ch 11 acids and bases student - pbworks

General, Organic, and Biological Chemistry: Structures of Life, 5/e

Karen C. Timberlake

© 2016 Pearson Education, Inc.

Karen C. Timberlake

Lecture Presentation

Chapter 11

Acids and Bases


Karen C. Timberlake


Clinical laboratory technicians prepare specimens for the detection of cancerous tumors and type blood samples for transfusions. They must also interpret and analyze the test results, which are then passed on to the physician.

Chapter 11 Acids and Bases


Karen C. Timberlake


Chapter 11 Readiness

Key Math Skills

• Solving Equations (1.4D)

• Writing Numbers in Scientific Notation (1.4F)


Karen C. Timberlake


Chapter 11 Readiness

Core Chemistry Skills

• Writing Ionic Formulas (6.2)

• Balancing a Chemical Equation (7.1)

• Using Concentrations as a Conversion Factor (9.4)

• Writing the Equilibrium Constant Expression (10.3)

• Calculating Equilibrium Concentrations (10.4)

• Using Le Châtelier’s Principle (10.5)


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Chapter 11 Acids and Bases

A soft drink contains phosphoric acid (H3PO4) and carbonic acid (H2CO3).

Learning Goal Describe and name acids and bases.


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Arrhenius Acids

Arrhenius acids

• produce hydrogen ions (H+) when they dissolve in water.

HCl(g) H+(aq) + Cl−(aq)

• are also electrolytes, because they produce H+ in water.

• have a sour taste.

• turn blue litmus red.

• corrode some metals.

H2O(l)


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Naming Acids

• Acids with a hydrogen ion (H+) and a nonmetal (or CN−) ion are named with the prefix hydro and end with ic acid.

HCl(aq) hydrochloric acid

• Acids with a hydrogen ion (H+) and a polyatomic ion are named by changing the end of the name of the polyatomic ion from

ate to ic acid or ite to ous acid

ClO3− chlorate ClO2

− chlorite

HClO3 chloric acid HClO2 chlorous acid


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Study Check

Select the correct name for each of the following acids:1. HBr A. bromic acid

B. bromous acid C. hydrobromic acid

2. H2CO3 A. carbonic acidB. hydrocarbonic acidC. carbonous acid

3. HBrO2 A. bromic acidB. hydrobromous acidC. bromous acid


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Arrhenius Bases

Arrhenius bases

• produce hydroxide ions (OH−)in water.

• taste bitter or chalky.• are also electrolytes, because

they produce hydroxide ions (OH−) in water.

• feel soapy and slippery.• turn litmus indicator paper • blue and phenolphthalein

indicator pink.

An Arrhenius base produces

cations and OH− anions in an aqueous solution.


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Naming Bases

Typical Arrhenius bases are named as hydroxides.

NaOH sodium hydroxide

KOH potassium hydroxide

Ba(OH)2 barium hydroxide

Al(OH)3 aluminum hydroxide

Calcium hydroxide, Ca(OH)2,

is used in the food industryto produce beverages, and in

dentistry as a filler for root canals.


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Characteristics of Acids and Bases


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Study Check

Match the formulas of acids and bases with their names.

1. ___ HNO2 A. iodic acid

2. ___ Ca(OH)2 B. sulfuric acid

3. ___ H2SO4 C. sodium hydroxide

4. ___ HIO3 D. nitrous acid

5. ___ NaOH E. calcium hydroxide


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11.2 Brønsted–Lowry Acids and Bases

According to the Brønsted–Lowry theory,

• an acid is a substance that donates H+.

• a base is a substance that accepts H+.

Learning Goal Identify conjugate acid–base pairs for Brønsted–Lowry acids and bases.


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NH3, a Brønsted–Lowry Base

In the reaction of ammonia and water,

• NH3 acts as the base that accepts H+.

• H2O acts as the acid that donates H+.

Because the nitrogen atom of NH3 has a stronger attraction for H+ than oxygen, water acts as an acid by donating H+.


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Study Check

In each of the following equations, identify the Brønsted–

Lowry acid and base in the reactants:

A. HNO3(aq) + H2O(l) H3O+(aq) + NO3

−(aq)

B. HF(aq) + H2O(l) H3O+ (aq) + F−(aq)


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Study Check

Identify each as a characteristic of

A. an acid or B. a base.

____ 1. has a sour taste

____ 2. produces OH− in aqueous solutions

____ 3. has a chalky taste

____ 4. is an electrolyte

____ 5. produces H+ in aqueous solutions


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Conjugate Acid–Base Pairs

In any acid–base reaction, there are two conjugate acid–

base pairs.

• Each pair is related by the loss and gain of H+.

• One pair occurs in the forward direction.

• One pair occurs in the reverse direction.

Acid and conjugate base pair 1

HA + B A− + BH+

Base and conjugate acid pair 2


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In this acid–base reaction,

• the first conjugate acid–base pair is HF, which donates H+ to form its conjugate base, F−.

