unit 8 chemical equilibrium focusing on acid-base systems

Unit 8 Chemical Equilibrium Focusing on

Acid-Base Systems

88unit

Chemical EquilibriumFocusing on Acid–BaseSystems

Chemical EquilibriumFocusing on Acid–BaseSystemsEquilibrium describes any condition or situation of balance. We recognize equilib-

rium in a chemical reaction system, oddly enough, by noticing nothing—we see no

change in any property of the system. The easiest conclusion to draw would be that

nothing is happening, but closer study reveals that, at the molecular level, a lot of

change is going on.

Chemical reaction equilibrium is always a dynamic balance between two opposing

changes, which are balanced because they are occurring at equal rates, within a closed

system. What we observe directly is the net effect—neither an increase nor a decrease

in any measurable property.

Chemistry involves the study of change in chemical substances. To predict and con-

trol chemical change, we must better understand the nature of the system at the

molecular level. For instance, to understand why and how bubbles of gas form or

dissolve in a liquid, we must take into account the nature of the gas, the nature of

the liquid, and the actions of their invisible entities—all at the same time.

This unit explores the nature of dynamic equilibrium in chemical systems. It

explains much more thoroughly and completely many of the chemical change con-

cepts you have already learned. You will examine some very important reactions—

those involving acids and bases in solution—at a higher conceptual level. This

knowledge will allow you to describe, explain, and predict many new chemical sys-

tems and situations. The continual exploration and improvement of concepts such

as these is a critical part of the nature of science.

As you progress through the unit, think about these focusing questions:

• What is happening in a system at equilibrium?

• How do scientists predict shifts in the equilibrium position of a system?

• How do Brønsted–Lowry acids and bases illustrate equilibrium?

670 Unit 8 NEL

Unit 8 - Ch 15 Chem30 11/1/06 1:17 PM Page 670

GENERAL OUTCOMESIn this unit, you will• explain that there is a dynamic balance

of opposing reactions in chemicalsystems at equilibrium

• determine quantitative relationships inacid–base ionization and other simplesystems at equilibrium

Chemical Equilibrium Focusing on Acid–Base Systems 671NEL

Unit 8


Unit 8ChemicalEquilibriumFocusing onAcid–BaseSystems

ARE YOU READY?

672 Unit 8 NEL

These questions will help you find out what you already know, and what you need toreview, before you continue with this unit.

Knowledge1. A large number of chemical reactions (most notably redox reactions) will only

occur in aqueous solution, at least at a rate great enough to be observable in alaboratory. In terms of collision–reaction theory (Figure 1), state (a) why a given reaction might only occur in solution(b) why collisions between entities do not necessarily result in reaction(c) the two effects that an increase in temperature has on collisions between

entities involved in a reaction(d) the effect that an increase in concentration of one kind of entity has on

collisions between all entities involved in a reaction

2. Assume that each of the following substances is placed in water. Rewrite theformula and the physical state to indicate whether it is very soluble or onlyslightly soluble in water at SATP. For those substances predicted to produce ionsin solution, write symbols for all aqueous ions present after the substancedissociates upon dissolving.(a) MgSO4·7H2O(s)(b) CH3COCH3(l)(c) CH2CH2(g)(d) CaCO3(s) (e) PbCl2(s)(f) FeCl3(s)(g) C3H5(OH)3(l)

3. Each of the following substances is mixed with water to form an aqueoussolution. Write an equation (using the modified Arrhenius theory) to explainthe acidity or basicity of the solution in terms of hydronium or hydroxide ions.Your equations should, when necessary, represent the “ionizing” of the substanceas a reaction with water.(a) HCl(g)(b) CH3COOH(l)(c) H2SO4(l)(d) HNO3(l)(e) NaCl(s)(f) NH3(g)(g) NaOH(s)

Concepts

• collision–reaction theory

• dissociation and ionization

• amount concentration

• ion concentration

• percent reaction

• stoichiometric calculation

• net ionic equations

• acids and bases

• indicators

Skills

• laboratory safety

• scientific problem solving

You can review prerequisiteconcepts and skills in theappendices and on theNelson Web site.

A Unit Pre-Test is alsoavailable online.

Prerequisites

An Ineffective Collision

Figure 1Collision–reaction theory considers the numbers, speeds, and orientation of colliding entities toexplain the progress of chemical reactions.

www.science.nelson.com GO


Unit 8


4. The concentration of chemical substances in solution can vary widely (Figure 2). Concentration affects solution properties such as colour,conductivity, freezing point, and viscosity. Concentration also affects thefrequency of particle (entity) collisions; and, thus, will usually affect theobserved rate of a reaction. Complete Table 1.

5. Chemical substances may also have widely varying concentrations in thegaseous state. Calculate the amount concentration of(a) 24.0 g of hydrogen in a 2.00 L container(b) 500 kPa of oxygen in a 10.0 L container at 0 °C(c) 4.40 mol of carbon dioxide at SATP(d) 0.227 mol of methane at STP

6. Understanding chemical equilibrium theory often involves using a net ionicequation. For each of the following combinations of reagents, predict theproduct(s), and write a net ionic reaction equation. Balance the equation withsimplest integer coefficients, and include physical states for all substances.(a) Copper(II) chloride and potassium carbonate solutions are mixed.(b) Ethene (ethylene) reacts with hydrogen chloride to form chloroethane.(c) Aluminium foil reacts with hydrochloric acid.(d) Ammonia undergoes simple decomposition.(e) Magnesium metal is placed in an aqueous solution of gold(III) chloride.(f) Mixing sulfurous acid solution with sodium hypochlorite solution results in

a spontaneous redox reaction.

7. The acidity of solutions often has a considerable effect on the type, rate, andextent of the reactions they will undergo. Acids may be classed as strong orweak, depending on the extent of their reaction with water. Complete Table 2,identifying the acids as strong or weak.

Table 1 Concentration of Entities and Quantities of Reagents in Solution

Mass Solution ConcentrationReagent dissolved (g) volume (L) Entity (mol/L)

NaOH(s) 1.74 0.500 OH–(aq)

Al2(SO4)3(s) 2.00 SO42–(aq) 0.100

Al2(SO4)3(s) 2.00 Al3+(aq) 0.100

CaCl2(s) 1.00 Cl–(aq) 0.00440

Table 2 Solution Acidity and Basicity

Strength [Acid] % Reaction [H3O�(aq)] Acid S/W (mol/L) in/with water pH (mol/L)

HCl(aq) 0.016 �99

HBr(aq) �99 0.024

HNO3(aq) �99 4.0

CH3COOH(aq) 0.100 1.3

HCN(aq) 0.200 5.0

HNO2(aq) 0.010 7.0

dilutesolution

concentratedsolution

Figure 2Concentration affects the rate ofchemical reaction as well as physicalproperties.


In this chapter

1515 Equilibrium Systems

chapter

Equilibrium Systems

Equilibrium in Chemical SystemsThe simplest equilibrium systems are static: Nothing is moving or changing to createthe balance. A textbook sitting on a level desktop is an example of static equilibrium. Itstays motionless because two equal and opposite forces act on it simultaneously. Thedownward pull of Earth (gravity) on the book is exactly balanced by the upward pushby the desktop. A laboratory balance is a common technology that uses this kind ofequilibrium.

Chemical equilibrium is also a balance between two opposing agents of change, butalways in a dynamic system. An expert juggler in performance (Figure 1) is similar to achemical system at equilibrium. The juggler’s act is a dynamic equilibrium, with some ballsmoving upward and some moving downward at any given moment. There is no netchange because the rates of upward movement and downward movement are equal atany given moment.

Chemical systems at equilibrium have constant observable properties. Nothing appearsto be happening because the internal movement involves entities that are too small to see.A critical task of chemical engineers is to disturb (unbalance) chemical equilibria inindustrial reactions. Production of specific desired products is controlled by manipulatingthe conditions under which reactions occur. Some general concepts apply to all chem-ical equilibrium systems; these concepts are the focus of this chapter.

674 Chapter 15 NEL

Exploration: Shakin’ theBlues

Investigation 15.1: TheExtent of a ChemicalReaction

Web Activity: EquilibriumState

Mini Investigation:Modelling DynamicEquilibrium

Lab Exercise 15.A: TheSynthesis of anEquilibrium Law

Web Activity: PaulKebarle

Lab Exercise 15.B:Determining anEquilibrium Constant

Web Activity: WritingEquilibrium Expressions

Investigation 15.2:Equilibrium Shifts(Demonstration)

Biology Connection: CO2 Transport

Investigation 15.3: TestingLe Châtelier’s Principle

Case Study: UreaProduction in Alberta

Lab Exercise 15.C: TheNitrogen Dioxide–Dinitrogen TetroxideEquilibrium

Investigation 15.4:Studying a ChemicalEquilibrium System

Web Activity: PoisonAfloat

Answer these questions as best you can with your current knowledge. Then, using theconcepts and skills you have learned, you will revise your answers at the end of thechapter.

1. What, precisely, is happening to the chemical entities involved in a reaction whileobservation shows that products are being formed?

2. Is anything happening to the chemical entities involved in a reaction whenobservation shows the reaction appears to have stopped, with no more productsbeing formed?

3. Why do some reactions seem to occur partially, and apparently stop while some of allof the reactants are still present, while in other reactions all of the limiting reagentappears to be consumed?

4. Can the chemical amount of product be predicted successfully for reactions that arenot quantitative?

STARTING Points

Career Connection:Food Science Technologist; Chemical Process Engineer

Unit 8 - Ch 15 Chem30 11/1/06 1:18 PM 74

Equilibrium Systems 675NEL

Figure 1How is a juggler similar to achemical system in equilibrium?

Figure 3How many times does this reactionhappen?

Exploration Shakin’ the Blues

When small pieces of zinc are added to a dilute solution ofexcess hydrochloric acid in an open beaker (Figure 2), avigorous reaction occurs with lots of gas and heat given off. Thezinc continues to react and, when it is completely consumed,the visible signs of reaction come to an end. After seeing manyreactions such as this one in your science studies, you mayhave come to think that all chemical reactions only go one way:from reactants to products. But do they always?

Materials: lab apron; eye protection; 400 mL flask and stopper; 250 mL water; 5.0 g potassiumhydroxide (KOH(s)); 3.0 g glucose or dextrose; 2% methyleneblue; stirring rod

• Pour 250 mL of water into the flask.• Add 6 drops of methylene blue and all of the potassium

hydroxide and glucose to the flask.• Stir the mixture with the stirring rod until the solids have

dissolved.• Stopper the flask and set it on the bench. Observe the

colour of the solution.• Shake the solution vigorously and note any changes

(Figure 3).• Set the flask on the table and leave it standing until another

change is noticed.• Repeat the previous two steps many times. Make

observations each time.

(a) Describe the reaction in the flask in relation to the discussion at the beginning of this activity.

(b) What evidence do you have to substantiate your answer toquestion (a)?

(c) Predict whether the colour changes will continue forever.• Test your prediction over a reasonable period of time.(d) Evaluate your prediction.

Potassium hydroxide is poisonous and corrosive.

Keep potassium hydroxide away from skin andeyes. Wear eye protection.

Figure 2Zinc reacts rapidlyand quantitatively withhydrochloric acid.


676 Chapter 15 NEL

15.115.1 Explaining Equilibrium SystemsScientists describe chemical systems in terms of empirical properties such as tempera-ture, pressure, volume, and amounts of substances present. Chemical systems are sim-pler to study when separated from their surroundings by a definite boundary. Thisseparation gives an experimenter control over the system so that no matter can enteror leave. Such a physical arrangement is called a closed system. A reaction in solutionin a test tube or a beaker can be considered a closed system, as long as no gas is used orproduced in the reaction. Systems involving gases must be closed on all sides by a solidcontainer. Separating a chemical reaction system from its surroundings makes studyingits properties, conditions, and changes much simpler. The use of controlled systems isan integral part of scientific study.

Closed Systems at EquilibriumOne example of a chemical system at equilibrium is a soft drink in a closed bottle—a closedsystem in equilibrium. Nothing appears to change, until the bottle is opened. Removingthe bottle cap and reducing the pressure alters the equilibrium state, as the carbondioxide is allowed to leave the system (Figure 1). Carbonated drinks that have gone“flat” because of the decomposition of carbonic acid can be carbonated again by theaddition of pressurized carbon dioxide to the solution to reverse the reaction, and thencapping the container to restore the original equilibrium.

Collision–reaction theory is fundamental to the study of chemical systems. As origi-nally introduced in this textbook to provide a basis for stoichiometric calculations, thistheory required us to initially assume, for simplicity, that reactions are always spontaneous,rapid, quantitative, and stoichiometric. Common experience, however, shows that thisassumption is not always true. Not all reactions are rapid; for example, corrosion of a carbody may take years. A study of oxidation–reduction reactions soon provides evidencethat many reactions are not spontaneous. This chapter will examine and test the assump-tion of quantitative reaction in detail to significantly increase your understanding ofchemical systems.

Figure 1When the pressure on this equilibrium system changes, theequilibrium is disturbed.

Purpose Design AnalysisProblem Materials Evaluation (1, 2, 3)Hypothesis ProcedurePrediction Evidence

To perform this investigation, turn to page 700.

The Extent of a Chemical ReactionIn Chapter 8, you performed experiments that produced evidencethat reactions are quantitative. In a quantitative reaction, thelimiting reagent is completely consumed. To identify the limitingreagent, you can test the final reaction mixture for the presence ofthe original reactants. For example, in a diagnostic test, you mighttry to precipitate ions from the final reaction mixture that werepresent in the original reactants.

PurposeThe purpose of this investigation is to test the validity of theassumption that chemical reactions are quantitative.

ProblemWhat are the limiting and excess reagents in the chemicalreaction of selected quantities of aqueous sodium sulfate andaqueous calcium chloride?

DesignSamples of sodium sulfate solution and calcium chloride solutionare mixed in different proportions and the final mixture is filtered.Samples of the filtrate are tested for the presence of excessreagents, using diagnostic tests.

INVESTIGATION 15.1 Introduction Report Checklist



Evidence obtained from many reactions contradicts the assumption that reactionsare always quantitative. In Investigation 15.1, there is direct evidence for the presence ofboth reactants after the reaction appears to have stopped. This apparent anomaly can beexplained, in terms of collision–reaction theory, by the idea that a reverse reaction canoccur: the products, calcium sulfate and sodium chloride, can react to re-form the orig-inal reactants. The final state of this chemical system can be explained as a competitionbetween collisions of reactants to form products and collisions of products to re-formreactants.

forwardNa2SO4(aq) � CaCl2(aq) 0 CaSO4(s) � 2 NaCl(aq)

reverse

This competition requires that the system be closed so that reactants and productscannot escape from the reaction container. The chemical system in Investigation 15.1 canbe considered a closed system, bounded by the volume of the liquid phase.

We assume that any closed chemical system with constant macroscopic properties(no observable change occurring) is in a state of equilibrium, usually classified, for con-venience, as one of three types. Phase equilibrium involves a single chemical substanceexisting in more than one phase in a closed system. Water placed in a sealed containerevaporates until the water vapour pressure (concentration of water in the gas phase)rises to a maximum value, and then remains constant (Figure 2). Solubility equilib-rium involves a single chemical solute interacting with a solvent substance, where excesssolute is in contact with the saturated solution (Figure 3). A chemical reaction equi-librium involves several substances: the reactants and products of a chemical reaction.All three types of equilibrium are explained by a theory of dynamic equilibrium—abalance between two opposite processes occurring at the same rate.

The terms forward and reverse are used to identify which process is being referred to,and are specific to a written equilibrium equation. When any equation is written witharrows to show that the change occurs both ways, the left-to-right change is called theforward reaction, and the right-to-left change is called the reverse reaction.

Section 15.1

Practice1. The evidence gathered in Investigation 15.1 may be classified as an anomaly—an

unexpected result that contradicts previous rules or experience.(a) Write the balanced equation for the double replacement reaction of sodium

sulfate and calcium chloride solutions.(b) Write a statement describing the anomaly that occurred, using chemical names

from the equation.(c) Write the net ionic equation for the reaction.(d) Use chemical names from the net ionic equation to write a statement about the

anomaly.(e) Which of the previous statements more accurately describes the chemical system,

according to collision–reaction theory?

2. When scientists first encounter an apparent anomaly, they carefully evaluate thedesign, procedure, and technological skills involved in an investigation. Oneimportant consideration is the reproducibility of the evidence. Compare your evidencein Investigation 15.1 with the evidence collected by other groups. Is there support forthe reproducibility of this evidence?

Anomalies—Signals forChangeAnomalies, or discrepant events,are important as scientists acquireand develop scientific knowledge.Sometimes these events havebeen ignored, discredited, orelaborately explained away byscientists who do not wish toquestion or reconsider acceptedlaws and theories. Investigatinganomalies sometimes leads to therestriction, revision, or replacementof scientific laws and theories.

DID YOU KNOW ??

Figure 2According to the theory of dynamicequilibrium, as long as the containerremains closed at constanttemperature, the rate at whichmolecules in the liquid stateevaporate is equal to the rate atwhich molecules in the gas statecondense.

H2O(l) 0 H2O(g)

H2O(g)

H2O(l)

0

Figure 3For excess solid copper(II) sulfate in equilibrium with its saturated aqueous solution, the rates ofdissolving and crystallization are equal.CuSO4(s) 0 Cu2�(aq) � SO4

2–(aq)


678 Chapter 15 NEL

Chemical Reaction EquilibriumChemical reaction equilibria are more complex than phase or solubility equilibria, dueto the variety of possible chemical reactions and the greater number of substancesinvolved. To explain chemical equilibrium systems, we need to combine ideas fromatomic theory, kinetic molecular theory, collision–reaction theory, and the concepts ofreversibility and dynamic equilibrium. Although this synthesis is successful as a descrip-tion, and also as an explanation, it has only limited application in predicting quantita-tive properties of an equilibrium system.

WEB Activity


Simulation—Equilibrium StateThis simulation illustrates the establishment of a simple dynamic equilibrium by showing howthe concentrations of entities change over time, beginning from an initial condition. A textbookcan only represent this process as a sequence of static (still) diagrams (such as those in Figure 4) or graphs (such as that in Figure 5).

mini Investigation Modelling Dynamic Equilibrium

This activity models the progress of a chemical reaction toequilibrium, representing concentrations of reactants andproducts as volumes of water. The establishment of equilibriumis graphed as volume versus number of transfers. This graphthen represents a typical concentration–time graph of achemical reaction to equilibrium.

Materials: two drinking straws of different diameters; two 25 mL graduated cylinders; graph paper; water; meniscus finder

• Label one 25 mL graduated cylinder R (for reactants), andfill with water to the 25.0 mL mark.

• Label the other cylinder P (for products) and leave it empty.

• Holding a straw in each hand, place one straw in eachgraduated cylinder so that each straw rests on the bottomof its cylinder. Use the larger straw in the R cylinder.

• Transfer water simultaneously from each cylinder to theother by placing an index finger over the open end of eachstraw (to seal it). Then lift the straws out of their originalcylinders, move the bottom of each straw over the othercylinder, and lift your index fingers to allow the water ineach straw to drain into the cylinder below. Replace thestraws in their original cylinders.

• After each transfer, measure and record the volume of waterin each cylinder to 0.1 mL. Also record the number of thetransfer.

• Repeat the transfer step until no significant change in watervolumes has occurred for at least three transfers.

• Graph the volume of water in each cylinder as a dependent(responding) variable against the number of transfers as theindependent (manipulated) variable.

• Repeat the activity, switching the straws used in eachcylinder. Plot the values on the same axes as for the firsttrial.

(a) How does the graph change, and how is it the same, whenthe smaller straw is initially in the R cylinder?

(b) How would doing the transfers more slowly affect the finalvolumes in each cylinder?

(c) How would replacing the larger straw with an even largerone affect the final volumes in each cylinder?

(d) Express the equilibria of the water transfer trials as percentyields; that is, the percentage of “reactant” water that isconverted to “product”.

(e) Predict the graph’s shape if both of the straws chosen hadthe same diameter.

Unit 8 - Ch 15 Chem30 11/3/06 9:42 AM Page 678


The Hydrogen–Iodine Reaction SystemChemists have studied the reaction of hydrogen gas and iodine gas extensively, becausethe molecules are simple in structure and the reaction takes place in the gas phase. Oncehydrogen and iodine are mixed, the reaction proceeds rapidly at first. The initial dark purple colour of the iodine vapour gradually fades, and then remains constant(Figure 4).

Section 15.1

I2(g)

H2(g)

Initially, hydrogen (in excess)and iodine are added to theflask. The colour of the iodinevapour is the only easilyobservable property.

Early in the reaction, hydrogenand iodine form hydrogeniodide faster than hydrogeniodide forms hydrogen andiodine. Overall, the amount ofiodine decreases, so the colourof the flask contents appearsto lighten. Both hydrogen andhydrogen iodide are colourless.

At equilibrium, analysis showsthat the flask contains allthree substances. The purplecolour shows that some iodineremains. The constancy of thecolour is evidence thatequilibrium exists. Forwardand reverse reactions areoccurring at equal rates.

I2(g)

H2(g)

HI(g)

I2(g)

H2(g)

HI(g)

Figure 4When hydrogen and iodine areadded to the flask, the colour of theiodine vapour is the only easilyobservable (empirical) property. Anequilibrium equation describes thisevidence theoretically.H2(g) � I2(g) 0 2 HI(g), t � 448 °C

Figure 5The graph of the quantity of eachsubstance against time shows thatthe rate of reaction of the reactantsdecreases as the number of reactantmolecules decreases, and the rateat which the product changes backto reactants increases as thenumber of product moleculesincreases. These two rates mustbecome equal at some point, afterwhich the quantity of eachsubstance present will not change.

Time (reaction progress)

Qua

ntit

y

Reaction of Hydrogenand Iodine

t = 448 °C

H2

HI

I2

Table 1 contains data from three experiments with the hydrogen–iodine system: onein which hydrogen and iodine are mixed; one in which hydrogen, iodine, and hydrogeniodide are mixed; and one in which only hydrogen iodide is present initially. At a tem-perature of 448 °C, the system quickly reaches an observable equilibrium each time.Chemists use evidence such as that in Table 1 to describe a state of equilibrium in twoways: in terms of percent reaction and in terms of an equilibrium constant. Percentreaction describes the equilibrium for one specific system example only, whereas anequilibrium constant describes all systems of the same reaction at a given temperature.Alternatively, you can draw a graph of the reaction progress, plotting quantity (or con-centration) of the reagents versus time (Figure 5).

Table 1 The Hydrogen–Iodine System at 448 °C

Initial system concentrations Equilibrium system concentrationsSystem (mmol/L) (mmol/L)

H2(g) I2(g) HI(g) H2(g) I2(g) HI(g)

1 5.00 5.00 0 1.10 1.10 7.80

2 0.50 0.50 1.70 0.30 0.30 2.10

3 0 0 3.20 0.35 0.35 2.5


680 Chapter 15 NEL

A percent yield is defined as the yield of product measured at equilibrium comparedwith the maximum possible yield of product. In other words, percent yield can be usefulfor communicating the position of an equilibrium. The maximum possible yield ofproduct is calculated using the method of stoichiometry, assuming a quantitative forwardreaction with no reverse reaction. Percent yield provides an easily understood way torefer to quantities of chemicals present in equilibrium systems. For example, analysisof the evidence in System 1, Table 1, shows that, at 448 °C, this particular hydrogen–iodinesystem reaches an equilibrium with a percent yield of 78.0% (Table 2).

Equilibrium arrows (0) communicate that an equilibrium exists. To communicate theextent of a reaction, a percent yield may be written above the equilibrium arrows in achemical equation. The following equation describes the position of a hydrogen–iodineequilibrium in System 1, Table 2, at 448 °C.

78%

H2(g) � I2(g) 0 2 HI(g) t � 448 °C

Scientists now think of all chemical reactions as occurring in both forward and reversedirections. Any reaction falls loosely into one of four categories. Reactions that favourreactants very strongly, that is, reactions that normally have a percent yield of much lessthan 1%, are simply observed as being nonspontaneous. In these reactions, mixing reac-tants has no observable result. Reactions producing observable equilibrium conditionsmay react less or more than 50%, favouring reactants or products respectively. Significantamounts of both reactants and products are always present. Finally, reactions that favourproducts very strongly, much more than 99%, are observed to be complete (quantitative).The chemical equations for quantitative reactions are generally written with a singlearrow to indicate that the effect of the reverse reaction is negligible. Table 3 shows howpercent yield may be used to classify equilibrium systems and how the classification maybe communicated in reaction equations.

Table 2 Percent Yield of the Hydrogen–Iodine System at 448 °C

Equilibrium [HI]* Maximum possible [HI]* Percent yieldSystem (mmol/L) (mmol/L) (%)

1 7.80 10.0 78.0

2 2.10 2.70 77.8

3 2.50 3.20 78.1

*Square brackets [ ] indicate amount concentration.

Learning TipWhen a reaction is shown to bequantitative (as written in theequation), it means that thereverse reaction happens solittle that it can be ignored forall normal purposes. Anotherway to think of a quantitativereaction is that if the productsshown (as written) were mixedtogether as reactants, therewould be no apparent reaction.A reaction that is quantitative inthe forward direction isnecessarily nonspontaneous inthe reverse direction. In thisunit, we will (arbitrarily) assumethat "quantitative" specifies areaction that, at equilibrium, ismore than 99.9% complete.Another way to think of it is thatless than one part per thousand(0.1%) of an original reactantremains unreacted, atequilibrium, in a quantitativereaction.

Table 3 Classes of Chemical Reaction Equilibria

Percent yield Description of equilibrium Position of equilibrium

negligible nonspontaneous(no apparent reaction)

� 50% reactants favoured � 50%

0

� 50% products favoured � 50%

0

� 99.9% quantitative →

DID YOU KNOW ??Lavoisier and Closed SystemsAntoine Laurent Lavoisier(1743–1794) is recognized as thefather of modern chemistry for manyreasons. He created the basicnomenclature system we still use forcompounds, demonstrated the lawof conservation of mass, andexplained and clarified the theory ofcombustion. Most importantly,perhaps, his successes convincedthe chemical community of thecritical importance of his methods ofcareful measurement, and ofcarrying out experiments in closedsystems—carefully accounting for,and preventing, invisible reactantsand products (notably, gases) fromescaping. When chemists began tounderstand the importance ofmaintaining closed systems in orderto draw correct conclusions aboutreactions, chemistry could finallymove forward as a true science.



Section 15.1

Consider the reaction equation for the formation of hydrogen iodide at 448 °C. Assume thereaction is begun with 1.00 mmol/L concentrations of both H2(g) and I2(g). Construct anICE table to determine equilibrium concentrations of the reagents. The equilibriumconcentration of I2(g) (determined by colour intensity) is 0.22 mmol/L.

Set up the ICE table as follows:

Table 4 The H2(g) � I2(g) 0 2 HI(g) Equilibrium

[H2(g)] [I2(g)] [HI(g)] Concentration (mmol/L) (mmol/L) (mmol/L)

Initial 1.00 1.00 0.00

Change

Equilibrium 0.22

Begin by calculating the change (decrease) in concentration of iodine.

(0.22 � 1.00) mmol/L � �0.78 mmol/L

The changes of the other concentrations may be calculated directly from this value, usingstoichiometric ratios from the balanced equation.

For hydrogen, the change is also a decrease (negative value) of

0.78 mmol/L � � �0.78 mmol/L

For hydrogen iodide, the change is an increase (positive value) of

0.78 mmol/L � � �1.6 mmol/L

Complete the ICE table, using these values to enter the concentrations at equilibrium.

Table 5 The H2(g) � I2(g) 0 2 HI(g) Equilibrium

[H2(g)] [I2(g)] [HI(g)] Concentration (mmol/L) (mmol/L) (mmol/L)

Initial 1.00 1.00 0.00

Change �0.78 �0.78 �1.6

Equilibrium 0.22 0.22 1.6

2�1

1�1

SAMPLE problem 15.1

Diabetes: Blood SugarEquilibriumFor people with diabetes, reactionequilibrium established by sugar inthe human body is criticallyimportant. Recent advances in thetechnology allow testing of bloodsugar concentration with personaldevices such as the One Touch®SureStep® blood-glucose meter(Figure 6). The meter displays theconcentration in mmol/L afteranalyzing a single drop of bloodextracted from a fingertip. Multiplereadings, including date and time,are stored electronically and canbe displayed at any time. The datacan be downloaded to a computer.

DID YOU KNOW ??

Figure 6This device helps diabetics monitortheir blood glucose levels.

When considering equilibrium systems, we cannot use the simple assumption ofquantitative reaction. When there is no limiting reagent for a reaction, and when wecannot assume complete reaction, stoichiometric calculations require a little morethought. Such calculations may conveniently be set up as an ICE table, meaning thatthe initial, change, and equilibrium values are arranged in tabular form.

In Sample Problem 15.1, notice that every substance in the reaction is a gas. Therefore,all of the stoichiometric calculations can use concentrations directly, rather than chem-ical amounts, because the volume must be the same for every gaseous substance in aclosed container. The volume is a common factor in the calculation step that uses the stoichiometric ratio. This same reasoning means that concentrations can also be useddirectly for stoichiometric calculation whenever every substance in a reaction is anaqueous entity dissolved in the same volume of solvent.

CAREER CONNECTION

Food Science TechnologistThe development and analysis offood products for individuals withdiabetes is crucial. Food sciencetechnologists carefully measureand conduct tests oncarbohydrates so that patients cancontrol their blood sugarequilibrium.

Learn more about the manyfood industries that employ foodscience specialists.



682 Chapter 15 NEL

Practice3. For a chemical system at equilibrium:

(a) What are the observable characteristics?(b) Why is the equilibrium considered “dynamic”?(c) What is considered “equal” about the system?

4. In a gaseous reaction system, 2.00 mol of methane, CH4(g), is initially added to 10.00 mol of chlorine, Cl2(g). At equilibrium the system contains 1.40 mol ofchloromethane, CH3Cl(g), and some hydrogen chloride, HCl(g). (a) Write a balanced reaction equation for this equilibrium and calculate the

maximum possible yield of chloromethane product.(b) Calculate the percent yield at this equilibrium and state whether products or

reactants are favoured.

5. Combustion reactions, such as the burning of methane, often favour products sostrongly that they are written with a single arrow. Assuming the forward reaction hasa very low activation energy (Chapter 12) and the reverse reaction has a very highactivation energy (to account for the difference in the tendency to occur), sketch apossible potential energy diagram representing the progress of such a reaction.

6. After 4.0 mol of C2H4(g) and 2.50 mol of Br2(g) are placed in a sealed container, thereaction

C2H4(g) � Br2(g) 0 C2H4Br2(g)

establishes an equilibrium. Figure 7 shows the concentration of C2H4(g) as itchanges over time at a fixed high temperature until equilibrium is reached.

(a) Sketch this graph. Draw lines on your copy to show how the concentration ofeach of the other two substances changes.

(b) Create an ICE table, using reagent amount concentrations.(c) What is the volume of the container?(d) Calculate the percent yield of dibromoethane.

7. Write a balanced net ionic equation for each of the following described reactions,showing appropriate use of equilibrium "arrow" symbols where appropriate, andindicating (with symbols) whether products or reactants are favoured.(a) An excess of solid copper reacts with virtually all of the silver ions in a sample

solution.(b) When a solution containing calcium ions is mixed with a solution containing a

large excess of sulfate ions, a precipitate forms, but tests indicate that a smallquantity of calcium ions remains in solution.

(c) When acetic acid is dissolved in water, the acetic acid molecules react with watermolecules to form hydronium and acetate ions. Careful pH testing shows thatabout 980 of every 1000 acetic acid molecules remain in their molecular form, atequilibrium.

Figure 7A graph of the reaction of ethene with bromine

Reaction of Ethene and Bromine5.0

4.0

3.0

2.0

1.0

0.0

Con

cent

rati

on (

mol

/L)

Time

C2H4

Learning TipChemists often refer to“homogeneous” reactionsystems. This term means thatevery entity involved in thereaction exists in the samephysical state. The reactionfrom Sample Problem 15.1,where all reactants andproducts are gases, is a typicalcase. The other common caseinvolves reactions where allentities involved are in aqueoussolution. A system that hasmore than one phase, such assolid copper reacting inaqueous silver nitrate solution,is a “heterogeneous” reactionsystem.



Section 15.1

Purpose Design AnalysisProblem Materials EvaluationHypothesis ProcedurePrediction Evidence

The Synthesis of an Equilibrium LawThe following chemical equation represents a chemical equilibrium:

Fe3�(aq) � SCN�(aq) 0 FeSCN2�(aq)

This equilibrium is convenient to study because the colour of thesystem characterizes the equilibrium position of the system(Figure 8).

PurposeThe purpose of this investigation is the synthesis of an equilibriumlaw. Complete the Analysis of the investigation report.

ProblemWhat mathematical formula, using equilibrium concentrations of reactants and products, gives a constant for the iron(III)–thiocyanate reaction system?

DesignReactions are performed using various initial concentrations ofiron(III) nitrate and potassium thiocyanate solutions. Theequilibrium concentrations of the reactants and the product aredetermined from the measurement and analysis of the colourintensity using a spectrophotometer. Possible mathematicalrelationships among the concentrations are tried and analyzed todetermine if the mathematical formula gives a constant value.

Evidence

Table 6 Iron(III)–Thiocyanate Equilibrium at SATP

[Fe3�(aq)] [SCN�(aq)] [FeSCN2�(aq)]Trial (mol/L) (mol/L) (mol/L)

1 3.91 � 10�2 8.02 � 10�5 9.22 � 10�4

2 1.48 � 10�2 1.91 � 10�4 8.28 � 10�4

3 6.27 � 10�3 3.65 � 10�4 6.58 � 10�4

4 2.14 � 10�3 5.41 � 10�4 3.55 � 10�4

5 1.78 � 10�3 6.13 � 10�4 3.23 � 10�4

AnalysisTest the following mathematical relationships for constancy:

1. [Fe3�(aq)][SCN�(aq)][FeSCN2�(aq)]

2. [Fe3�(aq)] � [SCN�(aq)] � [FeSCN2�(aq)]

3.

4.

5.[SCN�(aq)]��[FeSCN2�(aq)]

[Fe3�(aq)]��[FeSCN2�(aq)]

[FeSCN2�(aq)]��[Fe3�(aq)][SCN�(aq)]

LAB EXERCISE 15.A Report Checklist

Figure 8The two reactantscombine to form a darkred equilibrium mixture.The red colour of thesolution is due to theaqueous thiocyanate–iron(III) product,FeSCN2�(aq).

Fe3+(aq)

FeSCN2+(aq)

SCN–(aq)

ComputersScientists often use computers toanalyze numerical evidence inorder to establish mathematicalrelationships among experimentalvariables. The mathematicalformulas derived are useful inunderstanding chemical processesand in applying these processes totechnology.

DID YOU KNOW ??WEB Activity


Canadian Achievers—Paul KebarlePaul Kebarle (Figure 9) pioneered the measurements of gas-phase ion-molecule equilibria. Kebarle’s findings, now significantly expanded by other workers, constitute a central database that is of fundamentalimportance in many diverse fields of scientific research.

1. What fundamental data did Kebarle and his co-workers obtainfrom their research?

2. List three fields of research that Kebarle’s work has aided.Figure 9Paul Kebarle


The Equilibrium Constant, KcAnalysis of the evidence from many experiments such as those in Lab Exercise 15.A(page 683) reveals a mathematical relationship that provides a constant value for a chem-ical system over a range of amount concentrations. This constant value is called theequilibrium constant, Kc, for the reaction system. Evidence and analysis of many equi-librium systems have resulted in the following equilibrium law.

For the reaction a A � b B 0 c C � d D,

the equilibrium law expression is Kc � �[

[

C

A

]

]

c

a

[

[

D

B]

]b

d

�

In this mathematical expression, A, B, C, and D represent chemical entity formulas anda, b, c, and d represent their coefficients in the balanced chemical equation. The rela-tionship holds only when amount concentrations are observed to remain constant, in aclosed system, at a given temperature.

We use a balanced chemical equation with whole-number coefficients to write the math-ematical expression of the equilibrium law. The coefficients of the balanced equationbecome the exponents of the amount concentrations. If the equation were to be writtenin reverse, the equilibrium law expression would simply be the reciprocal of the expres-sion above, and the equilibrium constant would be the reciprocal of the one for thereaction as written here. Using the products over reactants convention results in a rela-tionship between the numerical value of Kc and the forward extent of the equilibrium thatis easier to visualize for the equation as written. The higher the numerical value of theequilibrium constant, the greater the tendency of the system to favour the forward direc-tion; that is, the greater the equilibrium constant, the more the products are favoured atequilibrium.

684 Chapter 15 NEL

Figure 10Cato Maximilian Guldberg(1836–1902) and Peter Waage(1833–1900) were related by morethan their interest in chemistry:They were brothers-in-law!

Learning TipIt is common practice(convention) to ignore unitsand list only the numericalvalue for an (amountconcentration) equilibriumconstant. The expression ofunits is often very complex forKc relationships. But, becauseeach entity concentration isalways entered with mol/Lunits, any entity concentrationwe calculate from a Kc valuewill always give an answerhaving mol/L units, so you needonly memorize this (simplifying)rule.

DID YOU KNOW ??Related InterestsTwo Norwegian chemists, CatoMaximilian Guldberg and PeterWaage, conducted detailed empiricalstudies of many equilibrium systemsin the mid-1800s (Figure 10). By1864, they had proposed amathematical description of theequilibrium condition that they calledthe “law of mass action.” Analyzingthe results of their experiments,Guldberg and Waage noticed that,when they arranged the equilibriumconcentrations into a specific form ofratio, the resulting value was thesame no matter what combinationsof initial concentrations were mixed.

Write the equilibrium law expression for the reaction of nitrogen monoxide gas withoxygen gas to form nitrogen dioxide gas.

Solution

2 NO(g) � O2(g) 0 2 NO2(g)

Kc �[NO2(g)]2

��[NO(g)]2[O2(g)]

COMMUNICATION example 1

The value of Kc for the formation of HI(g) from H2(g) and I2(g) is 40, at temperature t.Determine the value of Kc for the decomposition of HI(g) at the same temperature.

Solution

2 Hl(g) 0 H2(g) � I2(g)

Kc � � �410� = 0.025

[H2(g)][I2(g)]��

[Hl(g)]2


Note that the decomposition reaction equation is the reverse of the formation reactionequation, and the value of Kc for decomposition is the reciprocal of the Kc for formation.



Experiments have shown that the value of the equilibrium constant depends on tem-perature. The value is also affected by very large changes in the equilibrium concentra-tion of a reactant or a product. A moderate change in the concentration of any one ofthe reactants or products results in a change in the other concentrations, so that theequilibrium constant remains the same. The equilibrium constant provides only ameasure of the equilibrium position of the reaction; it does not provide any informationon the rate of the reaction. Because they hold for a significant range of different con-centrations, equilibrium constant expressions have been found to be very useful, andKc values for reactions are in common use throughout the scientific community.

Equilibrium constants are adjusted to reflect the fact that pure substances in solid orliquid (condensed) states have concentrations that are essentially fixed—the chemicalamount (number of moles) per unit volume is a constant value. For example, a litre ofliquid water at SATP has a mass of 1.00 kg (a chemical amount of 55.5 mol) and, thus,a fixed amount concentration of 55.5 mol/L. The concentration of condensed states isnot included in a Kc expression—we assume that these constant values become part ofthe expressed equilibrium constant. Substances in a gaseous or dissolved state have vari-able concentrations, and must always be shown in an equilibrium law expression.

Section 15.1

Using ConstantRelationshipsYou are already familiar with theusefulness of some other constantmathematical relationships aboutreal phenomena. Finding arelationship that is constant forequilibrium concentrations is justanother example.

If you examine circles ofdifferent sizes carefully, youdiscover that the distance aroundany circle divided by the distanceacross it (at the widest part)always gives the same (constant)answer, no matter how big or smallthe circle. This constant value,3.14159…, is so useful that it hasbeen given its own symbol, theGreek letter pi, �. This relationshipis most usefully expressed as C � �d because measuring adiameter is much easier thanmeasuring a circumference.

