4 Reactions in solution Contents

4-1 Some Important Definitions

4-2 Electrolytes

4-3 Reactions Between Ions in Solution

4-4 Ionic Equations

4-5 Single-Replacement Reactions

4-6 Some Uses of Reactions in Solution

4-7 Concentration Expressed as Percent

4.8 Molarity

4.9 Stiochiometry of Reaction in solution

4.10 Titration

Reactions in aqueous (water) solution are important

because :

(1) water is inexpensive and able to dissolve a vast number

of substances;

(2) in aqueous solution, many substances are dissociated

into ions, which can participate in chemical reactions;

(3) water is nontoxic and does not burn;

(4) aqueous solutions are found everywhere, from

seawater to living systems.

4-1 Some Important Definitions

A solution is a homogenous mixture of 2 or more substances.

The solute(s) is (are) the substance(s) present in the smaller a

mount(s).

The solvent is the substance present in the larger amount.

A saturated solution contains as much solute as will dissolve

at a given temperature in the presence of undissolved solute.

The quantity of solute that will dissolve in a given quantity of s

olvent or solution is called the solubility of the solute.

Unsaturated solution;

With some substances, it is possible to make a solution that

contains more solute than a saturated solution. A solution like

this is called a supersaturated solution.

A solid that separated from a solution is called a precipitate.

The concentration of a solution is the amount of solute

dissolved in a given quantity of solvent or solution.

A dilute solution contains only a low concentration of solute.

A concentrated solution contains a high concentration of solute.

• Aqueous solutions, solutions in water, have the potential to

conduct electricity.

• The ability of the solution to conduct electricity (conductiv

ity) depends on the number of ions in solution.

• There are three types of solution:

• Strong electrolytes: most of solute is in the form of ions.

• Weak electrolytes: most of solute is in the form of mole

cules.

• Nonelectrolytes: the solution do not conduct an observa

ble amount of electricity.

4-2 Electrolytes

Electrolytic Properties

Electrolytes are compounds that conduct electricity when dissolved or melted.

Molten salts are also strong electrolytes.– e.g. NaCl melts at 801°C, and the liquid is composed of mobile Na+ and Cl– ions.

Strong and Weak Electrolytes

• Strong electrolytes: most of the solute is in the form of ions.

For example:

• Weak electrolytes: most of the solute is in the form of molecules.

These ions exist in equilibrium with the unionized substance.

For example:

• When a substance dissolves, but little of the dissolved substa

nce exists as ions, it is called a “WEAK ELECTROLYTE”.

• e.g. The weak electrolyte HF (aq) is completely soluble in wat

er, but exists mainly as HF (aq) molecules in water.

• HF (aq) = H+ (aq) + F– (aq)

97% 3%

• e.g. Ammonia, NH3, and acetic acid, CH3CO2H, are two comm

on substances that exist mainly as molecules in solution.

• NH3 (aq) + H2O (l) = NH4+ (aq) + OH– (aq)

• HC2H3O2 (aq) = C2H3O2– (aq) + H+ (aq)

99.6% 0.4%

Non-Electrolytes

• When a substance dissolves, but none of the dissolved

substance exists as ions, it is called a “NON-ELECTROLYTE”.

• E.g. Many organic molecules that contain hydrogen-bonding

groups (like –OH or C=O) are very soluble in water, but they

do not ionize.

– Examples are alcohols and sugars like glucose & sucrose, and acetone.

Some General Terms• Acids - substances that able to ionize in solution to form hydrog

en ion (H+) and increase the concentration of H+ in the solution.• For example, HCl dissociate in water to form H+ and Cl- ions.

• Bases – a compound that increases the concentration of hydroxi

de ions when dissolved in water.

are substances that can react with or accept H+ ions.

• For example, OH- will accept H+ from HCl forming H2O.

• Salts –compounds of metallic cations (or polyatomic cations) an

d nonmetallic anions (or polyatomic anions).

• For example, NaCl instead of HCl. • The reaction of an acid with a base to form a salt and wat

er is called neutralization.

Identifying Strong and Weak Electrolytes

• Most salts are strong electrolytes (NaCl, CaCO3).

• Most acids are weak electrolytes. However, HCl, HBr, H

I, HNO3, H2SO4, HClO3, and HClO4 are strong acids.

• The common strong bases are the hydroxides, of the alkal

i metals (NaOH, etc) and the heavy alkaline earth metals

(from Ca(OH)2).

• Most other substances are nonelectrolytes.

Most chemical reactions in the general chemistry lab are

carried out in solutions. This is partly because mixing the

reactants in solution helps to achieve the close contact between

atoms, ions, or molecules necessary for a reaction to occur. one

component of a solution, called the solvent, determines whether

the solution exists as a solid, liquid, or gas.