• the other conjugate acid–base pair is H2O, which accepts H+ to form its conjugate acid, H3O

+.

• each pair is related by a loss and gain of H+.


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In the reaction of NH3 and H2O,

• one conjugate acid–base pair is NH3/NH4+.

• the other conjugate acid–base pair is H2O/H3O+.

Core Chemistry Skill Identifying Conjugate Acid–Base Pairs


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Study Check

1. Write the conjugate base for each of the following acids:

A. HBr

B. H2S

C. H2CO3

2. Write the conjugate acid of each of the following bases:

A. NO2−

B. NH3

C. OH−


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Study Check

Identify the sets that contain acid–base conjugate pairs.

1. HNO2, NO2−

2. H2CO3, CO32−

3. HCl, ClO4−

4. HS−, H2S

5. NH3, NH4+


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Amphoteric Substances

Substances that can act as both acids and bases are

amphoteric or amphiprotic.

For water, the most common amphoteric substance, the acidic or basic behavior depends on the other reactant.

• Water donates H+ when it reacts with a stronger base.

• Water accepts H+ when it reacts with a stronger acid.


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Guide to Writing Conjugate Acid–Base Pairs


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Study Check

Identify the conjugate acid–base pairs in the following

reaction:

HNO3(aq) + NH3(aq) NO3−(aq) + NH4

+(aq)


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Solution

Identify the conjugate acid–base pairs in the following reaction:


+(aq)

STEP 1 Identify the reactant that loses H+ as the acid.

In the reaction, HNO3 donates H+ to NH3.

STEP 2 Identify the reactant that gains H+ as the base.

In the reaction, NH3 gains H+ to form NH4+. Thus,

• NH3 is the base and NH4+ is its conjugate acid.

• HBr is the acid and Br− is its conjugate base.


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Solution

Identify the conjugate acid–base pairs in the following

reaction:


+(aq)

STEP 3 Write the conjugate acid–base pairs.

HBr/Br− is the acid and conjugate base pair.

NH3/NH4+ is the base and conjugate acid pair.


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11.3 Strengths of Acids and Bases

Weak acids only partially

dissociate in water.

Hydrofluoric acid, HF, is

the only halogen that

forms a weak acid.

Learning Goal Write equations for the dissociation of strong and weak acids; identify the direction of reaction.


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Strong and Weak Acids

A strong acid completely ionizes (100%) in aqueous solutions.

HCl(g) + H2O(l) H3O+(aq) + Cl−(aq)

A weak acid dissociates only slightly in water to form a few ions in aqueous solutions.

H2CO3(aq) + H2O(l) H3O+(aq) + HCO3

− (aq)


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Strong Acids

In water, the dissolved molecules

of HA, a strong acid,

• dissociate into ions 100%.

• produce large concentrations of H3O

+ and the anion (A−).

The strong acid HCl dissociates completely into ions:

HCl(g) + H2O(l)

H3O+(aq) + Cl−(aq)


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Weak Acids

In weak acids, only a few

molecules dissociate.

• Most of the weak acid remains as the undissociated(molecular) form of the acid.

• The concentrations of H3O+

and the anion (A−) are small.

H2CO3 is a weak acid:

H2CO3(aq) + H2O(l)

H3O+(aq) + HCO3

−(aq)


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Strong and Weak Acid Dissociation

• In an HCl solution,

the strong acid HCldissociates 100% to form H+ and Cl−.

• A solution of the weak acid HC2H3O2

contains mostly molecules of HC2H3O2 and a few ions of H+ and C2H3O2

−.


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Strong and Weak Acid Dissociation

Figure 11.2 ▶ After dissociation in water, (a) the strong acid HI has highconcentrations of H3O

+ and I–, and (b) the weak acid HF has a high

concentration of HF and low concentrations of H3O+ and F

–

.


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Diprotic Acids: Carbonic Acid

• Some weak acids, such as carbonic acid, are diprotic

acids that have two H+, which dissociate one at a time.

H2CO3(aq) + H2O(l) H3O+(aq) + HCO3

−(aq)

• Because HCO3− is also a weak acid, a second dissociation

can take place to produce another hydronium ion and the carbonate ion, CO3

2−.

HCO3−(aq) + H2O(l) H3O

+(aq) + CO32−(aq)


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Diprotic Acids: Sulfuric Acid

• Some strong acids, such as sulfuric acid, are

diprotic acids that have two H+, which dissociate

one at a time.

H2SO4(aq) + H2O(l) H3O+(aq) + HSO4

−(aq)

• Because HSO4− is a weak acid, a second

dissociation can take place to produce another H+

and the sulfate ion, SO42− .

HSO4−(aq) + H2O(l) H3O

+(aq) + SO42−(aq)


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Strong Bases

Strong bases as strong electrolytes

• are formed from metals of Groups 1A (1) and 2A (2).