Many other relationshipexpressions produce this sameconstant, including those usuallyused to calculate the area of acircle and the time period of apendulum’s swing.

DID YOU KNOW ??

Write the equilibrium law expression for the decomposition of solid ammonium chloride togaseous ammonia and gaseous hydrogen chloride.

Solution

NH4Cl(s) 0 NH3(g) � HCl(g)

Kc � [NH3(g)][HCl(g)]


Write the equilibrium law expression for the reaction of zinc in copper(II) chloride solution.

Solution

Zn(s) � Cu2�(aq) 0 Cu(s) � Zn2�(aq)

Kc �[Zn2�(aq)]��[Cu2�(aq)]


The concentration of solid NH4Cl(s) is omitted from the equilibrium law expression.The role of temperature in equilibrium constant expressions is critical, although the

temperature is not written in the expression directly. The value of the equilibrium con-stant, Kc, always depends on the temperature. Any stated numerical value for an equi-librium constant, or any calculation using an equilibrium constant expression, mustspecify the reaction temperature at equilibrium.

Since equilibrium depends on the concentrations of reacting substances, these sub-stances must be represented in the expression as they actually exist—meaning that ionsin solution must be represented as individual entities. Equilibrium constant expressionsare always written from the net ionic form of reaction equations, balanced with simplestwhole-number (integral) coefficient values unless otherwise specified.

Again, note that the solids, as well as the spectator ions (the chloride ions in this example),are omitted from the equilibrium law expression.

The Meaning of theEquilibrium ConstantTry this simulation to deepen yourunderstanding of the equilibriumconstant.


EXTENSION +


686 Chapter 15 NEL


Determining an Equilibrium ConstantDetermining the equilibrium constant for a reaction at specificconditions is often essential for an industrial chemist. Thisknowledge is necessary before adjusting the reaction conditionsin order to optimize production of the desired substances.Complete the Analysis of the investigation report.

PurposeThe purpose of this investigation is to use the equilibrium law todetermine the equilibrium constant at 200 °C for thedecomposition reaction of the molecular compound phosphoruspentachloride.

ProblemWhat is the value of the equilibrium constant for thedecomposition of phosphorus pentachloride gas to phosphorustrichloride gas and chlorine gas, at a temperature of 200 °C?

Evidenceequilibrium temperature � 200 °Cequilibrium concentrations:[PCl3(g)] � [Cl2(g)] � 0.014 mol/L[PCl5(g)] � 4.3 � 10�4 mol/L

LAB EXERCISE 15.B Report Checklist

Write an equilibrium law expression based on a balanced equation for the reactionsystem. Use single whole-number coefficients, written in net ionic form, and ignoreconcentrations of pure solid or liquid phases:

If: aA � bB 0 cC � dD

then: Kc �

An equilibrium constant value

• always depends on the system temperature

• is independent of the reagent concentrations

• is independent of any catalyst present

• is independent of the time taken to reach equilibrium

• is normally stated as a numerical value, ignoring any units

• is greater, the more the system favours the formation of products

[C]c[D]d

�[A]a[B]b

SUMMARY Writing Equilibrium Law Expressions

Predicting Final Equilibrium ConcentrationsFor simple homogeneous systems, it is possible to algebraically predict reagent concen-trations at equilibrium using a known value for Kc and initial reactant concentrationvalues. For more complex systems, the calculation becomes more difficult—such systemsare left for more advanced chemistry courses.

In a 500 mL stainless steel reaction vessel at 900 °C, carbon monoxide and water vapourreact to produce carbon dioxide and hydrogen. Evidence indicates that this reactionestablishes an equilibrium with only partial conversion of reactants to products. Initially,2.00 mol of each reactant is placed in the vessel. Kc for this reaction is 4.20 at 900 °C.What amount concentration of each substance will be present at equilibrium?

Write a balanced equation for the reaction equilibrium.

CO(g) � H2O(g) 0 CO2(g) � H2(g) Kc � 4.20 at 900 °C

SAMPLE problem 15.2



Section 15.1

Use the balanced equation to write the equilibrium law expression.

Kc � 4.20 �

The initial amount concentrations of the CO(g) and the H2O(g) are the same:

c � �20.0.5000m

Lol

� � 4.00 mol/L � [CO(g)] � [H2O(g)]

An ICE table makes it easier to keep track of amount concentration changes that occurduring a reaction, and to find the amount concentrations at equilibrium.

At equilibrium, let the final amount concentration of the product H2(g) be x mol/L (anyconvenient symbol could be used). Then [CO2(g)] must also be x mol/L, since the stoichiometric ratio of the reaction is 1:1:1:1. By this same reasoning, at equilibrium, the initial concentrations of CO(g) and H2O(g) must have decreased by x mol/L; so [CO(g)] � [H2O(g)] � (4.00 – x) mol/L. When you enter values for “Change” into the ICEtable, you must show increases as “�”, and decreases as “�”.

Substitute equilibrium concentrations in the equilibrium law expression.

4.20 � �

Since the right side of the equation is a perfect square, solving for x is quitestraightforward.

�4.20� � ��2.05 � �

4.00x� x�

x � (2.05)(4.00 � x)

� 8.20 – 2.05x

3.05x � 8.20

x � 2.69

Assign positive and negative signs and complete the ICE table.

At equilibrium, at 900 °C, [CO(g)] � [H2O(g)] � 1.31 mol/L and

[CO2(g)] � [H2(g)] � 2.69 mol/L

x2��(4.00 � x)2

x2��(4.00 � x)2

[CO2(g)][H2(g)]��[CO(g)][H2O(g)]

[CO2(g)][H2(g)]��[CO(g)][H2O(g)]

Table 7 The CO(g) � H2O(g) 0 CO2(g) � H2(g) Equilibrium

[CO(g)] [H2O(g)] [CO2(g)] [H2(g)]Concentration (mol/L) (mol/L) (mol/L) (mol/L)

Initial 4.00 4.00 0 0

Change �x �x �x �x

Equilibrium (4.00 � x) (4.00 � x) x x

Table 8 The CO(g) � H2O(g) 0 CO2(g) � H2(g) Equilibrium

[CO(g)] [H2O(g)] [CO2(g)] [H2(g)]Concentration (mol/L) (mol/L) (mol/L) (mol/L)

Initial 4.00 4.00 0 0

Change �2.69 �2.69 �2.69 �2.69

Equilibrium 1.31 1.31 2.69 2.69

Note that, if both initial reactant concentrations are not the same, solving the equationfor x is more complicated, requiring use of the quadratic formula. Questions in this textare restricted to examples that do not require the quadratic formula for solution.

Equilibrium Constant andReaction QuotientIs there any way of knowingwhether or not a reaction is atequilibrium? If we know theconcentrations of reactants andproducts, and the equilibriumconstant at the appropriatetemperature, we can use aconcept called the "reactionquotient" to predict which way thereaction will proceed, or if it isalready at equilibrium.


EXTENSION +


688 Chapter 15 NEL

Section 15.1 Questions1. Write a balanced equation with integer coefficients and the

expression of the equilibrium law for each of the followingreaction systems at fixed temperature. (a) Hydrogen gas reacts with chlorine gas to produce

hydrogen chloride gas in the industrial process thateventually produces hydrochloric acid.

(b) In the Haber process (Chapter 8), nitrogen reacts withhydrogen to produce ammonia gas.

(c) At some time in the future, industry and consumersmay make more extensive use of the combustion ofhydrogen as an energy source.

(d) When aqueous ammonia is added to an aqueousnickel(II) ion solution, the Ni(NH3)6

2�(aq) complex ion isformed (Figure 11).

(e) In the Solvay process for making washing soda(Chapter 7), one reaction involves heating solid calciumcarbonate (limestone) to produce solid calcium oxide(quicklime) and carbon dioxide.

(f) In Investigation 15.1, aqueous solutions of sodiumsulfate and calcium chloride are mixed. (Remember touse a net ionic equation.)

(g) In a sealed can of soda, carbonic acid, H2CO3(aq),decomposes to liquid water and carbon dioxide gas.

2. You can apply the empirical and theoretical concepts ofequilibrium to many different chemical reaction systems.Use the generalizations from your study of organicchemistry to predict the position of equilibrium for bromineplaced in a reaction container with ethylene at a hightemperature.

3. Interpret the graph in Figure 12 to answer the questionsabout the reaction. Hydrogen and iodine were placed in areaction vessel, which was then sealed, and heated to 450 °C.

(a) All three substances are gases. If the container has avolume of 2.00 L, what chemical amount of eachsubstance was present initially?

(b) What chemical amount of hydrogen iodide had formedat equilibrium? (Create an ICE table.)

(c) Describe the rate at which hydrogen is reacting fromthe moment the reactants are mixed to the time whenequilibrium has been established, in terms ofcollision–reaction theory.

4. For each of the following, write the chemical reaction equation with appropriate equilibrium arrow, as shown in Table 3 (page 680). (a) The Haber process is used to manufacture ammonia

fertilizer from hydrogen and nitrogen gases. Under less-than-desirable conditions, only an 11% yield ofammonia is obtained at equilibrium.

(b) A mixture of carbon monoxide and hydrogen, known aswater gas, is used as a supplementary fuel in manylarge industries. At high temperatures, the reaction ofcoke and steam forms an equilibrium mixture in whichthe products (carbon monoxide and hydrogen gases)are favoured. (Assume that coke is pure carbon.)

(c) Because of the cost of silver, many high school sciencedepartments recover silver metal from waste solutionscontaining silver compounds or silver ions. Aquantitative reaction of waste silver ion solutions with

WEB Activity

Simulation—Writing Equilibrium ExpressionsThis simulation allows you to select a reaction type and the initial reactant concentrations,which the program uses to plot the resulting equilibrium graph. You will then be guidedthrough a series of questions.

Time

H2

HI

I2

8.0

6.0

4.0

2.0

7.2

2.4

0.4

Con

cent

rati

on (

mol

/L)

Reaction of Hydrogen and Iodine t = 448 °C

Figure 12The progress of a hydrogen–iodine reaction

Figure 11A Ni2�(aq)solution is green.Ammonia reactswith the nickel(II)ion to form theintensely bluehexaammine-nickel(II) ion,Ni(NH3)6

2�(aq).

Ni2+(aq) Ni(NH3)62+(aq)




Section 15.1

copper metal results in the production of silver metaland copper(II) ions.

(d) One step in the industrial process used to manufacturesulfuric acid is the production of sulfur trioxide fromsulfur dioxide and oxygen gases. Under certainconditions, the reaction produces a 65% yield ofproducts.

5. Write the expression of the equilibrium law for thehydrogen–iodine–hydrogen iodide system at 448 °C. Usingthe evidence for System 1 as reported in Table 1 on page679, calculate the value of the equilibrium constant.

6. In the Haber process for synthesizing ammonia gas fromnitrogen and hydrogen, the value of Kc is 6.0 � 10�2 for thereaction at 500 °C. In a sealed container at equilibrium at500 °C, the concentrations of H2(g) and of N2(g) aremeasured to be 0.50 mol/L and 1.50 mol/L, respectively.Write the equilibrium law expression and calculate theequilibrium concentration of NH3(g).

7. At a certain constant (very high) temperature, 1.00 mol ofHBr(g) is introduced into a 2.00 L container. Decompositionof this gas to hydrogen and bromine gases quicklyestablishes an equilibrium, at which point the amountconcentration of HBr(g)is measured to be 0.100 mol/L. (a) Write a balanced equation for the reaction.(b) Write the equilibrium law expression.(c) Calculate the chemical amount of HBr(g) present at

equilibrium.(d) Calculate the chemical amount of HBr(g) that has

reacted to form H2(g) and Br2(g) products whenequilibrium is established.

(e) Calculate the chemical amounts of H2(g) and Br2(g)that have been produced, and, thus, are present, whenequilibrium is established.

(f) Calculate the amount concentration of all substancespresent at equilibrium.

(g) Calculate Kc for this reaction at this temperature.

8. To a heated reaction vessel with a volume of 1.00 L, a labtechnician adds 6.23 mmol H2(g), 4.14 mmol of I2(g), and22.40 mmol of HI(g). At equilibrium, a spectrophotometer isused to determine that the concentration of iodine vapouris 2.58 mmol/L. Construct an ICE table and find Kc for thereaction system H2(g) � I2(g) 0 2 HI(g).

9. Consider the system

CO2(g) � H2(g) 0 CO(g) � H2O(g)

Initially, 0.25 mol of water and 0.20 mol of carbon monoxideare placed in the reaction vessel. At equilibrium,spectroscopic evidence shows that 0.10 mol of carbondioxide is present. Construct an ICE table and find Kc forthis system.

10. Consider the system

2 HBr(g) 0 H2(g) � Br2(g)

Initially, 0.25 mol of hydrogen and 0.25 mol of bromine areplaced into a 500 mL electrically heated reaction vessel. Kc for the reaction at the temperature used is 0.020. (a) Find the concentrations of the substances at

equilibrium.(b) Calculate the chemical amount of each substance

present at equilibrium.

11. Explain briefly how atomic theory, kinetic molecular theory,collision–reaction theory, and the concepts of reaction rateand reversible reactions are all necessary to explainchemical reaction equilibrium observations.

Extension

12. In a very long-term sense, Earth may be considered aclosed system. One equilibrium of concern to scientists isthe same one involved in carbonation of soft drinks, on avastly larger scale. Scientists believe that over time, thecarbon dioxide gas in the atmosphere should be inequilibrium with carbon dioxide dissolved in the oceans.They also know that the concentration of CO2(g) in theatmosphere has been increased significantly (by about20%) in the last century, which, they believe, is mostly dueto the burning of fossil fuels. Concerns about theconsequences of global warming make it imperative thatscientists improve their theories about the various cycles,processes, and equilibria involving this greenhouse gas.Research and summarize currently accepted theory aboutcarbon dioxide dissolved in the oceans, and list some othercycles and systems involving reaction or production ofCO2(g).



690 Chapter 15 NEL

15.215.2Qualitative Change in EquilibriumSystemsObserving the effects of varying system properties on the equilibrium of systems con-tributes greatly to our understanding of the equilibrium state. From a technologicalperspective, controlling the extent of equilibrium by manipulating properties is verydesirable because control leads to more efficient and economic processes. From a scientificperspective, observing systems at equilibrium leads to improved theories that describe,explain, and predict the nature of equilibrium, thus increasing our understanding.

Equilibrium is an area of study where, historically, technology has led science. Reactionswere first manipulated in response to some human need, although the reactions’ responseswere not explained until much later by successive theories of increasing validity. Thissection of the chapter will examine equilibrium manipulation in the same way, withempirical descriptions of equilibrium manipulation given first, followed by theoreticalexplanations of the observed results.

According to Le Châtelier’s principle, when a chemical system at equilibrium is dis-turbed by a change in a property of the system, the system always appears to react inthe direction that opposes the change, until a new equilibrium is reached (Figures 1 to3). The application of Le Châtelier’s principle involves a three-stage process: an initial equi-librium state, a shifting non-equilibrium state, and a new equilibrium state.

Le Châtelier’s principle provides a method of predicting the response of a chemicalsystem to an imposed change. Using this simple and completely empirical approach,chemical engineers could produce more of the desired products, making technologicalprocesses more efficient and more economical. For example, Fritz Haber used Le Châtelier’s principle to devise a process for the economical production of ammoniafrom atmospheric nitrogen. (See the Haber process, Chapter 8, page 325.)

Figure 1Henri Louis Le Châtelier(1850–1936), French chemist andengineer, worked in chemicalindustries. To maximize the yield ofproducts, Le Châtelier usedsystematic trial and error. Aftermeasuring properties of equilibriumstates in chemical systems, hediscovered a pattern and stated it asa generalization. This generalizationhas been supported extensively byevidence and is now considered ascientific law. By convention, it isknown as Le Châtelier’s principle.

Figure 2Fe3�(aq) � SCN�(aq) 0

FeSCN2�(aq)The test tube on the left is atequilibrium, as shown by theconstant colour of the FeSCN2�(aq)ion. The equilibrium is disturbed bythe addition of Fe3�(aq) ions. Thesystem shifts and some of theadditional Fe3+(aq) reacts toproduce more FeSCN2�(aq), thusestablishing a new equilibrium state.When the shift is complete, theconcentration of Fe3+(aq) is higherthan before (only some of the addedions react), the concentration ofSCN�(aq) is lower, and theconcentration of FeSCN2�(aq) ishigher. The higher concentration ofFeSCN2�(aq) is evident from themore intense colour of the solutionobserved in the test tube on theright. Fe3�(aq) � SCN�(aq) 0 FeSCN2�(aq)



Le Châtelier’s Principle and Concentration ChangesLe Châtelier’s principle predicts that if the addition of a reactant to a system at equilibriumincreases the concentration of that substance, then that system will undergo an equilib-rium shift forward (to the right). The effect of the shift is that, temporarily, we observe thereactant concentration decreasing, as some of the added reactant changes to products.This period of change ends with the establishment of a new equilibrium state where, onceagain, there are no observable changes. The system has changed in such a way as to opposethe change introduced. For example, the production of freon-12, a CFC refrigerant,involves the following equilibrium reaction taking place at a fixed temperature:

CCl4(l) � 2 HF(g) 0 CCl2F2(g) � 2 HCl(g)freon-12

To improve the yield of the primary product, freon-12, more hydrogen fluoride is addedto the initial equilibrium system. The additional concentration of reactant disturbs theequilibrium state and the system shifts to the right, consuming some of the addedhydrogen fluoride by reaction with carbon tetrachloride. As a result, more freon-12 is pro-duced and a new equilibrium state is reached. In chemical reaction equilibrium shifts,an imposed concentration change is normally only partially counteracted, and the finalequilibrium state concentrations of the reactants and products are usually different fromthe values at the original equilibrium state. See Figure 3 for a graphic interpretation ofthe freon-12 equilibrium shift.

Note that adding more carbon tetrachloride, CCl4(l), would have no effect on theequilibrium state in the container. This reactant is (and stays) in liquid form, so its con-centration is constant and would not be increased by increasing the amount of CCl4(l)present.

Adjusting an equilibrium state by adding and/or removing a substance is by far the mostcommon application of Le Châtelier’s principle. For industrial chemical reactions, engi-neers strive to design processes where reactants are added continuously and products arecontinuously removed, so that an equilibrium is never allowed to establish. If the reac-tion is always shifting forward, the process is always making product (and, presumably,the industry is always making money).

Section 15.2

Figure 3The reaction establishes anequilibrium that is disturbed (at thetime indicated by the vertical dottedline) by the addition of HF(g). Someof the added HF reacts, decreasingin concentration, while theconcentration of both productsincreases until a new equilibrium isestablished and concentrationsbecome constant again. Note thatthe concentration of HF(g) at thenew equilibrium is greater than atthe original equilibrium, so theimposed change is only partlycounteracted. The initial Kc valueand final Kc value are the same.

[HF][HCI]

[CCI2F2]

CCI4(I) + 2 HF(g) 0CCI2F2(g) + 2 HCI(g)

Con

cent

rati

on

(mol

/L)

Time

Purpose Design AnalysisProblem Materials Evaluation (2, 3)Hypothesis ProcedurePrediction Evidence


Equilibrium Shifts (Demonstration)In this investigation, you will be looking at two equilibrium systems:

N2O4(g) � energy 0 2 NO2(g)colourless reddish brown

CO2(g) � H2O(l) 0 H�(aq) � HCO3�(aq)

The second equilibrium system, produced by the reaction ofcarbon dioxide gas and water, is commonly found in the humanbody and in carbonated drinks. A diagnostic test is necessary todetect shifts in this equilibrium. Bromothymol blue, an acid–baseindicator, can detect an increase or decrease in the hydrogen ionconcentration in this system. Bromothymol blue turns blue whenthe hydrogen ion concentration decreases, and yellow when thehydrogen ion concentration increases.

PurposeThe purpose of this demonstration is to test Le Châtelier’sprinciple by studying two chemical equilibrium systems: theequilibrium between two oxides of nitrogen, and the equilibriumof carbon dioxide gas and carbonic acid.

ProblemHow does a change in temperature affect the nitrogen dioxide–dinitrogen tetroxide equilibrium system? How does a change inpressure affect the carbon dioxide–carbonic acid equilibriumsystem?



692 Chapter 15 NEL

The removal of a product (if the removal decreases concentration as well as chemicalamount) will also shift an equilibrium forward, producing more product to counteractthe change imposed. The freon-12 reaction can be shifted forward by removing eithergaseous product, since decreasing the amount of a gas lowers its concentration in any reac-tion container of fixed size (Figure 4).

The following equation represents the final step in the production of nitric acid:

3 NO2(g) � H2O(l) 0 2 HNO3(aq) � NO(g)

In this industrial process, nitrogen monoxide gas is removed from the chemical systemby a reaction with oxygen gas. The removal of the nitrogen monoxide causes the systemto shift to the right—some nitrogen dioxide and water react, replacing some of theremoved nitrogen monoxide. As the system shifts, more of the desired product, nitric acid,is produced.

Although equilibrium systems are important in industrial chemical production, theyare even more vital in biological systems. A particularly important biological equilibriumis that of hemoglobin (a protein in red blood cells), oxygen, and oxygenated hemo-globin.

Hb � O2 0 HbO2

As blood circulates to the lungs, the high concentration of oxygen shifts the equilib-rium to the right and the blood becomes oxygenated (Figure 5). As the blood circulatesthroughout the body, cell reactions consume oxygen. This removal of oxygen shifts theequilibrium to the left and more oxygen is released.

Collision–Reaction Theory and Concentration ChangesCollision–reaction theory provides a simple explanation of the equilibrium shift thatoccurs when a reactant concentration is increased. We assume that the number of reac-tant entities per unit volume suddenly increases, so that collisions are suddenly much morefrequent for the forward reaction. The forward reaction rate, therefore, increases sig-nificantly. Since the reverse reaction rate is not changed, the opposing rates are no longerequal, and, for a time, the difference in rates results in an observed increase ofproducts.

Of course, as the concentration of products increases, so does the reverse reactionrate. At the same time, the new forward rate decreases as reactant is consumed, untileventually the two rates become equal to each other again. The rates at the new equilibriumare faster than those at the original equilibrium, because the system now contains alarger number of particles (and, therefore, a higher concentration) in dynamic equilib-rium. If a substance is removed, causing an equilibrium shift, the explanation is similarexcept that the initial effect is to suddenly decrease either the forward or the reverse rateby decreasing the concentration.

Addition or removal of a reagent present in pure solid or pure liquid state does notchange the concentration of that substance. The reaction of condensed phases (solidsand liquids) takes place only at an exposed surface—and if the surface area exposed is changed, it is always exactly the same change in available area for both forward and reverse reaction collisions. The forward and reverse rates change by exactly the sameamount if they change at all, so equilibrium is not disturbed and no shift occurs.

Figure 5Oxygenated blood from the lungs ispumped by the heart to bodytissues. The deoxygenated bloodreturns to the heart and is pumpedto the lungs. Shifts in equilibriumoccur over and over again as oxygenis picked up in the lungs andreleased throughout the body.

air

O2

O2 O2

O2

O2

O2

air-filledsacs inlungs

heart

O2

capillary blood vesselsand body cells

Figure 4The reaction establishes anequilibrium that is disturbed (at thetime indicated by the vertical dottedline) by the removal of HCl(g). Theequilibrium shifts forward,increasing the concentration of bothproducts while decreasing HF(g)concentration, until a newequilibrium is established. The initialKc value and the final Kc value arethe same.

CCI4(I) + 2 HF(g) 0CCI2F2(g) + 2 HCI(g)

Con

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(mol

/L)

Time

[HF][HCI][CCI2F2]



Le Châtelier’s Principle and Temperature ChangesThe heat energy in a chemical equilibrium equation is treated as though it were a reac-tant or a product.

reactants � energy 0 products (endothermic in the forward direction)

reactants 0 products � energy (exothermic in the forward direction)

Heating or cooling a system adds or removes heat energy from the system. In either sit-uation, the equilibrium shifts to minimize the change. If the system is cooled, the equi-librium shifts so that more heat energy is produced. If the system is heated, the equilibriumshifts in the direction in which heat energy is absorbed.

For example, in the salt–sulfuric acid process used to produce hydrochloric acid, thesystem is heated in order to increase the percent yield of hydrogen chloride gas.

2 NaCl(s) � H2SO4(l) � energy 0 2 HCl(g) + Na2SO4(s)

Adding heat energy shifts the system to the right, absorbing some of the added energy.In the production of sulfuric acid, the key reaction step is the equilibrium represented

by the following equation. Percent yield of the product is increased at low temperature.

2 SO2(g) � O2(g) 0 2 SO3(g) � energy

Removing heat energy causes the system to shift to the right. This shift yields more sulfurtrioxide while partially replacing the heat energy that was removed.

Collision–Reaction Theory and Energy ChangesCollision–reaction theory explains the equilibrium shift (that occurs when the heatenergy of a system at equilibrium is changed) as the result of an imbalance of reactionrates. Consider the previously mentioned reaction equation—a typical exothermic reac-tion. The reaction energy is shown this time in standard ∆rH notation.

2 SO2(g) � O2(g) 0 2 SO3(g) ∆rH � �198 kJ

We explain the result of cooling the system by assuming that both forward and reversereaction rates are slower at lower temperatures, because the particles move more slowlyand collide less frequently. The reverse rate decreases more than the forward rate, how-ever. While the rates remain unequal, the observed result is the production of moreproduct and the release of more heat energy. The shift causes concentration changesthat will increase the reverse rate and decrease the forward rate until they become equalagain, at a new, lower temperature (Figure 6).

Note that industrial exothermic equilibrium reactions are often carried out at high tem-peratures, even though adding heat energy shifts the equilibrium toward reactants (lowersthe percent yield). The Haber process (Case Study, Section 8.3) is a good example. Heatenergy is added because the forward and reverse reaction rates are too slow at lowertemperatures to allow the reaction to reach equilibrium in a reasonable time. Making largequantities of the marketable product in a short time is much more important to a man-ufacturer than creating a small increase in the yield of each batch. Whenever possible,chemical engineers try to design a continuous process for an industrial reaction—onethat shifts the reaction forward by constantly adding reactants, and constantly removingproducts. This system is no longer a closed system, so the reaction never establishesequilibrium. Such an industrial reaction may run continuously for months or even years.

Section 15.2

Figure 6The reaction establishes an equilib-rium that is disturbed (at the timeindicated by the vertical dotted line)by a decrease in temperature. Theequilibrium shifts forward, increasingthe concentration of SO3 productwhile decreasing the concentrationof both reactants, until a new equi-librium is established. Because thetemperature is changed, the final Kcvalue for this example is greaterthan the initial Kc value because theshift favours the forward reaction.

Con

cent

rati

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(mol

/L)

Time

2 SO2(g) + O2(g) 0 2 SO3(g)

[SO2][O2][SO3]


BIOLOGY CONNECTION

CO2 Transport

Many biological processes dependon equilibria. For example, thetransportation of carbon dioxidedepends on the CO2 0 H2CO3equilibrium system. In the tissuesof the body where carbon dioxideis produced, the equilibrium shiftsso that more carbonic acid isformed. In the lungs, the shift is inthe reverse direction, as carbondioxide is released. You will discover other examples ofequilibria in a biology textbook.

CAREER CONNECTION

Chemical Process EngineerWhat role do chemical engineersplay in designing systems andsequences of reactions for industrialchemical production? Research theeducation requirements, jobprospects, and work assignments ofpeople in this profession.



694 Chapter 15 NEL

Le Châtelier’s Principle and Gas Volume ChangesAccording to Boyle’s law, the amount concentration of a gas in a container is inverselyproportional to the volume of the container. Since the amount concentration of a gas isdirectly proportional to its pressure, we can predict the possible effect of containervolume change on the equilibrium position of homogeneous gaseous systems. Decreasingthe volume by half doubles the concentration of every gas in the container. To predictwhether a change in pressure will affect a system’s equilibrium, you must consider thetotal chemical amount of gas reactants and the total chemical amount of gas products.For example, in the equilibrium reaction of sulfur dioxide and oxygen, three moles ofgaseous reactants produce two moles of gaseous products.

2 SO2(g) � O2(g) 0 2 SO3(g)

If the volume is decreased, the overall pressure is increased. Increased pressure causes ashift to the right, which decreases the total number of gas molecules (three moles totwo moles) and, thus, reduces the pressure. If the volume is increased, the pressure isdecreased, and the shift is in the opposite direction. A system with equal numbers ofgas molecules on each side of the equation, such as the equilibrium reaction betweenhydrogen and iodine (page 679), is not affected by a change in volume. Similarly, systemsinvolving only liquids or solids are not affected by changes in pressure. Note that addinga gas that is not involved in the equilibrium (such as an inert gas) to the container willincrease the overall pressure in the container, but will not cause a shift in equilibrium.Adding or removing gaseous substances not involved in the reaction does not change theconcentrations of the reactant and product gases.

Collision–Reaction Theory and Gas Volume ChangesWhen a system involving gaseous reactants and products is changed in volume, theresulting equilibrium shift is again explained as an imbalance of reaction rates.

2 SO2(g) � O2(g) 0 2 SO3(g) � 198 kJ

Collision–reaction theory explains the result of decreasing the volume of this system byassuming that both forward and reverse reaction rates become faster because the con-centrations of reactants and products both increase. For this example, however, the for-ward rate increases more than the reverse rate because there are more particles involvedin the forward reaction. Consequently, the increase in the total number of collisions isgreater for the forward reaction process. Again, while the rates remain unequal, theobserved result is the production of more product. The shift causes concentration changesthat gradually increase the reverse rate and decrease the forward rate until they becomeequal again (Figure 7).

Catalysts and Equilibrium SystemsCatalysts are used in most industrial chemical systems. A catalyst decreases the timerequired to reach an equilibrium position, but does not affect the final position of equi-librium. The presence of a catalyst in a chemical reaction system lowers the activationenergy for both forward and reverse reactions by an equal amount (Chapter 12), so theequilibrium establishes much more rapidly but at the same position as it would withoutthe catalyst present. Forward and reverse rates increase equally. The final equilibrium con-centrations are reached in a shorter time compared with the same, but uncatalyzed,reaction. The value of catalysts in industrial processes is to decrease the time required forequilibrium shifts created by manipulating other variables, allowing a more rapid overallproduction of the desired product.

Learning TipSince increasing thetemperature of any exothermicreaction at equilibrium alwaysshifts the equilibrium left(toward reactants), an increasein temperature must decreasethe value of Kc for such areaction. Similarly, increasingthe temperature will increasethe value of Kc for anyendothermic reaction. You havealready learned that the valueof Kc is temperature dependent.No other change imposed on asystem at equilibrium changesthe numerical value of theequilibrium constant.

Figure 7The reaction equilibrium isdisturbed by a decrease in containervolume (at the time indicated by thevertical dotted line). The equilibriumshifts forward, increasing theconcentration of SO3 whiledecreasing the concentration ofreactants, until a new equilibrium isestablished. The initial Kc value andthe final Kc value are the same.

Con

cent

rati

on

(mol

/L)

Time

[SO2][O2][SO3]

2 SO2(g) + O2(g) 0 2 SO3(g)



Section 15.2

Variables Imposed Change Response of System

concentration increase shifts to consume some of the addedreactant or product

decrease shifts to replace some of the removed reactant or product

temperature increase shifts to absorb some of the added heat energy

decrease shifts to replace some of the removedheat energy

volume increase shifts toward the side with the larger(gaseous systems (decrease in pressure) total chemical amount of gaseous only) entities

decrease shifts toward the side with the smaller(increase in pressure) total chemical amount of gaseous entities

SUMMARY Variables Affecting Chemical Equilibria

Practice1. What three types of changes shift the position of a chemical equilibrium?

2. For each of the following chemical systems at equilibrium, use Le Châtelier’s principleto predict the effect of the change imposed on the chemical system. Indicate thedirection in which the equilibrium is expected to shift. For each example, sketch thegraph of concentrations versus time, plotted from just before the change to theestablished new equilibrium. (a) H2O(l) � energy 0 H2O(g)

The container is heated.(b) H2O(l) 0 H�(aq) � OH�(aq)

A few crystals of NaOH(s) are added to the container.(c) CaCO3(s) � energy 0 CaO(s) � CO2(g)

CO2(g) is removed from the container.(d) CH3COOH(aq) 0 H�(aq) � CH3COO�(aq)

A few drops of pure CH3COOH(l) are added to the system.

3. Much methanol is produced industrially by the exothermic reaction CO(g) � 2 H2(g) 0 CH3OH(l), carried out at high pressure (5–10 MPa) andtemperature (250 °C) in the presence of several catalyst substances. Methanol is lessflammable than gasoline, and so it is a safer fuel. It is the fuel used in open-wheelChamp Car racing, and also in the Indianapolis 500.(a) State, in terms of forward and reverse reaction rates, why using a very high

pressure of the reactant gases is economically desirable for the manufacturer .(b) State in which direction a high temperature will shift this reaction equilibrium. (c) Explain why using a high temperature is desirable, in terms of the time required

for the reaction to reach equilibrium. (d) Explain, in terms of equilibrium position and equilibrium shift, why this reaction is

done in an open system, where reactants are continually added to the pressurevessel and liquid product is continually removed.


696 Chapter 15 NEL



Testing Le Châtelier’s PrincipleThe equilibria chosen for this investigation involve chemicals thatprovide coloured solutions. The investigation tests predictionsabout equilibrium shifts (made using Le Châtelier’s principle) byobserving colour changes.

In order to complete the Prediction section of the report, youmust read the Design, Materials, and Procedure carefully. Thenmake a Prediction about the result of each change made in theProcedure.

PurposeThe purpose of this investigation is to test Le Châtelier’s principleby applying stress to four different chemical equilibria.

ProblemHow does applying changes to conditions of particular chemicalequilibria affect the systems?

DesignStresses are applied to four chemical equilibrium systems andevidence is gathered to test predictions made using Le Châtelier’sprinciple. Control samples are used in all cases.


Urea Production in AlbertaCanada has a vast wealth of natural resources that we can useto produce many chemicals with an incredible variety of uses.Because plants require nitrogen for growth, the primarypurpose of some of these chemicals is for agricultural use,such as nitrogen fertilizers. You have already learned thatAlberta produces large amounts of ammonia, which can beused directly as a fertilizer. Ammonia is stored as a liquid athigh pressure, and injected directly into the soil (see Section8.3, Figure 7), but it is toxic and corrosive to human tissue,which makes it dangerous to use. Special equipment and careare required. Many food producers prefer to use a high-nitrogen, nontoxic, solid compound.

Urea (Figure 8) is another simple molecular chemical—alsoa nitrogen fertilizer—that is inexpensive, simple to produce,easy to transport, and extraordinarily useful. This chemical isused in the millions of tonnes, for applications as diverse as

• wastewater plants (for treating effluent) • air transportation (for de-icing runways) • forestry and agriculture (for fertilizer)• livestock feeding (for a protein supplement)• woodworking (for making glues and resins) • construction supplies (for making insulation)• furniture (for making particle board and chipboard)• clothing (for making certain dyes)

Until the 1900s, the common source of this chemical was staleanimal urine. The body forms this compound as its principalmeans of removing excess nitrogen. In fact, until the early1800s, it was thought that this, or any other “organic” chemicalextracted from living things, always had to be produced by aliving organism. In 1828, however, Friedrich Wöhler, in afamous classic experiment, synthesized urea in his laboratory.His work forever changed the “organic” concept of chemistry.Early in the last century, an efficient industrial method was

Case StudyCase Study

Figure 8Urea is a small and simple molecule, but a critically importantnitrogen-containing compound. It has a very high nitrogenpercentage by mass, and is very soluble in water—both veryimportant points for a compound to be useful as a plant fertilizer. Three different representations of the urea moleculeare shown here.

H NC O

H

H N

Hurea

developed, and has since been used for mass production ofthis versatile and valuable substance.

Urea is produced through the reaction of ammonia, NH3(g),with carbon dioxide, CO2(g), at high temperature and



Section 15.2

pressure. Most of the reactants exist in liquid form at thepressures used. A hot concentrated solution, containing about80% urea by mass, results from this reaction. This hot solutionis further concentrated and cooled through evaporation of thewater content, to form either granules (angular crystals) orprills (small round pellets) of white solid urea (Figure 9). Theoverall reaction may be written as

CO2(l) � 2 NH3(l) 0 NH2CONH2(aq) � H2O(l) t � 150–200 °CP � 12–20 MPa

The science and technology for urea production developedin response to a strong demand for this compound. Theprocess depends on an understanding of reaction equilibrium,and of the effect (on equilibrium) of high temperature andpressure conditions. Also key was the design and constructionof reaction containers (vessels) able to withstand highpressures and temperatures.

Sometimes, as well as building on existing scientificknowledge, new technology results in scientific advances. Atechnology, such as equipment to create very high pressures,may allow great leaps forward in science, such as thesynthesis of completely new forms of crystalline solids. Newobservations then lead to new theories and sometimes evennew laws.

Case Study Questions

1. There are two phase equilibria that shift during thereaction of carbon dioxide and ammonia. Writeequilibrium equations to represent these phase changes,and use Le Châtelier’s principle to explain how they arecontinuously being shifted, both by pressure, and also bythe effect of the chemical reaction that is occurring in thevessel.

2. Crystallization of urea from aqueous solution can beexpressed as a solubility equilibrium, according to theequation

NH2CONH2(aq) 0 NH2CONH2(s)

(a) On which side of this equation would heat energy bewritten? Explain your reasoning.

(b) Based on molecular structure and bonding theory,explain why you would expect urea to be a very highlysoluble compound.

(c) Urea granules that are bagged and sold for fertilizeruse are quite uniform in size (Figure 10). Explainwhat physical process would likely be used to separate these granules from any larger or smallergranules that form during crystallization.

Extension

3. Research Friedrich Wöhler’s classic experiment of 1828.Write the balanced equation for the reaction heperformed.

4. Find out how much urea is produced annually in Alberta,where it is produced, and its current price per tonne.Assemble your findings into an attractive presentation.



Figure 9Prills form when sprayed droplets of very hot urea solution aremade to fall through air in a huge tower, cooling andevaporating to dryness on the way down.

Figure 10Urea fertilizer is a nitrogen sourcefor crops.


698 Chapter 15 NEL


The Nitrogen Dioxide–DinitrogenTetroxide EquilibriumComplete the Prediction, Analysis, and Evaluation sections of thereport.

PurposeThe purpose of this problem is to use Le Châtelier’s principle topredict the response of an equilibrium to an introduced change inconditions.

ProblemHow does increasing the pressure affect the nitrogendioxide–dinitrogen tetroxide equilibrium?

DesignA sample of nitrogen dioxide gas is compressed in a syringe andthe intensity of the colour is used as evidence to test thePrediction.

EvidenceThe orange-brown nitrogen dioxide gas colour increases inintensity when the plunger on the syringe is depressed, and thendecreases in intensity (Figure 11). The final colour is slightly moreintense than the original colour (before moving the plunger).

LAB EXERCISE 15.C Report Checklist

Figure 112 NO2(g) 0 N2O4(g) An increase in pressure on the nitrogen dioxide–dinitrogen tetroxide equilibrium in the closed system results initially in a more intensecolour followed by a decrease in colour intensity.



Studying a Chemical Equilibrium SystemFigure 2, page 690, shows the colours of aqueous solutions ofiron(III), thiocyanate, and iron(III) thiocyanate ions. Use yourknowledge of Le Châtelier’s principle to write a Problemstatement, and then design and carry out a simple investigationto determine whether the reaction as written is exothermic orendothermic.

PurposeThe purpose of this investigation is to use Le Châtelier’s principleto solve a problem concerning the effect of an energy change onthe following equilibrium system.

Fe3�(aq) � SCN�(aq) 0 FeSCN2�(aq)almost colourless colourless red


WEB Activity

Web Quest—Poison Afloat Have you ever considered becoming a crime scene investigator? In this Web Quest, you arepromoted to Chief Chem Crime Investigator. You are presented with a body and a series ofclues… The detecting is up to you. You will have a chance to use your knowledge of chemistryto solve this puzzle and gather the evidence to unravel what happened.