4-3 Reactions Between Ions in Solution

Predicting the products of reactions between electrolyte soluti

ons: The rule says that either possible products is insoluble

or a weak or nonelectrolyte, a reaction will take place.

Water solubility of common inorganic compounds: P117

4-4 Ionic Equations

Molecular equations: show all compounds as if they exist in solution in the form of molecules.

Ionic equations: compounds that exist mostly as ions in solution are shown as ions.

A net ionic equation: shows only the species that take part in the reaction.

Reactions between ions in solution to form insoluble substanc

es and weak or nonelectrolytes are sometimes called metathe

sis or double-replacement reactions.

________________________

________________________

Pb2+ + 2NO3- + 2Na+ + 2I- PbI2 (s) + 2Na+ + 2NO3

-

Na+ and NO3- are spectator ions.

PbI2

Pb(NO3)2 (aq) + 2NaI (aq) PbI2 (s) + 2NaNO3 (aq)

precipitate

Pb2+ + 2I- PbI2 (s)

4.2

Writing Net Ionic Equations

1. Write the balanced molecular equation.

2. Write the ionic equation showing the strong electrolytes

3. Determine precipitate from solubility rules

4. Cancel the spectator ions on both sides of the ionic equation

AgNO3 (aq) + NaCl (aq) AgCl (s) + NaNO3 (aq)

4.2

Write the net ionic equation for the reaction of silver nitrate with sodium chloride.

• A reactive metal will displace H2 from H2O :

2 Li (s) + 2 H2O (l) —> 2 LiOH (aq) + H2 (g)

• A less reactive metal will displace H2 from an acid :

Fe (s) + 2 H+ (aq) —> Fe2+(aq) + H2 (g)

4-5 Single-Replacement Reactions

Single-Replacement Reactions: one element takes the

place of another element in a compound.

A metal will displace another metal ion from solution :

Cu (s) + 2 Ag+ (aq) —> Cu2+ (aq) + 2 Ag (s)

• The reactant metal must be more reactive (a stronger reducing

agent or “less noble”) than the product metal.

• The order of this displacement gives the ‘activity series’ of the

metals.

The reactivity of metals in Groups IA and IIA appears to

decrease from left to right across a row in the period table.

The reactivity of metals in Groups IA and IIA increases from

top to bottom of a group.

For transition metals, reactivity decreases from top to bottom

of a column in the period table.

Nonmetals: the reactivity of nonmetals increases from left to ri

ght across a row of the period table.

Cl2 (aq) + Na2S (aq) = S (s) + 2NaCl (aq)

Nonmetals: the reactivity of nonmetals decreases from top to b

ottom of a group in the period table.

Cl2 (aq) + NaBr (aq) = Br2 (aq) + 2NaCl (aq)

4-6 Some Uses of Reactions in Solution

1. Dissolving insoluble compounds:

Fe2O3(s) + 6H+ —> 2Fe3+ + 3H2O

2. Synthesis of inorganic compounds:

AgNO3 (aq) + NaBr (aq) —> AgBr (s) + NaNO3 (aq)

4-7 Concentration Expressed as Percent

Mass % A = ×100Mass component A

Total mass solution

Volume % A = ×100Volume component A

Total volume solution

ppm (volume) = ×106Volume of soluts

Volume of solution

(Parts per Million)

The concentration, or molarity, of a solution is defined as the number of moles of solute per liter of solution:

molarity (M) =amount of solute (in moles)

volume of solution (in liters)

Molarity

4.8 Molarity

If 0.440 mol urea, CO(NH2)2, is dissolved in enough water to make 1.000L of solution, the solution concentration, or molarity, is

0.440 mol CO(NH2)2

1 L solu= 0.440 M CO(NH2)2

The symbol M is often used in place of the unit mol/L and the word “molar” in place of molarity. Thus, we might describe a solution that has

0.440 mol CO(NH2)2/L as 0.440 M CO(NH2)2 or 0.440 molar CO(NH2)2.

Calculating molarity from measured quantities. A solution is pre

pared by dissolving 25.0 mL ethanol, C2H5OH (d= 0.789 g/mL), i

n enough water to produce 250.0 mL solution. What is the mola

rtity of ethanol in the solution?

Example

Solution

To determine the molarity, we must first calculate how

many moles of ethanol are in the solution. To calculate the

number of moles of ethanol in a 25.0-mL sample,we need

conversion factors based on density and molar mass .

? Mol C2H5OH = 25.0 mL C2H5OH × ×

= 0.428 mol C2H5OH

Now,we note that 250.0mL = 0.2500L,and then we use the definition of molatity .