• include LiOH, NaOH, KOH, Ba(OH)2, Sr(OH)2, and Ca(OH)2.

• dissociate completely in water.

KOH(s) K+(aq) + OH−(aq)

• are found in household products used to remove grease and unclog drains.


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Weak Bases

Weak bases are weak electrolytes

• that are poor acceptors of H+ ions.

• produce very few ions in solution.

• include ammonia.

NH3(g) + H2O(l) NH4+(aq) + OH−(aq)

Ammonia Ammonium hydroxide


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Strong and Weak Bases

Strong Bases

Lithium hydroxide LiOHSodium hydroxide NaOHPotassium hydroxide KOHRubidium hydroxide RbOHCesium hydroxide CsOHCalcium hydroxide Ca(OH) 2*Strontium hydroxide Sr(OH) 2*Barium hydroxide Ba(OH)2*

*Low solubility, but they dissociate completely

Bases in Household Products

Weak BasesWindow cleaner, ammonia, NH3

Bleach, NaOClLaundry detergent, Na2CO3, Na3PO4

Toothpaste and baking soda, NaHC3

Baking powder, scouring powder, Na2CO3

Lime for lawns and agriculture, CaCO3

Laxatives, antacids, Mg(OH) 2, Al(OH)3

Strong BasesDrain cleaner, oven cleaner, NaOH


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Direction of Reaction

Strong acids have weak conjugate bases that do not readily

accept H+. • As the strength of the acid decreases, the strength of its

conjugate base increases.

In any acid–base reaction, there are two acids and two bases.• However, one acid is stronger than the other acid, and one

base is stronger than the other base.• By comparing their relative strengths, we can determine

the direction of the reaction.


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Direction of Reaction: H2SO4

Sulfuric acid, H2SO4, is a strong acid that readily gives up H+

to water.

H2SO4(aq) + H2O(l) H3O+(aq) + HSO4

−(aq)

Stronger Stronger Weaker Weaker

acid base acid base

• The hydronium ion H3O+ produced is a weaker acid than

H2SO4.• The conjugate base HSO4

− is a weaker base than water.


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Direction of Reaction: CO32−

The carbonate ion from carbonic acid, H2CO3, reacts with water.

• Water donates one H+ to carbonate, CO32− to form HCO3

−

and OH−.

• From Table 11.3, we see that HCO3− is a stronger acid than H2O.

• We also see that OH− is a stronger base than CO32−.

To reach equilibrium, the strong acid and strong base react in the

direction of the weaker acid and weaker base.

CO32− (aq) + H2O(l) OH−(aq) + HCO3

−(aq)

Weaker Weaker Stronger Stronger

acid base base acid


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Study Check

Identify each of the following as a strong or weak acid

or base:

A. HBr

B. HNO2

C. NaOH

D. H2SO4

E. Cu(OH)2


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Study Check

Using Table 11.3, identify the stronger acid in each pair.

A. HNO2 or H2S

B. HCO3− or HBr

C. H3PO4 or H3O+


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11.4 Dissociation Constants for Acids and Bases

HCHO2(aq) + H2O(l) H3O+(aq) + CHO2

−(aq)

Learning Goal Write the expression for the dissociation constant of a weak acid or weak base.


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Dissociation of a Weak Acid

Because the dissociation of strong acids in water is essentially complete, the reaction is not considered to be an equilibrium process.

• Weak acids partially dissociate in water as the ion products reach equilibrium with the undissociated weak acid molecules.

• Formic acid is a weak acid that dissociates in water to form hydronium ion, H3O

+, and formate ion, CHO2−.

HCHO2 (aq) + H2O(l) H3O+(aq) + CHO2

−(aq)


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Writing Dissociation Constants

As with other dissociation expressions,

• the molar concentration of the products is divided by the molar concentration of the reactants.

• water is a pure liquid with a constant concentration and is omitted.

• the expression is called acid dissociation constant, Ka.

HCHO2 (aq) + H2O(l) H3O+(aq) + CHO2

−(aq)


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Acid Dissociation Constant, Ka

When the value of the Ka

• is small, the equilibrium lies to the left, favoring the reactants.

• is large, the equilibrium lies to the right, favoring the products.

Weak acids have small Ka values, while strong acids have very large Ka values.


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Base Dissociation Constant, Kb

When the value of the Kb,

• is small, the equilibrium lies to the left, favoring the reactants.

• is large, the equilibrium lies to the right, favoring the products.

The stronger the base, the larger the Kb value.

CH3—NH2(aq) + H2O(l) CH3—NH3+(aq) + OH−(aq)

The concentration of water is omitted from the base dissociation constant expression.


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Study Check

Write the acid dissociation constant expression for nitrous acid, HNO2.


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11.5 Dissociation of Water

The equilibrium reached between the conjugate acid–base

pairs of water produces both H3O+ and OH−.