Section 15.2

Section 15.2 Questions1. The following equation represents part of the industrial

production of nitric acid. Predict the direction of theequilibrium shift for each of the following changes. Explainany shift in terms of the changes in forward and reversereaction rates.

4 NH3(g) � 5 O2(g) 0 4 NO(g) � 6 H2O(g) � energy

(a) O2(g) is added to the system.(b) The temperature of the system is increased.(c) NO(g) is removed from the system.(d) The pressure of the system is increased by decreasing

the volume.

2. The following chemical equilibrium system is part of theHaber process for the production of ammonia.

N2(g) � 3 H2(g) 0 2 NH3(g) � energy

Suppose you are a chemical process engineer. Use Le Châtelier’s principle to predict five specific changes thatyou might impose on the equilibrium system to increasethe yield of ammonia.

3. In a solution of copper(II) chloride, the followingequilibrium exists:

CuCl42�(aq) � 4 H2O(l) 0 Cu(H2O)4

2�(aq) � 4 Cl�(aq)dark green blue

For the following stresses put on the equilibrium, predictthe shift in the equilibrium and draw a graph ofconcentration versus time to communicate the shift.(a) Concentrated hydrochloric acid is added.(b) Saturated aqueous silver nitrate is added, causing a

precipitation reaction.

4. Identify the nature of the changes imposed on the followingequilibrium system at the four times indicated bycoordinates A, B, C, and D (Figure 12).

5. In which of the following cases would an increase intemperature increase the percent yield at equilibrium?(a) H2O(l) 0 H2O(g)(b) N2(g) � 3 H2(g) 0 2 NH3(g) rH � �91 kJ(c) KOH(s) 0 K�(aq) � OH�(aq) � heat(d) 2 C(s) � 2 H2(g) 0 C2H4(g) rH � �53 kJ

6. Chloromethane (methyl chloride) is manufactured by“chlorinating” methane. For this reaction system atequilibrium, explain the effect of each of the imposedchanges on the position of reaction equilibrium.

CH4(g) � Cl2(g) 0 CH3Cl(g) � HCl(g) rH is negative

(a) More methane is injected into the reaction vessel.(b) The container volume is increased.(c) The temperature is lowered.(d) A catalyst is introduced into the system.

7. Ethyne (acetylene) is manufactured by a high-temperaturecombustion of methane, using a large excess of methane.For this endothermic reaction system at equilibrium, explainthe effect of each of the imposed changes on the value ofthe equilibrium constant.

6 CH4(g) � O2(g) 0 2 C2H2(g) � 10 H2(g) � 2 CO(g)

(a) More methane is injected into the reaction vessel.(b) The container volume is decreased.(c) The temperature is lowered.(d) A catalyst is introduced into the system.

Extension

8. In a deep lake or in the ocean, the pressure that a humanconsiders “normal” has doubled by the time a diver reachesa depth of 10 m, and increases by about one atmospherefor every extra 10 m, to a maximum of about 100 MPa at thedeepest points in Earth’s oceans. This fact is of majorconcern to scuba divers for several reasons. Pressure insidea scuba diver’s lungs must constantly be adjusted to equaloutside water pressure, otherwise the lungs could collapseupon diving, and could explode upon rising. In fact, thedevelopment of the pressure regulator (Figure 13) was thetechnology that made scuba diving possible.

The gases in a diver’s lungs are dissolved to some extentin the blood that passes through. Pressure changes canchange several solubility equilibria, with some seriouseffects. Research the gases dissolved in human blood, andwhat equilibrium shifts cause the diving conditions callednitrogen narcosis, and the “bends.”

Figure 12Graph showing four disturbances to an equilibrium system

Figure 13The mouthpiece pressurefrom a diver’s tankautomatically adjusts forchanges in depth.

A B C D

C2H6

C2H4

H2

C2H4(g) + H2(g) 0 C2H6(g) + energy

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Chapter 15 INVESTIGATIONS

The Extent of a Chemical Reaction

In Chapter 8, you performed experiments that produced evidence that reactions are quantitative. In a quantitativereaction, the limiting reagent is completely consumed. Toidentify the limiting reagent, you can test the final reactionmixture for the presence of the original reactants. For example,in a diagnostic test, you might try to precipitate ions fromthe final reaction mixture that were present in the original reactants.

PurposeThe purpose of this investigation is to test the validity of theassumption that chemical reactions are quantitative.

ProblemWhat are the limiting and excess reagents in the chemicalreaction of selected quantities of aqueous sodium sulfate andaqueous calcium chloride?

DesignSamples of sodium sulfate solution and calcium chloridesolution are mixed in different proportions and the final mix-ture is filtered. Samples of the filtrate are tested for the pres-ence of excess reagents, using the following diagnostic tests:

Soluble barium compounds are toxic. Remember towash your hands before leaving the laboratory.

Barium nitrate in solid form is a strong oxidizingsubstance.

• If a few drops of Ba(NO3)2(aq) are added to the filtrateand a precipitate forms, then sulfate ions are present.Ba2�(aq) � SO4

2�(aq) → BaSO4(s)• If a few drops of Na2CO3(aq) are added to the filtrate and

a precipitate forms, then calcium ions are present.Ca2�(aq) � CO3

2�(aq) → CaCO3(s)

Materialslab aproneye protection25 mL of 0.50 mol/L

CaCl2(aq)25 mL of 0.50 mol/L

Na2SO4(aq)1.0 mol/L Na2CO3(aq) in

dropper bottlesaturated Ba(NO3)2(aq) in

dropper bottle

two 50 mL or 100 mLbeakers

two small test tubes10 mL or 25 mL graduated

cylinderfiltration apparatusfilter paperwash bottlestirring rod


INVESTIGATION 15.1 Report Checklist

Equilibrium Shifts (Demonstration)

In this investigation, you will be looking at two equilibrium systems:

N2O4(g) � energy 0 2 NO2(g)colourless reddish brown

CO2(g) � H2O(l) 0 H�(aq) + HCO3�(aq)

The second equilibrium system, produced by the reaction ofcarbon dioxide gas and water, is commonly found in thehuman body and in carbonated drinks. A diagnostic test isnecessary to detect some shifts in this equilibrium.Bromothymol blue, an acid–base indicator, can detect anincrease or decrease in the hydrogen ion concentration in

this system. Bromothymol blue turns blue when the hydrogenion concentration decreases, and yellow when the hydrogenion concentration increases.

Purpose The purpose of this demonstration is to test Le Châtelier’sprinciple by studying two chemical equilibrium systems: theequilibrium between two oxides of nitrogen, and the equi-librium of carbon dioxide gas and carbonic acid.



700 Chapter 15 NEL


Chapter 15

ProblemHow does a change in temperature affect the nitrogendioxide–dinitrogen tetroxide equilibrium system? How doesa change in pressure affect the carbon dioxide–carbonic acidequilibrium system?

Materialslab aproneye protectiontwo NO2(g)/N2O4(g) sealed flaskscarbon dioxide–hydrogen carbonate ion equilibrium

mixture (pH � 7)bromothymol blue indicator in dropper bottlesmall syringe with needle removed (5 to 50 mL)solid rubber stopper to seal end of syringebeaker of ice–water mixturebeaker of hot water

Procedure1. Place the sealed NO2(g)/N2O4(g) flasks in hot and

cold water baths (Figure 1) and record yourobservations.

2. Place two or three drops of bromothymol blueindicator in the carbon dioxide–hydrogen carbonateion equilibrium mixture.

3. Draw some of the carbon dioxide–hydrogencarbonate ion equilibrium mixture into the syringe,and then block the end with a rubber stopper.

4. Slowly move the syringe plunger and record yourobservations.

Figure 1Each of these flasks contains an equilibriummixture of dinitrogentetroxide and nitrogendioxide. Shifts inequilibrium can be seenwhen one of the flasks isheated or cooled.

hot water ice waterNO2(g) N2O4(g)0

Be careful with the flasks containing nitrogendioxide: this gas is highly toxic. Use in a fume hoodin case of breakage.

INVESTIGATION 15.2 continued

Testing Le Châtelier’s Principle

The equilibria chosen for this investigation involve chemi-cals that provide coloured solutions. This investigation testspredictions about equilibrium shifts (made using Le Châtelier’sprinciple) by observing colour changes. For example, if thecolour in Reaction 3 becomes more intensely red, the equi-librium has shifted right to produce more FeSCN2�(aq) ions,increasing their concentration. See Table 1.

In order to complete the Prediction section of the report,you must read the Design, Materials, and Procedure care-fully. Then, make a Prediction about the result of each changemade in the Procedure.

Purpose The purpose of this investigation is to test Le Châtelier’s prin-ciple by applying stress to four different chemical equilibria.

Table 1 Solution Colours

Ion Colour

CoCl42�(alc) blue

Co(H2O)62�(alc) pink

H2Tb(aq) red

HTb�(aq) yellow

Tb2�(aq) blue

Fe3�(aq) pale yellow

SCN�(aq) colourless

FeSCN2�(aq) red

Cu(H2O)42�(aq) pale blue

Cu(NH3)42�(aq) deep blue





702 Chapter 15 NEL

ProblemHow does applying stresses to particular chemical equilibriaaffect the systems?

Part I

CoCl42�(alc) � 6 H2O(alc) 0

Co(H2O)62�(alc) � 4 Cl�(alc) � energy

Part II

H2Tb(aq) 0 H�(aq) � HTb�(aq)

HTb�(aq) 0 H�(aq) � Tb2�(aq)

Part III

Fe3�(aq) � SCN�(aq) 0 FeSCN2�(aq)

Part IV

Cu(H2O)42�(aq) � 4 NH3(aq) 0

Cu(NH3)42�(aq) � 4 H2O(l)

DesignStresses are applied to four chemical equilibrium systems andevidence is gathered to test predictions made using Le Châtelier’s principle. Control samples are used in all cases.For example, before adding sodium hydroxide to a new equi-librium solution, split the solution into two samples in orderto have a control sample for colour comparison.

Part I Cobalt(II) ComplexesWater, saturated silver nitrate, and heat are added to, and heatis removed from, samples of the provided equilibrium mix-ture. Note: This reaction equilibrium is in solution using analcohol solvent, shown as (alc), so the concentration of wateris a variable in this system.

Part II Thymol Blue IndicatorHydrochloric acid and sodium hydroxide are added to sam-ples of the provided equilibrium mixture.

Part III Iron(III)–Thiocyanate EquilibriumIron(III) nitrate, potassium thiocyanate, and sodiumhydroxide are added to samples of the provided equilibriumsystem.

Part IV Copper(II) ComplexesAqueous ammonia and hydrochloric acid are added to sam-ples of the provided equilibrium mixture.

Materialslab apron eye protection100 mL beaker large waste beaker6 to 12 small test tubes test-tube rackdistilled water crushed ice

hot water bathcobalt(II) chloride equilibrium mixture in ethanol dropper bottles containing

0.2 mol/L AgNO3(aq) thymol blue indicator 0.1 mol/L HCl(aq) 0.1 mol/L NaOH(aq) iron(III) thiocyanate equilibrium mixture0.2 mol/L Fe(NO3)3(aq) 0.2 mol/L KSCN(aq) 6.0 mol/L NaOH(aq) 0.1 mol/L CuSO4(aq)1.0 mol/L NH3(aq) 1.0 mol/L HCl(aq)

The chemicals used may be corrosive orpoisonous, and may cause other toxiceffects. Exercise great care when using thechemicals and avoid skin and eye contact.Immediately rinse the skin if there is anycontact. If any chemicals get in the eyes,flush eyes for a minimum of 15 min andinform the teacher. Ethanol is flammable.Make sure there are no open flames in thelaboratory when using the ethanol solutionof cobalt(II) chloride.

ProcedurePart I Cobalt(II) Complexes

1. Obtain 25 mL of the equilibrium mixture with thecobalt(II) chloride complex ions.

2. Place a small amount of the mixture into each of fivesmall test tubes. Use the fifth test tube as a control forcomparison purposes.

3. Add drops of water to one test tube until a change isevident. Record the evidence.

4. Add drops of 0.2 mol/L silver nitrate to another testtube and record the evidence.

5. Heat another equilibrium mixture in a hot water bathand record the evidence.

6. Cool an equilibrium mixture in an ice bath andrecord the evidence.

Part II Thymol Blue Indicator

7. Add about 5 mL of distilled water to each of twosmall test tubes.

8. Add 1 to 3 drops of thymol blue indicator to thewater in each test tube to obtain a noticeable colour.Use one test tube of solution as a control.

9. Add drops of 0.1 mol/L HCl(aq) to the experimentaltest tube to test for the predicted colour changes.

10. Add drops of 0.1 mol/L NaOH(aq) to the same tubeto test for the predicted colour changes.



Chapter 15

Part III Iron(III)–Thiocyanate Equilibrium

11. Obtain about 20 mL of the iron(III)–thiocyanateequilibrium mixture.

12. Place about 5 mL of the equilibrium mixture in eachof three test tubes. Use one test tube as a control.

13. Add drops of Fe(NO3)3(aq) to one test tube until achange is evident.

14. Add drops of 6.0 mol/L NaOH(aq) to this newequilibrium mixture until a change occurs. (Iron(III)hydroxide has very low solubility.)

15. Add drops of KSCN(aq) to another equilibriummixture until a change is evident.

INVESTIGATION 15.3 continued Part IV Copper(II) Complexes

16. Obtain 2 mL of 0.1 mol/L CuSO4(aq) in a small testtube.

17. Add three drops of 1.0 mol/L NH3(aq) to establishthe equilibrium mixture.

18. Add more 1.0 mol/L NH3(aq) to the aboveequilibrium mixture and record the results.

19. Add 1.0 mol/L HCl(aq) to the equilibrium mixturefrom step 18 and record the results.

Dispose of the chemicals as directed by your teacher. Identifyeach as toxic (to be collected) or nontoxic (disposable in thesink).Ensure that all equipment and surfaces are clean and wash your hands thoroughly before leaving the laboratory.

Studying a Chemical EquilibriumSystem

Figure 1 shows the colours of aqueous solutions of iron(III),thiocyanate, and iron(III) thiocyanate ions. Use your knowl-edge of Le Châtelier’s principle to write a Problem statement,and then design and carry out a simple investigation to deter-mine whether the reaction as written is exothermic orendothermic.

Purpose The purpose of this investigation is to use Le Châtelier’s prin-ciple to solve a problem concerning the effect of an energychange on the following equilibrium system.

Fe3�(aq) � SCN�(aq) 0 FeSCN2�(aq)almost colourless colourless red

Iron(III) compounds are irritants. Thiocyanate ionsolutions are toxic. Avoid skin and eye contact. Ifthere is any skin or eye contact, immediately rinsewith plenty of water. Flush the eyes for at least 15 min and inform the teacher.




Figure 1The iron–thiocyanate reaction


Chapter 15 SUMMARY

704 Chapter 15 NEL

Outcomes

Knowledge

• define equilibrium and state the criteria that apply to achemical system in equilibrium (15.1)

• identify, write, and interpret chemical equations for systemsat equilibrium (15.1, 15.2)

• predict, qualitatively, using Le Châtelier’s principle, shifts inequilibrium caused by changes in temperature, pressure,volume, concentration, or the addition of a catalyst, anddescribe how these changes affect the equilibrium constant(15.2)

• define Kc and write equilibrium law expressions for givenchemical equations, using lowest whole-number coefficients(15.1)

• calculate equilibrium constants and concentrations forhomogeneous systems when concentrations at equilibriumare known, when initial concentrations and one equilibriumconcentration are known, and when the equilibrium constantand one equilibrium concentration are known (15.1)

STS

• state that the goal of science is knowledge about the naturalworld (15.1, 15.2)

• list the characteristics of empirical and theoreticalknowledge (15.2)

• state that a goal of technology is to solve practical problems(15.2)

Skills

• initiating and planning: predict variables that can cause ashift in equilibrium (15.2); design an experiment to showequilibrium shifts (15.2); describe procedures for safehandling, storage, and disposal of materials used in thelaboratory (15.1, 15.2)

• performing and recording: perform an experiment to test,qualitatively, predictions of equilibrium shifts (15.2)

• analyzing and interpreting: write the equilibrium lawexpression for a given equation (15.1); analyze, qualitatively,the changes in concentrations of reactants and productsafter an equilibrium shift (15.2); interpret data from a graphto determine when equilibrium is established, and determinethe cause of a stress on the system (15.2)

• communication and teamwork: work collaboratively inaddressing problems and communicate effectively (15.1, 15.2)

closed systemequilibriumphase equilibriumsolubility equilibriumchemical reaction equilibriumdynamic equilibrium

forward reactionreverse reactionICE tableequilibrium constant, Kcequilibrium law

15.2Le Châtelier’s principle equilibrium shift

Key EquationsFor the reaction a A � b B 0 c C � d D,

the equilibrium law is Kc �[C]c[D]d

�[A]a[B]b

Key Terms

15.1

MAKE a summary

1. Make a concept map, beginning with the word“Equilibrium” in the centre of a page. Link all of the KeyTerms from this chapter, together with points of yourown, to explain and illustrate how connections amongthese terms include the equilibrium law expression, ICEtable format, Le Châtelier’s principle, and points fromthe section Summaries.

2. Refer back to your answers to the Starting Pointsquestions at the beginning of this chapter. How hasyour thinking changed?

Go To

The following components are available on the Nelson Web site. Follow the links for Nelson Chemistry Alberta 20–30.

• an interactive Self Quiz for Chapter 15

• additional Diploma Exam-style Review questions

• Illustrated Glossary

• additional IB-related material

There is more information on the Web site wherever you seethe Go icon in this chapter.


Plague and the Little Ice AgeA Dutch researcher discusses his theory that the atmosphericcarbon dioxide equilibrium shifted when 40% of Europe’shuman population was killed by a plague in the 14th century.Could this have caused “the little ice age”?


EXTENSION +



Chapter 15Chapter 15 REVIEW

Many of these questions are in the style of the DiplomaExam. You will find guidance for writing Diploma Exams inAppendix H. Exam study tips and test-taking suggestionsare on the Nelson Web site. Science Directing Words usedin Diploma Exams are in bold type.

DO NOT WRITE IN THIS TEXTBOOK.

Part 11. A “closed” system for a chemical equilibrium means that

A. the reaction container must be solid and sealed, with afixed volume

B. no substance involved in the equilibrium must be ableto enter or leave

C. pressure and temperature in the reaction vessel mustbe constant

D. no chemical of any kind may be added to the reactionvessel

2. The possible equilibrium that is not dynamic is the onebetween A. the liquid and gas phases of octaneB. chromate and dichromate ions in aqueous solutionC. the downward force of gravity exerted by Earth on an

object, and the upward force of a balance exerted onthe object

D. the oxygen dissolved in water in a lake and the oxygendissolved in nitrogen in the atmosphere

3. In a reaction, 4.22 mol of product has formed but there areno more visible signs of change. Calculations show that asmuch as 6.00 mol of product could have formed from thechemical amounts of reactants used. The percent yield is__________ %.

4. Which is the correct form of the equilibrium law for thisreaction, as the equation is written?

A. Kc �

B. Kc � [SO2(g)]2[O2(g)][SO3(g)]2

C. Kc �

D. Kc �

5. The imposed condition that does not shift equilibriumtoward the product is A. adding the vanadium pentoxide catalystB. decreasing the temperatureC. decreasing the container volumeD. adding oxygen to the system container

6. If the equilibrium constant, Kc, for the above oxidationreaction at a given temperature is 2.4 � 103, then theconstant for the reverse reaction (written as thedecomposition of sulfur trioxide) is A. 2.4 � 103

B. 4.2 � 10�4

C. 2.4 � 10�3

D. 4.2 � 104

[SO3(g)]2

��[SO2(g)]2[O2(g)]

[SO2(g)]2[O2(g)]��

[SO3(g)]2

[SO2(g)]2[O2(g)]��

[SO3(g)]

Use this information to answer questions 4 to 6.

Sulfuric acid is the most common commercial acid, withmillions of tonnes produced each year (Figure 1). The secondstep in the “contact” process for industrial production ofsulfuric acid involves the oxidation of sulfur dioxide gascatalyzed by contact with V2O5(s) powder. The reactionequation is

2 SO2(g) � O2(g) 0 2 SO3(g) ∆H° � �198 kJ

Figure 1Most sulfuric acid is produced in plants such as this one by thecontact process, which includes two exothermic combustionreactions. Sulfur reacts with oxygen, forming sulfur dioxide; then,sulfur dioxide, in contact with a catalyst, reacts with oxygen,forming sulfur trioxide. Sulfur trioxide and water form sulfuric acid.


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706 Chapter 15 NEL

7. If excess copper reacts in a solution of silver nitrate, it iscorrect to state thatA. the reaction is not considered quantitativeB. the equilibrium constant at SATP will have a numerical

value between 1 and 100C. water is not written in the equilibrium law expression

because it is a spectator speciesD. silver nitrate is not written in the equilibrium law

expression because it is a species with a constantconcentration

8. This reaction is done in a laboratory autoclave (a stainlesssteel pressure vessel) and allowed to reach equilibrium.More methane is then injected into the autoclave. We canpredict that, when a new equilibrium is reached (at thesame temperature), the concentration of every reagent inthe equation will have increased, except that ofA. methane B. waterC. carbon dioxide D. hydrogen

9. When this gaseous reaction system at equilibrium isdisturbed by heating the autoclave, we theorize that A. both forward and reverse reaction rates increase, but

the forward rate increases moreB. the forward reaction rate does not change, but the

reverse reaction rate increasesC. the reverse reaction rate does not change, but the

forward reaction rate increasesD. both reaction rates increase equally, so the equilibrium

position is unchanged

10. A test reaction is done starting with only methane andexcess water in the autoclave. If the initial concentration ofmethane is 0.110 mol/L, and the methane concentration atequilibrium is 0.010 mol/L, then the equilibriumconcentration of hydrogen is __________ mol/L.

Part 211. Define chemical equilibrium empirically.

12. What main idea explains chemical equilibrium?

13. What phrase is used to describe a reaction equilibrium inwhich the proportion of reactants to products is quite high?

14. Describe and explain a situation in which a carbonatedsoft drink is in (a) a non-equilibrium state(b) an equilibrium state

15. Predict whether adding a catalyst affects a state ofequilibrium. What does the catalyst do?

16. For each of the following descriptions, write a chemicalequation for the system at equilibrium. Communicate theposition of the equilibrium with equilibrium arrows. Thenwrite a mathematical expression of the equilibrium law foreach chemical system. (a) A combination of low pressure and high temperature

provides a percent yield of less than 10% for theformation of ammonia in the Haber process.

(b) At high temperatures, the formation of water vapourfrom hydrogen and oxygen is quantitative.

(c) The reaction of carbon monoxide with water vapour toproduce carbon dioxide and hydrogen has a percentyield of 67% at 500 °C.

17. Scientists and technologists are particularly interested inthe use of hydrogen as a fuel. Interpret this reactionequation by predicting the relative proportions of reactantsand products in this system at equilibrium.

2 H2(g) � O2(g) 0 2 H2O(g) Kc � 1 � 1080 at SATP

18. Write the equilibrium law expression for a saturatedsolution of oxygen in water.

19. If the gas in the closed system is pure oxygen, the solubilityis higher than it is if the gas is air. Express the solubility ofpure oxygen in water at SATP as an amount concentration.

20. Express the concentration of pure oxygen gas at SATP as anamount concentration. (Recall that the molar volume ofgases at SATP is 24.8 L/mol.)

21. Use the answers to the previous two questions todetermine the value of Kc for an equilibrium of pureoxygen gas in contact with its saturated solution at SATP.

22. Predict whether the value of Kc for this equilibrium will bedifferent for an equilibrium of air in contact with water at 25 °C. Predict which system condition changes will, andwhich will not, change the value of an equilibrium constant.


Consider the following reaction to produce hydrogen, done asa first step in the industrial process to make ammonia.Methane (from natural gas) reacts with steam over a nickelpowder catalyst. The reaction equation is

CH4(g) � 2 H2O(g) � heat energy 0 CO2(g) � 4 H2(g)


The solubility of pure oxygen in contact with liquid water atSATP is very low: only about 42 ppm, or 42 mg/L. The solubilityequilibrium when water is in contact with air (21% oxygen) at SATP, is even lower: about 8.7 mg/L. The equilibriumequation is

O2(g) 0 O2(aq)

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Chapter 15

23. In terms of the equilibrium law, explain why more oxygendissolves in water when the gas above it is pure oxygenthan when the gas above it is air.

24. The quality of surface water in lakes and streams (Figure 2)is of critical importance to society. Dissolved oxygencontent of the surface fresh water in Canada averages 10 ppm. Explain what stress is placed upon fish and otherorganisms in streams, lakes, and wetlands if climate changeincreases the average temperature of the water.

25. In many processes in industry, engineers try to maximizethe yield of a product. Outline how concentration can bemanipulated in order to increase the yield of a product.

26. In a container at high temperature, ethyne (acetylene) andhydrogen react to produce ethene (ethylene). No ethene isinitially present. Later, at equilibrium, the concentration ofethene is 0.060 mol/L.

C2H2(g) � H2(g) 0 C2H4(g)

The initial concentrations of both acetylene and hydrogenare 1.00 mol/L. (a) Use an ICE table to determine the equilibrium

constant.(b) Sketch a reaction progress graph to show the change

in concentration values over time, from the beginningof the reaction to equilibrium.

27. H2(g) � Br2(g) 0 2 HBr(g) Kc � 12.0 at t °C(a) 8.00 mol of hydrogen and 8.00 mol of bromine are

added to a 2.00 L reaction container. Construct an ICEtable and use it to predict the concentrations atequilibrium.

(b) 12.0 mol of hydrogen and 12.0 mol of bromine areadded to a 2.00 L reaction container. Construct an ICEtable and use it to predict the concentrations atequilibrium.

28. CO(g) � H2O(g) 0 CO2 (g) � H2(g) Kc � 4.00 at 900 °C

In a container, carbon monoxide and water vapour react toproduce carbon dioxide and hydrogen. The equilibriumconcentrations are [H2O(g)] � 2.00 mol/L, [CO2(g)] � 4.00 mol/L, and [H2(g)] � 2.00 mol/L. Determine the equilibrium concentration of carbonmonoxide.

29. Write a statement of Le Châtelier’s principle.

30. What variables are commonly manipulated to shift achemical equilibrium system?

31. Describe how a change in volume of a closed system containing a gaseous reaction at equilibrium affects thepressure of the system.

32. In a sealed container, nitrogen dioxide is in equilibrium withdinitrogen tetroxide.

2 NO2(g) 0 N2O4(g) Kc � 1.15, t � 55 °C

(a) Write the mathematical expression for the equilibriumlaw applied to this chemical system.

(b) If the equilibrium concentration of nitrogen dioxide is0.050 mol/L, predict the concentration of dinitrogentetroxide.

(c) Predict the shift in equilibrium that will occur whenthe concentration of nitrogen dioxide is increased.

33. Predict the shift in the following equilibrium systemresulting from each of the following changes:

4 HCl(g) � O2(g) 0 2 H2O(g) � 2 Cl2(g) � 113 kJ

(a) an increase in the temperature of the system(b) a decrease in the system’s total pressure due to an

increase in the volume of the container(c) an increase in the concentration of oxygen(d) the addition of a catalyst

34. Chemical engineers use Le Châtelier’s principle to predictshifts in chemical systems at equilibrium resulting fromchanges in the reaction conditions. Predict the changesnecessary to maximize the yield of product in each of thefollowing industrial chemical systems: (a) the production of ethene (ethylene)

C2H6(g) � energy 0 C2H4(g) � H2(g)(b) the production of methanol

CO(g) � 2 H2(g) 0 CH3OH(g) � energy

Figure 2The Bow River begins as a cold, glacier-fed mountainstream with a higher-than-average oxygen content, and isworld famous for its trout fly fishery. It also supplies the cityof Calgary with water for a million people daily, as well asproviding water for agriculture in southern Alberta.

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35. Apply Le Châtelier’s principle to predict whether, and inwhich direction, the following established equilibriumwould be shifted by the change imposed:

2 CO(g) � O2(g) 0 2 CO2(g) � heat energy

(a) temperature is increased(b) vessel volume is increased(c) oxygen is added(d) platinum catalyst is added(e) carbon dioxide is removed

36. For each example, predict whether, and in which direction,an established equilibrium would be shifted by the changeimposed. Explain any shift in terms of changes in forwardand reverse reaction rates. (a) Cu2�(aq) � 4 NH3(g) 0 Cu(NH3)4

2�(aq) CuSO4(s) is added

(b) CaCO3(s) � energy 0 CaO(s) � CO2(g) temperature is decreased

(c) Na2CO3(s) � energy 0 Na2O(s) � CO2(g) sodium carbonate is added

(d) H2CO3(aq) � energy 0 CO2(g) � H2O(l) vessel volume is decreased

(e) KCl(s) 0 K�(aq) � Cl�(aq)AgNO3(s) is added

(f) CO2(g) � NO(g) 0 CO(g) � NO2(g) vessel volume is increased

(g) Fe3�(aq) � SCN�(aq) 0 FeSCN2�(aq) Fe(NO3)3(s) is added

37. Predict in which of the following equilibria a decrease intemperature favours the forward reaction.(a) Br2(l) 0 Br2(g)(b) N2(g) � 3 H2(g) 0 2 NH3(g) rH is negative(c) LiCl(s) 0 Li�(aq) � Cl�(aq) � heat(d) 6 C(s) � 3 H2(g) 0 C6H6(l) rH � �49 kJ(e) CaCO3(s) � energy 0 CaO(s) � CO2(g)

38. The scrubbing operation involves a gas mixture of low-molar-mass hydrocarbons, carbon dioxide, and hydrogensulfide. This mixture of gases is injected into the scrubberunit at high pressure, with the scrubber solution at about 40 °C. For hydrogen sulfide, first the toxic gas dissolves

H2S(g) 0 H2S(aq) (negative rH)

which is followed immediately by the chemical reaction

(C2H4OH)2NH(aq) + H2S(aq) 0(C2H4OH)2NH2

+(aq) + HS–(aq) (negative rH) (a) Draw a structural formula for diethanolamine.(b) Explain, using Le Châtelier’s principle, how the

relatively low temperature and high pressure act tohelp the scrubber solution “absorb” toxic hydrogensulfide.

39. When the unabsorbed hydrocarbon gases are removedfrom the scrubber unit, the sulfur atoms remain behind,trapped in solution as hydrogen sulfide ions. In the nextprocess step, the scrubber solution is “regenerated”: theabsorbed toxic compound is now emitted from solution andremoved from the reaction vessel. Use Le Châtelier’sprinciple to describe how conditions should be altered inthe scrubber vessel, to shift equilibrium positions to makethis process as efficient as possible.

708 Chapter 15 NEL

Use this information to answer questions 38 and 39.

Alberta’s petroleum industry has a chronic problem—all fossilfuels found in the province contain some sulfur. If notremoved, this sulfur will react upon burning to release SO2(g).Sulfur dioxide is very irritating to lung tissue and is highly corrosive; thus, it is a major contributor to air pollution, acidrain, and respiratory disease. Furthermore, sulfur impuritiesmay damage the fuel injection and anti-pollution systems ofmodern internal combustion engines if not removed fromgasoline and diesel fuels. For these reasons, recent Canadianlegislation requires that sulfur content in diesel fuels sold foron-road use must be less than 15 ppm (15 mg/kg) by July,2006.

In a refinery or bitumen upgrader, the sulfur is first removedfrom fossil fuel feedstock by cracking and/or hydrogenation,resulting in reaction of the sulfur to produce (extremely toxic)H2S(g). Standard industry technology to remove hydrogen sulfide gas from petrochemical gas stream mixtures involvesthe use of an amine scrubber unit, in a two-step process thatdepends on two kinds of equilibrium. For simplicity, assumethat an amine scrubber reaction vessel contains a 25%aqueous solution of diethanolamine (C2H4OH)2NH(aq), whichapproximates the actual solution used in the various oil sandsplants in Alberta.

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Chapter 15


40. When carbon dioxide gas “dissolves” in water, the processis more correctly thought of as being an exothermicchemical reaction with water, to form aqueous carbonicacid. Soft drink beverages are “carbonated” in this way, athigh pressure. Write and balance an equilibrium equationfor this reaction.

41. Carbonic acid, H2CO3(aq), then reacts exothermically withbasic aqueous diethanolamine in essentially the same waythat hydrosulfuric acid, H2S(aq), does. Write and balancethe equilibrium equation for this reaction, and use it toexplain whether the process conditions for scrubbingH2S(g) should work to remove CO2(g) as well.

Extension

42. Work cooperatively to research, assemble, and present amore complete and accurate summary of the operation of atypical Alberta industrial amine “scrubber” system. Your presentation should include• a description of the role of chemical equilibrium in the

system• information on applications of this process throughout

Alberta• the use of the best features of any available word

processing or slideshow software

43. A halogen light bulb contains a tungsten (wolfram)filament, W(s), in a mixed atmosphere of a noble gas and ahalogen; for example, Ar(g) and I2(g) (Figure 3). Theoperation of a halogen lamp depends, in part, on theequilibrium system

W(s) � I2(g) 0 WI2(g)

Research and describe the role of temperature in theoperation of a halogen lamp. For example, how is itpossible for a halogen lamp to operate with the filament at2700 °C when the tungsten normally would not last verylong at this high temperature? Why is such a hightemperature desirable?

44. When the Olympic Games were held in Mexico in 1968,many athletes arrived early to train in the higher altitude(2.3 km) and lower atmospheric pressure of Mexico City.Exertion at high altitudes, for people who are notacclimatized, may make them dizzy or “lightheaded” fromlack of oxygen. Explain this observation. Your explanationshould include• the theory of dynamic equilibrium• Le Châtelier’s principle• a description of how people who normally live at high

altitudes are physiologically adapted to their reduced-pressure environment


Figure 3Halogen light bulbs are more efficient than ordinaryincandescent bulbs, but they burn so hot they may requirespecial fixtures, and must be treated with extra care.


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In this chapter

Career Connection: Environmental Engineer; Chemistry Researcher; Microbiologist

1616 Equilibrium inAcid–Base Systems

chapter

Equilibrium inAcid–Base Systems

Exploration: Salty Acid orAcidic Salt?

Lab Exercise 16.A: TheChromate–DichromateEquilibrium

Web Activity: EdgarSteacie

Investigation 16.1:Creating an Acid–BaseStrength Table

Lab Exercise 16.B:Predicting Acid–BaseEquilibria

Web Activity: PoolChemistry

Lab Exercise 16.C:Aqueous Bicarbonate IonAcid–Base Reactions

Investigation 16.2: TestingBrønsted–Lowry ReactionPredictions

Lab Exercise 16.D:Creating an Acid–BaseTable

Case Study: ChangingIdeas on Acids andBases—The Evolution of aScientific Theory

Web Activity: Titration ofPolyprotic Acids andBases

Biology Connection:Homeostasis

Web Activity: Preparationof Buffer Solutions

Web Activity: MaudMenten

Investigation 16.3: Testinga Buffer Effect

710 Chapter 16 NEL

Answer these questions as best you can with your current knowledge. Then, usingthe concepts and skills you have learned, you will revise your answers at the end ofthe chapter.

1. How can some substances neutralize both acids and bases?

2. Can acid–base reactions and their products be predicted? Explain.

3. Can pH curves for titrations of weak acids and weak bases predict equivalence pointsas strong acid–strong base pH curves do? Explain.

4. How is the buffering of some medications related to stomach fluid acidity?

STARTING Points

The nature of science involves constant questioning and testing of theories—and so itis with theories about acids and bases. A great many chemical reaction systems involveacids in some way, including the one that begins the decomposition (digestion) of thefood you eat, and, thus provides you with energy. Many other types of aqueous reac-tion systems have rates that are easily controlled by adjusting the level of acidity. Becausesuch systems are commonly found both in nature and in industry, scientists seek towork with the most complete and successful acid–base concepts—which in turn meansseeking new ways to test previously accepted theories. Theories that do not describe,explain, and predict the chemistry of acids and bases well enough will initially be restrictedto only those situations where they work. As a result of further testing, scientists willeither revise the theory, or replace it altogether.

This chapter presents new hypotheses, evidence, and analyses to help you develop amore comprehensive understanding of both the aqueous reaction environment, andthe activity of acids and bases within that environment. Your knowledge of chemicalequilibrium (Chapter 15) allows you to explore these questions from a new perspec-tive, and, in turn, will allow you to form a more complete and less restrictive theory ofacids and bases. In fact, few topics in chemistry illustrate this scientific principle ofongoing theory testing and development so well.

These underlying principles—that theories must be supported by evidence, and thatunderstanding is increased by always questioning and testing existing knowledge—arethe basis of the uniquely productive “way of knowing” about the natural world that wecall science. To these principles, and to the enormous accumulation of knowledge theyhave made possible, we owe most aspects of our present technological civilization.


Equilibrium in Acid–Base Systems 711NEL

Exploration Salty Acid or Acidic Salt?

In Chapter 5, you learned about saturated solutions as examplesof dynamic equilibrium, and prepared a saturated sodiumchloride solution. The strong acid, HCl(aq), shares half of itschemical formula with NaCl(aq). How might the two solutionsinteract?

Materials: two 10 mL graduated cylinders; saturated sodiumchloride solution, NaCl(aq); concentrated hydrochloric acid,HCl(aq)

(a) Write and balance a chemical reaction equation for anyreaction you can predict when sodium chloride solution andconcentrated hydrochloric acid are mixed.

(b) Write an equation to express the nature of a saturatedsodium chloride aqueous solution as a dynamic equilibrium.

(c) Write the equilibrium law expression for a saturatedaqueous sodium chloride solution. Identify the substancewith a constant concentration.

Hydrochloric acid is corrosive. Wear appropriateeye protection, lab gloves, and a lab apron.

• Measure approximately 8 mL of saturated sodium chloridesolution into the larger graduated cylinder.

• Measure approximately 2 mL of concentrated hydrochloricacid into the smaller graduated cylinder.

• Pour the concentrated hydrochloric acid into the cylindercontaining the saturated sodium chloride solution. Record your observations.

• Dispose of all substances down the sink, using lots of water.(d) Explain what has apparently happened to the saturated

NaCl(aq) equilibrium. (e) Explain which ion concentration in the large cylinder

must have changed, whether it was increased ordecreased, and what principle you use to “know” thisanswer.

(f) Which initial solution was more concentrated? Use Le Châtelier’s principle to explain how you “know” thisanswer.

(g) Was the initial HCl(aq) a saturated solution? Was it atequilibrium? Was it at equilibrium before being removedfrom its closed storage container? Explain how you canuse the basic principles of equilibrium to “know” theseanswers.

Figure 1Testing your concepts is a continual part of “doing” science.


712 Chapter 16 NEL

16.116.1Water Ionization and Acid–Base Strength

In many of the preceding units, you have studied examples of chemical reactions and sys-tems that in some way depend on the nature of acids and bases and/or the pH of solu-tion. For example, aqueous permanganate ions are powerful oxidizing agents, but can actto oxidize other reagents only in the presence of hydrogen (hydronium) ions. Almost anysoluble R–COOH organic compound will make an aqueous solution with a pH below7, because the hydrogen atom of such a group is relatively easily removed. The electro-lysis of potassium iodide solution produces a solution that is strongly basic. Understandingthese connections involves considering equilibrium effects, as shown by the chro-mate–dichromate aqueous ion system (Figure 1).

You have already learned several concepts about the strengths and properties of acidsand bases (see Chapter 6 Summary, page 262). If we now combine equilibrium con-cepts with these acid–base concepts, we can develop a much more comprehensive under-standing of acids and bases. This understanding, in turn, will allow you to better explainand predict how acids and bases behave. For instance, the reaction examined in LabExercise 16.A is an easily explained example of how the acidity of a solution can directlyaffect other ion equilibria.

Figure 1The chromate—dichromate aqueousion system, in equilibria at high andlow pH

CrO42_

(aq) Cr2O72_

(aq)


The Chromate–DichromateEquilibriumIn an aqueous solution, chromate ions are in equilibrium withdichromate ions (Figure 1).

