chapter 9

Chapter 9

Aqueous Solutionsand Solubility Equilibria

Solutions for Practice ProblemsStudent Textbook page 424

1. ProblemPredict whether an aqueous solution of each salt is neutral, acidic, or basic.(a) NaCN(b) LiF(c) Mg(NO3)2

(d) NH4I

What Is Required?Predict whether each aqueous solution is acidic, basic, or neutral.

What Is Given?The formula of each salt is given.

Plan Your StrategyDetermine whether the cation is from a strong or weak base, and whether the anion isfrom a strong or weak acid. Ions derived from weak bases or weak acids react with waterand affect the pH of the solution.

Act on Your Strategy(a) Sodium cyanide, NaCN, is the salt of a strong base (NaOH) and a weak acid

(HCN). Only the cyanide ions react with water. The solution is basic.(b) Lithium fluoride, LiF, is the salt of a strong base (LiOH) and a weak acid (HF).

Only the fluoride ions react with water. The solution is basic.(c) Magnesium nitrate, Mg(NO3)2, is the salt of a strong base (Mg(OH)2) and a

strong acid (HNO3(aq)). Neither ion reacts with water, so the solution is neutral.(d) Ammonium iodide, NH4I, is the salt of a weak base (NH3(aq)) and a strong acid

(HI(aq)). Only the ammonium ions react with water. The solution is acidic.

Check Your SolutionCheck that you have correctly identified whether the cation is from a strong or weakbase, and whether the anion is from a strong or weak acid.

2. ProblemIs the solution of each salt acidic, basic, or neutral? For solutions that are not neutral,write equations that support your predictions.(a) NH4BrO4

(b) NaBrO4

(c) NaOBr(d) NH4Br

What Is Required?Predict whether each aqueous solution is acidic, basic, or neutral. Write equations thatrepresent the solutions that are acidic or basic.

158Chapter 9 Aqueous Solutions and Solubility Equilibria • MHR

CHEMISTRY 12

What Is Given?The formula of each salt is given.

Plan Your StrategyDetermine whether the cation is from a strong or weak base, and whether the anion isfrom a strong or weak acid. Ions derived from weak bases or weak acids react with waterand affect the pH of the solution.

Act on Your Strategy(a) Ammonium perbromate, NH4BrO4, is the salt of a weak base (NH3(aq)) and a

strong acid (HBrO4). Only the ammonium ions react with water.NH4

+(aq) + H2O(�) ⇀↽ NH3(aq) + H3O+

(aq)

The solution is acidic.(b) Sodium perbromate, NaBrO4, is the salt of a strong base (NaOH) and a strong acid

(HBrO4). Neither ion reacts with water, so the solution is neutral.(c) Sodium hypobromite, NaOBr, is the salt of a strong base (NaOH) and a weak acid

(HOBr(aq)). Only the hypobromite ions react with water.OBr−

(aq) + H2O(�) ⇀↽ HOBr(aq) + OH−(aq)

The solution is basic.(d) Ammonium bromide, NH4Br, is the salt of a weak base (NH3(aq)) and a strong

acid (HBr(aq)) .Only the ammonium ions react with water.NH4

+(aq) + H2O(�) ⇀↽ NH3(aq) + H3O+

(aq)

The solution is acidic.

Check Your SolutionThe equations that represent the reactions with water support the prediction thatNH4BrO4 and NH4Br dissolve to form an acidic solution and NaOBr dissolves toform a basic solution. Sodium hypobromite is the salt of a strong base-strong acid, so neither ion reacts with water and the solution is neutral.

3. ProblemKa for benzoic acid, C6H5COOH, is 6.3 × 10−5 . Ka for phenol, C6H5OH, is1.3 × 10−10. Which is the stronger base, C6H5COO−

(aq) or C6H5O−(aq)?

Explain your answer.

What Is Required?You must determine which ion, C6H5COO−

(aq) or C6H5O−(aq), is the stronger base.

What Is Given?The Ka of each conjugate acid is given.

Plan Your StrategyFor a conjugate acid-base pair, KaKb = 1.0 × 10−14 . Therefore, a small value of Ka foran acid results in a large value of Kb for the conjugate base.

Act on Your StrategyBecause Ka for phenol is smaller than Ka for benzoic acid, Kb for the phenolate ion must be larger than Kb for the benzoate ion. Consequently, C6H5O−

(aq), is the stronger base.