H2O(l) + H2O(l) H3O+(aq) +

OH−(aq)

Learning Goal Use the water dissociation constant to calculate the [H3O

+] and [OH−] in an aqueous solution.

+ ++ -


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Dissociation Constant of Water, Kw

Water is amphoteric—it can act as an acid or a base. In water,

• H+ is transferred from one H2O molecule to another.

• one water molecule acts as an acid, while another acts as a base.

• equilibrium is reached between the conjugate acid–base pairs.


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Writing the Dissociation Constant, Kw

In the equation for the dissociation of water, there is both a

forward and a reverse reaction.

H2O(l) + H2O(l) H3O+(aq) + OH−(aq)

• In pure water, the concentrations of H3O+ and OH− at 25

°C are each 1.0 × 10−7 M.

[H3O+] = [OH−] = 1.0 × 10–7 M

Kw = [H3O+] [OH−]

Kw = (1.0 × 10−7 M) (1.0 × 10−7 M) = 1.0 × 10–14 at 25 °C

Base Acid Conjugate Conjugate

acid base


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Dissociation Constant, Kw

The ion product constant for water, Kw, is defined as

• the product of the concentrations of H3O+ and OH−.

• equal to 1.0 × 10−14 at 25 °C (the concentration units

are omitted).

When

• [H3O+] and [OH−] are equal, the solution is neutral.

• [H3O+] is greater than the [OH−], the solution is acidic.

• [OH−] is greater than the [H3O+], the solution is basic.


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Using Kw to Calculate [H3O+] and [OH−]

• If we know the [H3O+] of a solution, we can use the

Kw to calculate the [OH−].

• If we know the [OH−] of a solution, we can use the Kw to calculate the [H3O

+].


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Pure Water Is Neutral

[H3O+] = 1.0 × 10−7 M

[OH−] = 1.0 × 10−7 M

[H3O+] = [OH−]

Pure water is neutral.

In pure water, the ionization of water molecules produces small but equal quantities of H3O

+ and OH− ions.


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Acidic Solutions

Adding an acid to pure water

• increases the [H3O+].

• causes the [H3O+] to exceed

1.0 × 10−7 M.

• decreases the [OH−].

[H3O+] > [OH−]

The solution is acidic.


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Basic Solutions

Adding a base to pure water

• increases the [OH−]

• causes the [OH−] to exceed 1.0 × 10−7 M

• decreases the [H3O+]

[H3O+] < [OH−]

The solution is basic.


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Comparison of [H3O+] and [OH−]


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© 2016 Pearson Education, Inc.General, Organic, and Biological Chemistry: Structures of Life, 5/e

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Guide to Calculating [H3O+] and [O–] in

Aqueous Solutions


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Calculating [H3O+]

What is the [H3O+] of a solution if [OH−] is 5.0 × 10−8 M?

STEP 1 State the given and needed quantities.

STEP 2 Write the Kw for water and solve for the unknown [H3O

+].

ANALYZE Given Need Know

THE [OH−] = 5.0 × 10−8 M [H3O+] Kw = [H3O

+][OH−]

PROBLEM = 1.0 × 10−14


THE [OH−] = 5.0 × 10−8 M [H3O+] Kw = [H3O

+][OH−]

PROBLEM = 1.0 × 10−14


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Calculating [H3O+]

What is the [H3O+] of a solution if [OH−] is 5.0 × 10−8 M?

STEP 3 Substitute in the known [H3O+] or [OH−] and

calculate.

Because the [H3O+] of 2.0 × 10–7 M is larger than the

[OH−] of 5.0 × 10–8 M, the solution is acidic.


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Study Check

If lemon juice has [H3O+] of 2.0 × 10–3 M, what is the [OH−] of

the solution?

A. 2.0 × 10−11 M

B. 5.0 × 10−11 M

C. 5.0 × 10−12 M


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11.6 The pH Scale

Learning Goal Calculate the pH from [H3O+]; given the

pH, calculate [H3O+] and [OH−] of a solution.

The pH scale is used to describe the acidity of solutions.

A dipstick is used to measure the pH of urine.


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The pH Scale

The pH of a solution

• is used to indicate the acidity of a solution.

• has values that usually range from 0 to 14.

• is acidic when the values are less than 7.

• is neutral at a pH of 7.

• is basic when the values are greater than 7.


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The pH Scale

The pH of a solution is commonly measured using

• a pH meter in the laboratory.

• pH paper, an indicator that turns specific colors at a specific pH value.

The pH of a solution is found by comparing the colors of indicator paper to a chart.


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pH Measurement

The pH of a solution can be determined using (a) a pH meter,

(b) pH paper, and (c) indicators that turn different colors corresponding to different pH values.


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pH of Common Substances

On the pH scale,

values below 7.0 are acidic, a value of 7.0 is neutral, and values above 7.0 are basic.


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Study Check

Identify each solution as acidic, basic, or neutral.