2 CrO42�(aq) � 2 H�(aq) 0 Cr2O7

2�(aq) � H2O(l)

Complete the Prediction and Design (including diagnostic tests)of the investigation report.

PurposeThe scientific purpose of this investigation is to test a Design forvarying the acidity of an equilibrium.

ProblemHow does changing the hydrogen ion concentration affect thechromate–dichromate equilibrium?

HypothesisThe position of this equilibrium depends on the acidity of thesolution.

LAB EXERCISE 16.A Report Checklist

For simplicity, a great many chemical reaction equations use H�(aq) to represent anaqueous hydrogen ion. This representation often works very well, as in the redox equa-tions used in Unit 7, or the titration analysis equations used for stoichiometric calcula-tions in Chapter 8. For other purposes, however, it is necessary to represent this ion moreaccurately in order to understand and explain the theoretical nature of the reaction. Asyou learned in Chapter 6, it is more consistent with evidence to think of this entity as ahydronium ion, and to represent it as H3O

�(aq). Acid–base equilibrium theory necessarilyinvolves collision–reaction theory for a variety of entities. This chapter will, therefore, usethe hydronium ion convention almost exclusively. Before aqueous acidic or basic solu-tion equilibrium can be investigated further, the equilibrium nature of the ions of the sol-vent (water) must first be examined, understood, and taken into consideration.



The Water Ionization Constant, KwEven highly purified water has a very slight conductivity that is only observable if meas-urements are made with very sensitive instruments (Figure 2). According to Arrhenius’theory, conductivity is due to the presence of ions (Figure 3). Therefore, the conduc-tivity observed in pure water must be the result of ions produced by the ionization of somewater molecules into hydronium ions and hydroxide ions. Because the conductivity is soslight, the equilibrium at SATP must greatly favour the water molecules.

Kc � or Kc � [H3O�(aq)] [OH�(aq)]

Evidence indicates that, at 25 °C, at any given moment, fewer than two of every billionmolecules in pure liquid water exist in ionized form! When we write an equilibriumconstant expression for this ionization, the value of Kc is an extremely small number.Recall that, because pure liquid water (or the water in any dilute aqueous solution) hasan essentially constant concentration, it does not appear in the expression; it is simplyincorporated into the equilibrium constant value.

The water ionization equilibrium relationship is so important in chemistry that thisparticular Kc constant is given its own special symbol and name. This new constant is calledthe ion product or ionization constant for water, Kw.

Kw � [H3O�(aq)][OH�(aq)] � 1.00 � 10�14 at SATP

The equilibrium equation for the ionization of water shows that hydronium ions andhydroxide ions form in a 1:1 ratio. Therefore, the concentration of hydronium ions andhydroxide ions in pure water must be equal. This equality must also be true for any neu-tral aqueous solution. Using the mathematical expression for Kw, and the value of Kw atSATP, the concentrations of H3O�(aq) and OH–(aq) can be calculated by taking thesquare root of the Kw value. Recall (from Chapter 15) that in all entity concentrationcalculations from any Kc value, the entity concentration is simply assumed to have unitsof mol/L.

[H3O�(aq)] � [OH�(aq)] � �1.00 �� 10�14� � 1.00 � 10�7 mol/L

The ionization of water is especially important in the empirical and theoretical studyof acidic and basic solutions. Recall from Chapter 6 that, according to the modifiedArrhenius theory, an acid is a substance that reacts with water to produce hydronium ions.The additional hydronium ions provided by the acid increase the hydronium ion concentration. Since the hydronium ion concentration is greater than 10–7 mol/L, the solu-tion is acidic. A basic solution is one in which the hydroxide ion concentration is greater

[H3O�(aq)][OH� (aq)]

��[H2O(l)][H2O(l)]

Section 16.1

Figure 2A sensitive multimeter is required todetect the electrical conductivity ofthe highly purified water that istypically used in a chemistrylaboratory. (See Appendix C.3 forguidance on using a multimeter.)

H

O

H

O

H

H

H

O

H

O

H

HH

O

+–

H

HH

O

Figure 3Collisions of water molecules very occasionally produce this cation–anion pair.

�10�6%H2O(l) � H2O(l) 0 H3O

�(aq) � OH�(aq)

Successful CollisionsA collision that successfully formshydronium and hydroxide ions isvery rare. This is not becausecollisions are rare—each watermolecule collides with others tensof trillions of times every second!But, the chance that any givencollision will both have sufficientenergy, and also be at exactly theright orientation, is very, very smallindeed. Ordinarily, an equilibriumso strongly favouring the reversereaction would just be ignored—thought of as not happening atall—but, in this case, the ions areuniquely important because of theeffects they have on all reactions inaqueous solution.

Conversely, you already knowthat the reverse reaction(hydronium ions with hydroxideions) is quantitative. Almost everysuch ion collision will be effectivebecause the required energy isvery low, and the attraction ofopposite ion charges acts to orientthe entities correctly.

Modelling water molecules thisway—as space-filling models withsuperimposed atomic symbols andlines to represent bonds—gives anoverall representation of theprocess that is logically consistentwith experimental evidence.

DID YOU KNOW ??


714 Chapter 16 NEL

than 10�7 mol/L. Basic solutions are produced in two ways: either by the complete dis-sociation (upon dissolving in water) of an ionic hydroxide, or by partial reaction ofsome weak base entity (ion or molecule) with water to produce hydroxide ions.

The most important point about Kw is that it applies to pure water, and also to any solu-tion that is mostly water. This means that this ionization equilibrium will be involved inany other reaction going on in aqueous solution, if that reaction involves hydroniumions or hydroxide ions in any way. As an example, consider the equilibrium reactionjust studied in Lab Exercise 16.A. Since that reaction involves hydronium (hydrogen)ions, the chromate–dichromate equilibrium can be controlled (shifted) easily, simplyby adjusting either the hydronium ion or the hydroxide ion concentration; which reallyjust means deliberately shifting the water ionization equilibrium. A great many chem-ical reactions are dependent in this way on the water ionization equilibrium. Many of them(those containing H3O

�(aq) or OH–(aq) ions) are evident in the Relative Strengths ofOxidizing and Reducing Agents table (Appendix I).

Note that, since the mathematical relationship is simple, we can easily use Kw to cal-culate either the hydronium ion amount concentration or the hydroxide ion amountconcentration in an aqueous solution, if the other concentration is known.

Since [H3O�(aq)][OH�(aq)] � Kw

then [H3O�(aq)] �

and [OH�(aq)] �

In ordinary dilute aqueous acidic or basic solutions, the presence of substances otherthan water decreases the certainty of the Kw value at 25 °C to two significant digits.

All questions and examples in this text assume temperatures of 25 °C, and aqueous solu-tions that are not highly concentrated, with a Kw value of 1.0 � 10�14, unless specificallystated otherwise.

Kw��[H3O

�(aq)]

Kw��[OH�(aq)]

Learning TipKeep in mind that the value forKw is subject to the samerestrictions as any otherequilibrium constant: one beingthat it will change if thetemperature changes. Forexample, in pure water at 20 °C,Kw has a numerical value of6.76 � 10�15, whereas at 30 °Cits value is 1.47 � 10�14.

Kw will also change enoughto be invalid in any aqueoussolution with a very high soluteconcentration—because thenthe assumption that theconcentration of the watersolvent is at or near a constantvalue (55.5 mol/L) no longerholds true.

A 0.15 mol/L solution of hydrochloric acid at 25 °C is found to have a hydronium ionconcentration of 0.15 mol/L. Calculate the amount concentration of the hydroxide ions.

Solution

HCl(aq) � H2O(l) 0 H3O�(aq) � Cl�(aq)

[OH�(aq) �

� (entity concentration units assumed to be mol/L)

� 6.7 � 10�14 mol/L

Using the Kw relationship, the hydronium ion concentration is 6.7 � 10�14 mol/L.

1.0 � 10�14

��0.15 mol/L

Kw��[H3O

�(aq)]


CAREER CONNECTION

Environmental EngineerMonitoring and protecting waterquality is an essential part of thework of environmental engineers.They design systems andtreatment processes that controlpH levels, temperature, andamounts of dissolved oxygen forindustries that use water.

Would you like to become anenvironmental engineer? Look intosalaries, employmentopportunities, and educationalprograms at at least two differenteducational institutions.

Learning TipThese communication examplesshow the application of theusual units convention forsimplifying calculations fromequilibrium constants. Units forthe constant are ignored, andother concentration units arealways entered in mol/L. Then,since the units for thecalculated value are alwaysmol/L, they are just written inwith the answer.




Section 16.1

Calculate the amount concentration of the hydronium ion in a 0.25 mol/L solution ofbarium hydroxide.

Solution

Ba(OH)2(s) → Ba2�(aq) � 2 OH�(aq)

[OH�(aq)] � 2 � [Ba(OH)2(aq)]

� 2 � 0.25 mol/L

� 0.50 mol/L

[H3O�(aq)] �

�

� 2.0 � 10�14 mol/L

Using the Kw relationship, the hydronium ion concentration is 2.0 � 10�14 mol/L .

1.0 � 10�14

��0.50 mol/L

Kw��[OH�(aq)]


Determine the hydronium ion and hydroxide ion amount concentrations in 500 mL of anaqueous solution for home soap-making containing 2.6 g of dissolved sodium hydroxide.

Solution

nNaOH � 2.6 g � � 0.065 mol

[NaOH(aq)] � � 0.13 mol/L

NaOH(s) → Na�(aq) � OH�(aq)

[OH�(aq)] � [NaOH(aq)] � 0.13 mol/L

[H3O�(aq)] �

�

� 7.7 � 10�14 mol/L

Using the Kw relationship, the hydronium ion concentration is 7.7 � 10�14 mol/L, and thehydroxide ion concentration is 0.13 mol/L.

1.0 � 10�14

��0.13 mol/L

Kw��[OH�(aq)]

0.065 mol��

0.500 L

1 mol�40.00 g


Warnings on PackagingMany households, in previouscenturies, kept supplies of lye(sodium hydroxide) on hand formaking soap. Because it lookedlike sugar, it was occasionallyswallowed by curious children,causing terrible injuries to theirthroats. A prominent Americanphysician, Dr. Chevalier Jackson(1865–1958) realized that warningson the packaging wouldencourage parents to keep thisdangerous substance out of thereach of their children. This is oneof the earliest instances of warninglabelling on packaging. Partiallybecause of Dr. Jackson's efforts,the United States Congress passedthe Federal Caustic Labelling Actin 1927. In Canada, we have theConsumer Chemicals andContainers Regulations.

DID YOU KNOW ??


716 Chapter 16 NEL

Communicating Concentrations: pH and pOHRecall (Chapter 6) that the enormous range of aqueous solution hydronium ion con-centrations is more easily expressed using the logarithmic pH scale (Figure 5).Mathematically,

pH � � log [H3O�(aq)] and, inversely, [H3O

�(aq)] = 10�pH

For basic solutions, it is sometimes more useful to use a scale based on the amount con-centration of hydroxide ions. Recall that the definition of pOH follows the same formatand the same certainty rule as pH.

pOH � �log[OH�(aq)] and, inversely, [OH�(aq)] � 10�pOH

Practice1. The hydronium ion concentration in an industrial effluent is 4.40 mmol/L. Determine

the concentration of hydroxide ions in the effluent.

2. The hydroxide ion concentration in a household cleaning solution is 0.299 mmol/L.Calculate the hydronium ion concentration in the cleaning solution.

3. Calculate the hydroxide ion amount concentration in a solution prepared bydissolving 0.37 g of hydrogen chloride in 250 mL of water.

4. Calculate the hydronium ion amount concentration in a saturated solution of calciumhydroxide (limewater) that has a solubility of 6.9 mmol/L.

5. What is the hydronium ion amount concentration in a solution made by dissolving20.0 g of potassium hydroxide in water to form 500 mL of solution?

6. Calculate the percent ionization of water at SATP. Recall that 1.000 L of water has amass of 1000 g.

Calculate the amount concentration of hydronium ions in a 0.100 mol/L aqueous solutionof ammonia that is used in a spray bottle for window cleaning solution (Figure 4). Areference states that there is 2.1% reaction of dissolved ammonia with water (ionization) atthis concentration, at SATP.

Solution

NH3(aq) � H2O(l) 0 NH4�(aq) � OH–(aq)

[OH�(aq)] � � 0.100 mol/L

� 0.0021 mol/L

[H3O�(aq)] �

�

� 4.8 � 10�12 mol/L

Using the Kw relationship, in 0.100 mol/L aqueous ammonia, the hydronium ionconcentration is 4.8 � 10–12 mol/L.

1.0 � 10�14

��0.0021 mol/L

Kw��[OH�(aq)]

2.1�100


Figure 4Many solutions sold for cleaningwindows contain ammonia.



The mathematics of logarithms allows us to express a simple relationship between pHand pOH. According to the rules of logarithms,

log(ab) � log(a) � log(b)

Using the equilibrium law for the ionization of water,

[H3O�(aq)][OH�(aq)] � Kw

log[H3O�(aq)] � log[OH�(aq)] � log(Kw)

(�pH) � (�pOH) � �14.00

pH � pOH � 14.00 (at SATP)

This relationship allows for a quick conversion between pH and pOH values.

Section 16.1

Figure 5The pH scale

20 1 3 4 5 6 7 8 9 10 11 12 13 14

14 13 12 11 10 9 8 7 6 5 4 3 2 1 0

pH

lyehouseholdammonia

antacidsolution

seawater

bloodnormalrain

softdrink

vinegarbatteryacid

increasing acidity pH < 7 increasing basicity pH > 7neutral (pure water)

pOH

Kw � [H3O�(aq)][OH�(aq)] � 1.0 � 10�14

pH � �log[H3O�(aq)] pOH � �log[OH�(aq)]

[H3O�(aq)] � 10�pH [OH�(aq)] � 10�pOH

pH � pOH � 14.00

SUMMARYWater Ionization Conversions and Values (at SATP)


718 Chapter 16 NEL

Acid Strength as an Equilibrium PositionIn Chapter 6, you learned that acidic solutions of different substances at the same con-centration do not possess acid properties to the same degree. The pH of a 1.00 mol/L solu-tion of an acid can vary anywhere from a value of nearly 7 to a value of nearly 0, dependingon the specific acid in the solution. Other properties can also vary. For example, aceticacid does not conduct an electric current nearly as well as hydrochloric acid of equalconcentration (Figure 6). When we observe chemical reactions of these acids, it isapparent that acetic acid, although it reacts in the same manner and amount ashydrochloric acid, does not react as quickly. The concepts of strong and weak acids weredeveloped to describe and explain these differences in properties of acids.

An acid is described as weak if its characteristic properties are less than those of acommon strong acid, such as hydrochloric acid. Weak acids are weaker electrolytes andreact at a slower rate than strong acids do; the pH of solutions of weak acids is closer to7 than the pH of strong acids of equal concentration.

In Chapter 6, strong acids were explained as ionizing quantitatively by reacting withwater to form hydronium ions, whereas weak acids were explained as ionizing only par-tially (usually �50%). The empirical distinction between strong and weak acids can beexplained much more completely now by combining the modified Arrhenius theorywith equilibrium theory. A strong acid is explained as an acid that reacts quantitativelywith water to form hydronium ions. For example, the reaction of dissolved hydrogenchloride (hydrochloric acid) with water is virtually complete. Even though the equa-tion could be written with double equilibrium arrows and the extent shown with a �99.9% note, it is simpler, and much more common, to just use a single arrow to showthat the reaction is quantitative.

HCl(aq) � H2O(l) → H3O�(aq) � Cl�(aq)

A weak acid is an acid that reacts partially with water to form hydronium ions.Measurements of pH indicate that most weak acids react less than 50%. For example, aceticacid reacts only 1.3% in solution at 25 °C and 0.10 mol/L concentration.

Learning TipRecall this convenient “rule ofthumb” for keeping track ofsignificant digits in(logarithmic) calculations of pHor pOH.

The number of digits followingthe decimal point in the pH orpOH (logarithm) value should beequal to the number of signifi-cant digits shown in the amountconcentration of the ion.

For example, a hydronium ion concentration of2.7 � 10�3 mol/L is expressedas a pH of 2.57, and a pOH of4.3 is expressed as a hydroxideion concentration of5 � 10�5 mol/L.

Practice7. Food scientists and dieticians measure the pH of foods when they devise recipes and

special diets.

(a) Copy and complete Table 1.

(b) Based on pH only, predict which of the foods would taste most sour.

8. To clean a clogged drain, 26 g of sodium hydroxide is added to water to make 150 mLof solution. What are the pH and pOH values for the solution?

9. What mass of potassium hydroxide is contained in 500 mL of solution that has a pHof 11.5? Comment on the degree of certainty of your answer.

Table 1 Acidity of Foods

[H3O�(aq)] [OH�(aq)]Food (mol/L) (mol/L) pH pOH

oranges 5.5 � 10�3

asparagus 5.6

olives 2.0 � 10�11

blackberries 10.6

Learning TipFor any aqueous solution of anacid, percent reaction of thatacid with water is alsocommonly called its percentionization in chemistryreferences, because the neteffect is the same as if the acidmolecules simply ionize, as wasonce (simplistically) assumed inArrhenius’ original theory.

You should consider “reactionwith water” and “ionization inwater” to be equivalent termsfor acid–base solution theory,and may expect to see eitherterm used routinely in text andin questions.



Recall, from Chapter 15, that any equilibrium position depends on concentration(s)as well as on temperature; so, this 1.3% ionization value for acetic acid is only valid fora 0.10 mol/L solution at 25 °C. Laboratory pH experiments show that the higher theconcentration of a weak acid solution, the lower its percent ionization becomes.

1.3%

CH3COOH(aq) � H2O(l) 0 H3O�(aq) � CH3COO�(aq)

The hydronium ion concentration of any acid solution can be calculated by multi-plying the percent reaction by the initial amount concentration of the acid solute. Forexample, in HCl(aq) solution, virtually 100% of the HCl molecules react with watermolecules at equilibrium.


[H3O�(aq)] � �

1

1

0

0

0

0� � 0.10 mol/L

� 0.10 mol/L

There are six acids ordinarily classed as “strong”: hydrochloric, nitric, sulfuric, hydro-bromic, hydroiodic, and perchloric acid solutions. Only the first three are common.

For any aqueous strong acid, we can simply assume that the concentration of hydro-nium ions in solution is equal to the initial concentration of the acid dissolved. For weakacids (the majority of examples), we always need to calculate the hydronium ion con-centration in solution, because only a small proportion of the initial acid concentrationwill be converted to ions at equilibrium.

Section 16.1

Figure 6In solutions of equal concentration,a weak acid such as acetic acidconducts electricity to a lesserextent than does a strong acid suchas hydrochloric acid.


720 Chapter 16 NEL

As explained in Chapter 6, we can easily compare the strengths of different acids bycomparing the measured pH values for aqueous solutions of equal concentration. Thelower the pH, the higher the hydronium ion concentration, the greater the percent reac-tion, and thus the stronger the acid. Furthermore, we can find the percent reaction forionization of any weak acid solution from the measured pH of a solution of known ini-tial concentration.

In a 0.10 mol/L solution of acetic acid, only 1.3% of the CH3COOH molecules have reactedat equilibrium to form hydronium ions. Calculate the hydronium ion amountconcentration.

Solution

1.3%CH3COOH(aq) � H2O(l) 0 H3O

�(aq) � CH3COO�(aq)

[H3O�(aq)] � � 0.10 mol/L

� 1.3 � 10�3 mol/L

The hydronium ion concentration in 0.10 mol/L acetic acid is 1.3 � 10�3 mol/L.

1.3�100


The pH of a 0.10 mol/L methanoic acid solution is 2.38. Calculate the percent reaction forionization of methanoic acid.

Solution

[H3O�(aq)] � 10�pH

� 10�2.38 mol/L

� 4.2 � 10�3 mol/L

[H3O�(aq)] �

p � � 100

� � 100

� 4.2%

The percent ionization of 0.10 mol/L aqueous methanoic acid is 4.2%.

4.2 � 10�3 mol/L��

0.10 mol/L

[H3O�(aq)]

��[HCOOH(aq)]

p��100 � [HCOOH(aq)]

COMMUNICATION example 6Learning Tip

Percent ionization:

p � � 100

Rearrange the equation to solvefor hydronium ionconcentration:

[H3O�(aq)] � � [HA(aq)]

where p � percent ionizationand

[HA(aq)] � initial concentrationof weak acid

p�100

[H3O�(aq)]

��[HA(aq)]



Section 16.1

Section 16.1 Questions1. How does the hydronium ion concentration compare with

the hydroxide ion concentration if a solution is (a) neutral?(b) acidic?(c) basic?

2. What two diagnostic tests can distinguish a weak acid froma strong acid?

3. According to Arrhenius’ original theory, what do all baseshave in common?

4. Hydrocyanic acid is a very weak acid.

(a) Write an equilibrium reaction equation for the ionizationof 0.10 mol/L HCN(aq). The percent ionization at SATP is7.8 � 10�3%.

(b) Calculate the hydronium ion concentration and the pHof a 0.10 mol/L solution of HCN(aq).

5. At 25 °C, the hydronium ion concentration in vinegar is 1.3 mmol/L. Calculate the hydroxide ion concentration.

6. At 25 °C, the hydroxide ion concentration in normal humanblood is 2.5 � 10�7 mol/L. Calculate the hydronium ionconcentration and the pH of blood.

7. Acid rain has a pH less than that of normal rain. Thepresence of dissolved carbon dioxide, which forms carbonicacid, gives normal rain a pH of 5.6. What is the hydroniumion concentration in normal rain?

8. If the pH of a solution changes by 3 pH units as a result ofadding a weak acid, by how much does the hydronium ionconcentration change?

9. If 8.50 g of sodium hydroxide is dissolved to make 500 mLof cleaning solution, determine the pOH of the solution.

10. What mass of hydrogen chloride gas is required to produce250 mL of a hydrochloric acid solution with a pH of 1.57?

11. Determine the pH of a 0.10 mol/L hypochlorous acidsolution, which has 0.054% ionization at 25 °C.

12. Calculate the pH and pOH of a hydrochloric acid solutionprepared by dissolving 30.5 kg of hydrogen chloride gas tomake 806 L of solution. What assumption is made whendoing this calculation?

13. Acetic (ethanoic) acid is the most common weak acid usedin industry. Determine the pH and pOH of an acetic acidsolution prepared by dissolving 60.0 kg of pure, liquidacetic acid to make 1.25 kL of solution. The percent reactionwith water at this concentration is 0.48%.

14. Determine the mass of sodium hydroxide that must bedissolved to make 2.00 L of a solution with a pH of 10.35.

15. Write an experimental design for the identification of fourcolourless solutions: a strong acid solution, a weak acidsolution, a neutral molecular solution, and a neutral ionicsolution. Write sentences, create a flow chart, or design atable to describe the required diagnostic tests.

16. Sketch a flow chart or concept map that summarizes theconversion of [H3O

�(aq)] to and from [OH�(aq)], pH, andpercent reaction (ionization) of acid solute. Make your flowchart large enough that you can write the procedurebetween the quantity symbols in the diagram.

WEB Activity


Canadian Achievers—Edgar SteacieEdgar Steacie (Figure 7) was an internationally acclaimed research scientist and a senioradministrator of the National Research Council.

1. What was Steacie’s main area of research?

2. Why was Steacie known as a statesman of science for Canada?

3. What is a Steacie Fellowship?

Figure 7Edgar Steacie (1900–1962)


722 Chapter 16 NEL

16.216.2The Brønsted–Lowry Acid–BaseConceptBy now, our much-revised acid and base theory includes concepts of hydronium ions,reaction with water, reaction equilibrium, and the ionization equilibrium of water. Ourmodified theory is, thus, much more comprehensive, and much better at describing,explaining, and predicting acid–base reactions, than the original theory proposed byArrhenius. We find, however, that there are still problems, and our theory must still beconsidered too restrictive. There is no provision in it for reactions that do not occur inaqueous solution. In addition, we find that there are some substances that seem to haveboth acid and base properties.

As a common example, sodium hydrogen carbonate (sodium bicarbonate, bakingsoda) forms a basic solution (raises the pH) in water, but the same compound will partlyneutralize (lower the pH of) a sodium hydroxide (lye) solution. When we observe thatNaHCO3(s) forms a basic aqueous solution, we conclude that this happens becausesome hydrogen carbonate ions react with water molecules to produce hydroxide ions. Weknow from many other observations that sodium ions have no acidic or basic proper-ties. The following reaction equation seems to explain our observation easily:

HCO3�(aq) � H2O(l) 0 H2CO3(aq) � OH�(aq)

But, if adding sodium hydrogen carbonate makes a (strongly basic) sodium hydroxidesolution less basic, the concepts we are using lead us to conclude that something mustbe reacting to decrease the concentration of the hydroxide ions. It must be the hydrogencarbonate ions because there are no other entities in this system except sodium ionsand water molecules. We can easily write an equation to explain this observation, aswell:

HCO3�(aq) � OH�(aq) 0 CO3

2�(aq) � H2O(l)

This equation seems to explain how the hydroxide ions can be partially consumed. But,if it is correct, it begs the question, “Is sodium bicarbonate a base, or is it an acid?”Clearly, our theory still needs some work! We need a broader, more comprehensive con-cept to successfully explain these kinds of observations.

The Proton Transfer ConceptNew theories in science usually result from looking at the evidence in a way that hasnot occurred to other observers. In 1923, two European scientists independently devel-oped a new approach to acids and bases (Figure 1). These scientists focused on the roleof an acid and a base in a reaction rather than on the acidic or basic properties of theiraqueous solutions. An acid, such as hydrogen chloride, functions in a way opposite to abase, such as ammonia. According to the Brønsted–Lowry concept, hydrogen chloride,upon dissolving, donates a proton to a water molecule:


and ammonia, upon dissolving, accepts a proton from a water molecule.

NH3(aq) � H2O(l) 0 OH�(aq) � NH4�(aq)

Figure 1Johannes Brønsted (1879–1947) (a)and Thomas Lowry (1874–1936) (b)independently created newtheoretical definitions for acids andbases, based upon proton transferduring a reaction.

(a)

(b)

H�

H�

acid

base



Water does not have to be one of the reactants. For example, hydronium ions present ina hydrochloric acid solution can react directly with dissolved ammonia molecules.

H3O�(aq) � NH3(aq) → H2O(l) � NH4

�(aq)

We can describe this reaction as NH3 molecules removing protons from H3O� ions.Hydronium ions act as the acid, and ammonia molecules act as the base. Water is presentas the solvent but not as a primary reactant. In fact, water does not even have to bepresent, as evidenced by the reaction of hydrogen chloride and ammonia gases(Figure 2).

HCl(g) � NH3(g) → NH4Cl(s)

According to the Brønsted–Lowry concept, a Brønsted–Lowry acid is a protondonor and a Brønsted–Lowry base is a proton acceptor. A Brønsted–Lowry neutralizationis a competition for protons that results in a proton transfer from the strongest acidpresent to the strongest base present. A Brønsted–Lowry reaction equation is an equa-tion written to show an acid–base reaction involving the transfer of a proton from oneentity (an acid) to another (a base).

The Brønsted–Lowry concept does away with defining a substance as being an acidor base. Only an entity that is involved in a proton transfer in a reaction can be definedas an acid or base—and only for that particular reaction. This last point is extremelyimportant: protons may be gained in a reaction with one entity, but lost in a reaction withanother entity. For example, in the reaction of HCl with water shown on the previouspage, water acts as the base; whereas, in its reaction with NH3, water acts as the acid. Asubstance that appears to act as a Brønsted–Lowry acid in some reactions and as aBrønsted–Lowry base in other reactions is called amphoteric, or sometimes (incorrectly)amphiprotic. The empirical term, amphoteric, properly refers to a chemical substancewith the ability to react as either an acid or a base. The theoretical term, amphiprotic,describes an entity (ion or molecule) having the ability to either accept or donate aproton. The hydrogen carbonate ion in baking soda (Figure 3), like every other hydrogenpolyatomic ion, is amphiprotic, as shown by the following reactions:

HCO3�(aq) � H2O(l) 0 OH�(aq) � H2CO3(aq) Kc � 2.2 � 10�8 mol/L

base acid

HCO3�(aq) � H2O(l) 0 H3O

�(aq) � CO32�(aq) Kc � 4.7 � 10�11 mol/L

acid base

When bicarbonate ions are in aqueous solution, some react with the water molecules byacting as an acid, and some react by acting as a base. Kc values given for these two equi-libriums show that one of them predominates—so much so that the other reaction issimply inconsequential. The number of ions reacting as a base is over 2000 times morethan the number reacting as an acid. As you might expect, the resulting solution is basic.

We can get a clearer idea of the amphoteric nature of baking soda (caused by theamphiprotic nature of the hydrogen carbonate ion) by noting the evidence expressed inthe next two equations. Note that bicarbonate ions partly neutralize a strong acid, butthey can also partly neutralize a strong base.

HCO3�(aq) � H3O

�(aq) 0 H2CO3(aq) � H2O(l) (raises the pH of a strong base acid acid solution)

HCO3�(aq) � OH�(aq) 0 CO3

2�(aq) � H2O(l) (lowers the pH of a acid base strong base solution)

Section 16.2

H�

acid base

H�

acid base

Figure 2Invisible fumes of ammonia gas andhydrogen chloride gas mix and reactabove these beakers, producing tinyvisible particles of solid ammoniumchloride that are suspended in theair.

Figure 3Baking soda is a commonhousehold chemical, but it requiresan uncommonly sophisticatedtheory to describe, explain, orpredict all of its properties.


724 Chapter 16 NEL

Scientists consider the Brønsted–Lowry concept to be a theoretical definition. It fallsshort of being a comprehensive theory because it does not explain why a proton isdonated or accepted, and cannot predict theoretically which reaction will occur for agiven entity in any given new situation. The advantage of the Brønsted–Lowry definitionsis that they enable us to define acids and bases in terms of chemical reactions rather thansimply as substances that form acidic and basic aqueous solutions. A definition of acidsand bases in terms of chemical reactions allows us to describe, explain, and predict agreat many more reactions, whether they take place in aqueous solution, in solution insome other solvent, or between two undissolved entities in pure chemical states.

Figure 4White vinegar is a 5% acetic(ethanoic) acid solution. The amountconcentration is 0.83 mol/L. Such asolution is only about 0.43% ionizedat 25 °C, making acetic acid atypical weak acid, by definition.

DID YOU KNOW ??Superacids

In aqueous solution, all strong acidsare equal in strength, since they allreact instantly and completely withwater to form the strongest possibleacid entity that can exist in water—the hydronium ion. Scientists havelong suspected, however, that thewell-known strong acids, HCl andH2SO4, may not be equal in strengthwhen no water is present and thatmuch stronger acids could exist. Dr. Ronald Gillespie of McMasterUniversity has done extensiveresearch on non-aqueous strongacids. His definition of a superacid—one that is stronger than puresulfuric acid—is now generallyaccepted by scientists. Perchloricacid, the only common superacid,easily loses protons to H2SO4(l)molecules. Fluorosulfonic acid,HSO3F(l), is more than onethousand times stronger thanH2SO4(l). It is the strongestBrønsted–Lowry acid known.

Practice1. Theories in science develop over a period of time. Illustrate this development by

writing a theoretical definition of an acid, using the following concepts. Begin youranswer with, “According to [authority], acids are substances that….”(a) Arrhenius’ original theory(b) the modified Arrhenius theory(c) the Brønsted–Lowry concept

2. How does the definition of a base according to the modified Arrhenius theorycompare with the Brønsted–Lowry definition?

3. Classify each reactant in the following equations as a Brønsted–Lowry acid or base. (a) HF(aq) � SO3

2�(aq) 0 F�(aq) � HSO3�(aq)

(b) CO32�(aq) � CH3COOH(aq) 0 CH3COO�(aq) � HCO3

�(aq)(c) H3PO4(aq) � OCl�(aq) 0 H2PO4�(aq) � HOCl(aq)(d) HCO3

�(aq) � HSO4�(aq) 0 SO4

2�(aq) � H2CO3(aq)

4. Evidence indicates that the hydrogen sulfite ion is amphiprotic. A sodium hydrogensulfite solution can partly neutralize either a sodium hydroxide spill or a hydrochloricacid spill. (a) Write a net ionic equation for the reaction of aqueous hydrogen sulfite ions with

the hydroxide ions in solution. Label the reactants as acids or bases.(b) Write a net ionic equation for the reaction of hydrogen sulfite ions with the

hydronium ions from a hydrochloric acid solution. Label the reactants as acids orbases.

5. What restrictions to acid–base reactions do the Brønsted–Lowry definitions remove?

6. Why is the Brønsted–Lowry concept labelled a theoretical definition rather than atheory?

Conjugate Acids and BasesAccording to the Brønsted–Lowry concept, acid–base reactions involve the transfer of aproton. These reactions are universally reversible and always result in an acid–base equi-librium.

In a proton transfer reaction at equilibrium, both forward and reverse reactions involveBrønsted-Lowry acids and bases. For example, in an acetic acid solution (Figure 4), theforward reaction is explained as a proton transfer from acetic acid to water molecules,and the reverse reaction is a proton transfer from hydronium to acetate ions.

base acid

CH3COOH(aq) � H2O(l) 0 CH3COO�(aq) � H3O�(aq)

acid base

H�

H�



This equilibrium is typical of all acid–base reactions. There will always be two acids (inthe example, CH3COOH and H3O

�) and two bases (in the example, H2O and CH3COO�)in any acid–base reaction equilibrium. Furthermore, the base on the right (CH3COO�)is formed by removal of a proton from the acid on the left (CH3COOH). The acid on theright (H3O

�) is formed by the addition of a proton to the base on the left (H2O). A pairof substances with formulas that differ only by a proton is called a conjugate acid–basepair. An acetic acid molecule and an acetate ion are a conjugate acid–base pair. Acetic acidis the conjugate acid of the acetate ion, and the acetate ion is the conjugate base of aceticacid. The equation shows that there must always be two conjugate acid–base pairs inany proton transfer reaction. The hydronium ion and water are the second conjugateacid–base pair in this equilibrium. Conjugate acid–base pairs appear opposite each otherin a table of acids and bases, such as that in Appendix I.

1.3%

CH3COOH(aq) � H2O(l) 0 CH3COO�(aq) � H3O�(aq) (0.10 mol/L at 25 °C)

At equilibrium, only 1.3% of the CH3COOH molecules have reacted with water in a0.10 mol/L solution at SATP. It appears that the ability of the CH3COO� part of theacetic acid molecule to keep its proton (H�) is much greater than the ability of H2O toattract the proton away (Figure 5). This means that CH3COO� is a stronger base (it hasa greater attraction for protons) than H2O.

When HCl molecules react with water (Figure 6), the Cl� of each HCl molecule has amuch weaker attraction for its proton (H�) than any colliding water molecule has. Thewater molecules “win” this “competition” for protons overwhelmingly. At equilibrium, essen-tially all of the HCl molecules have lost protons to water molecules. In this case, thetransfer of protons is quantitative, and, because its molecules have an extremely weakattraction for their protons, HCl(aq) is called (somewhat confusingly) a strong acid.

Remember: The stronger the base, the more it attracts another proton. The strongeran acid, the less it attracts its own proton.

Section 16.2

Figure 5This pictorial analogy is used to explain why acetic acid is a weak acid in aqueous solution. Webelieve that the proton (H�) of the carboxyl group is attracted more strongly by the rest of theacetic acid molecule than it is by the water molecule. We infer this conclusion because, atequilibrium, pH evidence indicates that very few of the acetic acid molecules have lost protonsin their collisions with water molecules.

Figure 6When gaseous hydrogen chloridedissolves in water, the HClmolecules are thought to collide andreact quantitatively with watermolecules to form hydronium andchloride ions.

CH3COO–

H+

HO

H

HH

HH

HH ClCl

O O+

+ –

+

conjugate pair

conjugate pair


726 Chapter 16 NEL

The terms strong acid and weak acid can be explained in terms of the Brønsted–Lowryconcept, and also by comparing the reactions of different acids with the same base—for example, water. Using HA as the general symbol for any acid and A� as its conjugatebase, the empirically derived Relative Strengths of Aqueous Acids and Bases table(Appendix I) lists the position of equilibrium of aqueous solutions of many different acids.They are ordered according to how much they ionize in (react with) the water solvent.

HA(aq) � H2O(1) 0 H3O�(aq) � A�(aq)

The extent of the proton transfer between HA and H2O determines the relative strengthof HA(aq). In Brønsted–Lowry terms, when a strong acid reacts with water, an almostcomplete transfer of protons results for the forward reaction and almost no transfer ofprotons occurs for the reverse reaction; a nearly 100% reaction with water. The equilibriumconstant value is found to be very large. Theoretically, a strong acid holds its protonvery weakly, and easily loses the proton to any base, even very weak bases such as water.This result leads to the interpretation that the conjugate base, A�, of a strong acid musthave a very weak (negligible) attraction for protons. A useful generalization regarding therelative strengths of an acid–base conjugate pair is: the stronger an acid, the weaker its con-jugate base; and conversely, the weaker an acid, the stronger its conjugate base. (See theRelative Strengths of Aqueous Acids and Bases table in Appendix I.)

Chemists have no simple explanation, in terms of forces or bonds, for the differing abil-ities of acids to donate protons or of bases to accept them. The inability to predict acidand base strengths for an entity not already included in an empirically determined tableof acids and bases is a major deficiency of all acid–base theories.

Learning TipAcids were studied muchearlier than bases were, andsometimes the old terminologyused to describe acid–basesituations can be misleading—because it always focuses onthe nature of the acid. Strongacids are very reactive, but webelieve it is because they arevery weak proton attractors. It iscommon to speak of acids“donating” protons, and ofbases “accepting” them, butthis terminology gives anunrealistic view of what webelieve is really happening. It islike saying that a bank robber“accepts” money “donated” by abank teller. The tug-of-waranalogy in Figure 5 is a morerealistic way to think of protontransfer reactions—as acompetition between twobases, both attracting the sameproton.

• An acid is a proton donor and a base is a proton acceptor, in a specific reaction.

• An acid–base reaction involves a single proton transfer from one entity (the acid) toanother (the base).

• An amphiprotic entity (amphoteric substance) is one that acts as a Brønsted–Lowryacid in some reactions and as a Brønsted–Lowry base in other reactions.

• A conjugate acid–base pair consists of two entities with formulas that differ only bya proton.

• A strong acid has a very weak attraction for protons. A strong base has a very strongattraction for protons.

• The stronger an acid, the more weakly it holds its proton. The stronger a base, themore it attracts another proton.

• The stronger an acid, the weaker is its conjugate base. The stronger a base, the weakeris its conjugate acid.

SUMMARY Brønsted–Lowry Definitions

Practice7. Use the Brønsted–Lowry definitions to identify the two conjugate acid–base pairs in

each of the following acid–base reactions. (a) HCO3

�(aq) � S2�(aq) 0 HS�(aq) � CO32�(aq)

(b) H2CO3(aq) � OH�(aq) 0 HCO3�(aq) � H2O(l)

(c) HSO4�(aq) � HPO4

2�(aq) 0 H2PO4�(aq) � SO4

2�(aq)(d) H2O(l) � H2O(l) 0 H3O


8. Some ions can form more than one conjugate acid–base pair. List the two conjugateacid–base pairs involving a hydrogen carbonate ion in the reactions in question 7.



Predicting Acid–Base Reaction EquilibriaThe Brønsted–Lowry concept unfortunately does not include any theoretical explana-tion about why any given entity attracts a proton more or less strongly. To predict the out-come of any acid–base combination, we must rely on empirical evidence, gained bymeasuring and recording the relative strengths of acid and base entities. Predictionsmust be restricted to only those acid–base combinations for which we already have data.To help us, we can now look for a simple generalization that might allow us to predictthe approximate position of equilibrium in an acid–base proton transfer.

Section 16.2

Purpose Design AnalysisProblem Materials Evaluation (1)Hypothesis ProcedurePrediction Evidence


Creating an Acid–Base Strength TableAn acid–base table organizes common acids (and their conjugatebases) in order of decreasing acid strength. Acid strength can betested several ways, including by a carefully designed use ofindicators. Predict the order of strengths using the RelativeStrengths of Aqueous Acids and Bases table (Appendix I). Usethe indicators provided to create a valid and efficient Design, inwhich you clearly identify the relevant variables. Evaluate theDesign (only), and suggest improvements if any problems areidentified.