Molarity = = 1.71 M C2H5OH

0.789 g C2H5OH

1 mL C2H5OH 46.07g C2H5OH

0.428 mol C2H5OH

0.2500 L soln

1 mol C2H5OH

Calculating the mass of solution in a solution of known molarity. We want to prepare exactly 0.2500L(250.0mL) of an 0.250 M K2CrO4 solution in water.What mass of K2CrO4 should we use ?

Example

As just stated, we can use solution molarity to convert between volume of solution and amount of solution.The other conversion factor we need is the molar mass of K2CrO4.

(L soln mol K2CrO4 g K2CrO4)

? g K2CrO4 = 0.2500 L soln × ×

= 12.1 g K2CrO4

0.250 mol K2CrO4

1L soln 1 mol K2CrO4

Solution

194.2 g K2CrO4

This statement and the definition of molarity are all that you need to work out dilution problems. You may, however, prefer a method based expression :

molarity (M) =volume of solution (in liters)

amount of solute (in moles)

Suppose we refer to a solution concentration (molarity) as M, a solution volume, in liters, as V, and the amount of solution, in moles, as n. Then, expression above becomes, M=n/V, which we can rearrange to

n=MV

When a solution is diluted, the amount of solution remains constant between the initial (i) solution taken and the final (f) solution produced. That is, MiVi = ni = nf = MfVf

or Mi × Vi = Mf ×Vf

Solution Dilution

Example

Preparing a solution by dilution. A particular analytical chemistry procedure requires 0.0100 M K2CrO4. What volume of 0.250M K2CrO4 must be diluted with water to prepare 0.250L of 0.0100M K2CrO4 ?

solution

To solve this problem,start with the final solution because we known the most about it.Then,determine something about the initial solution.First,calculation the amount of solute that must be present in the final solution.

? Mol K2CrO4 = 0.250 L soln× 0.0100 mol K2CrO4

1L soln

Because all the solution in the final,diluted solution comes from the initial,more concentrated solution,we must answer the question: What volume of 0.250M K2CrO4 contains 0.00250 mol K2CrO4?

?L soln = 0.00250 mol K2CrO4× 1 L soln

0.250 mol K2CrO4

= 0.0100 L soln

Alternatively, we can use equation Mi × Vi = Mf ×Vf. We know

n the volume of solution that we wish to prepare (Vf=250.0 mL)

and the concentrations of the final (0.0100M) and initial (0.250M)

solutions.We must solve for the initial volume, Vi. Note that, alth

ough in deriving equation above we expressed volumes in liters,

in applying it we can use any volume unit as long as we use the

same unit for both Vi and Vf (milliliters in the present case). The

term needed to convert volumes to liters would appear on each

side of the equation and cancel out.

Vi = Vf × = 250.0 mL × = 10.0 mL 0.0100 M

0.250 M

Mf

Mi

4.9 Stoichiometry of reactions in Solution

When we measure the volume of a solution of know molarity,

we obtain a fixed number of moles of solute. Therefore, becaus

e stoichiometry calculations require us to work in moles, we ca

n use molarities and volumes instead.

Relating the mass of a product to the volume and molarity of a reactan

t solution. A 25.00-mL pipetful of 0.250 mol•L － 1 K2CrO4(aq) is added t

o an excess of AgNO3(aq). What mass of Ag2CrO4(s) will precipitate fr

om the solution?

K2CrO4(aq) + 2 AgNO3(aq) Ag2CrO4(s) + 2KNO3(aq)

Example

Solution

(1)Determine the amount of K2CrO4 that reacts; (2) use a stoichiometric fac

tor from the chemical equation to the amount (in moles) of Ag2CrO4 (s) p

roduced; (3) convert the amount of Ag2CrO4 (s) to its mass in grams. Thi

s convert pathway is mL K2CrO4 mol K2CrO4 mol Ag2Cr

O4 g Ag2CrO4. We can combine these steps into a single setup.

1 L1000 mL? g Ag2CrO4= 25.00mL× ×

× ×

1 L 0.250 mol K2CrO4

1 mol Ag2CrO4

1 mol K2CrO4

331.7g Ag2CrO41 mol Ag2CrO4

= 2.07 g Ag2CrO4(s)

• There are two different types of units: – laboratory units (macroscopic units: measure in

lab);– chemical units (microscopic units: relate to moles).

• Always convert the laboratory units into chemical units

first.– Grams are converted to moles using molar mass.– Volume or molarity are converted into moles using

M = mol/L.• Use the stoichiometric coefficients to move between

reactants and product.

4.10 Titration

Titrations

• Suppose we know the molarity of a NaOH solution and

we want to find the molarity of an HCl solution.

• We know:– molarity of NaOH, volume of HCl.