A. ___ HCl with a pH = 1.5

B. ___ pancreatic fluid, [H3O+] = 1 × 10−8 M

C. ___ Sprite soft drink, pH = 3.0

D. ___ pH = 7.0

E. ___ [OH−] = 3 × 10−10 M

F. ___ [H3O+ ] = 5 × 10−12


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Calculating the pH of Solutions

The pH scale

• is a logarithmic scale that corresponds to the [H3O+] of

aqueous solutions.

• is the negative logarithm (base 10) of the [H3O+].

pH = −log[H3O+]

To calculate the pH, the negative powers of 10 in the molar concentrations are converted to positive numbers. If [H3O

+] is 1.0 × 10−2 M,

pH = −log[1.0 × 10−2 ] = −(−2.00) = 2.00

Key Math Skill Calculating pH from [H3O+]


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pH: Significant Figures

To determine the number of significant figures in the pH value,

• the number of decimal places in the pH value is the same as the number of significant figures in the coefficient of [H3O

+].

• the number to the left of the decimal point in the pH value is the power of 10.


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pH Scale and [H3O+]

Because pH is a log scale,

• a change of one pH unit corresponds to a tenfold change in [H3O

+].

• pH decreases as the [H3O+] increases.

pH 2.00 is [H3O+] = 1.0 × 10−2 M

pH 3.00 is [H3O+] = 1.0 × 10−3 M

pH 4.00 is [H3O+] = 1.0 × 10−4 M


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Guide to Calculating pH of Solutions

The pH of a solution is calculated from the [H3O+] by using the

log key in your calculator and changing the sign.


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pH Calculation

Aspirin, which is acetylsalicylic acid, was the first nonsteroidal

anti-inflammatory drug used to alleviate pain and fever. If a solution of aspirin has a [H3O

+] = 1.7 × 10−3 M, what is the pH of the solution?


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pH Calculation

If a solution of aspirin has a [H3O+] = 1.7 × 10−3 M, what is

the pH of the solution?



THE [H3O+] = 1.7 × 10−3 M pH of solution pH = −log[H3O

+]

PROBLEM


THE [H3O+] = 1.7 × 10−3 M pH of solution pH = −log[H3O

+]

PROBLEM


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pH Calculation

If a solution of aspirin has a [H3O+] = 1.7 × 10−3 M, what is the

pH of the solution?

STEP 2 Enter the [H3O+] into the pH equation and

calculate.

pH = −log[H3O+] = −log[1.7 × 10−3]

Calculator Procedure:

1.7 3

Calculator Display:

EE or Exp +/−log =

2.769551079


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pH Calculation

If a solution of aspirin has a [H3O+] = 1.7 × 10−3 M, what is the

pH of the solution?

STEP 3 Adjust the number of SFs on the right of the decimal point.

Coefficient Power of ten

1.7 × 10–3

Two SFs Exact

pH = −log[1.7 × 10−3] = 2.77


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Study Check

Find the pH of a solution with a [H3O+] of 4.0 × 10−5.


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Calculating [H3O+] from pH

Given the pH of a solution, we can reverse the calculation to

obtain the [H3O+].

• For whole number pH values, the negative pH value is the power of 10 in the [H3O

+] concentration.

[H3O+] = 10−pH

• For pH values that are not whole numbers, the calculation requires the use of the 10x key, which is usually a 2nd function key.

Key Math Skill Calculating [H3O+] from pH.


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Guide to Calculating [H3O+] from pH


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Study Check

Determine the [H3O+] for a solution that has a pH

of 3.42.


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11.7 Reactions of Acids and Bases

Gastric acid contains

HCl and is produced by parietal cells that line the stomach. When protein enters the stomach, HClis secreted until the pH reaches 2, the optimum pH for digestion.

Learning Goal Write balanced equations for reactions of acids with metals, carbonates, and bases.


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Reactions of Acids

Acids react with

• metals to produce salt and hydrogen gas.

• bases to produce a salt and water.

• bicarbonate and carbonate ions to produce carbon dioxide gas.

A salt is an ionic compound that does not have H+ as the cation or OH− as the anion.


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Acids and Metals

Acids react with metals to produce hydrogen gas and the salt

of the metal.

2K(s) + 2HCl(aq) 2KCl(aq) + H2(g)

Metal Acid Salt Hydrogen

Zn(s) + 2HCl(aq) ZnCl2(aq) + H2(g)

Metal Acid Salt Hydrogen

Magnesium reacts

rapidly with acid and

forms H2 gas and a

salt of magnesium.


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Acids, Carbonates, and Bicarbonates

Acids react with carbonates and hydrogen carbonates to produce

carbon dioxide gas, a salt, and water:

2HCl(aq) + CaCO3(s) CO2(g) + CaCl2(aq) + H2O(l)

Acid Carbonate Carbon Salt Water

dioxide

HCl(aq) + NaHCO3(s) CO2(g) + NaCl(aq) + H2O(l)

Acid Bicarbonate Carbon Salt Water

dioxide

Core Chemistry Skill Writing Equations for Reactions of Acids

and Bases


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Acids and Hydroxides: Neutralization

In a neutralization reaction,

• an acid reacts with a base to produce salt and water.