PurposeThe purpose of this investigation is to test an experimental designfor using indicators to create a table of relative strengths of acidsand bases.

ProblemCan the indicators available be used to rank the acids and basesprovided in order of strength?



Predicting Acid–Base EquilibriaIs it possible to predict how far a reaction will proceed? Use thetable of Relative Strengths of Aqueous Acids and Bases inAppendix I and the evidence of position of equilibrium tocomplete the Analysis of the investigation report.

PurposeThe purpose of this investigation is to develop a generalization forpredicting the position of acid–base equilibria.

ProblemHow do the positions of the reactant acid and base in theacid–base table relate to the position of equilibrium?

�50%

1. CH3COOH(aq) � H2O(l) 0 H3O�(aq) � CH3COO�(aq)

�99%

2. HCl(aq) � H2O(l) 0 H3O�(aq) � Cl�(aq)

�50%

3. CH3COO�(aq) � H2O(l) 0 CH3COOH(aq) � OH�(aq)�50%

4. H3PO4(aq) � NH3(aq) 0 H2PO4�(aq) � NH4

�(aq)�50%

5. HCO3�(aq) � SO3

2�(aq) 0 HSO3�(aq) � CO3

2�(aq)

6. H3O�(aq) � OH�(aq) → H2O(l) � H2O(l)

LAB EXERCISE 16.B Report Checklist

WEB Activity

Web Quest—Pool ChemistryA swimming pool can be an enjoyable place to spend a hot summer afternoon. Most of us arefamiliar with the dangers of pools. This Web Quest introduces an unfamiliar risk: pool gas. Whatis pool gas and how does it form? What are the dangers of pool gas, and how can they bereduced?



728 Chapter 16 NEL

Predicting Acid–Base ReactionsWhen making complex predictions, scientists often combine a variety of empirical andtheoretical concepts. We need to use a combination of concepts to predict both theproducts and the extent of acid–base reactions. According to the collision–reactiontheory, a proton transfer may result from a collision between an acid and a base. In a systemthat is a mixture of several different acid and/or base entities, there are countless randomcollisions of all the entities present. So, in theory, in any such system, there are manydifferent possible acid–base reactions, and all of these reactions occur (to some extent)all the time. Evidence indicates that one reaction predominates: it occurs to an extent somuch greater than the other reactions that it is the only observable reaction. We will beable to explain which reaction will predominate if we use a combination of the colli-sion–reaction theory and the Brønsted–Lowry concept.

According to collision–reaction theory, in an acid–base system, collisions of all enti-ties present are constantly occurring. According to the Brønsted–Lowry concept, a protonwill only transfer if an acid entity collides with a base entity that is a better protonattractor than itself. A proton could theoretically transfer several times (if there wereseveral different acids and bases in the system), transferring each time to a strongerproton attractor. Once gained by an entity of the strongest base present, a proton willremain there, because there is no other base present in the system that can attract thatproton strongly enough to remove it. By the same logic, once any entity of the strongestacid present has lost its proton, its (remaining) conjugate base cannot gain one backfrom any other entity present, because it is a weaker proton attractor than anything elsein the system. Overall, theory suggests that the predominant acid–base reaction shouldbe the one that involves proton transfer from the strongest acid to the strongest basepresent in the system. Experimental evidence indicates that this explanation is correct.Other possible proton transfer reactions occur to such a small extent that they have a neg-ligible effect on the reactants and products, so they are normally ignored.

We theorize that the only significant reaction in any acid–base system involves protontransfer from the strongest acid present to the strongest base present. For an aqueous solu-tion system, we first must list all the entities present as they exist in aqueous solution.Table 1 summarizes how to represent the entities that are present.

The process for determining the nature and extent of the predominant proton transferreaction in an aqueous acid–base system can be thought of as five distinct steps, for con-venience, as shown in the following Sample Problem.

Table 1 Predominant Entities Present in Aqueous Solution

Substance dissolved Kinds of entities Predominant entities (example) in solution present (example)

Ionic compounds cations, anions Ca2�(aq) and HCO3�(aq)

Ca(HCO3)2(aq)

Ionic oxides cations, hydroxide ions Na�(aq), OH�(aq) orNa2O(aq), CaO(aq) Ca2�(aq), OH�(aq)

Strong acids H3O�(aq), conjugate base H3O

�(aq), Cl�(aq) orHCl(aq) or HNO3(aq) H3O

�(aq), NO3�(aq)

Weak acids molecules or ions HF(aq) or HSO3�(aq)

HF(aq) or HSO3�(aq)

Weak bases molecules or ions NH3(aq) or HCO3�(aq)

NH3(aq) or HCO3�(aq)

CAREER CONNECTION

Chemistry ResearcherFarideh Jalilehvand (Figure 7) isan associate professor ofchemistry at the University ofCalgary. Her research involves alot of X-ray absorptionspectroscopy. Find out how aciditydue to sulfur accumulation inwater-logged wood is connectedto ancient shipwrecks by visitingher information site, and check outher curriculum vitae (CV). Scienceis international in scope!


Figure 7Farideh Jalilehvand



Section 16.2

Figure 8The drain cleaner shown is aconcentrated solution of a verystrong base. A spill would be highlycorrosive and quite hazardous.Excess weak acid, such as thevinegar (5% acetic acid), is the bestchoice to “neutralize” a spilledstrong base. The final solution willnot actually be neutral, but it will beonly mildly acidic, and much safer tohandle.

Learning TipThe relative positions of thestrongest acid and the strongestbase on an acid–base table canbe used to approximatelydetermine the position of anacid–base equilibrium.

Products Favoured

SA

�

SB

Reactants Favoured

SB

�

SA

�50%0

What will be the predominant reaction if spilled drain cleaner (sodium hydroxide) solutionis neutralized with vinegar (Figure 8)? Are reactants or products favoured in this reaction?

Step 1: List all entities present as they exist in aqueous solution. Refer to Table 1 ifnecessary. The entity list for this situation is

Na�(aq) OH�(aq) CH3COOH(aq) H2O(l)

Step 2: Use the entity lists of the Relative Strengths of Aqueous Acids and Bases table toidentify and label each entity present as a Brønsted–Lowry acid or base. Amphiproticentities are labelled for both possibilities. Conjugate bases of strong acids are not includedor labelled as bases because they cannot act as bases in aqueous solution. Metal ions aretreated as spectators.

A ANa�(aq) OH�(aq) CH3COOH(aq) H2O(l)

B B

Step 3: Use the order of the entities in the Relative Strengths of Aqueous Acids andBases table to identify and label the strongest Brønsted–Lowry acid (the highest one onthe table) and the strongest Brønsted–Lowry base (the lowest one on the table) that arepresent in the solution.

SA ANa�(aq) OH�(aq) CH3COOH(aq) H2O(l)

SB B

Step 4: Write a balanced equation to show a proton transfer from the strongest acid tothe strongest base, assuming that their respective conjugates are the reaction products.

H�

CH3COOH(aq) � OH�(aq) 0 CH3COO�(aq) � H2O(l)

Step 5: Predict the position of equilibrium using the generalization developed in LabExercise 16.B and illustrated in the margin Learning Tip. For this reaction, the strongestacid is positioned higher on the table than the strongest base, so products are favoured.Under certain assumed conditions (see the following discussion), the equilibrium percentreaction may be labelled as greater than 50%.

�50%CH3COOH(aq) � OH�(aq) 0 CH3COO�(aq) � H2O(l)

SAMPLE problem 16.1

�50%0

The nature of water complicates the prediction of outcomes of acid–base reactions inaqueous solution. We need to consider the specific restrictions that apply when pre-dicting proton transfer reactions in aqueous solution, because they are much morecommon than acid–base reactions in any other environment.

• Hydronium ion is the strongest acid entity that can exist in aqueous solution. If astronger acid than hydronium ion is dissolved in water, it reacts instantly andcompletely with water molecules to form hydronium ions. For this reason, the sixstrong acids are all written as H3O

�(aq) when in aqueous solution.

• Hydroxide ion is the strongest base entity that can exist in aqueous solution. If astronger base than hydroxide ion is dissolved in water, it reacts instantly andcompletely to form hydroxide ions. The only common example is the dissolving ofsoluble ionic oxide compounds such as Na2O(s). In such cases, the oxide ion iswritten as OH�(aq) when in aqueous solution.


730 Chapter 16 NEL

• No entity in aqueous solution can react as a base if it is a weaker base than water.For this reason, the conjugate bases of the six strong acids are not considered asbases, in aqueous solution.

Even though you find that products are favoured (as in Sample Problem 16.1), forthe predominant reaction, you cannot accurately predict the actual equilibrium positionof all cases of any specific acid–base reaction by this method. It simply means you knowthat the forward reaction happens more readily than the reverse reaction, all thingsbeing equal. But, recall (from Chapter 15) that the initial concentration of entities playsa very large part in determining the percent reaction at equilibrium. To be able to predict something meaningful about the position of an acid–base reaction equilibrium,we must restrict the reaction conditions. Unless you are specifically informed otherwisein a question, assume in this text that, for the predominant reaction in an acid–basesystem,

• the equation represents a single proton transfer between two entities, neither ofwhich is water, and has a stoichiometric ratio of 1:1:1:1

• the strongest acid and the strongest base present in the reaction system are bothpresent in significant chemical amounts, in approximately equal concentrations

Under these circumstances (and only then), you may assume that a Brønsted–Lowryacid–base reaction where products are favoured has a percent reaction greater than 50%,or that a reaction where reactants are favoured may be labelled as “� 50%.” It is alsocorrect (given these restrictions) to state that the Kc value for any such reaction is � 1if products are favoured, and � 1 if reactants are favoured. It is not possible to be moreprecise about the equilibrium position, however, at our current level of theory.

Note that the above restrictions exclude any reaction equation written to describe theionization (reaction with water) of any single acid or base entity dissolved to make anaqueous solution. In these cases, water is acting as one of the reactants, and the amountconcentration of the water is very much greater than that of the other entity.

Finally, recall that the reaction of hydronium ions with hydroxide ions is always quan-titative. This equation should always be written with a single arrow. There are manyother acid–base reactions that are quantitative, particularly those where either hydroniumor hydroxide ions are involved, but it is not possible to easily predict whether any acid–basereaction will be quantitative from the table of Relative Strengths of Aqueous Acids andBases. In this text, other quantitative reactions will always be identified for you, as in thefollowing Communication Example.

Ammonium nitrate fertilizer is produced by the quantitative reaction of aqueous ammoniawith nitric acid. Write a balanced acid–base equilibrium equation.

Solution

A SANH3(aq), H2O(l), H3O

�(aq)SB B

NH3(aq) � H3O�(aq) → NH4

�(aq) � H2O(l)

COMMUNICATION example

Learning TipSince hydroxide ion is thestrongest possible base inaqueous solution, reaction ofany acid with this ion inaqueous solution isautomatically one that favoursproducts. Similarly, the reactionof any base with the strongestpossible acid, hydronium ion,must favour products. Ofcourse, the reaction ofhydroxide ion with hydroniumion is always quantitative.



Section 16.2

Figure 9Bottles of household bleach displaya warning against mixing the bleach(aqueous sodium hypochlorite) withacids. Does your prediction of thereaction between vinegar andhypochlorite ions provide any cluesabout the reason for the warning?

1. List all entities (ions, atoms, or molecules including H2O(l)) initially present as theyexist in aqueous solution. (Refer to Table 1, page 728.)

2. Identify and label all possible aqueous acids and bases, using the Brønsted–Lowrydefinitions.

3. Identify the strongest acid and the strongest base present, using the table of RelativeStrengths of Aqueous Acids and Bases (Appendix I).

4. Write an equation showing a transfer of one proton from the strongest acid to thestrongest base, and predict the conjugate base and the conjugate acid to be the products.

5. Predict the approximate position of the equilibrium, using the generalization devel-oped in Lab Exercise 16.B on page 727 and the table of Relative Strengths of AqueousAcids and Bases (Appendix I).

SUMMARYA Five-Step Method for Predicting thePredominant Acid–Base Reaction

PracticeUse the five-step method to predict the predominant reactions in the following chemicalsystems:

9. Hydrofluoric acid and an aqueous solution of sodium sulfate are mixed to test thefive-step method of predicting acid–base reactions.

10. Strong acids, such as perchloric acid, have been shown to react quantitatively withstrong bases, such as sodium hydroxide.

11. Methanoic acid is added to an aqueous solution of sodium hydrogen sulfide.

12. A student mixes solutions of ammonium chloride and sodium nitrite in a chemistrylaboratory.

13. Empirical work has shown that nitric acid reacts quantitatively with a sodium acetatesolution.

14. A consumer attempts to neutralize an aqueous sodium hydrogen sulfate cleaner witha solution of lye. (See Appendix J if you do not remember what lye is.)

15. Can ammonium nitrate fertilizer, added to water, be used to neutralize a muriatic acid(hydrochloric acid) spill?

16. Predict the acid–base reaction of bleach with vinegar (Figure 9).

17. Commercial laundry bleach (Figure 9) is made by reacting chlorine gas with a sodiumhydroxide solution, according to the equation

Cl2(g) � 2 OH�(aq) 0 OCl�(aq) � Cl�(aq) � H2O(l)

As sold, the pH of laundry bleach solutions is always well above 8. You know that theelement chlorine has very low solubility in pure water. Explain, using Le Châtelier’sprinciple, why bleach bottle labels warn so strongly against mixing bleach with othercleaning agents, such as acidic toilet bowl cleaners.

Note: Household bleach also produces toxic chloramines if mixed with basicammonia cleaning solutions. The best rule is, NEVER mix bleach directly with anyother cleaning powder or solution. It is arguably the most dangerous commonhousehold chemical.


732 Chapter 16 NEL

Figure 10The versatility of baking soda is demonstrated by its use inextinguishing fires, in baking biscuits, and in neutralizingexcess stomach acid. It is also used as a medium for localanesthetics—baking soda reduces stinging sensations byneutralizing the acidity of the anesthetic, with the result thatthe speed and efficiency of the anesthetic are improved. Thebroad range of uses for baking soda results, in part, from itsamphiprotic character.


Aqueous Bicarbonate Ion Acid–BaseReactionsAn acid–base table organizes common acids (and their conjugatebases) in a way that enables us to predict predominant acid–basereactions. Evaluate the reliability of the five-step method (only)using information from this investigation, by comparing youranalysis of the evidence with your predictions.

PurposeThe purpose of this investigation is to test the five-step methodfor predicting reactions in acid–base systems.

ProblemWhat are the products and position of the equilibrium for sodiumhydrogen carbonate (Figure 10) with stomach acid, vinegar,household ammonia, and lye, respectively?

Evidence

LAB EXERCISE 16.C Report Checklist



Testing Brønsted–Lowry ReactionPredictionsWhen predicting products for this investigation, list all entitiespresent as they normally exist in an aqueous environment. Forthose reactants that are added in solid state, assume that theywill dissolve. Use the resulting entities for prediction. Evaluate thepredictions, the Brønsted–Lowry concept, and the five-stepmethod for acid–base reaction prediction.

PurposeThe purpose of this investigation is to test the Brønsted–Lowryconcept and the five-step method for reaction prediction from atable of relative acid–base strength.

ProblemWhat reactions occur when various pairs of substances aremixed?

DesignA prediction is made for each of eleven pairs of substances. Theprediction is then tested using one or more diagnostic tests,complete with controls. Additional diagnostic tests increase thecertainty of the evaluation.


Table 2 The Addition of Baking Soda to Various Solutions

Reactant Bubbles Odour pH

HCl(aq) yes none increases

CH3COOH(aq) yes disappears increases

NH3(aq) no remains decreases

NaOH(aq) no none decreases



Section 16.2


Creating an Acid–Base TableComplete the Analysis of the investigation report, including ashort table of the four acids and bases involved. Use entityposition generalizations in the acid–base table for reactions thatfavour products or reactants. The evidence for reaction 1 isinterpreted as:

PurposeThe purpose of this investigation is to test an experimental designfor using equilibrium position to create a table of relativestrengths of acids and bases.

ProblemWhat is the order of acid strength for the first four members of thecarboxylic acid family?

Evidence�50%

CH3COOH(aq) � C3H7COO�(aq) 0CH3COO�(aq) � C3H7COOH(aq)

�50%

HCOOH(aq) � CH3COO�(aq) 0 HCOO�(aq) � CH3COOH(aq)�50%

C2H5COOH(aq) � C3H7COO�(aq) 0C2H5COO�(aq) � C3H7COOH(aq)

�50%

C2H5COOH(aq) � HCOO�(aq) 0C2H5COO�(aq) � HCOOH(aq)

LAB EXERCISE 16.D Report Checklist

CH3COOH(aq)

C3H7COO–(aq)

Acids Bases

> 50%Changing Ideas on Acids andBases—The Evolution of a ScientificTheoryHistorically, chemists have known the empirical properties ofsubstances long before any theory was developed to explainand predict those properties. For example, chemists werefamiliar with the distinguishing properties of several acids andbases, and used them routinely, by the middle of the 17thcentury. Early attempts at an acid–base theory tended to focuson acids and to ignore bases. Over time, several theoriespassed through cycles of formulation, testing, acceptance,further testing, and eventual rejection. Following is a briefhistorical summary of acid–base theories and the evidencethat led to their revision.

• Antoine Lavoisier (1743–1794) (Figure 11) believed thatthe properties of acids could be traced back to a singlesubstance. Lavoisier studied the combustion of phosphorusand sulfur and determined that these elements combinewith something in the atmosphere to produce compoundsthat form acidic solutions when dissolved in water. WhenJoseph Priestley identified the component of theatmosphere that actively supports combustion, Lavoisier(1777) named the gas oxygen, meaning “acid maker.”Lavoisier assumed that oxygen was the substanceresponsible for the generic properties of acids. There weresoon problems with this theory when it was found that

Case StudyCase Study

several acids, such as muriatic acid (HCl), do not containoxygen. Furthermore, many substances that form basicsolutions (such as lime, CaO) were found to contain oxygen.This evidence led to the rejection of the oxygen theory, but itis historically important because it is the first systematicattempt to chemically characterize acids and bases. Thegeneralization that nonmetallic oxides form acidic solutionsis still useful in chemistry.

• Sir Humphry Davy (1778–1829) (Figure 12) conducted theexperiments that demonstrated the absence of oxygen inmuriatic acid (HCl), which led to the rejection of Lavoisier’stheory. Davy (1810) advanced his own theory that the

Figure 11Antoine Lavoisier

Figure 12Humphry Davy


734 Chapter 16 NEL

presence of hydrogen gives compounds acidic properties.This theory, however, did not explain why many compoundscontaining hydrogen have neutral properties (for example,CH4) or basic properties (for example, NH3). Justus vonLiebig (1803–1873) revised Davy’s idea to define acids assubstances in which the hydrogen could be replaced by ametal. This revision meant that acids could be thought of asionic compounds in which hydrogen had replaced the metalion. Liebig had no corresponding theoretical definition forbases, which were still identified empirically as substancesthat neutralized acids.

• Svante Arrhenius(1859–1927) (Figure 13)developed a theory in 1887that provided the first usefultheoretical definitions of acidsand bases. He defined acidsas substances that ionize inaqueous solution to formhydrogen ions, H�(aq).Similarly, he defined bases assubstances that dissociate toform hydroxide ions, OH�(aq),in solution. This theoryexplained the process ofneutralization as thecombination of hydrogen ionsand hydroxide ions to form water

H�(aq) � OH�(aq) → H2O(l)

Arrhenius explained the strengths of acids in terms ofdegrees of ionization. While these ideas were a majordevelopment in chemists’ understanding of the properties ofacids and bases, there were some problems with Arrhenius’theory.

• In Arrhenius’ theory, an acid is expected to be an acid inany solvent, which was found not to be the case. Forexample, when dissolved in water, HCl supposedly breaksup into hydrogen ions and chloride ions, but when it isdissolved in benzene, the HCl remains as intactmolecules. The nature of the solvent, therefore, had toplay a critical role in acid–base properties of substances.

• The need for hydroxide to always be the base ledArrhenius to propose formulas such as NH4OH(aq) for theformula for ammonia in water, which led to themisconception that NH4OH(aq) was the base, notNH3(aq).

• According to Arrhenius’ theory, all salts (ionic solids)should produce neutral solutions, but many do not. Forexample, solutions of ammonium chloride are acidic andsolutions of sodium acetate are basic.

• Bonding theory suggested that it was very unlikely that asingle proton could exist in aqueous solution withoutbeing bonded to at least one water molecule.

• Johannes Brønsted (1879–1947) (Figure 14) and Thomas Lowry (1874–1936) (Figure 15) independentlypublished (1923) essentially the same concept about howacids and bases behave. These scientists focused on the

role of an acid and a base in a reaction rather than on theproperties of their aqueous solutions. According to theBrønsted–Lowry concept, an acid is a proton (H�) donor anda base is a proton acceptor. The solvent has a central role inthe Brønsted–Lowry concept. Water can be considered anacid or a base since it can lose a proton to form a hydroxideion (OH�) or accept a proton to form a hydronium ion(H3O

�). In their view, neutralization is a competition forprotons that results in a proton transfer from the strongestacid present to the strongest base present. For example, inthe reaction

HSO4�(aq) � NH3(aq) → SO4

2�(aq) � NH4�(aq)

the hydrogen sulfate ion acts as the acid (because itdonates a proton) and ammonia acts as the base (becauseit accepts a proton). A weakness of the Brønsted–Lowryconcept is its limitation to solutions (gaseous or liquid).

• Gilbert Lewis (1875–1946)(Figure 16) developed anacid–base theory in 1923 thatincludes all previous theoriesand definitions of acids andbases. He viewed acids andbases in terms of the covalentbond, a theory he haddeveloped in 1916. Lewisdefined acids as electron–pairacceptors and bases aselectron–pair donors. It isimportant to note that inLewis acid–base theory, nohydrogen ion and no solventneed be involved. The Lewisdefinition is broader than all previous definitions andexplains more inorganic and organic reactions (Figure 17).For example, in the reaction

BF3(g) � NH3(g) → H3NBF3(g)

boron trifluoride acts as a Lewis acid because it accepts(forms a bond with) a pair of electrons from the ammonia,which acts as the Lewis base.

Figure 13Svante Arrhenius

Figure 14Johannes Brønsted

Figure 15Thomas Lowry

Figure 16Gilbert Lewis



Section 16.2

In science, it is unwise to assume that any scientificconcept is complete. Whenever scientists assume that theyunderstand a concept, two things usually happen. First,conceptual knowledge tends to remain static for a whilebecause little conflicting evidence exists, or becauseconflicting evidence (being somewhat discomfiting) isignored. Second, when enough conflicting evidenceaccumulates, a change in thinking occurs within thescientific community in which the current theory isdrastically revised or entirely replaced. Revolutionaryconcepts most often tend to be formed in a moment ofinsight, usually following a long period of incredibly hardwork done by a great many people.

Case Study Questions

1. Identify a word or phrase that describes the central idea ineach of the theories described above.

2. For each theory, describe the weakness that led to itsbeing replaced.

3. What was the first theoretical definition of a base, andwho developed it?

4. Write Lewis formulas for each substance in the reaction ofboron trifluoride with ammonia, to illustrate the “electronpair transfer” concept.

Figure 17This Venn diagram shows how comprehensive some different acid–base theoriesare considered to be.

Section 16.2 Questions1. Aqueous solutions of nitric acid and nitrous acid of the

same concentration are prepared. (a) Predict how their pH values compare.(b) Explain your answer using the Brønsted–Lowry concept.

2. Briefly state the five steps involved in predicting thepredominant reaction and the approximate position ofequilibrium in an acid–base reaction system.

3. According to the Brønsted–Lowry concept, whatdetermines the position of equilibrium in an acid–basereaction?

4. What generalization from the table of Relative Strengths ofAqueous Acids and Bases (Appendix I) can be used topredict the position of an acid–base equilibrium?

5. State two examples of conjugate acid–base pairs, eachinvolving the hydrogen sulfite ion.

6. Predict, with reasoning, including Brønsted–Lowryequations, whether each of the following chemical systemswill be acidic, basic, or neutral. (a) aqueous hydrogen bromide(b) aqueous potassium nitrite(c) aqueous ammonia(d) aqueous sodium hydrogen sulfate(e) carbonated beverages(f) limewater(g) vinegar

7. Write an experimental design to test each of the predictionsin question 6.

8. Write two experimental designs to rank a group of bases inorder of strength.

Brønsted-Lowry

Lewis

Arrhenius

hydrogenoxygen


736 Chapter 16 NEL

9. Ammonia molecules and hydronium ions have remarkablysimilar shapes and structures (Figure 18). Both have threehydrogen atoms bonded in a pyramidal structure to a smallcentral atom, which has one lone pair of electrons. Oxygenand nitrogen atoms both have very high electronegativities,and they are adjacent on the periodic table. Explain why anammonia molecule is a good proton attractor, whereas ahydronium ion is an extremely poor proton attractor.

instructions, or you may find the cleaner also reacting withthe item you are trying to clean, or even with you! Thesecleaners have a variety of formulations, but most containhydroxyacetic acid (glycolic acid), CH2OHCOOH(aq). A 0.10 mol/L glycolic acid solution is 3.9% ionized at 25 °C,compared to 0.10 mol/L acetic acid, which is 1.3% ionized at the same temperature. (a) Write a Brønsted–Lowry equation for the reaction of

aqueous glycolic acid with carbonate ions (from hardwater “scum”), and label both conjugate acid–basepairs.

(b) Explain whether the equilibrium would favour productsmore, or less, if acetic acid (vinegar) were used to dothe same cleaning job. Is it likely that any differencewould be significant? State your reasoning.

(c) Explain whether glycolic acid would react with a givenchemical amount of rust more quickly or more slowlythan an equal volume of equally concentrated aceticacid solution.

(d) Explain whether glycolic acid would react with agreater, or lesser, chemical amount of rust than anequal volume of equally concentrated acetic acidsolution.

11. Since ancient times, people have known that strongheating of natural limestone (calcium carbonate) willproduce quicklime (calcium oxide). This very usefulsubstance is both reactive and corrosive with many organicsubstances because it has a high affinity for water. Addingwater to quicklime causes the formation of slaked lime(calcium hydroxide)—the common name was derived fromthe idea that the lime was “slaking” its thirst. This reactionalso produces a very considerable quantity of heat! (a) Write a balanced chemical equation for the

decomposition of calcium carbonate to carbon dioxideand calcium hydroxide.

(b) Write a balanced chemical equation for the addition ofwater to calcium oxide to produce calcium hydroxide.

(c) Write a Brønsted–Lowry equation for the instantaneousand overwhelmingly quantitative (and highlyexothermic) reaction of aqueous oxide ions with water.

Figure 19Commercial rust removing solutions must be usedwith care, but can be a very effective illustration ofthe old TV slogan, “better living, throughchemistry.”

Figure 18Evidence indicates that the ammoniamolecule, NH3, modelled here, has the samepyramidal shape as the hydronium ion,H3O

�.

10. Many household cleaning solutions claim to “dissolveaway” rust stains and hard water “lime” deposits (Figure 19). These cleaners are simply acidic solutions (asis vinegar) that react with slightly soluble substances suchas Fe2O3(s) and CaCO3(s). The word “react,” however,sounds dangerous to consumers, and is hardly ever used inadvertising. These cleaning solutions must be usedcarefully, and you should always read all of the label



16.316.3Acid–Base Strength and the

Equilibrium LawModifying Arrhenius’ original concept of acids and bases allowed us to explain thebehaviour of many substances in solution that could not be explained by his originaltheory. Then, problems with varying degrees of properties led to considering reactionsof acids and bases as examples of equilibrium, which allowed a much more compre-hensive understanding of relative acid or base strengths. Apparent conflicts in the def-inition of acids and bases then led us to define them by what they do in a reaction (in termsof proton transfer), rather than by what they are (in terms of some typical structure). Thisprocess of continually testing (identifying problems) and then expanding and improvingconcepts is typical of science—with the goal always being the formation of a more com-plete and comprehensive theory. Science assumes that no theory ever formed by thehuman mind can be “perfect,” and that any theory, no matter how well supported byevidence, must always be questioned and tested. We also realize that every step moves uscloser to a more complete understanding. Testing, and then restricting, revising, and(perhaps) replacing concepts, and then testing again is the constant cycle of the scien-tific method. Next, we explore the possible value of using the equilibrium law to fur-ther increase our understanding of acid–base behaviours.

The Acid Ionization Constant, KaOne important way that chemists communicate the strength of any weak acid is by usingthe equilibrium constant expression for a Brønsted–Lowry reaction equation showingthe acid ionizing in (reacting with) water. This expression is another very importantspecial case of equilibrium. The constant is given its own unique name and symbol: theacid ionization constant, Ka. See the Ka values listed in the Relative Strengths of AqueousAcids and Bases data table (Appendix I).

Note that in this table, the first six acids have Ka values given only as “very large.”These acids are collectively called the strong acids because they all react quantitatively (�99.9%) with water to form hydronium ions. Because no acid stronger than H3O

�(aq)can exist in aqueous solution, all of these acids are considered equivalent. For each of them,the actual acid species present is H3O

�(aq).All of the other acids listed on the table are considered to be weak acids, and they vary

greatly in extent of reaction with water at equilibrium. To empirically rank any weakacid as stronger or weaker than any other, we normally compare the ionization of solu-tions of the same concentration, to see which is more strongly acidic (has a lower pH). Thisis how Appendix I was created.

By considering the equilibrium established in solutions of weak acid to be a Brønsted–Lowry proton transfer, we gain a more complete understanding about what is actuallygoing on among the entities reacting. By using Ka equilibrium constants, we can deriveand predict quantities that are very useful when working with acid–base reactions.Consider a solution of hydrofluoric acid, HF(aq). The equation for the reaction in water(ionization) of this weak acid is

HF(aq) � H2O(l) 0 H3O�(aq) � F�(aq)

The equilibrium law expression is

Ka �

Note that the concentration of liquid water is omitted from the Ka equilibrium lawexpression. We make an assumption that this value will remain essentially constant for

[H3O�(aq)][F�(aq)]

��[HF(aq)]

Learning TipUse of ionization terminology(strong and weak) can create aproblem, caused by thecommon use of the word“strong.” For an acid, strengthrefers only to the ease withwhich a base can remove itsproton, and has nothing to dowith the quantity present in thesolution. Concentration refers tothe chemical quantity per unitvolume of solution, and hasnothing to do with strength.

Both the strength and theconcentration of an acid affecthow rapidly, and to what extent,that acid will react.

Acid–Base ReactionsThis quick simulation lets youexplore the percent reaction whenvarious acids and bases reacttogether. The acid ionizationconstants are provided for twostrong and three weak acids.


EXTENSION +


738 Chapter 16 NEL

aqueous solutions that are not highly concentrated. This assumption is not exactly true,because the water is not a separate liquid phase in an aqueous solution. However, aslong as the amount of water solvent is much greater than the amount of acid solute,making this assumption will cause negligible change in (and will not add uncertainty to)the answers for any calculations.

All Ka values are calculated by making this assumption. For this reason (and for othermeasurement reasons too involved to explain here), Ka values are necessarily somewhatinaccurate, and become less accurate as the acid concentration increases. It is impor-tant to know that Ka values given in tables are normally given to only two significantdigits because they are usually only known to a certainty of about ± 5%.

Two calculations involving the Ka constant are common for weak acid solutions:

• calculating a Ka value from measured (empirical) amount concentration data

• using a Ka value to predict a concentration of hydronium ions for an aqueoussolution where the initial weak acid amount concentration is known

Calculating Ka from Amount Concentrations

Learning TipThe great advantage of Kavalues over percent ionizationvalues is that, once determined,Ka values are valid over a widerange of acid solutionconcentrations. A percentionization value is useful onlyfor one specific concentrationof one specific weak acid.

The pH of a 1.00 mol/L solution of acetic acid is carefully measured to be 2.38 at SATP.What is the value of Ka for acetic acid?



Ka �

First, find the equilibrium concentration of aqueous hydronium ion from the pH.

[H3O�(aq)] � 10�pH

� 10�2.38

� 0.0042 mol/L

Next, use the hydronium ion concentration and the balanced chemical equation tocalculate the concentration of the acetate ion. Since one acetate ion forms for eachhydronium ion, the equilibrium concentration of acetate ion must be identical to that ofthe hydronium ion. The stoichiometric ratio is 1:1.

[CH3COO�(aq)] � [H3O�(aq)] � 0.0042 mol/L

An ICE table is useful to find the equilibrium concentration of acetic acid molecules.

Ka �

� 0.000 017

Regardless of numerical size, Ka values are usually expressed in scientific notation.The calculated Ka for acetic acid is 1.7 � 10�5.

(0.0042 mol/L)(0.0042 mol/L)��

(1.00 mol/L)

[H3O�(aq)][CH3COO�(aq)]

��[CH3COOH(aq)]

SAMPLE problem 16.2

Table 1 ICE Table for CH3COOH(aq) � H2O(l) 0 H3O�(aq) � CH3COO�(aq)

[CH3COOH(aq)] [H3O�(aq)] [CH3COO�(aq)]Concentration (mol/L) (mol/L) (mol/L)

Initial 1.00 0 0

Change � 0.0042 � 0.0042 � 0.0042

Equilibrium 1.00 0.0042 0.0042

Learning TipBecause the value for the initialacid concentration is preciseonly to two decimal places, thecalculation of the equilibriumconcentration of the acid (1.00 mol/L � 0.0042 mol/L)rounds to 1.00 mol/L. (See thePrecision Rule for Calculations,Appendix F.3.) In other words,the decrease in the initial acidconcentration is negligible, inthis particular case. A negligiblechange in concentration is quiteoften the case for Kacalculations for weak acids, butnot always.

Recall that amountconcentrations must be used tocalculate any Kc value and, byconvention, units for theequilibrium constant value aresimply ignored.



Section 16.3

A student measures the pH of a 0.25 mol/L solution of carbonic acid to be 3.48. Calculate the Ka for carbonic acid from this evidence.

Solution

H2CO3(aq) � H2O(l) 0 H3O�(aq) � HCO3

�(aq)

Ka �

At equilibrium,[H3O

�(aq)] � 10�pH

� 10�3.48

� 3.3 � 10�4 mol/L

[HCO3�(aq)] � [H3O

�(aq)] � 3.3 � 10�4 mol/L

[H2CO3(aq)] � (0.25 � 0.000 33) mol/L

� 0.25 mol/L

Ka � � 0.000 000 44

According to the equilibrium law, the Ka for carbonic acid is 4.4 � 10�7.

(0.000 33 mol/L)(0.000 33 mol/L)��

(0.25 mol/L)

[H3O�(aq)][HCO3

�(aq)]��

[H2CO3(aq)]


Table 2 ICE Table for H2CO3(aq) � H2O(l) 0 H3O�(aq) � HCO3

�(aq)

Amount [H2CO3(aq)] [H3O�(aq)] [HCO3�(aq)]

concentration (mol/L) (mol/L) (mol/L)

Initial 0.25 0 0

Change � 0.000 33 � 0.000 33 � 0.000 33

Equilibrium 0.25 0.000 33 0.000 33

The pH of a 0.400 mol/L solution of sulfurous acid is measured to be 1.17.

Calculate the Ka for sulfurous acid from this evidence.

Solution

H2SO3(aq) � H2O(l) 0 H3O�(aq) � HSO3

�(aq)

Ka �

At equilibrium,

[H3O�(aq)] � 10�pH

� 10�1.17

� 0.068 mol/L

[HSO3�(aq)] � [H3O

�(aq)] � 0.068 mol/L

[H2SO3(aq)] � (0.400 � 0.068) mol/L

� 0.332 mol/L

[H3O�(aq)][HSO3

�(aq)]��

[H2SO3(aq)]


Learning TipBecause a Brønsted-Lowryequation is written to show asingle proton transfer, thestoichiometric ratio will alwaysbe 1:1:1:1. Thus, the units for Kavalues will always be mol/L.Because we can make thisassumption, units are often notincluded with Ka values inreference tables or in finalanswer statements.


740 Chapter 16 NEL

Calculating [H3O�(aq)] from KaThe second type of calculation involving a Ka value allows us to predict the acidity ofany weak acid solution. We can calculate the concentration of hydronium ions from theinitial acid concentration and the Ka value as follows.

Predict the [H3O�(aq)] and pH for a 0.200 mol/L aqueous solution of methanoic acid.

Look up the value of Ka for methanoic (formic) acid in the Relative Strengths of AqueousAcids and Bases data table (Appendix I). The value is given as 1.8 � 10�4 mol/L.


HCOOH(aq) � H2O(l) 0 H3O�(aq) � HCOO�(aq)

Ka � 1.8 � 10�4 �

Use x to represent the numerical value for the equilibrium amount concentration ofaqueous hydronium ions.

[H3O�(aq)] � x mol/L

An ICE table is useful to keep track of the concentration values.

Using the 1:1:1:1 stoichiometric ratio, calculate concentration increases and decreases forthe reaction to equilibrium, and note them in the ICE table.

Substituting into the equilibrium law expression, ignoring units,

1.8 � 10�4 �

�(x) (x)

��(0.200 � x)

[H3O�(aq)][HCOO�(aq)]

��[HCOOH(aq)]

[H3O�(aq)][HCOO�(aq)]

��[HCOOH(aq)]

SAMPLE problem 16.3

Table 4 ICE Table for HCOOH(aq) � H2O(l) 0 H3O�(aq) � HCOO�(aq)

[HCOOH(aq)] [H3O�(aq)] [HCOO�(aq)]Concentration (mol/L) (mol/L) (mol/L)

Initial 0.200 0 0

Change � x � x � x

Equilibrium (0.200 � x)* x x

Ka � � 0.014

According to the equilibrium law, the Ka for sulfurous acid is 1.4 � 10�2.

(0.068 mol/L)(0.068 mol/L)��

(0.332 mol/L)

Table 3 ICE Table for H2SO3(aq) � H2O(l) 0 H3O�(aq) � HSO3

�(aq)

Amount [H2SO3(aq)] [H3O�(aq)] [HSO3�(aq)]


Initial 0.400 0 0

Change � 0.068 � 0.068 � 0.068

Equilibrium 0.332 0.068 0.068



Section 16.3

Note that the equation to be solved contains a squared value.

0.000 18 �

Multiplying both sides by (0.200 � x), collecting terms, and equating to zero gives x2 � 0.000 18x � 0.000 036 � 0 (expressed in the form of a quadratic equation).

Notice what happens when we test our present example for the simplifying assumption.

� �0.

00.02000

18� � 1111, which is greater than 1000

The test allows us to use the assumption (initial acid concentration � equilibrium acidconcentration, or [HA(aq)]initial � [HA(aq)]equilibrium). In this specific case, so little acid hasreacted with water at equilibrium that the initial acid concentration is decreased by anegligible amount—less than the uncertainty of the initial value.

Numerically, for this example, we may assume that (0.200 � x) � 0.200, and we can thensimplify the equation to

0.000 18 � , which makes solving for x quick and easy.

Isolate x and take the square root of both sides of the equation:

x � �0.000 1�8 � 0.�200�� 0.0060

so, [H3O�(aq)] � 0.0060 mol/L or 6.0 � 10�3 mol/L

and pH � �log [H3O�(aq)]

� �log (0.0060)

� 2.22

x2

�0.200

[HCOOH(aq)]initial��Ka

x2

��(0.200 � x)

* Note: Technically, solving for x (the amount concentration of hydronium ion) from this

expression requires the use of the quadratic formula, x � ,

which is quite tedious, unless you have a calculator that is preprogrammed tosolve such formulas.

Conveniently, for many (but not all) weak acid solutions, we can use anapproximation that greatly simplifies solving for x. If the percent reaction of theweak acid is quite small, we can assume that the initial concentration of the aciddecreases by so little that the numerical value of the acid amount concentrationwill remain effectively unchanged at equilibrium.