• What do we want?– Molarity of HCl.

• What do we do?– Take a known volume of the HCl solution, measure the mL of

NaOH required to react completely with the HCl.

Titrations

• What do we get?– Volume of NaOH. We know molarity of the NaOH, we can c

alculate moles of NaOH.

• Next step?

– We also know HCl + NaOH NaCl + H2O. Therefore, we kn

ow moles of HCl.

• Can we finish?– Knowing mol(HCl) and volume of HCl (20.0 mL above), we c

an calculate the molarity.

Titration is an important method for determining the amount of a substance

present in solution.

A solution of known concentration is called standard solution.

Ideally, addition is stopped when the quantity of reactant called for by the

equation for the reaction has been added. This point is called equivalence

point.

The point at which the color change takes place is called the end point of the

titration.

A substance whose change in color shows that the end point has been reached

is called an indicator.

5-2 Precipitation Reactions

Many chemical reactions occur in aqueous solution. A precipitation reaction is one that occurs in solution and results in the formation of an insoluble product. For example, an aqueous solution of the soluble ionic compound, sodium sulfate,

can be mixed with an aqueous solution of the soluble ionic compound, barium hydroxide. The result is the formation

of a precipitate, the insoluble ionic compound barium sulfate.

Note that only one of the products of the reaction is a solid. Sodium hydroxide is soluble. Many combinations of such solutions will not result in a precipitation because they produce no insoluble product. Whether or not an ionic product of a reaction in solution will precipitate from the solution can be predicted using a set of solubility guidelines

To predict the outcome when combining aqueous solutions of ionic compounds: 1. Identify the cation and anion in each reactant. 2. Combine the cation from the first reactant with the anion

of the second to get the first product. 3. Combine the anion from the first reactant with the cation of

the second to get the second product.4. Determine whether the products are aqueous or solid by

consulting the solubility guidelines

Having learned that substances such as sodium sulfate, barium hydroxide, and sodium hydroxide are strong electrolytes and exist entirely as ions in solution, we are equipped to write these chemical equations in a way that better represents what actually happens in solution.

The equation above is a molecular equation in which none of the species is represented as ionized. A molecular equation shows the

complete chemical formulas for the reactants and products. We know, however, that several of the species in the equation dissociate completely in solution. We convert the equation to

a complete ionic equation by identifying the strong electrolytes and representing them as separated ions. The equation above becomes

The ionic equation reveals that two of the species in solution (sodium ion and hydroxide ion) do not undergo any change in the course of the reaction. Ions that are

present but play no role in the reaction are called spectator ions. Eliminating the spectator ions from both sides of the equation

gives the net ionic reaction

The net ionic equation for the combination of aqueous sodium sulfate andaqueous barium hydroxide is the same as the

net ionic equation for any combination of a soluble sulfate and a soluble barium compound.

To write a net ionic equation:1.Write a balanced molecular equation. 2.Rewrite the equation to show the ions that form in solution when each soluble strong electrolyte dissociates or ionizes into its component ions. Only dissolved strong electrolytes are written in ionic form. 3.Identify and cancel spectator ions that occur on both sides of the equation.

Example

Using solubility rules to predict precipitation reactions.Predict whether a reaction will occur in each of the following cases.If so,write a net ionic equation for the reaction.

(a)NaOH(aq) + MgCl2(aq) ?

(b)BaS(aq) + CuSO4(aq) ?

(c)(NH4)2SO4(aq) + ZnCl2(aq) ?

Solution

(a)Because all common Na compounds are water soluble,Na+ remains in solution.The combination of Mg2+ and OH-produces insoluble Mg(OH)2.with this information.we can write

2 Na+ (aq) + 2 OH-(aq) + Mg2+(aq) + 2 Cl-(aq)

Mg(OH)2 (s) + 2 Na+ (aq) + 2 Cl-(aq)

Write the elimination of spectator ions,we obtain

2 OH-(aq) + Mg2+(aq) Mg(OH)2 (s)

(b) From the solubility rules,we conclude that insoluble combinations formed when Ba2+,S2-,Cu2+,and SO4

2- are found together in solution are BaSO4(s) and CuS(s).The net ionic equation is

Ba2+(aq) + S2-(aq) + Cu2+(aq) + SO42-(aq) BaSO4(s) + CuS(s)

(c) A careful review of the solubility rules shows that all the possible ion combinations lead to water-soluble compounds.

2 NH4+(aq) + SO4

2-(aq) + Zn2+(aq) + 2 Cl-(aq) no reaction

Keep in mind that a solution made by combining ZnSO4(aq) and

NH4Cl(aq) is indistinguishable from the combination of ZnCl2(aq) and

(NH4)2SO4(aq).