• the salt formed is the anion from the acid and the cationfrom the base.

HCl(aq) + NaOH(aq) NaCl(aq) + H2O(l)

Acid Base Salt Water


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Acids and Hydroxides: Neutralization

In neutralization reactions, one H+ always reacts with

one OH−.

• If we write the strong acid and strong base as ions,

HCl(aq) + NaCl(aq) NaOH(aq) + H2O(l)

we see that H+ reacts with OH− to form water, leaving the ions Na+ and Cl− in solution:

H+(aq) + Cl−(aq) + Na+(aq) + OH−(aq)

Na+(aq) + Cl−(aq) + H2O(l)

• The overall reaction occurs as the H+ from the acid and OH−

from the base form water:

H+(aq) + OH−(aq) H2O(l) Net ionic equation


Karen C. Timberlake


Guide to Balancing an Equation for Neutralization


Karen C. Timberlake


Balancing Neutralization Reactions

Write the balanced equation for the neutralization of solid magnesium hydroxide and nitric acid.

STEP 1 Write the reactants and products.

Mg(OH)2(s) + HNO3(aq) salt + H2O(l)

STEP 2 Balance the H+ in the acid with the OH− inthe base.

Mg(OH)2(s) + 2HNO3(aq) salt + H2O(l)


Karen C. Timberlake


Balancing Neutralization Reactions

Write the balanced equation for the neutralization of solid magnesium hydroxide and nitric acid.

STEP 3 Balance the H2O with H+ and the OH−.

Mg(OH)2(s) + 2HNO3(aq) salt + 2H2O(l)

STEP 4 Write the salt from the remaining ions.

Mg(OH)2(s) + 2HNO3(aq) Mg(NO3)2(aq) + 2H2O(l)


Karen C. Timberlake


Study Check

Write a balanced equation for the following reaction:

Mg(OH)2(s) + HBr(aq)


Karen C. Timberlake


Study Check

Select the correct group of coefficients for each of the

following neutralization equations:

1. HCl(aq) + Al(OH)3(aq) AlCl3(aq) + H2O(l)

A. 1, 3, 3, 1 B. 3, 1, 1, 1 C. 3, 1, 1, 3

2. Ba(OH)2(aq) + H3PO4(aq) Ba3(PO4)2(s) + H2O(l)

A. 3, 2, 2, 2 B. 3, 2, 1, 6 C. 2, 3, 1, 6


Karen C. Timberlake


Solution

Select the correct group of coefficients for each of the

following neutralization equations:

STEP 4 Write the salt from the remaining ions.

1. 3HCl(aq) + Al(OH)3(aq) AlCl3(aq) + 3H2O(l)

The answer is C, 3, 1, 1, 3.

2. 3Ba(OH)2(aq) + 2H3PO4(aq)

Ba3(PO4)2(s) + 6H2O(l)

The answer is B, 3, 2, 1, 6.


Karen C. Timberlake


Chemistry Link to Health: Antacids

Antacids are substances that

• are used to neutralize excess stomach acid.• are made of aluminum hydroxide and magnesium

hydroxide mixtures.

These hydroxides are not very soluble in water, so the levels of available OH−

are not damaging to theintestinal tract.


Karen C. Timberlake


11.8 Acid–Base Titration

The titration of an acid. A known volume of an acid is placed in

a flask with an indicator and titrated with a measured volume of a base solution, such as NaOH, to the neutralization endpoint.

Learning Goal Calculate the molarity or volume of an acid or base

from titration information.


Karen C. Timberlake


Acid–Base Titration

Titration

• is a laboratory procedure used to determine the molarity of an acid.

• uses a base such as NaOH to neutralize a measured volume of an acid.

• requires a few drops of an indicator such as phenolphthalein to identify the endpoint.

Core Chemistry Skill Calculating Molarity or Volume of an Acid or

Base in a Titration


Karen C. Timberlake


Acid–Base Titration

In the following titration, a specific volume of acidic solution is titrated to the endpoint with a known concentration of NaOH.

Base �

NaOH

Acid �

Solution


Karen C. Timberlake


Indicator

The indicator phenolphthalein

• is added to identify the

endpoint.

• turns pink when the solution is neutralized.


Karen C. Timberlake


Endpoint of Titration

At the endpoint of the titration,

• the moles of base are equal to the moles of acid in the solution.

• the concentration of the base is known.

• the volume of the base used to reach the endpoint is measured.

• the molarity of the acid is calculated using the neutralization equation for the reaction.