The problem, of course, is deciding when a percent reaction is small enough toallow this assumption to be used. In this text, we will apply the mathematicallysimplest (and most convenient) “rule of thumb” that is commonly used to makethis decision.

If the initial amount concentration of the acid is numerically at least 1000times its Ka value, then you may assume that the initial and equilibrium acidamount concentrations are numerically equal.

This limiting assumption is consistent with the �5% certainty of most Ka values,meaning it ignores those amount concentration changes that would change theKa value by less than the uncertainty that already exists.

�b��b2 � 4�ac��

2a

Rule of ThumbThe very old English phrase “ruleof thumb,” meaning a rule for anyquick way to approximate aquantity without measuring,probably comes directly from useof human thumbs. Since the widthof an adult’s thumb isapproximately one inch, and thelength of a thumb is approximatelyfour inches, you can make a fairlygood estimate of the length ofsomething (in Imperial units) bysimply placing your thumbs alongthe object, one after the other.What is the length of this textbookin (your) thumb widths? What doequestrians mean when they say ahorse stands 16 “hands” high?

A practical rule of thumb forAmericans driving their cars intoMexico or Canada is to multiplythe posted speed by 6 and dropthe last digit, in order to quicklyconvert the km/h speeds on roadsigns into a fairly close miles-per-hour equivalent.

DID YOU KNOW ??


Note: No Practice, Section, or Review question in this text will require you to use the quad-ratic formula to solve acid ionization constant problems. However, you may be asked tostate whether a given problem would require use of this formula for solution (in otherwords, whether the ionization assumption test fails). You should develop the habit ofalways initially testing the assumption and making a qualifying statement before com-pleting the solution to any Ka question where this assumption holds.

742 Chapter 16 NEL

Predict the [H3O�(aq)] and pH for a 0.500 mol/L aqueous solution of hydrocyanic acid.

SolutionTest the simplifying assumption:

� � 8.1 � 107 (greater than 1000)

The assumption holds, and may be used.

Ka � 6.2 � 10�10 �

At equilibrium:

Let x � [H3O�(aq)] � [CN�(aq)]

Then [HCN(aq)] � (0.500 � x) � 0.500 (using the assumption)

6.2 � 10�10 �

x � �6.2 ��10�10 �� 0.500� � 1.8 � 10�5

so, [H3O�(aq)] � 1.8 � 10�5 mol/L

and pH � �log [H3O�(aq)] � �log (1.8 � 10�5) � 4.75

According to the equilibrium law, the hydronium ion amount concentration of0.500 mol/L hydrocyanic acid is 1.8 � 10�5 mol/L, and the pH of the solution is 4.75.

x2

�0.500

[H3O�(aq)][CN�(aq)]

��[HCN(aq)]

0.500��6.2 � 10�10

[HCN(aq)]initial��Ka


Table 5 ICE Table for HCN(aq) � H2O(l) 0 H3O�(aq) � CN�(aq)

Amount [HCN(aq)] [H3O�(aq)] [CN�(aq)]concentration (mol/L) (mol/L) (mol/L)

Initial 0.500 0 0


Equilibrium (0.500 � x) � 0.500 x x

As shown in the next Communication Example, when the negligible ionization (per-cent reaction) assumption holds true, predictions of [H3O

�(aq)] using Ka values becomesimple and straightforward. Note that you can perform the 1000:1 ratio test for the sim-plifying assumption before starting the problem, which is very convenient.

DID YOU KNOW ??pKaFor purposes of easy comparison,the Ka values of different acids aresometimes expressed as pKa values.A pKa value is the negativelogarithm of the Ka. A reported pKavalue of 12 would represent an acidwith a Ka of 1 � 10�12, a very weakacid. In comparing any two weakacids, the one with the lower pKavalue is the stronger of the twoacids.

Unit 8 - Ch 16 Chem30 11/2/06 11:09 AM 2


Section 16.3

Practice1. (a) Write a theoretical definition for the strength of an acid.

(b) State the empirical properties that provide evidence for differing acid strengths. (c) Explain the difference in meaning between strength and concentration, as these

terms are used to refer to aqueous acids. (d) Does the stronger of two acids, when dissolved, necessarily make a more acidic

aqueous solution? Explain.

2. Refer to Appendix I for required information to make the following predictions. Foreach case, first decide and state whether a solution would require use of thequadratic formula. For all cases where use of the quadratic formula is not required,communicate the full solution. (a) Predict [H3O

�(aq)] for 0.20 mol/L hydrobromic acid. (b) Predict [H3O

�(aq)] for 0.20 mol/L hydrofluoric acid. (c) Predict [H3O

�(aq)] for 0.20 mol/L ethanoic acid. (d) Predict the pH of 2.3 mmol/L nitric acid.(e) Predict the pH of 2.3 mmol/L nitrous acid. (f) Predict the pH of 2.3 mmol/L hydrosulfuric acid.

3. For all of the cases (a–f) in question 2 where a prediction could be calculated, rank theacid solutions in order of decreasing acidity.

4. For all of the cases (a–f) in question 2, rank the acids in order of decreasing acidstrength. Which two acids have essentially the same strength?

5. The hydronium ion concentration in 0.100 mol/L propanoic acid is determined (from apH measurement) to be 1.16 � 10�3 mol/L. (a) Calculate the percent reaction (ionization) of this particular weak acid solution.(b) Calculate Ka for aqueous propanoic acid. (c) Is this Ka value constant for propanoic acid? Explain.

6. A 0.10 mol/L solution of lactic acid, a weak acid found in milk, has a measured pH of2.43. The chemical name for this very common organic compound is 2-hydroxypropanoic acid. (a) Find the percent reaction of this lactic acid solution.(b) Calculate the Ka value for aqueous lactic acid. (c) Compare the Ka values for 2-hydroxypropanoic acid and for propanoic acid. What

effect does adding an OH group to the central carbon atom of this molecule haveon the ability of the COOH group to attract a proton?

7. Unlike all other aqueous hydrogen halides, hydrofluoric acid is not a strong acid. Itdoes, however, have a special chemical property: it reacts with glass. This property isused to etch frosted patterns on glassware (Figure 1). (a) Write the Ka expression for hydrofluoric acid.(b) Calculate the hydronium and fluoride ion amount concentrations, the pH, and the

percent reaction in a commercial 2.0 mol/L HF(aq) solution at 25 °C.

8. Phosphoric acid is used in rust-remover solutions. The aqueous acid is available forpurchase by high schools in concentrations of about 15 mol/L. (a) Predict the hydronium ion amount concentration, the pH, and the percent

reaction of a 10 mol/L solution of phosphoric acid. (b) Suggest a reason (having to do with Ka values and solution concentrations) why

you might well suspect that these values could be inaccurate.

9. Ascorbic acid is the chemical name for Vitamin C (Figure 2). A student prepares a0.200 mol/L aqueous solution of ascorbic acid, tests its pH, and reads a value of 2.40from the pH meter. (a) From the student’s evidence, calculate the Ka for ascorbic acid.(b) Compare your result to the value listed in the table of Relative Strengths of

Aqueous Acids and Bases (Appendix I). The Ka value for ascorbic acid increaseswith increasing temperature. State whether the student’s aqueous solution waslikely warmer or colder than standard temperature (25 °C) when the pH wastested.

Figure 1Hydrofluoric acid is a weak acid but it etches glass. The etching isnot due to the presence ofhydronium ions, because, as youknow, even the strongest acids areroutinely stored in glass containers.

Figure 2Vitamin C (ascorbic acid) is presentin many fresh foods, particularlycitrus fruits. It is essential to thebody to promote healing of injuriesand to fight infection.

In the winter of 1535, the men ofJacques Cartier’s expedition weresuffering from scurvy (vitamin Cdeficiency) in their fort at Stadacona(now Montreal). Cartier lost 25 menbefore learning of a simple curefrom the Iroquois. The Aboriginalpeople prepared a tea made fromwhite cedar needles—rich inascorbic acid—that cured suffererswithin days.


744 Chapter 16 NEL

Base Strength and the Ionization Constant, KbThe Brønsted–Lowry concept specifies that the strongest base possible in aqueous solu-tion must be hydroxide ions. So, hydroxide ion is considered to be “the” strong base—somewhat parallel to the strong acids. A difference is that many ionic substances containhydroxide ions initially, before being dissolved; whereas hydronium ions are alwaysformed by the reaction of some entity with water, after a substance dissolves. Recall thatfor solutions of all ionic hydroxides, such as NaOH(aq) or Ca(OH)2(aq), we assumethat the compound dissociates completely upon dissolving. Therefore, finding thehydroxide ion concentration does not involve any reaction with water or any reaction equi-librium. Rather, in these cases, we find the hydroxide ion concentration directly from adissociation equation, as shown in the next Communication Example.

Find the hydroxide ion concentration of a 0.064 mol/L solution of barium hydroxide.

Solution

Ba(OH)2(aq) → Ba2�(aq) � 2 OH�(aq) (complete dissociation)

[OH�(aq)] � [Ba(OH)2(aq)] � �21

�

� 0.064 mol/L � �21

�

� 0.13 mol/L

According to the stoichiometric ratio, the hydroxide ion concentration is 0.13 mol/L.


Other entities that act as bases are collectively called weak bases, because empiricalevidence indicates that they attract protons less than hydroxide ions do. Turn to thetable of Relative Strengths of Aqueous Acids and Bases (Appendix I), and observe that,to the right of every acid entity formula listed, there is a formula for an entity that isidentical to the acid’s formula, except that one proton (represented as H�) is missing. Asyou have learned, such an entity is called the conjugate base of the acid. While the list ofthese entities (reading upward) can be taken to be a list of bases and their (decreasing)relative strengths, note that there is no corresponding value given to show the extent ofreaction of these entities with water. This convention is common in chemistry—infor-mation about bases in solution must be derived from a table of acid values.

Bases vary in strength much as acids do, and the system used to identify, classify, andrank the strength of bases is quite similar to the one you have just learned for acids.Note that the table in Appendix I shows that the weaker an acid is, the stronger its con-jugate base (and vice versa). This observation is common sense. If it is very easy toremove the proton from an acid, then the entity remaining (the conjugate base) must notattract protons very well.

As you know, weak bases react only partially (usually much less than 50%) with waterin aqueous solution to produce hydroxide ions. We can communicate the strength ofany weak base using the equilibrium constant expression for its reaction with water indilute aqueous solution. This equilibrium constant for weak bases is another specialcase, where the Kc constant is called the base ionization constant, Kb.

Consider a solution of sodium citrate, Na3C6H5O7(aq). We assume that in aqueoussolution, any ionic compound dissociates completely into its component ions. In this case,

DID YOU KNOW ??Sailors and ScurvyVitamin C deficiency (from a dietarylack of fresh fruits and vegetables)causes scurvy, a disease where thebody cannot fight infections, andcuts, sores, and bruises do not heal.Until the last half of the 18thcentury, this disease was a terribleproblem for sailors, and a majorcause of death on long voyages. Forexample, on his famous voyage of1498, Vasco da Gama lost 100 of160 crewmen to scurvy. The diet ofEuropean sailors was primarilysalted meat and hard biscuits—itemsselected for their resistance tospoilage.

In 1747, James Lind, a Scottishphysician, wrote that feeding citrusfruit to scurvy victims effected anamazingly rapid cure. Captain Cooktried this remedy on his crew duringhis famous voyages of exploration inthe 1770s, and reported that he wasastounded to have lost only oneman on a three-year voyage!

By 1795, the British Royal Navy(after decades of deliberation)formally adopted the practice ofproviding lemon juice (incorrectlyreferred to as lime juice) for allhands. Scurvy in the fleet was wipedout, and British sailors have beencalled “limeys” ever since.

Note that this story illustratesanother case of technology leadingscience: The cure for scurvy wasknown long before the cause of thedisease was explained.

Interestingly, sailors on the longsea voyages undertaken by theChinese during the Ming Dynasty(1368–1644) had no problems withscurvy because their traditional dietincluded fresh germinated soyabeans, the shoots of which arenaturally rich in Vitamin C.



Section 16.3

the ions are aqueous sodium and citrate ions: Na�(aq) and C6H5O73�(aq). Sodium ions

are “spectators,” with no apparent acidic or basic properties. The citrate ions, however,are found on the Relative Strengths of Aqueous Acids and Bases data table (Appendix I)in the conjugate bases column. So the citrate ion is a weak Brønsted–Lowry base andreacts with water to form a basic solution at equilibrium.

The reaction equilibrium equation for aqueous sodium citrate is

C6H5O73�(aq) � H2O(l) 0 HC6H5O7

2�(aq) � OH�(aq)

The equilibrium law expression is

Kb �

Just as with Ka values, there are two kinds of weak base solution calculations thatinvolve the Kb constant:

• calculating a Kb value from empirical (measured) amount concentration data

• using a Kb value to predict an amount concentration of hydroxide ions for anaqueous solution where the initial weak base concentration is known

Calculating Kb from Amount ConcentrationsCalculating Kb for a weak base uses essentially the same method as calculating Ka for aweak acid. Often, an extra (simple) calculation step must be included because meas-ured pH values may need to be converted to find the equilibrium hydroxide ion con-centration, as shown in the following Communication Example.

[HC6H5O72�(aq)][OH�(aq)]

��[C6H5O7

3�(aq)]

A student measures the pH of a 0.250 mol/L solution of aqueous ammonia and finds it tobe 11.32.

Calculate the Kb for ammonia from this evidence. Show the establishment of the reactionequilibrium using an ICE table.

Solution

NH3(aq) � H2O(l) 0 NH4�(aq) � OH�(aq)

Kb =

At equilibrium:

[H3O�(aq)] � 10�pH

� 10�11.32

� 4.8 � 10�12 mol/L

Kw � [H3O�(aq)][OH�(aq)]

[OH�(aq)] �

�

� 0.0021 mol/L

1.0 � 10�14

��4.8 � 10�12 mol/L

Kw��[H3O

�(aq)]

[NH4�(aq)][OH�(aq)]

��[NH3(aq)]


Learning TipThe concentration of liquidwater is omitted from Kbequilibrium law expressions forthe same reasons that it isomitted from Ka expressions. Aswith Ka values, Kb values arealso only certain to about ±5%.For weak base Kb values, wealso assume that aqueoussolutions are not highlyconcentrated, just as we do forweak acids. Just as with Kavalues, Kb value units areignored, by convention.


Calculating [OH�(aq)] from KbThe second type of calculation involving a Kb value is the prediction of the basicity of anyweak base solution. We can calculate the concentration of hydroxide ions from the ini-tial weak base concentration and the Kb value. There is an automatic problem, however,because tables of relative acid–base strengths do not normally list Kb values. We use theautomatic relationship between conjugate acid–base pairs to deal with this problem.

The Ka�Kb Relationship for Conjugate Acid–Base Pairs To develop the next useful concept, we use the common weak acid, acetic acid,CH3COOH(aq), and its conjugate base, the acetate ion, CH3COO�(aq). Of course, itwould be just as correct to state that we will use the weak base, acetate ion, and its con-jugate acid, acetic acid.

746 Chapter 16 NEL

[NH4�(aq)] � [OH�(aq)]

� 0.0021 mol/L

[NH3(aq)] � (0.250 � 0.0021) mol/L

� 0.248 mol/L

Kb � = 0.000 018

From this evidence, and according to the equilibrium law, the Kb for aqueous ammoniais 1.8 � 10�5.

[0.0021 mol/L][0.0021 mol/L]��

[0.248 mol/L]

Table 6 ICE Table for NH3(aq) � H2O(l) 0 NH4�(aq) � OH�(aq)

[NH3 (aq)] [NH4�(aq)] [OH�(aq)]

Concentration (mol/L) (mol/L) (mol/L)

Initial 0.250 0 0

Change � 0.0021 � 0.0021 � 0.0021

Equilibrium 0.248 0.0021 0.0021

Figure 3Dyes with molecular structuresbased on aniline make possible avariety of bright, stable colours forfabrics and leathers.

Learning TipThe use of chemical (IUPAC)and common names can beconfusing if you have notmemorized a few examples. Inthis textbook, the normalpractice is to give the chemicalname, followed by a commonname in parentheses. A fewsubstances are so widely used,however, that the commonname is ubiquitous (usedeverywhere, even by chemists).Baking soda (sodiumbicarbonate, sodium hydrogencarbonate) and vinegar (aceticacid, ethanoic acid) are twoexamples of such chemicals;the common name is oftengiven, rather than the IUPACname. You are expected to befamiliar with such examples,and others listed in Appendix J.

Practice10. List some empirical properties that would be useful when distinguishing strong bases

from weak bases.

11. For each of the following weak bases, write the chemical equilibrium equation andthe equilibrium law expression for Kb. (a) CN�(aq) (b) SO4

2�(aq)

12. The hydroxide ion concentration in a 0.157 mol/L solution of sodium propanoate,NaC2H5COO(aq), is found to be 1.1 � 10�5 mol/L. Calculate the base ionizationconstant for the propanoate ion.

13. Aniline, C6H5NH2, is used to make a wide variety of drugs and dyes (Figure 3). It hasthe structure of an ammonia molecule, with a phenyl group substituted for onehydrogen, and, like ammonia, acts as a weak base. If the pH of a 0.10 mol/L anilinesolution was found to be 8.81, what is its Kb?



Section 16.3

The equilibrium reaction and law expression for aqueous acetic acid are


Ka �

For the weak base, aqueous acetate ion,

CH3COO�(aq) � H2O(l) 0 CH3COOH(aq) � OH�(aq)

Kb �

Notice what happens when we multiply these equilibrium constant expressions:

Ka � Kb � �

� [H3O�(aq)][OH�(aq)] (Does this expression look familiar?)

� Kw

So, for any conjugate acid–base pair, Kw � KaKb and Kb � �K

Kw

a

�.

This last equation is particularly useful because now, to find a Kb value for any baselisted on the Relative Strengths of Aqueous Acids and Bases table (Appendix I), you needonly identify its conjugate acid, and then divide Kw, 1.0 � 10�14, by the Ka value givenfor that conjugate acid.

[CH3COOH(aq)][OH�(aq)]��

[CH3COO�(aq)]


��[CH3COOH(aq)]

[CH3COOH(aq)][OH�(aq)]��

[CH3COO�(aq)]


��[CH3COOH(aq)]

Solid sodium benzoate forms a basic solution. Determine Kb for the weak base present.

NaC6H5COO(s) → Na�(aq) � C6H5COO�(aq) (complete dissociation)

First identify, from the Relative Strengths of Aqueous Acids and Bases table, and theentities present in solution, which entity is reacting as a base.

The benzoate ion must be the weak base entity. Its equilibrium reaction with water is

C6H5COO�(aq) � H2O(l) 0 C6H5COOH(aq) � OH�(aq)

From the equilibrium equation, the conjugate acid of the benzoate ion is identified as benzoic acid. From the acid–base table,

Ka for C6H5COOH(aq) � 6.3 � 10�5

Using the Kw relationship for conjugate acid–base pairs, find Kb for C6H5COO�(aq).

Kb �

�

� 1.6 � 10�10

According to the Kw relationship, the benzoate ion has a Kb value of 1.6 � 10�10.

1.0 � 10�14

��6.3 � 10�5

Kw�Ka

SAMPLE problem 16.4

Why Acidity?It should be obvious by now thatpH values are used much morethan pOH values, and Ka constantsare listed in tables rather than Kbconstants. You might wonder whywe choose to emphasize thehydronium ion properties ofaqueous solutions—after all,hydroxide ions play exactly theopposite (and an equallyimportant) role. But historically,chemists have always been muchmore concerned with acidicproperties than with basicproperties. Acids dissolve (reactwith) many metals to formcommon ionic compounds andhydrogen. Hydrogen gas was afascinating product to earlychemists, both because it is lighterthan air and because it is violentlyexplosive. The stronger acids tendto have painfully sharp odours andflavours, and react destructivelywith human tissue—all propertiesthat make acids quite noticeable.Even when it became clearlyunderstood that acids and baseswere opposite aspects of the sameconcept, it seemed easier tochoose acid properties as theinitial basis for studying acid–baserelationships. We talk about the“acidity” of something routinely; itis a common term, found in anypocket dictionary. What is theequivalent term for how basicsomething is? There is anaccepted word for it, but you’llneed a really good (very big)dictionary to find it. Or a goodChemistry text, of course...

DID YOU KNOW ??


Now we can easily predict how basic a solution of known concentration will be fromits Kb value. Predicting basicity always involves calculating the hydroxide ion concentration,and may also involve calculating a value for pOH, pH, and/or percent reaction. The firststep may be determining the Kb value by using a table of Ka values. The form of the cal-culation then follows the same format as the equivalent type of question for weak acids,with a similar assumption made, and the same restrictions.

• Assume that the initial weak base aqueous concentration decreases so little that it isnumerically unchanged at equilibrium, but only if the initial amount concentrationof the weak base is at least 1000 times greater than its Kb value.

• For questions in this text, you are not required to do calculations using the quad-ratic formula.

The Sample Problem that follows shows the format for calculations of this type, andalso an example of each conversion that may be required.

Recall, from Section 16.2, that, in aqueous solution, the amphiprotic ion HCO3�(aq)

reacts with water as a base to a very much greater extent than it does as an acid. Youmay, therefore, assume that the solution is basic because the ion reacts as a base, andignore its negligible reaction extent as an acid.

748 Chapter 16 NEL

Calculate Kb for the weak base aqueous ammonia, commonly used in commercial windowcleaning solutions. Write the equation for the equilibrium.

SolutionNH3(aq) � H2O(l) 0 NH4


For NH3(aq), NH4�(aq) is the conjugate acid:

Ka � 5.6 � 10�10

Kb �

�

� 1.8 � 10�5

According to the table of Relative Strengths of Aqueous Acids and Bases, and the Ka–Kbrelationship, the Kb value for aqueous ammonia is 1.8 � 10�5.

1.0 � 10�14

��5.6 � 10�10

Kw�Ka


Find the hydroxide ion amount concentration, pOH, pH, and percent reaction (ionization)of a 1.20 mol/L solution of baking soda.

The compound is sodium hydrogen carbonate, NaHCO3(s). The weak base is theHCO3

�(aq) ion, produced by dissolving NaHCO3(s) in water.

For HCO3�(aq), H2CO3(aq) is the conjugate acid:

Ka � 4.5 � 10�7

Find the Kb value using the Ka–Kb–Kw relationship.

Kb �

�

� 2.2 � 10�8

1.0 � 10�14

��4.5 � 10�7

Kw�Ka

SAMPLE problem 16.5



Section 16.3

Write the equation for the reaction and the equilibrium expression for Kb for hydrogencarbonate ion.

HCO3�(aq) � H2O(l) 0 H2CO3 (aq) � OH�(aq)

Kb �

At equilibrium, let x be the numerical value of [OH�(aq)] and also [H2CO3(aq)].

Then [HCO3�(aq)] � (1.20 � x)

Test the assumption that (1.20 � x) is numerically equal to 1.20.

� � 5.5 � 107 (much greater than 1000)

The assumption holds, so substitute into the Kb expression and solve for x.

2.2 � 10�8 �

x � �2.2 ��10�8 �� 1.20�� 1.6 � 10�4

[OH�(aq)] � 1.6 � 10�4 mol/L

Use [OH�(aq)] to find the value of pOH.

pOH � �log [OH�(aq)]

� �log (1.6 � 10�4)

� 3.80

Use the pOH value to find pH.

pH � 14.00 � pOH

� 14.00 � 3.80

� 10.20

Percent reaction is the percent ratio of the equilibrium concentration of producedhydroxide ions to the initial concentration of hydrogen carbonate ions.

percent reaction � � 100%

� � 100%

� 0.013%

A 1.20 mol/L sodium hydrogen carbonate solution has a hydroxide ion concentration of1.6 � 10�4 mol/L, a pOH of 3.80, a pH of 10.20, and a percent reaction of 0.013%.

1.6 � 10�4 mol/L��

1.20 mol/L

[OH�(aq)]equilibrium��[HCO3

�(aq)]initial

x2

�1.20

1.20��2.2 � 10�8

[HCO3�(aq)]initial��Kb

[H2CO3(aq)][OH�(aq)]��

[HCO3�(aq)]

Table 7 ICE Table for HCO3�(aq) � H2O(l) 0 H2CO3(aq) � OH�(aq)

Amount [HCO3�(aq)] [H2CO3 (aq)] [OH�(aq)]


Initial 1.20 0 0


Equilibrium (1.20 � x) x x

The Effect of Amphoteric EntitiesFinally, we deal with the problem of amphoteric entities. If an entity can react as eithera Brønsted–Lowry acid or base, how do you know which will be the predominant reac-tion in its aqueous solution? You can determine the answer with one simple calculation.

Visualizing Reaction ExtentsFrom any percent reaction value,you can get a better idea of thereaction extent at equilibrium byconverting mentally to a whole-number ratio. If bicarbonate ionsreact with water 0.013% atequilibrium, the ratio is 0.013:100,or 13:100 000. This means that atany given instant, about 13 ofevery 100 000 initial bicarbonateions have changed to carbonicacid molecules, but 99 987 of themhave not. The few ions that havechanged are responsible for all thebasicity of the solution!

If you really want to test yourpowers of imagination, try tovisualize that for every 100 000bicarbonate ions in this aqueoussolution, there are approximately 5 500 000 water molecules. Nowthink about all of these entitiesconstantly moving at speedsaveraging about 1500 km/h, andthen imagine every one of themcolliding with another molecule orion at a rate of approximately 7 000 000 000 times everysecond…

DID YOU KNOW ??


750 Chapter 16 NEL

Read the Ka value for the entity from the table of Relative Strengths of Aqueous Acidsand Bases. Calculate the Kb value for the entity from Kw and the Ka value (from the table)of its conjugate acid, as shown in Sample Problem 16.5. The higher of the “K” valuestells you which reaction predominates, and, therefore, whether the entity reacts as anacid or as a base in aqueous solution.

Which reaction predominates when NaHSO3(s) is dissolved in water to produceHSO3

�(aq) solution?

SolutionKa � 6.3 � 10�8

The conjugate acid is H2SO3(aq), with Ka � 1.4 � 10�2.

So, for HSO3�(aq),

Kb �

�

� 7.1 � 10�13

The Ka value for HSO3�(aq) far exceeds its calculated Kb value.

According to the Kw relation and the equilibrium law, an aqueous solution of thissubstance will be acidic because a hydrogen sulfite ion will react predominately as aBrønsted–Lowry acid.

1.0 � 10�14

��1.4 � 10�2

Kw�Ka


Section 16.3 Questions1. Codeine, an ingredient in these migraine pills (Figure 4),

has a Kb of 1.73 � 10�6. Calculate the pH of a 0.020 mol/Lcodeine solution (use Cod as a chemical shorthand symbolfor this complex weak base entity.).

2. What is the pH of a 0.18 mol/L cyanide ion solution?

3. Acetylsalicylic acid (ASA) is a painkiller used in manyheadache tablets. This drug forms an acidic solution thatattacks the digestive system lining. The Merck Index lists itsKa at 25 °C to be 3.27 � 10�4. Explain the difficulty incalculating the pH of a saturated 0.018 mol/L solution ofacetylsalicylic acid, C6H4COOCH3COOH(aq).

4. Boric acid is used for weatherproofing wood andfireproofing fabrics. Very dilute aqueous boric acid is usedin eyewash solution as a preservative. Predict the pH of a0.50 mol/L solution of boric acid.

5. Salicylic acid (2-hydroxybenzoic acid, C6H5OHCOOH(s)) isan active ingredient of medications, such as Clearasil®,that are used to treat acne. Since the Ka for this acid wasnot listed in any convenient references, a student tried todetermine the value experimentally. She kept adding waterslowly to a 1.00 g sample, with constant stirring, until thecrystals all dissolved. The volume of solution was 460 mL.She found that the pH of this solution of salicylic acid was2.40 at 25 °C. Calculate the ionization constant for this acid.

6. Sodium ascorbate is the sodium salt of ascorbic acid and isused as an antioxidant in food products. The pH of a 0.15 mol/L solution of the ascorbate ion, HC6H6O6

�(aq), is8.65. Calculate the Kb of the ascorbate ion.

7. Sodium hypochlorite is a strong oxidizing agent that is afire hazard when in contact with organic materials.Solutions of sodium hypochlorite are used as bleach anddisinfectant. Determine the hydroxide ion amountconcentration of a sodium hypochlorite solution sold ashousehold bleach. The bleach bottle label reads “5.25% (bymass) when packed.”

8. Write the equilibrium law expression for acetic acid reacting(ionizing) in water. Use Le Châtelier’s principle to explain, interms of equilibrium shift, why dilute acetic acid solutionshave a higher percent reaction than more concentratedsolutions. Consider that diluting a solution containingreacting aqueous entities is very similar to increasing thevolume of a container of reacting gaseous entities.

9. Predict and write the predominant Brønsted–Lowryreaction in aqueous solutions of the following substances:(a) K2HPO4(s) (b) NaH2PO4(s) (c) Na2HC6H5O7(s)

10. Calculate the pH of 5.0 � 10�2 mol/L solutions of each ofthe reagents in question 9.

Figure 4Codeine (methylmorphine,C18H21NO3) is available byprescription as an ingredient insome analgesic (pain relief)medicines. This narcotic alkaloidwas initially extracted from opium.Other alkaloids include nicotine,quinine, cocaine, strychnine, andcaffeine. All are common plant-derived organic compounds basedon nitrogen-containing ringstructures.



16.416.4Interpreting pH CurvesFor many acid–base reactions, the appearance of the products resembles that of thereactants, so we cannot directly observe the progress of a reaction. Also, acids cannoteasily be distinguished from bases except by measuring pH. As you first learned inChapter 8, a graph showing the continuous change of pH during an acid–base titra-tion, continued until the titrant is in great excess, is called a pH curve (titration curve)for the reaction. The pH values and changes provide important information about thenature of acids and bases, the properties of conjugate acid–base pairs and indicators,and the stoichiometric relationships in acid–base reactions.

Buffering Regions, Endpoints, and IndicatorsThe pH curves for acid–base reactions have characteristic shapes. You have learned thata common reason for plotting a pH curve is to determine the value of the solution’s pHwhen an acid–base reaction reaches its equivalence point. We can then use this informationto select an appropriate indicator—one that will produce an easily seen endpoint whena solution reaches this pH value. Then we can use the same acid–base reaction, done asa titration to that indicator’s endpoint, for titration analysis.

In all titration analyses, the critical measured quantity is the titrant volume requiredto reach the titration endpoint. For accurate analysis, this value needs to be as close aspossible to the theoretically exact value needed to reach the reaction equivalence point.Analysis titrations, like all empirical work, necessarily involve some uncertainty. We callany variation between endpoint values and equivalence point values the titration errorof an experiment. Good technique and careful application of knowledge can ensure thatthis variation is as small as possible—preferably negligible.Remember:

• Endpoint refers to that point in a titration analysis where the addition of titrant isstopped. The endpoint is defined (empirically) by the observed colour change of anindicator.

• Equivalence point refers to that point in any chemical reaction where chemicallyequivalent amounts of the reactants have been combined. The equivalence point isdefined (theoretically) by the stoichiometric ratio from the reaction equation.

Endpoints are easily detectable because pH changes a great deal, and very abruptly, as thereaction solution changes (at the equivalence point) from a tiny excess of acid to a tinyexcess of base (or vice versa). But this effect begs the question: Why, for most of a titra-tion, does the pH hardly change at all?

To better understand the information that a pH curve provides, we will use a familiarreaction about which we already know a great deal. The pH curve for the strongacid–strong base reaction (Figure 2) has three regions of interest. In the course of thisexcess titration, the pH first changes very slowly, then very rapidly, and finally very slowlyagain, as titrant is steadily added. Notice that the addition of acid titrant to the basesample initially has very little effect on the pH of the solution in the flask. In fact, the highpH has not changed much even after enough acid has been added to react with 90% ofthe original amount of the base sample. This nearly level region on a pH curve identi-fies a buffering region. Buffering is the property of some solutions (often called buffersolutions) of resisting (counteracting) any significant change in pH when an acid or abase is added. In this particular case, buffering occurs because any strong acid addedimmediately reacts with excess hydroxide ions. Consider that before any acid is added,the sample solution is primarily water molecules and hydroxide ions.

Learning TipWhen you began to study acidsand bases, the term “neutralize”was understood to refer to anacid–base reaction that had thesame final pH as neutral purewater: a pH of 7. Later, youlearned that the pH at theequivalence point for an acid–base reaction is almost never 7,except for the strong acid–strong base reaction of hydro-nium ions with hydroxide ions.

As commonly used, the word“neutralize” is taken to have amore general meaning: that onesubstance can completely orpartially lessen the acidic orbasic properties of anothersubstance. When we saysodium carbonate can be usedto neutralize spilled nitric acid(Figure 1), we mean thatadding the base brings the pHvalue up much closer to neutral.

Figure 1Soda ash (sodium carbonate)is blown onto nitric acid thathas spilled from some railwaytank cars, to neutralize thehighly reactive and dangerousacid.


752 Chapter 16 NEL

As long as a large amount of hydroxide ions is present, any added acid is immediatelyconverted to water, producing a solution that still consists primarily of water moleculesand hydroxide ions. The hydroxide ion concentration decreases a little, but that affectsthe solution pH value only very slightly. This pH “levelling effect” finally fails near theequivalence point—when the hydroxide ions in the solution become almost completelyconsumed. But, until then, the solution maintains a (nearly) constant pH. This bufferingproperty turns out to be critically important for a great number of reactions in appliedchemistry and biology. Note that once excess acid has been added, the solution consistsprimarily of water molecules and hydronium ions. Again, the pH remains stable becauseadding more hydronium ions now does not change the nature of the solution; it onlyincreases the hydronium ion concentration slightly. Another buffering region is estab-lished—this time at a low pH. You will learn more about the causes of buffering regionsand the nature of buffer solutions later in this section.

Following the first buffering region there is a very rapid change in pH for a very smalladditional volume of the titrant. The inflection point on the pH curve represents theequivalence point of this reaction: At this point, we know from theory that the pH mustbe 7. We also know from experience (Chapter 8) that this reaction can be used for titra-tion analysis, if an indicator is selected that changes colour in solution at a pH very closeto 7. As shown in Figure 3, bromothymol blue indicator has a colour change pH rangewith a middle value very close to (just slightly below) pH 7, which makes it an appropriateindicator for this particular reaction. The values below show that the (theoretical) titrantvolume required for complete reaction, as predicted by calculation, should be 24 mL.Figure 3 shows that if the colour change of bromothymol blue to a green intermediatecolour is the endpoint, then the titrant volume actually measured at the endpoint (24 mL)is negligibly different from the theoretical predicted value. So, this titration analysis givesvery accurate results.

OH�(aq) � H3O�(aq) → 2 H2O(l)

25 mL � 0.48 mol/L 24 mL � 0.50 mol/L

12 mmol 12 mmol

Figure 2This pH curve for the continuous and excess addition of 0.50 mol/L HCl(aq) to a 25 mL sampleof 0.48 mol/L NaOH(aq) helps chemists to understand the nature of acid–base reactions.

25.0 mL of 0.48 mol/L NaOH(aq) Titratedwith 0.50 mol/L HCl(aq)

14

12

10

8

6

4

2

5 10 15 20 25 30 35 40 45 50

Volume of HCl (mL)

pH

00

buffering region

buffering region

the curve inflection pointrepresents the reactionequivalence point

Learning TipRecall from Unit 4 that theinflection point of any plottedcurve is the point where thecurvature of the line changesdirection. For instance, on thestrong acid–strong base pHcurve in Figure 2, the linecurves downward until about 24 mL of acid has been added.At some observable point, theline curvature changes toupward.

For this specific titration, thechange in pH is so large and sorapid that the pH plot becomesa nearly vertical line. In such acase the equivalence point canalso be identified as the “mid-point” of the steep drop in theplotted line.

Also recall that, for thisparticular reaction (only), wealready know (from memory)that the equivalence point pHmust be 7.0 (at SATP), withoutactually needing to plot atitration curve: The reaction ofhydroxide and hydronium ionsis highly quantitative, and theonly product is water.



Acid–Base Indicator EquilibriumAcid–base indicators may be better understood and explained by using Brønsted–Lowrydefinitions. Any acid–base indicator is really two entities for which we use the samename: a Brønsted–Lowry conjugate acid–base pair. At least one (most often both) ofthe entities is visibly coloured, so you can tell simply by looking when it forms or is con-sumed in a reaction. Phenolphthalein is a common example (Figure 4), where the con-jugate base form is bright red and the conjugate acid form is colourless. Typically,common indicators such as methyl red (Figure 5) have quite complex molecular formulas,so we use a (very simplified) shorthand to identify them, for convenience. For example,we symbolize the (invisible) acid form of phenolphthalein as HPh, and the (bright red)base form as Ph�. When an indicator equilibrium is shifted to the point where equalconcentrations of both forms exist, an intermediate colour may be observed. For example,equal concentrations of the blue and yellow forms of bromothymol blue (at a pH of6.8) appear green to the human eye.

The explanation of the behaviour of acid–base indicators depends, in part, on both theBrønsted–Lowry concept and the equilibrium concept. An indicator is a conjugate weakacid–weak base pair formed when an indicator dye dissolves in water. Using HIn(aq) to

Section 16.4

14

12

10

8

6

4

2

5 10 15 20 25 30 35 40 45 50

Volume of HCl (mL)

pH

25.0 mL of 0.48 mol/L NaOH(aq) Titratedwith 0.50 mol/L HCl(aq)

alizarin yellow

bromothymol blue

orange IV

00

Figure 3Alizarin yellow is not a suitableindicator because it will changecolour long before the equivalencepoint of this strong acid–strong basereaction, which theoretically has apH of exactly 7. Orange IV is alsounsuitable; its colour change wouldoccur too late. The pH at the middleof the colour change range forbromothymol blue is 6.8, which veryclosely matches the equivalencepoint pH; so, a titration analysisendpoint for this reaction, asindicated by bromothymol blue,should give accurate results.

Figure 4Sodium hydroxide solution has beenadded to hydrochloric acidcontaining phenolphthaleinindicator, which is colourless inacids and red in bases. The redcolour, indicating the temporarypresence of some unreacted sodiumhydroxide, is rapidly disappearing asthe flask contents swirl and mix.

Figure 5A few common acid–base indicatorsare shown here. Each indicator hasits own pH range over which itchanges colour from the acid form(HIn) at the lower pH value to thebase form (In�) at the higher pHvalue. Material Safety Data Sheetsare available for these chemicals.


754 Chapter 16 NEL

represent the acid form and In�(aq) to represent the base form of any indicator, we canwrite the following equilibrium equation. (Litmus colours are given below the equa-tion as an example.)

HIn(aq) � H2O(l) 0 In�(aq) � H3O�(aq)

acid base base acid

red (litmus colour) blue

According to Le Châtelier’s principle, an increase in the hydronium ion concentra-tion will shift the equilibrium to the left. Then more indicator will change to the colourof the acid form (HIn(aq)). This change happens, for example, when litmus is addedto an acidic solution. Similarly, in basic solutions the hydroxide ions remove hydroniumions by reacting with them to form water, with the result that the equilibrium shifts tothe right. Then the base colour of the indicator (In�) predominates. Since different indi-cators have different acid strengths, the acidity or pH of the solution at which an indi-cator changes colour varies (Figures 5 and 6). These pH values have been measured andare reported in the table of acid–base indicators on the inside back cover of this book.

Figure 6The visible colour of methyl red indicator depends on the equilibrium proportions of its twocoloured forms at the pH of the solution in which it is placed. Methyl red exists predominantlyin its red (acid) form at pH values less than 4.8, and in its yellow (base) form at pH valuesgreater than 6.0. Between pH values of 4.8 and 6.0, intermediate orange colours occur, as bothforms of the indicator are present in detectable quantities.

N

H3C

H3C

N N COO– + H3O+

pH < 4.8

pH > 6.0

HMr

Mr–

N

H3C

H3C

N N COOH + H2O

conjugate pair

Learning TipTypically, acid–base indicatorssuch as methyl red (Figure 6)are large molecules with quitecomplex formulas. Smaller mol-ecules and ions are less likelyto interact with light waves inthe visible spectrum.