Karen C. Timberlake


Guide to Calculating Boiling Point Elevation, Freezing Point Lowering


Karen C. Timberlake


Acid–Base Titration Calculations

What is the molarity of an HCl solution if 18.5 mL of

0.225 M NaOH is required to neutralize 0.0100 L of HCl?


STEP 1 State given and needed quantities and concentrations.

ANALYZE Given Need Equation

THE 18.5 mL of molarity of HCl(aq) + NaOH(aq) �

PROBLEM 0.225 M NaOH HCl solution NaCl(aq) + H2O(l)

0.0100 L HCl

ANALYZE Given Need Equation

THE 18.5 mL of molarity of HCl(aq) + NaOH(aq) �

PROBLEM 0.225 M NaOH HCl solution NaCl(aq) + H2O(l)

0.0100 L HCl


Karen C. Timberlake



What is the molarity of an HCl solution if 18.5 mL of

0.225 M NaOH is required to neutralize 0.0100 L of HCl?


STEP 2 Write a plan to calculate molarity or volume.

mL NaOH

solution

Metric

factorL NaOH

solutionMolarity

moles of

NaOH

Mole–Mole

factor

moles of

NaOH

moles of

HClDivide by

liters

molarity of

HCl solution


Karen C. Timberlake



What is the molarity of an HCl solution if 18.5 mL of 0.225 M

NaOH is required to neutralize 0.0100 L of HCl?


STEP 3 State equalities and conversion factors including concentration.


Karen C. Timberlake






STEP 4 Set up the problem to calculate the needed quantity.


Karen C. Timberlake


Study Check





Karen C. Timberlake


11.9 Buffers

A buffer solution maintains the pH by neutralizing small amounts

of added acid or base.

An acid must be present to react with any OH− added, and a

base must be present to react with any H3O+ added.

Learning Goal Describe the role of buffers in maintaining

the pH of a solution; calculate the pH of a buffer.


Karen C. Timberlake


Buffers

When an acid or a base

is added to water, the pH changes drastically.

In a buffer solution, the pH is maintained; pH does not change when acids or bases are added.


Karen C. Timberlake


How Buffers Work

Buffers work because

• they resist changes in pH from the addition of an acid or a base.

• in the body, they absorb H3O+ or OH− from foods and

cellular processes to maintain pH.

• they are important in the proper functioning of cells and blood.

• they maintain a pH close to 7.4 in blood.

A change in the pH of the blood affects the uptake of oxygen and cellular processes.


Karen C. Timberlake


Components of a Buffer

A buffer solution

• contains a combination of acid–base

conjugate pairs, a weak acid and a salt of its

conjugate base, such as

HC2H3O2(aq) and C2H3O2−(aq)

• has equal concentrations of a weak acid and

its salt.


Karen C. Timberlake


How Buffers Work

In the buffer with acetic acid (HC2H3O2) and sodium

acetate (NaC2H3O2),

• the salt produces acetate ions and sodium ions.

NaC2H3O2(aq) C2H3O2−(aq) + Na+(aq)

• the salt is added to provide a higher concentration of the conjugate base C2H3O2

− than from the weak acid alone.

HC2H3O2(aq) + H2O(l) C2H3O2−(aq) + H3O

+(aq) Large amount Large amount


Karen C. Timberlake


Function of a Weak Acid in a Buffer

If a small amount of base is added to this same buffer

solution, it is neutralized by the acetic acid, HC2H3O2, which shifts the equilibrium in the direction of the products acetate ion and water.

HC2H3O2(aq) + OH−(aq) C2H3O2−(aq) + H2O(l)

Equilibrium shifts in the direction of the products.


Karen C. Timberlake


Function of Conjugate Base in a Buffer

When a small amount of acid is added, the additional H3O+

combines with the acetate ion, C2H3O2−, causing the equilibrium to

shift in the direction of the reactants, acetic acid and water.

The acetic acid produced contributes to the available weak acid.

HC2H3O2(aq) + H2O(l) C2H3O2− (aq) + H3O

+(aq)

Equilibrium shifts in the direction of the reactants.


Karen C. Timberlake


Working Buffers

The buffer described here consists of about equal concentrations of

acetic acid (HC2H3O2) and its conjugate base, acetate ion (C2H3O2−).

• Adding H3O+ to the buffer reacts with the salt, C2H3O2

−, whereas

adding OH− neutralizes the acid HC2H3O2.

• The pH of the solution is maintained as long as the added

amounts of acid or base are small compared to the

concentrations of the buffer components.


Karen C. Timberlake


By rearranging the Ka expression to give [H3O+], we can obtain

the ratio of the acetic acid/acetate buffer and calculate the pH.

Solving for H3O+ gives

Weak acid

Conjugate base

Core Chemistry Skill Calculating the pH of a Buffer

Calculating the pH of a Buffer


Karen C. Timberlake


Study Check

The Ka for acetic acid, HC2H3O2, is 1.8 × 10–5. What is the pH of a buffer prepared with 1.0 M HC2H3O2 and 1.0 M C2H3O2

−?


+(aq)


Karen C. Timberlake


Solution


−?


+(aq)


ANALYZE Given Need

THE [HC2H3O2] = 1.0 M pH of solution

PROBLEM [C2H3O2−] = 1.0 M

Equation


+(aq)

ANALYZE Given Need

THE [HC2H3O2] = 1.0 M pH of solution

PROBLEM [C2H3O2−] = 1.0 M

Equation


+(aq)


Karen C. Timberlake


Solution


−?


+(aq)

STEP 2 Write the Ka expression and rearrange for [H3O+].


Karen C. Timberlake


Solution

The Ka for acetic acid, HC2H3O2, is 1.8 × 10–5. What is the pH

of a buffer prepared with 1.0 M HC2H3O2 and 1.0 M C2H3O2−?


+(aq)

STEP 3 Substitute [HA] and [A−] into the Ka expression.


Karen C. Timberlake


Solution

The Ka for acetic acid, HC2H3O2, is 1.8 × 10–5. What is the pH

of a buffer prepared with 1.0 M HC2H3O2 and 1.0 M C2H3O2−?


+(aq)

STEP 4 Use [H3O+] to calculate pH.


Karen C. Timberlake


Calculating the pH of a Buffer

Because Ka is a constant at a given temperature,

• the [H3O+] is determined by the [HC2H3O2]/[C2H3O2

−] ratio.

• the addition of small amounts of either acid or base changes the ratio of [HC2H3O2]/[C2H3O2

−] only slightly.

• the changes in [H3O+] will be small and the pH will be

maintained.

• the addition of a large amount of acid or base may exceed the buffering capacity of the system.


Karen C. Timberlake


Buffers and pH Changes

Buffers can be prepared from conjugate acid–base

pairs such as H2PO4−/HPO4

2− and HPO42−/PO4

3−,

HCO3−/CO3

2−, or NH4+/NH3.

The pH of the buffer solution will depend on the

conjugate acid–base pair chosen.


Karen C. Timberlake



Using a common phosphate buffer for biological specimens,

we can look at the effect of using different ratios of [H2PO4

−/HPO42−] on the [H3O

+] and pH. The Ka of H2PO4−

is 6.2 × 10−8.

The equation and the [H3O+] are written as follows:

H2PO4−(aq) + H2O(l) H3O

+(aq) + HPO42−(aq)


Karen C. Timberlake




Karen C. Timberlake


Chemistry Link to Health: Buffers in Blood Plasma

The arterial blood plasma has a normal pH of 7.35 to 7.45. If

changes in H3O+ lower the pH below 6.8 or raise it above 8.0, cells

cannot function properly and death may result.

In our cells, CO2

• is continually produced as an end product of cellular

metabolism.

• is carried to the lungs for elimination, and the rest dissolves in

body fluids such as plasma and saliva, forming carbonic acid,

H2CO3.

As a weak acid, carbonic acid dissociates to give bicarbonate,

HCO3−, and H3O

+.


Karen C. Timberlake



Karen C. Timberlake


Kidneys also supply more of the bicarbonate anion, HCO3−,

setting up an important buffer system in the body fluid:

CO2(g) + H2O(l) H2CO3(aq) HCO3−(aq) + H3O

+(aq)

Excess H3O+ entering the body fluids reacts with the HCO3

−, and

excess OH− reacts with the carbonic acid.

H2CO3(aq) + H2O(l) HCO3−(aq) + H3O

+(aq)




Karen C. Timberlake



Karen C. Timberlake




Karen C. Timberlake


Kidneys also supply more of the bicarbonate anion, HCO3−,

setting up an important buffer system in the body fluid:

CO2(g) + H2O(l) H2CO3(aq) HCO3−(aq) + H3O

+(aq)

• Excess H3O+ entering the body fluids reacts with the HCO3

−:

H2CO3(aq) + H2O(l) HCO3−(aq) + H3O

+(aq)


• Excess OH– entering the body fluids reacts with the H2CO3:

H2CO3(aq) + OH−(aq) H2O(l) + HCO3−(aq)

Equilibrium shifts in the direction of the products.


Karen C. Timberlake




Karen C. Timberlake


To maintain the normal blood plasma pH (7.35 to 7.45),

• the ratio of [H2CO3]/[HCO3−] needs to be about 1 to 10.

• concentrations of 0.0024 M H2CO3 and 0.024 M HCO3−

work to maintain that pH.


Karen C. Timberlake




Karen C. Timberlake


In the body, the concentration of carbonic acid is closely associated with the partial pressure of CO2, PCO2

.

• If the CO2 level rises, increasing H2CO3, the equilibrium shifts to produce more H3O

+, which lowers the pH. This condition is called acidosis.

• A lowering of the CO2 level leads to a high blood pH, a condition called alkalosis.


Karen C. Timberlake


Concept Map

ch 11 acids and bases student - pbworks

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