The shorthand symbolism weuse to represent indicators ispartly for convenience, but alsopartly to reinforce the conceptthat indicators are conjugateacid–base pairs. Consider theactual formulas for the methylred entities, which we representas HMr for the acid form, andMr� for the base form. Thissimplified symbolism makes theproton transfer, which causesthe colour change, moreobvious.

Practice1. The shape of a pH curve is interpreted to describe the change of properties

throughout the course of an acid–base reaction. (a) In terms of curve shape, describe the characteristics of a buffering region.(b) In terms of pH change and titrant volume, explain what a buffering region

represents.(c) In terms of curve shape, describe the characteristics near an equivalence point.(d) In terms of pH change and titrant volume, explain what an equivalence point

represents.

2. For each indicator following, write a Brønsted–Lowry equilibrium equation for thereaction of the acid form of the indicator with water, to form a hydronium ion and the


According to the Brønsted–Lowry concept, acid–base indicators are simply colouredconjugate acid–base pairs. As for all other weak acids, the conjugate acid forms ofthese indicators must differ in strength for different entities.



Polyprotic Entities and Sequential ReactionsExpanding on what you learned in Chapter 8, and adding the Brønsted–Lowry concept,we now examine some acids and bases that can lose (or gain) more than one proton.Polyprotic acids can lose more than one proton, and polyprotic bases can gain morethan one proton, in Brønsted–Lowry transfers. If more than one proton transfer actu-ally occurs in the course of a titration, chemists believe the process occurs as a series ofsingle-proton transfer reactions. Typical pH curves for reactions of diprotic or triproticacids and bases differ from those of monoprotic acids and bases—and can provide usefulinformation about the reactions going on.

We observe (Figure 7) that the pH curve for the addition of HCl(aq) to Na2CO3(aq)shows two equivalence points—two significant changes in pH. Chemists interpret suchcurves as indicating how many quantitative reactions have occurred, sequentially, forthat particular acid–base titration. Here, for example, two successive quantitative reac-tions have occurred. The two equivalence points evident in Figure 7 can be explained bytwo different proton transfer equations.

Section 16.4

25.0 mL of 0.50 mol/L Na2CO3(aq) Titratedwith 0.50 mol/L HCl(aq)

12

10

8

6

4

2

5 10 15 20 25 30 35 40 45 55 60 65 7050

Volume of HCl (mL)

pH

75

methyl orange

00

first reactionequivalence point

second reactionequivalence point

Figure 7A pH curve for the addition of0.50 mol/L HCl(aq) to a 25.0 mLsample of 0.50 mol/L Na2CO3(aq)can be used to select anappropriate indicator for thisreaction done as a titration analysis.The colour change of methyl orange(from pH 4.4 to pH 3.2) means it willshow an endpoint that correspondsclosely to the second reactionequivalence point.

base form of the indicator. Use the same indicator symbolism as used in the table ofAcid–Base Indicators on the inside back cover of this book. Identify both conjugateacid–base pairs for each equation, and label all entities “acid” or “base.”(a) methyl orange(b) indigo carmine(c) thymol blue (the entity with two “H’s” in the symbolized formula)(d) thymol blue (the entity with one “H” in the symbolized formula)(e) bromothymol blue

3. Which indicator entity, in your answers to question 2, can behave as both an acid anda base? What chemical term is used to describe this property of an entity?

4. (a) Refer to the table of Acid–Base Indicators to determine which of the five conjugate acid forms for the indicators in question 2 is the strongest acid. (Notethe colour change pH ranges.) Then rank the other four indicators beneath it inorder of decreasing strength, as in a relative strengths of acids table.

(b) Suppose that five identical samples of an aqueous hydroxide ion solution eachhave four drops of a different indicator (from question 2) added. If each sample isnow titrated with the same HCl(aq) titrant, in which indicator/sample combinationwill the least amount of titrant cause a colour change?


756 Chapter 16 NEL

First, protons transfer from hydronium ions to carbonate ions, the strongest basepresent in the initial sample. The first significant drop in pH is evidence that the first reac-tion is quantitative. We write the Brønsted–Lowry equation in the usual way, starting withthe acid and base entities present in the sample solution, and also listing the H3O

�(aq)that is being added.

SA A

H3O�(aq), Cl�(aq), Na�(aq), CO3

2�(aq), H2O(l)SB B

H3O�(aq) � CO3

2�(aq) → H2O(l) � HCO3�(aq)

Because the first reaction is quantitative, essentially all the carbonate ions will have beenconsumed at the equivalence point, and an equal quantity of (new) hydrogen carbonateions will have been formed. Then, in a second reaction, protons transfer from additional(continuosly added) hydronium ions to the hydrogen carbonate ions that were formedin the first reaction.

SA A A

H3O�(aq), Cl�(aq), Na�(aq), H2O(l), HCO3

�(aq)B SB

H3O�(aq) � HCO3

�(aq) → H2O(l) � H2CO3(aq)

Because the second reaction is also quantitative, all of the hydrogen carbonate ions havereacted at the equivalence point, and only newly formed carbonic acid molecules remainin the solution. Continuing to add more hydronium ion after this point will cause no sig-nificant further change (no third reaction), because the strongest base in the systemnow is water.

The carbonate ion is a diprotic base because it can accept a total of two protons. Otherpolyprotic bases include sulfide ions and phosphate ions. The conjugate entities formedwhen these bases gain successive protons are:

S2�(aq) — HS�(aq) — H2S(aq)

PO43�(aq) — HPO4

2�(aq) — H2PO4�(aq) — H3PO4(aq)

Other polyprotic acids include oxalic acid and phosphoric acid (Figure 8).

HOOCCOOH(aq) — HOOCCOO�(aq) — OOCCOO2�(aq)

H3PO4(aq) — H2PO4�(aq) — HPO4

2�(aq) — PO43�(aq)

Evidence from pH measurements clearly shows that, for every proton transferred bya polyprotic entity, the strength of the new acid or base entity formed greatly decreases.The new entity may be anywhere from 100 to 100 000 times weaker! Chemists believethat electric charge and electrostatic attraction explain this effect. For example, it shouldlogically be much easier to pull a positively charged proton away from a neutral oxalic acidmolecule, than it is to pull one away from a negatively charged hydrogen oxalate ion.

Figure 9 shows the pH curve for phosphoric acid titrated with sodium hydroxide.Phosphoric acid is triprotic, so three reactions should be possible. However, note that onlytwo equivalence points are evident. We describe the process as before. At the first equiv-alence point (pH 4), equal chemical amounts of H3PO4(aq) and OH�(aq) have beencombined in solution. As the last of the H3PO4(aq) reacts, the pH rises abruptly, becausewhen the reaction moves past the equivalence point, only newly formed H2PO4

�(aq)is now present, and it is a much weaker acid than H3PO4(aq).

Figure 8For oxyacids, the protons thattransfer in Brønsted–Lowryreactions are covalently bonded tooxygen atoms (identified by a redcolour in the models above). Ofthese three examples, themonoprotic and diprotic acids arestrong acids, and the triprotic acid isa weak acid.

Acid formulas were originallyderived from stoichiometricevidence. Arguably, it would bemore informative to write themolecular formulas differently, asNO2OH, SO2(OH)2, and PO(OH)3, toclarify the bonding concept. In bothscience and society, however, onceconventions are established, theyare difficult to change.

sulfuric acid, H2SO4(aq)

nitric acid, HNO3(aq)

phosphoric acid, H3PO4(aq)



To write the reaction for the transfer of the first proton, start by listing the entities.

SA A

Na�(aq), OH�(aq), H3PO4(aq), H2O(l)SB B

OH�(aq) � H3PO4(aq) → H2O(l) � H2PO4�(aq)

Since all the H3PO4(aq) has reacted when the reaction has passed the first equivalencepoint, the second plateau must represent the reaction of OH�(aq) with H2PO4

�(aq).The second equivalence point (pH 9) corresponds to the completion of the reaction ofH2PO4

�(aq) with additional OH�(aq) solution. As the last of the H2PO4�(aq) reacts, the

pH begins to rise abruptly again because, when the reaction has passed this (second)equivalence point, only newly formed HPO4

2�(aq) is left, and it is an even weaker acid.

A SA

Na�(aq), OH�(aq), H2O(l), H2PO4(aq)SB B B

OH�(aq) � H2PO4�(aq) → H2O(l) � HPO4

2�(aq)

No pH endpoint is apparent for the possible reaction of HPO42�(aq) with additional

OH�(aq). A clue to this lack of a third endpoint can be obtained from the table of acidsand bases. The hydrogen phosphate ion is an extremely weak acid and apparently doesnot quantitatively lose protons to OH�(aq). For the third reaction, an equilibrium isestablished that gradually shifts right as more base is added. There is no pH increase ata third equivalence point because the reaction never goes to completion.

A SA

Na�(aq), OH�(aq), H2O(l), HPO42�(aq)

SB B B

OH�(aq) � H2PO4�(aq) 0 H2O(l) � HPO4

2�(aq)

>50%

HPO42�(aq) � OH�(aq) 0 PO4

3�(aq) � H2O(l)

As a general rule, only quantitative reactions produce detectable equivalence points in anacid–base titration.

Section 16.4

25.0 mL of 0.50 mol/L H3PO4(aq) Titratedwith 0.48 mol/L NaOH(aq)

12

10

8

6

4

21

5 10 15 20 25 30 35 40 45 55 60 65 7050

Volume of NaOH (mL)

11

9

7

5

3

pH

75 800

0

Figure 9A pH curve for the addition ofNaOH(aq) to a sample of H3PO4(aq) displays only two rapid changes inpH. This result is interpreted asindicating that there are only twoquantitative proton transferreactions of phosphoric acid withhydroxide ions. The third protontransfer reaction never goes tocompletion, but instead establishesan equilibrium.

Learning TipWhen any weak polyprotic acidor base is initially dissolved inwater, we always assume thatthe ionization equilibrium onlyinvolves the first protontransfer—to or from water. Thisis essentially true because thetendency of any entity that isformed by a first ionization tolose or gain a (second) protonis very much less than that ofthe original polyprotic entitydissolved; so much so that itcan safely be considerednegligible.


758 Chapter 16 NEL

Sulfuric acid, H2SO4(aq), is a unique polyprotic acid because it is the only commonone for which the first proton loss is already quantitative in aqueous solution; that is, itis the only strong acid that is polyprotic. Sulfuric acid acts like any of the other fivecommon strong acids, except that its 100% ionization produces hydrogen sulfate ions,HSO4

�(aq), along with hydronium ions, H3O�(aq). Recall that of all the anions of this

type (like HCO3�(aq), or HPO4

2�(aq)), the hydrogen sulfate ion is the only one that isa weaker base than water, so it cannot react as a base in aqueous solution. When reactingas an acid, however, hydrogen sulfate ion is one of the stronger weak acids, and usuallyreacts quantitatively with bases (except for very weak bases, of course). So sulfuric acidwill usually, but not always, react with a base (assuming the base is in excess) in twocomplete proton transfer reactions, one after the other. When you first began a studyof reactions of acids and bases, you would have written a complete reaction of sulfuricacid with sodium hydroxide as

H2SO4(aq) � 2 NaOH(aq) → Na2SO4(aq) � 2 H2O(l)

This reaction equation, if you think about it now, implies that you assumed that both ofthis acid’s “hydrogens” (protons) were “replaced” (transferred). You had no informa-tion at that time, however, about how this transfer occurs. For simple stoichiometriccalculations based on this reaction, this limited understanding (and simplified equa-tion) worked perfectly well. Explaining and predicting other reactions of sulfuric acid,however, is a different story. To do so, understanding how the neutralization processoccurs (as two distinct and sequential reactions) is necessary, because both reactionsmay not be quantitative when a different (weaker) base is involved, and stoichiometriccalculation can only be done for a reaction that is quantitative.

There are almost no overall reactions of common polyprotic acids or bases that havemore than two definite endpoints. For any such reactions that are actually done asanalysis titrations, an indicator can be selected so that the titration can be stopped ateither chosen equivalence point. For the titration of sodium carbonate solution withhydrochloric acid, this would mean titrating through two sequential quantitative reac-tions to the second equivalence point, as shown in Figure 7. For this titration, the secondequivalence point involves a greater pH change, and thus is easier to detect (more accu-rate). As we found earlier, methyl orange indictor is suitable for detecting this second equiv-alence point.

For purposes of stoichiometric calculation, we can combine sequential quantitativeBrønsted–Lowry reaction equations into a single “overall” reaction equation. This exampleis for the titration of sodium carbonate solution with hydrochloric acid.

1. H3O�(aq) � CO3

2�(aq) → HCO3�(aq) � H2O(l) followed by

2. H3O�(aq) � HCO3

�(aq) → H2CO3(aq) � H2O(l)“totals” to

2 H3O�(aq) � CO3

2�(aq) → H2CO3 � 2 H2O(l) (overall reaction)

For purposes of stoichiometric calculation, this equation is equivalent to writing

2 HCl(aq) � Na2CO3(aq) → H2CO3(aq) � 2 NaCl(aq)

because, either way, the reactant acid–base stoichiometric ratio is found to be 2:1.Chemists normally use the simplest representation that will be useful, so Brønsted–Lowryequations are more often used for correctly predicting products and equilibria, whereasstandard chemical formula equations are more often used for “doing” stoichiometry as,for example, in titration analyses.

DID YOU KNOW ??Sulfuric AcidSulfuric acid (Figure 10) is probablythe world’s most importantindustrial chemical. It is used for somany things that the industrialdevelopment and standard of livingof a country may be thought of asbeing proportional to its sulfuricacid production. Canada producesapproximately four million tonnesevery year. U.S. production (theworld’s largest) is about 10 times asmuch. The main direct consumeruse of sulfuric acid is the 4.5 mol/Lsolution inside every standard carand truck battery.

Sulfuric acid plants (Figure 11)can be quite compact, and areassembled from “off-the-shelf”technology. They involve only a fewsimple reactions, starting with sulfur(in Alberta, mostly extracted fromfossil fuels) and oxygen.

Figure 10A model of sulfuric acid, showingrelative atom size, bonding, andmolecular shape

Figure 11More sulfuric acid is manufacturedin North America than any otherchemical.



Section 16.4

Empirical pH curves provide a wealth of information:

• initial pH of sample solutions

• pH when excess titrant is added

• number of quantitative reactions

• non-quantitative (equilibrium) reactions

• equivalence point(s) for indicator selection

• buffering regions

SUMMARY pH Curve Reaction Information

Practice5. How is buffering action displayed on a pH curve?

6. How are quantitative reactions displayed on a pH curve?

7. How is a pH curve used to choose an indicator for a titration?

8. An acetic acid sample is titrated with sodium hydroxide (Figure 12.)

(a) Based on Figure 12, estimate the pH at the equivalence point.(b) Choose an appropriate indicator for this titration.(c) Write a Brønsted–Lowry equation for this reaction.(d) At the very beginning of the titration, before the curve levels off, it rises. Explain

this rise in terms of entities present in the mixture before and after beginning thetitration.

9. A sodium phosphate solution is titrated with hydrochloric acid (Figure 13 on nextpage). (a) Why are only two equivalence points evident?(b) Write three Brønsted–Lowry equations for the sequential reactions shown on the

pH curve in Figure 13. Communicate the position of equilibrium for each of thethree reactions.

10. Oxalic acid reacts quantitatively in a two-step overall reaction with a sodiumhydroxide solution. Assuming that an excess of sodium hydroxide is added, sketch apH curve (without any numbers) for all possible reactions.

Figure 12The pH curve for the addition of 0.48 mol/L NaOH(aq) to 25.0 mL of 0.49 mol/LCH3COOH(aq) illustrates pH changes during the reaction of a weak acid with astrong base.

CH3COOH(aq) Titrated with NaOH(aq)

14

12

10

8

6

4

25 10 15 20 25 30 35 40 45 50

Volume of NaOH (mL)

0

pH


760 Chapter 16 NEL

pH Curve Shape versus Acid and Base StrengthYou have learned that the reaction of the strongest possible acid in aqueous solution,H3O�(aq), and the strongest possible base in aqueous solution, OH�(aq), is over-whelmingly quantitative, and always results in an equivalence point pH of 7.00 at 25 °C. Not all titrations involve the hydronium and the hydroxide ions, however.

pH curves show that some weak acids (such as acetic acid) react quantitatively withOH�(aq) (Figure 12). The reactions of some weak bases (such as PO4

3�(aq) andHPO4

2�(aq)) with H3O�(aq), can also produce a quantitative reaction (Figure 13). The

strongest of the weak acids also react quantitatively with the strongest of the weak bases.For example, the reaction of HSO4

�(aq) with CO32�(aq) is a quantitative reaction. The

farther apart the reacting acid and base are on the Relative Strengths of Aqueous Acidsand Bases table (i.e., the larger the difference in Ka), the more likely it is that the reactionwill be quantitative. After plotting many pH curves, chemists have developed the gen-eral pH curve reference diagram shown in Figure 14.

Weak and Strong Acids and Bases

Volume of base

pH

SAWA

<7

7

>7 WA — SB

SB

WB

SA — SB

SA — WB

SA strong acidWA weak acidSB strong baseWB weak base

Figure 14Laboratory evidence is generalizedhere to show approximations of thepH curves for quantitative reactionsof weak and strong acids and bases.Note the initial change in pH forcurves involving “weak” entities, dueto the immediate change of thekinds of entities present in solutionwhen the titration begins.

25.0 mL of 0.51 mol/L Na3PO4(aq) Titratedwith 0.50 mol/L HCl(aq)

14

12

10

8

6

4

2

10 20 30 40 50 60 70 80 90 100

Volume of HCl (mL)

pH

00

Figure 13The pH curve for the addition of HCl(aq) to Na3PO4(aq) can be interpretedusing the Brønsted–Lowry acid–base concept.

Learning TipRecall that, for the examplesyou have studied, bromothymolblue has been an appropriateindicator for the SA–SBreaction, because it detects anendpoint with a pH of 7 verywell. Methyl orange detectsendpoint pH values around 4well, and is, therefore, useful formany SA–WB reactions. Finally,phenolphthalein detectsendpoints with pH valuesaround 10 well, and is,therefore, useful for manyWA–SB titrations.



Section 16.4

The (stoichiometric) equivalence point pH for a strong acid–strong base (SA–SB)reaction is seven (7). The equivalence point pH of any quantitative strong acid–weakbase (SA–WB) reaction is less than seven (�7), whereas that of any quantitative weakacid–strong base reaction (WA–SB) is greater than seven (�7).

The composite curves in Figure 14 are not acceptable (do not predict well) for any reac-tions between a weak acid and a weak base. Weak acid–weak base reactions often do nothave a detectable equivalence point because they are usually not quantitative. For allintents and purposes, analysis reactions and stoichiometric calculations are not donefor weak acid–weak base reactions, so a pH curve for such a reaction is simply irrelevant.

The pH of a solution at the equivalence point of an acid–base reaction may be explainedby considering the nature of those entities that are present in significant quantities at thatpoint. The entities present at an equivalence point can be predicted either by followingthe five-step Brønsted–Lowry method, or simply by converting a chemical substanceformula equation into a net ionic equation.

Now let’s examine a case of each of the common acid–base reaction types, and explainthe equivalence point pH in each case. Each of the examples used is a quantitative reaction.

Strong Acid–Strong Base ReactionFor example, nitric acid reacts with potassium hydroxide: SA–SB pH � 7.

SA A

H3O�(aq) , NO3

�(aq), K�(aq), OH�(aq), H2O(l)SB B

H3O�(aq) � OH�(aq) → 2 H2O(l)

or

HNO3(aq) � KOH(aq) → H2O(l) � KNO3(aq)

H�(aq) � NO3�(aq) � K�(aq) � OH�(aq) → H2O(l) � K�(aq) � NO3

�(aq)

H�(aq) � OH�(aq) → H2O(l)

At the equivalence point, water is the only acid or base entity present, which producesa neutral solution (pH � 7).

Strong Acid–Weak Base ReactionFor example, hydrochloric acid reacts with aqueous ammonia: SA–WB pH � 7.

SA A

H3O�(aq), Cl�(aq), NH3(aq), H2O(l)

SB B

H3O�(aq) � NH3(aq) → H2O(l) � NH4

�(aq)

orHCl(aq) � NH3(aq) → NH4Cl(aq)

H�(aq) � Cl�(aq) � NH3(aq) → NH4�(aq) � Cl�(aq)

H�(aq) � NH3(aq) → NH4�(aq)

At the equivalence point, the only acid or base entity present (other than water) is theammonium ion, which is a weak acid, resulting in an acidic solution (pH �7).

Learning TipRecall that entities common toboth sides of the equation arecancelled out and thecoefficients of the remainingentities are reduced to thesimplest ratio, in a net ionicequation.


762 Chapter 16 NEL

Weak Acid–Strong Base ReactionFor example, acetic acid reacts with barium hydroxide: WA–SB pH � 7.

SA A

CH3COOH(aq), Ba2�(aq), OH�(aq), H2O(l) SB B

CH3COOH(aq) � OH�(aq) → H2O(l) � CH3COO�(aq)

or

2 CH3COOH(aq) � Ba(OH)2(aq) → 2 H2O(l) � Ba(CH3COO)2(aq)

2 CH3COOH(aq) � Ba2�(aq) � 2 OH�(aq) →2 H2O(l) � Ba2�(aq) � 2 CH3COO�(aq)

CH3COOH(aq) � OH�(aq) → H2O(l) � CH3COO�(aq)

At the equivalence point, the only acid or base entity present (other than water) is theacetate ion, which is a weak base, resulting in a basic solution (pH >7).

• Strong acid–strong base reactions are quantitative (100%) and have an equivalence pointpH � 7.

• Strong acid–weak base quantitative reaction equivalence points have a pH �7.

• Weak acid–strong base quantitative reaction equivalence points have a pH �7.

• Polyprotic entity samples produce sequential reactions in titrations, each of whichmay or may not be quantitative.

SUMMARY Titration Generalizations

WEB Activity

Simulation—Titration of Polyprotic Acids and BasesThese two simulations enable you to explore the pH curves that result from various titrations.Of course, polyprotic acids and bases each have more than one K value, so the shapes of theirtitration curves will reflect this. Predict the shapes of the curves before working through theexercises.


Learning TipIn an equation for a titration, asingle arrow, →, is used. Thereaction represented by theequation must be quantitativein order for any stoichiometriccalculations based on theequation to be valid. In suchcases, chemical formulaequations are often the mostconvenient form.

Practice11. For the first quantitative reaction in each of the following acid–base titrations, predict

(where possible) whether the equivalence point pH will be greater than, less than, orequal to 7. (a) hydroiodic acid � aqueous sodium hydrogen phosphate →(b) boric acid � aqueous sodium hydroxide →(c) aqueous sodium hydrogen sulfate � aqueous potassium hydroxide →(d) hydrochloric acid � solid magnesium hydroxide →(e) hydrosulfuric acid � aqueous sodium hydrogen carbonate →(f) sulfuric acid � aqueous ammonia →



pH Curve Buffering Regions and Buffer SolutionsA titration pH curve for any quantitative acid–base reaction shows at least one regionwhere buffering action occurs between the beginning of the titration and the equivalencepoint. If the titration is continued much past the equivalence point, it enters anotherregion of buffering. Such a region represents a solution in which partial reaction hasoccurred, and, therefore, contains significant amounts of both entities of a conjugateacid–base pair. The term buffer refers specifically to the combination of any weak acidwith its conjugate base, in the same solution. Buffer solutions have a specific pH due tothe nature and concentration of the entities present, and change pH very little whenother acids or bases are added—provided that the quantity added is much less than thequantities of the conjugate pair entities present in the buffer.

For example, consider the titration of acetic acid with sodium hydroxide (Figure 15).The solution pH is approximately 4.8 when 10 mL of sodium hydroxide solution hasbeen added. This volume represents one-half of the (calculated) equivalence pointvolume of titrant. At this point in the reaction, one-half of the acetic acid originallypresent will have reacted (changed to acetate ions), according to the following equation:

OH�(aq) � CH3COOH(aq) → H2O(l) � CH3COO�(aq)

The mixture, therefore, now contains approximately equal amounts of the remainingunreacted weak acid, CH3COOH, and of the conjugate base, CH3COO�, produced in thereaction. Chemists would describe this mixture as an acetic acid–acetate ion buffer.

Section 16.4

12. Assume only three indicator solutions are available in your lab: methyl orange,bromothymol blue, and phenolphthalein. Choose the most suitable indicator fordetecting the equivalence point of each of the following acid–base titrations (for thefirst quantitative reaction only). (a) HBr(aq) � Ca(OH)2(s) →(b) HNO3(aq) � Na2CO3(aq) →(c) KOH(aq) � HNO2(aq) →

13. The following formulas each represent a solution at an acid–base reactionequivalence point. From the entities present in each solution, predict the observed pH as being greater than, less than, or equal to 7. (a) NH4Cl(aq)(b) Na2S(aq)(c) KNO3(aq)(d) NaHSO4(aq)

14. Use the five-step Brønsted–Lowry method to predict the overall reaction net ionicequation when the following chemicals are mixed. (a) solutions of perchloric acid and sodium carbonate (A pH curve shows two

protons are transferred quantitatively in successive reactions.)(b) solutions of nitrous acid and potassium hydroxide(c) solutions of phosphoric acid and sodium hydroxide (A pH curve shows two

protons are transferred quantitatively in successive reactions.)

15. Write a standard chemical formula equation, and also the net ionic equation, for thepredominant reaction that occurs when each of the following pairs of reagents mix. (a) Stomach acid is neutralized by solid magnesium hydroxide.(b) Aqueous ammonia is added to sulfurous acid.(c) A solution of sulfuric acid is neutralized by sodium hydroxide. (A pH curve

indicates two quantitative reactions.) Learning TipYou have observed (Figure 2,page 752) that the strong acid—strong base reaction pH curveshows that the conjugate pairH3O

�(aq)—H2O(l) will maintaina stable (very low) pH; and thatthe H2O(l)—OH�(aq) pair willstabilize pH at a very highvalue.

Even though these solutionscertainly do produce abuffering effect, they are nottechnically considered to bebuffers, because in each caseone of the two conjugate pairentities (water) is also thesolution solvent, and is presentin great excess.

A buffer is normally thoughtof as an aqueous solutionmixture of a weak acid and itsconjugate base, with bothentities of the conjugate pairpresent in significant quantity,but with both quantities muchless than the quantity of thesolution solvent (water) present.


764 Chapter 16 NEL

Buffering can be readily explained using Brønsted–Lowry equations. Suppose a smallamount of NaOH(aq) is added to the acetic acid–acetate ion buffer. Using the five-stepmethod for predicting the predominant acid–base reaction (page 731), the followingequation is obtained:

SA A

Na�(aq), OH�(aq), CH3COOH(aq), CH3COO�(aq), H2O(l)SB B B

OH�(aq) � CH3COOH(aq) → H2O(l) � CH3COO�(aq)

Figure 15 shows that this reaction is quantitative. A small amount of OH� would con-vert a small amount of acetic acid to acetate ions. The overall effect will be just a smalldecrease in the ratio of acetic acid to acetate ions in the buffer and a slight increase in thepH. The very small change in concentration of each entity of the acid–base conjugate pairpresent, and the complete consumption of the added hydroxide ions in the process,explains why the pH change is small. This buffer would work equally well if a smallamount of a strong acid, such as HCl(aq), were to be added. Evidence indicates that thereaction that occurs in that case is also quantitative.

SA A A

H3O�(aq), Cl�(aq), CH3COOH(aq), CH3COO�(aq), H2O(l)

SB B

H3O�(aq) � CH3COO�(aq) → H2O(l) � CH3COOH(aq)

Added hydronium ion is consumed and the mixture then has a slightly higher ratio ofacetic acid to acetate ions and a slightly lower pH. Figure 16 illustrates the concept ofbuffer capacity—the limit of the ability of a buffer to maintain a pH level. When the entityof the conjugate acid–base pair that reacts with an added reagent is completely con-sumed, the buffering fails and the pH changes dramatically.

The ability of buffers to maintain a relatively constant pH is important in many bio-logical processes where certain chemical reactions can only occur at a specific pH value.Many aspects of cell functions and metabolism in living organisms are very sensitive topH changes. For example, each enzyme carries out its function optimally over a small pHrange. One essential buffer operates to maintain a stabilized pH in the internal fluid ofall living cells. This critically important buffer is a mixture of dihydrogen phosphateions and hydrogen phosphate ions. In mammals, cellular fluid has a pH in the range

Volume of NaOH(aq)

(a)

pH

buffercapacity

Acetic Acid Buffer

Figure 16When an initial chemical amount ofa buffer is eventually depleted, thepH then changes very quickly.

(b)

pH

Volume of HCI(aq)

buffercapacity

Acetic Acid Buffer

Figure 15The highlighted plateau shows aneffective buffering region during thistitration of aqueous acetic acid withsodium hydroxide. At the pointhalfway to the equivalence point,the solution is a 1:1 ratio mix of anacetic acid–acetate ion buffer.

pH

0

Volume of NaOH(aq) (mL)

321

5 15 2010

4

14

25 30 35 40 45 50

765

89

10111213

15

Titration Curve for Titrating 0.300 mol/L CH3COOH(aq)with 0.300 mol/L NaOH(aq)

equivalence point

Learning TipBuffers are normally preparedas mixtures of a weak acid witha solution of some salt of thatweak acid. Thus, dissolvingsolid sodium acetate in asolution of acetic acid wouldproduce a buffer solution withproperties the same as those ofthe highlighted buffering regionof Figure 15.



Section 16.4

of 6.9 to 7.4. The H2PO4�(aq)–HPO4

2�(aq) buffer system has a pH of 7.2 when theconcentrations of these two conjugate entities are equal. With small variations in con-centration, this buffer is effective in maintaining optimum pH levels for the innumer-able reactions going on within any given cell. The major buffer system in the blood andother body fluids (except the cytoplasm within cells) is the conjugate acid–base pairH2CO3(aq)–HCO3

�(aq). Blood plasma has a remarkable buffering ability, as shown bythe empirical results in Table 1. Since production of the entities of this buffer in thebody is a continuous process, the buffer is not a limiting reagent, and you do not ordi-narily have to worry about using up your buffering capacity.

Table 1 Buffering Action of Neutral Saline Solution and of Blood Plasma

Final pH after addingSolution (1.0 L) Initial pH of mixture 1 mL of 10 mol/L HCI

neutral saline 7.0 2.0

blood plasma 7.4 7.2

WEB Activity


Simulation—Preparation of Buffer SolutionsThis simulation shows you how to select an appropriate acid–base conjugate pair to make abuffer of the desired pH.

Human blood plasma normally has a pH of about 7.4. Any change of more than 0.4 pHunits, induced by poisoning or disease, can be lethal. If the blood were not buffered, theacid absorbed from a glass of orange juice would probably be fatal.

BIOLOGY CONNECTION

Homeostasis

Many reactions in the human bodyproduce or consume carbonic acidor hydrogen carbonate ions. Whenone of these entities is depleted inthe blood, Le Châtelier’s principlecomes into effect, and more of thatentity is produced by anequilibrium shift to keepeverything in balance.

The carbonic acid in solution inblood is really in equilibrium withdissolved carbon dioxide:

CO2(aq) � H2O(l) 0 H2CO2(aq)

The enzyme carbonic anhydraseacts as a catalyst to allow thisequilibrium to establish rapidly, orreestablish quickly, once shifted. Itis possible to make your bloodtemporarily too basic bydeliberately breathing heavily for along time, thereby causing yourbody to lose too much carbondioxide. Doctors call this conditionrespiratory alkalosis.

There are countless equilibriumreactions interconnected in thisway in and around living cells.Biologists refer to thisinterdependent network ofreaction equilibria as an exampleof homeostasis—the condition ofautomatic continual readjustmentof cells, systems, and wholeorganisms to very specificconditions. Biology courses will tellyou a lot more about homeostasisand the factors that affect it.


Buffers are also important in many consumer, commercial, and industrial applica-tions. Fermentation and the manufacture of antibiotics require buffering to optimizeyields and to avoid undesirable side reactions. The production of various cheeses, yogurt,and sour cream are very dependent on controlling pH levels, since an optimum pH isneeded to control the growth of micro-organisms and to allow enzymes to catalyze fer-mentation processes. Sodium nitrite and vinegar are widely used to preserve food; partof their function is to prevent the fermentation that takes place only at certain pH values.

The CRC Handbook of Chemistry and Physics provides recipes for preparing buffersolutions. For example, a buffer with a pH of 10.00 can be prepared by mixing 50 mL of0.050 mol/L NaHCO3(aq) with 10.7 mL of 0.10 mol/L NaOH(aq). The reaction to estab-lish the buffer is described in the same way as all other acid–base reactions.

SA A

Na�(aq), HCO3�(aq), OH�(aq), H2O(l)

B SB B

HCO3�(aq) � OH�(aq) → H2O(l) � CO3

2�(aq)

The buffer preparation recipe reaction converts about �35� of the initial hydrogen car-

bonate ions into carbonate ions. An aqueous mixture of a 3:2 ratio of these two acid–baseconjugates at the concentrations specified has been found empirically to have an equi-librium pH of precisely 10.00 at 25 °C. Buffer mixtures like this one—with highly precise pH values—are routinely used to calibrate pH meters for accuracy.

The Hydrogen CarbonateBuffer SystemThe pH of your blood is maintainedat a constant level, courtesy of aseries of equilibrium reactions,including a buffer system. Thisextension outlines the reactions,and gives an example of howclimbing at high altitude can upsetthis delicate balance.

EXTENSION +



766 Chapter 16 NEL

Practice16. Give both an empirical and a theoretical definition of a buffer.

17. List two buffers that help maintain a normal pH level in your body.

18. Use the five-step method to predict the quantitative reaction of a carbonicacid–hydrogen carbonate ion buffer(a) when a small amount of HCl(aq) is added(b) when a small amount of NaOH(aq) is added

19. What happens if a large (excess) amount of a strong acid or base is added to a buffer?

20. Use Le Châtelier’s principle to predict what will happen to a benzoic acid–benzoateion buffer when a small amount of each of the following substances is added:(a) HCl(aq) (b) NaOH(aq)

21. Which of the following solution pairs, when mixed in equal quantities, will not form aneffective buffer? (a) HNO3(aq) and NaNO3(aq) (c) NH3(aq) and NH4Cl(aq)(b) C6H5COOH(aq) and NaC6H5COO(aq) (d) HCl(aq) and NaOH(aq)



Testing a Buffer EffectReferences provide “recipes” for preparing standard buffersolutions of any desired pH from 1.0 to 13.0. The one used in thisinvestigation has a pH of precisely 7.0, and might be used tocalibrate a pH meter, for example. If our theory of buffers iscorrect, this solution should resist significant change in pH upongradual addition of outside acid or base entities.

PurposeThe purpose of this investigation is to test our concept of buffers.Write the Design, Materials, and table of evidence to match theProcedure that is provided. The Materials list should include thesize of the equipment used. The buffer is prepared by a reactioncommunicated by the following chemical equation:

H2PO4�(aq) � OH�(aq) → HPO4

2�(aq) � H2O(l)excess limiting (base part(acid part reagent of buffer)of buffer)

ProblemHow does the pH change when a strong acid and a strong baseare slowly added separately to an H2PO4

�(aq)—HPO42�(aq)

buffer?


The five-step Brønsted–Lowry method to explain and predict acid–base reactions is a pre-ferred, acceptable method because it works for all quantitative and non-quantitativereactions studied so far:

• neutralization reactions • buffer reactions • excess reactions

• indicator reactions • polyprotic reactions • titration reactions

SUMMARY Brønsted–Lowry Is a Unifying Concept

DID YOU KNOW ??Commercial BuffersBuffers are fairly common,commercially. You can buy buffermixtures to adjust the pH ofaquarium water to suit the type offish you have. Buffer mixes are soldto add to hot tubs to control the pHof the water. Many food recipes mixingredients that form natural bufferswithin the dough, batter, or sauce.

WEB Activity


Canadian Achievers—Maud MentenMaud Menten (Figure 17) was a world-renowned pioneer in explaining the chemical action ofenzymes. What key concept was her chief contribution to understanding enzyme activity?

Figure 17Maud Menten (1879–1960)

CAREER CONNECTION

MicrobiologistMicrobiologists study, test for, andisolate microorganisms such asbacteria, fungi, and viruses. Aspart of a medical or researchteam, their work is essential forpublic health and safety.




Section 16.4

Section 16.4 Questions1. Draw a pH curve to illustrate the following information

about acid–base reaction systems. (a) What is buffering?(b) Where does buffering appear on a pH curve?(c) How are quantitative reactions represented on a pH

curve?(d) Define endpoint and equivalence point.(e) How is a suitable indicator chosen for a titration?(f) Do non-quantitative reactions have an endpoint?

Explain your answer briefly.

2. Sketch and label generalized pH curves on a single set ofaxes to illustrate the addition of a strong and a weak baseto a strong and a weak acid.

3. Sketch a generalized pH curve for the addition of a strongbase to a weak acid. Label the approximate equivalencepoint pH, and suggest an indicator that could be suitablefor titration analysis using this reaction.

4. If the pH of a solution is 6.8, what is the colour of each ofthe following indicators in this solution? (a) methyl red (d) phenolphthalein(b) chlorophenol red (e) methyl orange(c) bromothymol blue

5. Use Figure 18 to answer the following questions. (a) How many quantitative reactions have occurred?(b) Write the chemical equation for each quantitative

reaction.(c) State the equivalence point pH for each quantitative

reaction.(d) Choose a suitable indicator to correspond to the

equivalence point pH value(s).(e) Identify the buffering region(s) and state the chemical

formulas for the entities present in each region.

6. Write a net ionic equation to represent the followingacid–base reactions. (Use any method that you findconvenient to derive the net ionic equation.) (a) Oxalic acid is titrated with aqueous sodium hydroxide.(b) Sodium phosphate is titrated with hydrochloric acid.

(The titration is stopped at the first equivalence point.)(c) Sodium hydrogen phosphate is titrated with

hydrochloric acid. (An indicator is chosen to detect thefirst equivalence point.)

(d) Nitric acid is titrated with aqueous barium hydroxide.(e) A sulfuric acid spill is neutralized by adding excess lye.

7. Write a formula (non-ionic) equation, and also thecorresponding Brønsted–Lowry equation, to represent thefollowing acid–base reactions. (a) Acetic acid is titrated with aqueous sodium hydroxide.(b) Nitrous acid is titrated with aqueous barium hydroxide.(c) Sodium carbonate is titrated with hydrobromic acid.

(The titration is stopped after the first quantitativereaction.)

(d) Carbonic acid is titrated with aqueous sodiumhydroxide. (An indicator endpoint is used to stop thereaction with a stoichiometric ratio of 1:1.)

(e) A sulfuric acid spill is neutralized by adding excess lye(caustic soda).

8. State two different applications of buffers.

9. Suggest a compound that could be dissolved in a sulfurousacid solution to make an effective buffer.

10. Assume that you have a hydrogen carbonate ion solution.Suggest substances to add to make(a) a buffer with a lower pH than the original solution(b) a buffer with a higher pH than the original solution

11. Complete the Design of the following investigation report.

PurposeThe purpose of this investigation is to test buffer conceptsby quantitatively measuring equilibrium shifts for a buffersystem.

ProblemDoes the Brønsted–Lowry concept, as applied to buffers,predict changes in equilibrium concentration of H3O

�(aq)in a standard buffer solution, when strong acids or basesare added?

Materials pH 7.00 H2PO4

�/HPO42�

buffer solutionpH meter1.00 mol/L HCl(aq)1.00 mol/L NaOH(aq)

Extension

12. From a CRC Handbook or any other reference, locate andcopy at least one “recipe” for preparing a buffer solutionthat will have a pH of precisely 5.00 at 25 °C. Figure 18

pH curve for the titration of sodium sulfite solution withhydrochloric acid

10

8

6

4

2

10 20 30 40 50 60 70 80

Volume of HCl (mL)

00

25.0 mL of 0.46 mol/L Na2SO3(aq) Titratedwith 0.50 mol/L HCl(aq)

pH


Chapter 16 INVESTIGATIONS

NEL

Creating an Acid–Base StrengthTable

An acid–base table organizes common acids (and their con-jugate bases) in order of decreasing acid strength (Figure 1).Acid strength can be tested several ways, including by a care-fully designed use of indicators. Predict the order of strengthsusing the Relative Strengths of Aqueous Acids and Bases Table(Appendix I). Use the indicators provided to create a validand efficient Design, in which you clearly identify the rele-vant variables. Evaluate the Design (only), and suggestimprovements if any problems are identified.

PurposeThe purpose of this investigation is to test an experimentaldesign for using indicators to create a table of relative strengthsof acids and bases.

ProblemCan the indicators available be used to rank the acids andbases provided in order of strength?

Materials

0.10 mol/L Na2CO3(aq)0.10 mol/L NaOH(aq)indicators, for example:

methyl orange, methylviolet, bromothymolblue, phenolphthalein

(6) 13 � 100 mm test tubesfor each indicator or spotplate (microchem)

test-tube rack

Purpose Design AnalysisProblem Materials Evaluation (1)Hypothesis ProcedurePrediction Evidence


Chemicals used include toxic, corrosive, and irritantmaterials. Avoid eye and skin contact. If you spill anyof the chemical solutions on your skin, immediatelyrinse the area with lots of cool water. In the unlikelysituation of getting some of the chemicals in youreye, immediately rinse your eye for at least 15 minand inform your teacher.

SAH3O

+(aq)

H2O(l)WA

Acid–Base TableWB

H2O(l)

OH—(aq)SB

Figure 1Simplified acid–base table

Testing Brønsted–Lowry ReactionPredictionsWhen predicting products for this investigation, since at leastone reactant is always in solution, list all entities present as theynormally exist in an aqueous environment. For those reac-tants that are added in solid state, assume that they will dis-solve. Use the resulting entities for prediction. Evaluate thepredictions, the Brønsted–Lowry concept, and the five-stepmethod for acid–base reaction prediction.

PurposeThe purpose of this investigation is to test the Brønsted–Lowryconcept and the five-step method for reaction predictionfrom a table of relative acid–base strength.

ProblemWhat reactions occur when the following substances aremixed? (Hints for diagnostic tests are in parentheses.)

1. ammonium chloride and sodium hydroxide solutions(odour)

2. hydrochloric acid and sodium acetate solutions(odour)

3. sodium benzoate and sodium hydrogen sulfate solu-tions (benzoic acid has low solubility)

4. hydrochloric acid and aqueous ammonium chloride(odour)



lab aproneye protection0.10 mol/L HCl(aq)

0.10 mol/L NaHSO4(aq)0.10 mol/L CH3COOH(aq)0.10 mol/L NaHSO3(aq)

768 Chapter 16


Chapter 16


5. solid sodium chloride added to water (litmus)

6. solid aluminium sulfate added to water (litmus)

7. solid sodium phosphate added to water (litmus)

8. solid sodium hydrogen sulfate added to water(litmus)

9. solid sodium hydrogen carbonate added tohydrochloric acid (pH)

10. solid sodium hydrogen carbonate added to sodiumhydroxide solution (pH)

11. solid sodium hydrogen carbonate added to sodiumhydrogen sulfate solution (pH)

DesignA prediction is made for each of eleven pairs of substances.The prediction is then tested using one or more diagnostictests, complete with controls. Additional diagnostic testsincrease the certainty of the evaluation.

Chemicals used include toxic, corrosive, and irritantmaterials. Avoid eye and skin contact. If you spill anyof the chemical solutions on your skin, immediatelyrinse the area with lots of cool water. In the unlikelysituation of getting some of the chemicals in youreye, immediately rinse your eye for at least 15 minand inform your teacher. Remember to detect odourscautiously by wafting air toward your nose from thecontainer.

Testing a Buffer Effect

References provide “recipes” for preparing standard buffersolutions of any desired pH from 1.0 to 13.0. The one used inthis investigation has a pH of precisely 7.0, and might be usedto calibrate a pH meter, for example. If our theory of buffersis correct, this solution should resist significant change in pHupon gradual addition of outside acid or base entities.

PurposeThe purpose of this investigation is to test our concept ofbuffers. Write the Design, Materials, and table of evidence tomatch the Procedure that is provided. The Materials list shouldinclude the size of the equipment used. The buffer is pre-pared by a reaction communicated by the following chem-ical equation:

H2PO4�(aq) � OH�(aq) → HPO4

2�(aq) � H2O(l)excess limiting (base part(acid part reagent of buffer)of buffer)

ProblemHow does the pH change when a strong acid and a strong baseare slowly added separately to an H2PO4

�(aq)–HPO42�(aq)

buffer?

Procedure1. Obtain 50 mL of 0.10 mol/L KH2PO4(aq) and 29 mL of

0.10 mol/L NaOH(aq) in separate graduated cylinders.

2. Pour the KH2PO4(aq) and then the NaOH(aq) into abeaker to prepare a buffer with a pH of 7.

3. Pour an equal volume of the buffer into two test tubes.

4. Add 0.10 mol/L NaCl(aq) as a control into a thirdand a fourth test tube.

5. Add two drops of bromocresol green to one buffertest tube and one control test tube.

6. Add and count drops of 0.10 mol/L HCl(aq) until thecolour changes.

7. Repeat steps 5 and 6 with phenolphthalein and 0.10 mol/L NaOH(aq) using the other two test tubes.

8. Dispose of all solutions down the drain with runningwater.




Acids and bases are corrosive and toxic. Avoid skinand eye contact. If you spill any of the chemicalsolutions on your skin, immediately rinse the areawith lots of cool water. In the unlikely situation ofgetting some of the chemicals in your eye,immediately rinse your eye for at least 15 min andinform your teacher.


Chapter 16 SUMMARY

770 Chapter 16 NEL

Outcomes

Knowledge

• describe Brønsted–Lowry acids as proton donors and basesas proton acceptors (16.2)

• write Brønsted–Lowry equations and predict whetherreactants or products are favoured for acid–base equilibriumreactions (including indicators and polyprotic acids andbases) (16.2, 16.4)

• identify polyprotic acids, polyprotic bases, conjugate pairs,and amphiprotic entities (16.2, 16.4)

• define a buffer as relatively large amounts of a conjugateacid–base pair in equilibrium that maintain a relativelyconstant pH when small amounts of acid or base are added(16.4)

• sketch and qualitatively interpret titration curves ofmonoprotic and polyprotic acids and bases, identifyingequivalence points and regions of buffering for weakacid–strong base, strong acid–weak base, and strongacid–strong base reactions (16.4)

• define Kw , Ka , and Kb and use them to determine pH, pOH,[H3O

�], and [OH–] of acidic and basic solutions (16.1, 16.3)

• calculate equilibrium constants and concentrations forhomogeneous systems and Brønsted–Lowry acids and bases(excluding buffers) when concentrations at equilibrium areknown, when initial concentrations and one equilibriumconcentration are known, and when the equilibrium constantand one equilibrium concentration are known (16.1, 16.3)

STS

• state that the goal of science is knowledge about the naturalworld (16.1, 16.2, 16.3, 16.4)

• state that a goal of technology is to solve practical problems(16.2, 16.3, 16.4)

Skills

• initiating and planning: design an experiment to showquantitative equilibrium shifts in concentration under a givenset of conditions (16.4); describe procedures for safehandling, storage, and disposal of materials used in thelaboratory, with reference to WHMIS and consumer productlabelling information (16.2, 16.4)

• performing and recording: prepare a buffer to investigate therelative abilities of a buffer and a control to resist a pHchange when a small amount of strong acid or strong base isadded (16.4)

• analyzing and interpreting: use experimental data tocalculate equilibrium constants (16.3)

• communication and teamwork: work cooperatively inaddressing problems and apply the skills and conventions ofscience in communicating information and ideas and inassessing results (16.2, 16.4)

Key Terms

16.1ionization constant for water, Kw

16.2Brønsted–Lowry concept

Brønsted–Lowry acid

Brønsted–Lowry base

Brønsted–Lowry reaction equation

amphiprotic

amphoteric

conjugate acid–base pair

16.3acid ionization constant, Ka

base ionization constant, Kb

16.4pH curve

buffering

buffer

buffer capacity

Key Equations (All at SATP)

Kw � [H3O�(aq)][OH�(aq)] � 1.0 � 10�14 (16.1)

pH � pOH � 14.00 (16.1)

For any aqueous acid, HA(aq), Ka = (16.3)

For any aqueous base, B(aq), Kb = (16.3)

For any conjugate acid–base pair, KaKb � Kw (16.3)

[HB�(aq)][OH�(aq)]��

[B(aq)]

[H3O�(aq)][A�(aq)]

��[HA(aq)]



Chapter 16

MAKE a summary

1. Use a blank page to create a “star” (radial) chart forquantities and calculations related to Brønsted–Lowryacid–base reaction systems. Begin by writing H3O

�(aq)and OH�(aq) symbols about 10 cm apart horizontally,centred on the page and connected by a line. Drawlines from each of these symbols to other quantity symbols that are calculated from them (and from eachother, in some cases). Along each connecting line, indicate what information the calculation requires andhow it is performed. Include Ka, Kb, pH, and pOHquantities, but not percent ionization (because it is notspecific to acid–base reactions).

2. Revisit your answers to the Starting Points questions atthe beginning of this chapter. How would you answerthe questions differently now? Why?

Go To

The following components are available on the Nelson Web site. Follow the links for Nelson Chemistry Alberta 20–30.

• an interactive Self Quiz for Chapter 16

• additional Diploma Exam-style Review questions

• Illustrated Glossary

• additional IB-related material

There is more information on the Web site wherever you seethe Go icon in this chapter.


Lost Treasures of TibetWhat does acid–base chemistry have to do with ancient worksof art? Watch this video to discover how archeologists andconservators made use of chemical reactions to restore thebrilliance to Tibetan temple paintings.


EXTENSION +


Chapter 16 REVIEW

772 Chapter 16 NEL



Part 11. A hydrated proton is referred to as a

A. hydronium ionB. hydroxide ionC. hydroxyl groupD. Brønsted–Lowry base

2. The hydroxide ion concentration in a solution of windowcleaner is 2.1 mmol/L. The hydronium ion concentration inthis solution is calculated to beA. 2.1 � 10�15 mol/LB. 4.8 � 10�15 mol/LC. 2.1 � 10�12 mol/LD. 4.8 � 10�12 mol/L

3. A pH meter indicates that the pH of a soft drink is 3.46. ThepOH of the soft drink is calculated to beA. 10.46B. 10.54C. 14.46D. 14.54

4. The recipe for preparing a cleaning solution calls fordissolving 5.0 g of sodium hydroxide in 4.0 L of water. ThepH of the resulting solution is calculated to beA. 1.51B. 12.49C. 13.10D. 13.90

5. Acid–base theories have developed over the last twocenturies. Which of the following chemists did not make asignificant contribution to acid–base theory?A. Gilbert LewisB. Svante ArrheniusC. Ernest RutherfordD. Johannes Brønsted

6. Listed from strongest to weakest, the acids are ___ ___ ___ ___.

7. Listed from strongest to weakest, the bases are ___ ___ ___ ___.

8. The nearly level region on a pH curve representsA. the endpointB. a region of bufferingC. an indicator point D. the equivalence point

9. The titration of a weak acid with a strong base has anequivalence point pH of 9.0. Which of the followingindicators would be most suitable for this titration, done asan analysis?A. litmusB. methyl redC. phenolphthaleinD. alizarin yellow R

10. Which of the following combinations of ions does notrepresent a conjugate acid–base pair?A. H3PO4(aq) and H2PO4

�(aq)B. H2PO4

�(aq) and HPO42�(aq)

C. HPO42�(aq) and PO4

3�(aq)D. H3PO4(aq) and HPO4

2�(aq)

11. Which of the following statements about acid–basetitrations is not true?A. Strong acid–strong base reactions have an

equivalence point pH of 7.B. Strong acid–weak base reaction equivalence points

have a pH � 7.C. Weak acid–strong base reaction equivalence points

have a pH � 7.D. Strong acid–strong base reactions are quantitative.

12. Which of the following statements about buffers is nottrue?A. Buffers maintain a solution pH at approximately 7.B. Buffers can be formed by partially neutralizing a weak

acid with a strong base.C. Buffers maintain a relatively constant pH when small

amounts of acid or base are added.D. Buffers contain relatively large amounts of a conjugate

acid–base pair in equilibrium.


Use the names given and Appendix G to answer questions 6 and 7. Note that some of the ions listed are amphiprotic.1. phenol2. sulfide ion 3. cyanide ion4. ammonium ion5. hydrogen sulfate ion6. hydrogen carbonate ion

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Part 213. Define each of the following substances, according to the

Brønsted–Lowry theory.(a) an acid(b) a base(c) an acid–base reaction(d) an amphiprotic substance(e) a strong acid(f) a strong base

14. Write net ionic equations for each of the followingreactions in aqueous solution. Label the reactants asBrønsted–Lowry acids or bases. Identify any amphiproticions.(a) Solutions of sodium hydrogen sulfate and sodium

carbonate are mixed.(b) Aqueous ammonia is added to a solution of potassium

hydrogen sulfite.(c) Solutions of sodium hydrogen phosphate and acetic

acid are mixed.(d) Aqueous sodium hydroxide is added to a solution of

sodium hydrogen phosphate.

15. Identify all acids, bases, and conjugate pairs, and predictthe position of equilibrium (reactants or products favoured)in each of the following reactions.(a) CH3COOH(aq) � CN�(aq) 0 CH3COO�(aq) �

HCN(aq)(b) HSO3

�(aq) � HPO42�(aq) 0 SO3

2�(aq) �H2PO4

�(aq)(c) NH4

+(aq) � CO32�(aq) 0 NH3(aq) � HCO3

�(aq)

16. Write equilibrium law expressions for each of the followingequilibrium situations.(a) HF(aq) � H2O(l) 0 H3O

�(aq) � F�(aq)(b) NH3(aq) � H2O(l) 0 OH�(aq) � NH4

�(aq)(c) H2SO3(aq) � H2O(l) 0 H3O

�(aq) � HSO3�(aq)

(d) Ca(OH)2(aq) � H2O(l) 0 2OH�(aq) � Ca2�(aq)

17. Refer to Appendix G for required information to predict[H3O

�(aq)] and pH for each of these 0.10 mol/L solutions.(a) hydroiodic acid(b) methanoic acid(c) hydrosulfuric acid

18. A student measures the pH of a 0.25 mol/L solution ofpotassium hydrogen citrate to be 3.42. (a) Determine the Ka of the hydrogen citrate ion from

this evidence. (b) How could a student find the pH of a solution without

access to a pH meter? Design a procedure, involvingthree indicators, that would establish the approximatepH of this solution. Criticize your design.

Chapter 16

19. If a sample of acid rain has a pH of 5.0, predict the colourof each of the following indicators in this solution.(a) litmus(b) methyl red(c) methyl orange(d) phenolphthalein

20. Aqueous 0.10 mol/L solutions of potassium sulfate andpotassium benzoate are prepared. Predict how the pHvalues of the solutions would compare, using theBrønsted–Lowry proton transfer concept to justify youranswer.

21. Separate samples of a household cleaning solution weretested with two indicators. Indigo carmine was blue andphenolphthalein was red in the solution. Estimate theapproximate pH and approximate hydroxide ionconcentration in the solution.

22. Determine the percent reaction, pH, pOH, and hydroxideconcentration in a 0.015 mol/L solution of sodium acetate.

23. Sketch a pH (titration) curve for each of the followingtitrations, assuming all the acids and bases have aconcentration of 0.10 mol/L, and all reaction protontransfers are quantitative.(a) A diprotic acid is titrated with sodium hydroxide.(b) A diprotic base is titrated with hydrochloric acid.

24. Design an experiment to determine the relative strengthsof four weak acids.

25. Use the five-step Brønsted–Lowry method to predict thenet ionic equation for the overall reaction when thefollowing chemicals are mixed.(a) solutions of sodium sulfate and potassium benzoate(b) aqueous ammonium nitrate fertilizer and aqueous

sodium phosphate(c) solutions of citric acid and sodium hydrogen

carbonate

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Unit 8 REVIEW

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Part 11. The following aqueous solutions are each at equilibrium in

sealed containers that are half-full of liquid. For which oneof these may the volume of the liquid solution bereasonably considered to be the boundary of the closedsystem for the equilibrium? A. HCl(aq)B. NaOH(aq)C. H2CO3(aq)D. NH3(aq)

2. Basicity of solutions has been explained by a variety oftheories, each one more comprehensive than itspredecessors. The Brønsted–Lowry concept defines a baseas an entity that is a A. proton attractor in a specific acid–base reactionB. proton donor in a specific acid–base reactionC. hydronium attractor in any acid–base reactionD. hydroxide producer in any acid–base reaction

3. For which of the following aqueous solution titrationswould you expect to measure a pH of 7 at the equivalencepoint, at SATP? A. acetic acid titrated with sodium hydroxideB. nitric acid titrated with potassium carbonateC. hydrochloric acid titrated with lithium hydroxideD. ammonia titrated with hydrochloric acid

4. An industrial reaction process designer has a choice ofseveral different equilibrium reactions, all of which producethe same desired chemical substance. These reactionseach have different reaction characteristics. The designerwishes to produce product as fast as is safely possible. Thebest choice from an economic perspective is the reactionthat A. has a high rate (speed) of reaction and a low percent

yieldB. has a low rate (speed) of reaction and a high percent

yieldC. requires extreme high pressure to improve the percent

yieldD. requires an expensive catalyst to react noticeably

5. What energy condition must be met to maintain a dynamicequilibrium system? A. The temperature must be constant.B. The forward reaction must be exothermic.C. The container must not conduct heat.D. The activation energy must be high.

6. The conjugate base of arsenic acid in the ionizationreaction shown isA. H3AsO4(aq)B. H2AsO4

�(aq)C. HAsO4

2�(aq)D. AsO4

3�(aq)

7. Which is the correct form of the equilibrium law for thisacid ionization reaction?

A. Ka �

B. Ka �

C. Ka �

D. Ka �

8. An empirical pH curve for a titration of a sample ofaqueous arsenic acid with a standardized sodiumhydroxide solution shows two distinct endpoints. From thisevidence, it can be inferred that A. hydrogen arsenate ion is a very weak acid, with a

relatively strong conjugate baseB. arsenate ion is a very weak base, with a strong

conjugate acidC. an arsenic acid molecule can lose two protons

simultaneously to one hydroxide ionD. the third sequential proton transfer reaction is

quantitative

9. A student discovers that a pure liquid substance that issold as a common agricultural insecticide forms an acidicsolution in water. The measured pH of a 1.00 mol/Laqueous solution of this weak acid is 5.22. The calculatedKa value for this acid at this temperature may be expressednumerically as a.b � 10�cd. The values (in order) of a, b, c,and d are _____, _____, _____, and _____.

[H3O�(aq)][H2AsO4

�(aq)]��[H3AsO4(aq)][HAsO4

2�(aq)]

[H3O�(aq)]

��[H3AsO4(aq)][H2AsO4

�(aq)]

[H3O�(aq)][H2AsO4

�(aq)]��

[H3AsO4(aq)][H2O(aq)]

[H3O�(aq)][H2AsO4

�(aq)]��

[H3AsO4(aq)]


Pure solid arsenic acid, also called orthoarsenic acid, is ahydrated substance at room temperature with a formula thatmay be written as H3AsO4•�

12�H2O(s), or alternatively as

(H3AsO4)2•H2O(s). This substance is used in commercialglassmaking and in making wood-preservative solutions.Arsenic acid is soluble in water and ionizes according to thefollowing equation:

H3AsO4(aq) � H2O(l) 0 H3O�(aq) � H2AsO4

�(aq)Ka � 5.6 � 10�3 (18 °C)


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10. In writing an equilibrium law expression, condensed (puresolid and liquid) phases are ignored becauseA. forward and reverse reaction rates must be equal if

they occur at the surface of a solid or liquidB. solids and liquids are completely separate physically

from the rest of the reaction components, which aredissolved in each other

C. solid or liquid pure substances have essentiallyconstant amount concentration values

D. other entities cannot move between molecules ofsolids or liquids to collide with their entities

11. Consider the reaction N2(g) � 3 H2(g) 0 2 NH3(g), atequilibrium. If the system container volume were to besuddenly decreased, the concentration of nitrogen gas inthe system would immediately A. increase, then decrease to a new constant valueB. remain unchanged, because the reaction is at

equilibriumC. increase, then remain steady at a new constant valueD. decrease, then remain steady at a new constant value

Unit 8

12. The strong acid titration curve does not increase noticeablywhen the very first addition of NaOH(aq) is made, because A. OH�(aq) was already present in the original sample

solutionB. OH�(aq) does not initially react with entities present in

the sample solutionC. reaction of the added OH�(aq) creates no new entities

in the sample solutionD. hydronium ions resist any change to their structure

13. During most of these titrations, the pH curves remainnearly level. The conjugate acid–base pair that isresponsible for creating this buffering region during thestrong acid–strong base titration is A. H3O

�(aq)�H2O(l)B. H3O

�(aq)�CH3COOH(aq)C. CH3COOH(aq)�CH3COO�(aq)D. CH3COO�(aq)�OH�(aq)

14. The incorrect statement about the weak acid–strong basetitration is:A. Initial addition of OH�(aq) produces a new entity in

the sample solution.B. The titrant volume at the reaction equivalence point is

the same as for the strong acid–strong base titration.C. The solution pH at the reaction equivalence point will

be higher than for the strong acid–strong basetitration.

D. At the equivalence point, the solution pH will becontrolled by the concentration of hydroxide ion,which is the strongest base present.

Part 215. Formal concepts of acids have existed since the 18th

century. Explain the main idea and the limitations of eachof the following: the oxygen concept; the hydrogenconcept; Arrhenius’ concept; and the Brønsted–Lowryconcept of acids.

16. What happens when scientists find a theory, such asArrhenius’ original theory of acids, to be unacceptable?

17. Describe two main ways in which a theory or a theoreticaldefinition may be tested, in terms of what an acceptabletheory is required to do.

18. According to modern evidence, what is the nature of a“hydrogen ion” in aqueous solution?

19. Briefly outline how the theoretical definition of a base haschanged from Arrhenius’ original concept to the modifiedArrhenius concept, and subsequently to theBrønsted–Lowry concept.

20. Aqueous solutions of sodium sulfite and of sodiumcarbonate of equal concentrations are prepared. Explainhow the pH values would compare, using theBrønsted–Lowry proton transfer concept to justify youranswer.


Titration of acid samples with hydroxide ion solution producesvery different pH curves depending on the strength of theacid sample (Figure 1). These curves represent data from twoseparate titrations plotted on the same axes. The two acidsamples were standardized hydrochloric acid and acetic acidsolutions, both with the same initial concentration andvolume. The same sodium hydroxide titrant was used in bothtitrations.

Figure 1Titration curves for HCl(aq) and CH3COOH(aq) with NaOH(aq)

Volume of NaOH (mL)

Titration of Two Acids with NaOH(aq)

pH


776 Unit 8 NEL

21. Use the If [procedure], and [evidence], then [analysis]format (Appendix C.4) to describe a diagnostic test thatcould be used to determine which of two solutions (ofequal concentration) is sodium benzoate, and which issodium hydroxide.

22. Identify two different examples of conjugate acid–basepairs, each involving the dihydrogen phosphate ion.

23. Which of the following statements are necessarily truewhen products are strongly favoured in an acid–baseequilibrium, assuming equal amount concentrations andchemical amounts of the initial reactants? (a) The stronger of the two Brønsted–Lowry bases is a

product. (b) The equilibrium constant is greater than one. (c) The forward reaction is exothermic. (d) The stronger Brønsted–Lowry acid is a reactant. (e) The percent reaction is greater than 50%. (f) The pH of the final solution is greater than 7. (g) The reactant acid is above the reactant base in an

acid–base table.

24. From the equation information provided, predict all of thesystem changes you might introduce that, according to LeChâtelier’s principle, will act to shift the equilibria tomaximize the percent yield of the specified product. (a) production of propene (propylene)

C3H8(g) � energy 0 C3H6(g) � H2(g)(b) production of iodine

5 Sn2�(aq) � 2 IO3�(aq) � 12 H3O

�(aq) 05 Sn4�(aq) � I2(s) � 18 H2O(l)

25. Consider this system at equilibrium. PCl5(g) 0 PCl3(g) � Cl2(g); Kc � 0.40 at 170 °C(a) One mole of phosphorus pentachloride was initially

placed into a 1.0 L container. Once equilibrium hadbeen reached, it was found that the equilibriumconcentration of PCl5 was 0.54 mol/L. Figure 2describes the change in [PCl5] over time. Copy thegraph. Sketch lines to indicate the changingconcentrations of PCl3 and Cl2 over the same timeperiod.

(b) 0.50 mol of PCl3 and 0.30 mol of Cl2 were placed into a1.0 L container. Once equilibrium had been reached, itwas found that the equilibrium concentration of PCl3was 0.36 mol/L. Figure 3 describes the change in[PCl3] over time. Copy the graph. Sketch how theconcentrations of PCl5 and Cl2 change over the sametime period.

26. Consider the equilibrium reaction,CO(g) � H2O(g) 0 CO2(g) � H2(g) Kc � 5.0 at 650 °CIn a rigid 1.00 L laboratory reaction vessel, a technicianplaces 1.00 mol of each of the four substances involved inthis equilibrium. The vessel is heated to 650 °C. Determine the equilibrium amount concentrations of eachsubstance, organizing your values in an ICE table.

27. Write chemical formulas and net ionic equations for thefollowing overall reactions. (If necessary, refer to Appendix J.)(a) Vinegar is used to neutralize a drain cleaner spill. (b) Baking soda solution is used to neutralize spilled rust

remover containing oxalic acid. (c) An antacid tablet (flavoured calcium carbonate) is

used to neutralize excess stomach acid.

28. Identify all acids, bases, and conjugate pairs, and predictthe position of equilibrium (reactants or products favoured)in each of the following reactions:(a) HCOOH(aq) � CN�(aq) 0 HCOO�(aq) � HCN(aq)(b) HPO4

2�(aq) � HCO3� (aq) 0

H2PO4�(aq) � CO3

2�(aq)(c) Al(H2O)6

3�(aq) � H2O(l) 0H3O

�(aq) � Al(H2O)5OH2�(aq)Ka � 1 � 10�5

(d) C6H5NH2(aq) � H2O(l) 0C6H5NH3

�(aq) � OH�(aq)Kb � 4 � 10�10

Figure 2Reaction progress: PCl5(g) 0 PCl3(g) � Cl2(g)

0.8

0.6

0.4

0.2

Time

Con

cent

rati

on o

f PC

l 5(g

)�

0.1

0.3

0.5

0.7

0.9

1.0

Figure 3Reaction progress: PCl3(g) � Cl2(g) 0 PCl5(g)

0.8

0.6

0.4

0.2

Time

Con

cent

rati

on o

f PC

l 3(g

)�

0.1

0.3

0.5

0.7

0.9

1.0



29. Separate samples of an unknown solution were tested withtwo indicators. Congo red was red and chlorophenol redwas yellow in the solution. Predict the approximate pHand approximate hydronium ion concentration of thesolution.

30. The test for acceptability of any theory is its ability toexplain and predict a wide range of phenomena: to be aunifying theory. Test the Brønsted–Lowry concept by usingthe five-step procedure to write chemical equationsdescribing and predicting each of the following acid–basereactions. Design one diagnostic test that could be usedto verify each prediction. (a) the addition of hydrofluoric acid to a solution of

potassium sulfate(b) the addition of a solution of sodium hydrogen sulfate

to a solution of sodium hydrogen sulfide(c) the titration of methanoic acid with sodium hydroxide

solution(d) the addition of a small amount of a strong acid to a

hydrogen phosphate ion–phosphate ion buffer solution(e) the addition of colourless phenolphthalein indicator,

HPh(aq), to a strong base(f) sodium sulfate is dissolved in water(g) the addition of blue bromothymol blue, Bb�(aq), to

vinegar(h) the addition of washing soda to water(i) the addition of baking soda to water

31. Many acid–base phenomena can be described and/orpredicted by using Le Châtelier’s principle. Evaluate thisgeneralization's ability to describe or predict a variety ofreactions.

Your response should include• reference to the following three examples:

NaOH(aq) is added to the following buffer equilibrium:NH3(aq) � H2O(l) 0 NH4

�(aq) � OH�(aq)NaOH(aq) is added to a bromothymol blue indicatorsolution:

HBb(aq) � H2O(l) 0 H3O�(aq) � Bb�(aq)

NaOH(aq) titrant is added to a vinegar sample:CH3COOH(aq) � H2O(l) 0

H3O�(aq) � CH3COO�(aq)

• descriptions of diagnostic tests that would provideevidence for your predictions

32. One way to evaluate a theory is to test predictions withnew substances. Sodium methoxide, NaCH3O(s), isdissolved in water. Predict whether the final solution willbe acidic, basic, or neutral. Explain your answer using anet ionic equation. (Hint: Think of the methoxide ion as ahydroxide ion, with a methyl group substituted for thehydrogen.)

33. In an experimental investigation of amphoteric substances,samples of baking soda were added to a solution ofsodium hydroxide and to a solution of hydrochloric acid.The pH of the sodium hydroxide changed from 13.0 to 9.5after the addition of the baking soda. The pH of the

Unit 8

hydrochloric acid changed from 1.0 to 4.5 after the additionof baking soda. Explain these results.

Your response should include• chemical reaction equations to describe the reactions• identification of the amphiprotic entity

34. Each of seven unlabelled beakers was known to containone of the following 0.10 mol/L solutions: CH3COOH(aq),Ba(OH)2(aq), NH3(aq), C2H4(OH)2(aq), H2SO4(aq), HCl(aq),and NaOH(aq). Describe diagnostic test(s) required todistinguish the solutions and label the beakers. Use the “If ____, and ____, then ____” format (Appendix C.4), a flowchart, or a table to communicate your answer.

35. Use Figure 4 to answer the following questions. (a) Infer how many quantitative reactions have occurred.(b) Write the Brønsted–Lowry equation for each

successive proton transfer reaction, with appropriate“arrows” to indicate the extent of each reaction.

(c) Determine the titrant volume at each quantitativereaction equivalence point.

(d) Identify a suitable indicator to correspond to theequivalence point pH values.

(e) Identify the buffering region(s), and state thechemical formulas for the entities present in solutionin each buffering region.

36. Because chlorine–oxygen compounds are toxic to micro-organisms and react readily with organic materials in foodstains, they find wide application in disinfectants andbleaches. The reaction of hypochlorous acid with moleculesof coloured substances in stains often produces colourlessproducts, making the stain “disappear.” Hypochlorous acidmay be produced by the following reaction:H2O(g) � Cl2O(g) 0 2 HOCl(g) Kc � 0.090 (25 °C)Determine the concentrations of each reagent at equilibrium at 25 °C if the initial concentrations of bothwater vapour and chlorine monoxide were 4.0 mol/L.

Figure 4Graph for question 35

pH

Volume of HCl(aq) added (mL)

2

10 30 4020

4

50 60 70 80 90 100

6

8

10

12

25.0 mL of 0.50 mol/L Na3PO4(aq) Titrated with 0.50 mol/L HCl(aq)

0

14

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37. Ethyl acetate (ethyl ethanoate) is an ester with a greatmany different uses as an organic solvent—for everythingfrom paint to perfume. It is a product of the followingequilibrium reaction equation: CH3COOH(os) � C2H5OH(os) 0

CH3COOC2H5(os) � H2O(os) The equation shows the reaction taking place at 25 °C, with all components dissolved in a complex non-reacting liquid organic solvent, which we have symbolizedas “os” for convenience. Determine Kc for this reaction if, atequilibrium, the reagent concentrations are [CH3COOH(os)] � 2.5 mol/L[C2H5OH(os)] � 1.7 mol/L[CH3COOC2H5(os)] � 3.1 mol/L[H2O(os)] � 3.1 mol/L

38. Phenolphthalein indicator was used for a titration of several10.00 mL samples of hypochlorous acid with 0.350 mol/Lbarium hydroxide solution. An average titrant volume of12.6 mL was required to reach the observed endpoint ofthese trials. According to this evidence, predict theamount concentration of the hypochlorous acid solution.

39. In a titration, 0.20 mol/L HCl(aq) titrant is to be graduallyadded to a 20.0 mL sample of 0.10 mol/L NaOH(aq). (a) Sketch a theoretical (calculated) pH curve. (b) Include the following information as labels on your

sketch. Show all relevant calculations.(i) the equivalence point pH and titrant volume for

the reaction(ii) the initial pH of the sodium hydroxide sample

solution(iii) the pH after adding 5.0 mL of HCl(aq) (treat this

part as a limiting reagent calculation)(iv) the entities present at the equivalence point(v) the pH after adding 9.0 mL of HCl(aq)(vi) the pH after adding 11.0 mL of HCl(aq)

(c) Suggest a suitable indicator for an endpointdetermination for this titration. Indicate the pH valuesfor the indicator colour change range on your pHcurve.

40. The solubility of calcium hydroxide is expressed in thefollowing equation, representing a saturated solutionequilibrium at 25 °C: Ca(OH)2(s) 0 Ca2�(aq) � 2 OH�(aq) Ksp � 1.3 � 10�6

(a) Predict whether calcium hydroxide is highly soluble oronly slightly soluble in water.

(b) What is the common name for a saturated calciumhydroxide solution, and what diagnostic test can beperformed with it?

(c) Explain how the solubility makes this compound safeto handle, even though hydroxide ion is the strongestbase possible in aqueous solution.

(d) Explain why “liming” of a (moist) garden soil will onlyraise the pH by a small amount, but then will continueto keep it elevated for months.

41. In a concentrated commercial hydrochloric acid solution,about one of every four molecules is HCl, as calculatedfrom the mass percent printed on the label. Predict whatfraction of these HCl molecules is actually in the “ready toreact” form of H3O

�(aq).

42. When neutralizing this acid spill with calcium hydroxide,identify which compound you would want to be in excess,and explain why.

43. Write a chemical equation and a Brønsted–Lowry equationfor this neutralization.

Hydrochloric acid has also been in common use for a verylong time. It is still sold under its historical name, muriatic acid,usually as a solution of about 30–35% HCl. It is very useful as apowerful rust remover, or to adjust pH levels in swimmingpools. It will also “etch” the surface of concrete by reactingwith carbonate ions in the solid mixture, with the result thatpaint can adhere to the clean, rough concrete surface. A spillof this very corrosive acid is always a serious problem (Figure 5).

Figure 5Horticultural lime (calcium hydroxide) is quickly sprinkled on aspill of concentrated muriatic (hydrochloric) acid.


Lime is a very simple substance, and very easily prepared bystrongly heating natural chalk. It has been used throughoutrecorded history, and by now the word “lime” appears incommon names for many substances. Lime is calcium oxide,CaO(s), which is often sold as quicklime, or unslaked lime. Thisoxide is hazardous to handle because it reacts very readily andrapidly with water, releasing a lot of heat. When “slaked” withwater, it forms calcium hydroxide, Ca(OH)2(s), which is muchless dangerous to handle. This compound is widely sold ingarden stores as agricultural, or horticultural, lime—meaning arather impure form. It is commonly added to soils to raise thesoil pH value, because the absorption by plants of somenutrients and trace elements depends heavily on soil pH levels.

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44. A series of experiments with a non-aqueous solventdetermined that the products are highly favoured in eachof the following acid–base equilibria, as written. (C6H5)3C

� � C4H4NH 0 (C6H5)3CH � C4H4N�

CH3COOH � HS�0 CH3COO� � H2S

O2� � (C6H5)3CH 0 OH� � (C6H5)3C�

C4H4N� � H2S 0 C4H4NH � HS�

(a) Identify the Brønsted–Lowry acids, bases, andconjugate acid–base pairs in each of these chemicalreactions.

(b) Arrange the acids in these four chemical reactions inorder of decreasing acid strength (in the solventused), as a standard table of acids and conjugatebases.

45. The hydronium ion concentration of a 0.100 mol/Ln-butanoic (butyric) acid solution was measured to be 1.24 � 10�3 mol/L.

Determine the percent reaction (ionization) of thisparticular weak acid solution, and a Ka value for aqueousn-butanoic acid at this ambient temperature.

46. A 0.10 mol/L solution of the essential amino acidtryptophan (1-a-amino-3-indolepropanoic acid) has ameasured pH of 5.19 at 25 °C. Predict the percent reactionof this tryptophan solution, and the Ka value for aqueoustryptophan. The molecular formula for tryptophan may bewritten as C10H11N2COOH.

47. Glycine, H2NCH2COOH(aq), is a nonessential amino acid,with the simplest structure of all the amino acids, and hasa Ka � 4.5 � 10�7 at 25 °C. Calculate the hydronium ionconcentration, the pH, and the percent reaction in a 0.050 mol/L aqueous solution of glycine at 25 °C.

48. Thioacetic acid, CH3COSH(aq), has a Ka � 4.7 � 10�4

at 25 °C. (a) Write the Ka expression for thioacetic acid. (b) Calculate the hydronium and thioacetate ion

concentrations, the pH, and the percent reaction in a2.00 mol/L CH3COSH(aq) solution at 25 °C.

(c) Calculate Kb for the conjugate base, the thioacetateion.

49. Predict the percent reaction, pH, pOH, and hydroxide ionconcentration of a 0.012 mol/L solution of sodiumbenzoate.

50. A 0.100 mol/L laboratory solution of sodium propanoate,NaC2H5COO(aq), has a measured pH of 8.95 at 25 °C.Calculate Kb for the propanoate ion.

Unit 8

51. Chloro-substituted acetic acids are used in organicsynthesis, cleaners, and herbicides. These acids areprepared by the chlorination of acetic acid in the presenceof small amounts of phosphorus. This process is known asthe Hell–Volhard–Zelinsky reaction. As is typical of organicsynthesis reactions, at equilibrium, the reaction vesselcontains a mixture of all the different possible reactionproducts as well as some unreacted acetic acid andchlorine. When these acids were studied separately, thedata in Table 1 were obtained.

(a) Calculate acid ionization constants for eachchloroacetic acid. Write chemical equations todescribe the reactions of these acids with water.

(b) Suggest a theoretical explanation for the relativestrengths of this series of acids. (Hint: Would addingchlorines to the other end of the molecule make the —COOH end of the molecule more, or less, negative?)

(c) A chemical technician is assigned the design of anacid–base titration to determine the amounts of eachacid present in the mixture. She knows fromexperience that the reaction of acetic acid with astrong base is quantitative. Sketch a simplified pHcurve for titration of a mixture of the four acidsproduced by chlorinating acetic acid. How could thetechnician determine relative chemical amounts of theacids present in the mixture?

52. Liquid ammonia can be used as a solvent for acid–basereactions. (a) Predict the strongest acid species that could be

present in this solvent. (Consider the parallel withliquid water, H2O(l). Also consider the likely reaction ofa very strong proton donor, such as hydrogenchloride, when it dissolves and reacts quantitatively inliquid ammonia.)

(b) The ionization equilibrium of pure liquid ammonia issimilar to that of pure liquid water. Write theequilibrium equation for the ionization of liquidammonia.

(c) Predict the strongest base that could be present inliquid ammonia solution.

(d) Sketch a titration curve for the addition of thestrongest acid in ammonia solution to the strongestbase. Suggest what value, instead of pH, might beused on the vertical axis of your graph.

53. Review the focusing questions on page 670. Using theknowledge you have gained from this unit, briefly outline aresponse to each of these questions.

Chemical Equilibrium Focusing on Acid–Base Systems 779

O

HO C

C

C

C

Table 1 Comparison of 0.100 mol/L Solutions of AceticAcid and the Chloroacetic Acids

Substance Formula pH

acetic acid CH3COOH(aq) 2.89

chloroacetic acid CH2ClCOOH(aq) 1.94

dichloroacetic acid CHCl2COOH(aq) 1.30

trichloroacetic acid CCl3COOH(aq) 1.14

DE

DE

DE

DE


unit 8 chemical equilibrium focusing on acid-base systems

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