Check Your SolutionUsing the equation KaKb = 1.0 × 10−14 , you can calculate the Kb for each conjugatebase. Kb for C6H5COO−

(aq) is 1.6 × 10−10. Kb for C6H5O−(aq) is 7.7 × 10−5 .

This supports the reasoning that C6H5O−(aq) is the stronger base.


CHEMISTRY 12

4. ProblemSodium hydrogen sulfite, NaHSO3, is a preservative that is used to prevent the discolouration of dried fruit. In aqueous solution, the hydrogen sulfite ion can act as either an acid or a base. Predict whether NaHSO3 dissolves to form an acidic solution or a basic solution. (Refer to Appendix E for ionization data.)

What Is Required?You must decide whether NaHSO3 dissolves to form an acidic solution or a basic solution.

What Is Given?Ka for H2SO3(aq) = 1.4 × 10−2

Ka for HSO3−

(aq) = 6.3 × 10−8

Plan Your StrategyThe hydrogen sulfite ion is amphoteric. Compare the equilibrium constants Ka and Kb

for HSO3−

(aq) acting as an acid and a base, to determine which reaction goes farthestto completion.

Act on Your StrategyThe ion reactions with water areHSO3

−(aq) + H2O(�) ⇀↽ SO3

2−(aq) + H3O+

(aq) Ka = 6.3 × 10−8

HSO3−

(aq) + H2O(�) ⇀↽ H2SO3(aq) + OH−(aq) Kb = ?

Kb can be calculated using the value for Ka of H2SO3(aq).

Kb = KwKa

= 1.0 × 10−14

1.4 × 10−2

= 7.1 × 10−13

The equilibrium constant (Ka) for the hydrogen sulfite ion acting as an acid is greaterthan the equilibrium constant (Kb) for the ion acting as a base. Therefore, the solutionis acidic.

Check Your SolutionIt is a common difficulty in this type of problem to identify correctly the base reactionand the corresponding value for Kb. Kb must be calculated using the Ka value forH2SO3(aq), because HSO3

−(aq) is the conjugate base of H2SO3(aq).


5. ProblemAfter titrating sodium hydroxide with hydrofluoric acid, a chemist determined that thereaction had formed an aqueous solution of 0.020 mol/L sodium fluoride. Determinethe pH of the solution.

What Is Required?You need to calculate the pH of the solution.

What Is Given? [NaF] = 0.020 mol/LFrom Appendix E, Ka for HF(aq) = 6.3 × 10−4

Plan Your StrategyStep 1 Decide which ion reacts with water. Write the equation that represents the

hydrolysis reaction.


CHEMISTRY 12

Step 2 Determine the equilibrium constant for the ion involved in the hydrolysis reaction.

Step 3 Divide the ion concentration by the appropriate equilibrium constant todetermine whether the change in concentration of the ion can be ignored.

Step 4 Set up an ICE table for the ion that is involved in the reaction with water.Let x represent the change in the concentration of the ion that reacts.

Step 5 Write the equilibrium expression. Substitute the equilibrium concentrationsinto the expression, and solve for x.

Step 6 Use the value of x to determine [H3O+]. Then calculate the pH of the solution.

Act on Your StrategyStep 1 Sodium fluoride is the salt of a strong base (NaOH) and a weak acid (HF).

Thus, only the fluoride ion reacts with water.F−

(aq) + H2O(�) ⇀↽ HF(aq) + OH−(aq)

Step 2 Kb for the fluoride ion can be calculated using Ka for HF(aq) .

Kb = KwKa

= 1.0 × 10−14

6.3 × 10−4

= 1.6 × 10−11

Step 3 NaF(aq) ⇀↽ Na+(aq) + F−

(aq)

∴[F−] = 0.020 mol/L

F−

Kb= 0.020

1.6 × 10−11

= 1.2 × 109

Because this value is much greater than 500, the change in concentration ofF−

(aq) can be ignored compared with its initial concentration.

Step 4

Step 5 Kb = [HF][OH−][F−]

1.6 × 10−11 = (x)(x)0.020

x =√

3.2 × 10−13

= 5.6 × 10−7 mol/LStep 6 x = [OH−] = 5.6 × 10−7 mol/L

pOH = −log[OH]−

= −log(5.6 × 10−7)= 6.25

pH = 14.00 − pOH= 14.00 − 6.25= 7.75

Check Your SolutionThe pH of the solution is weakly basic. This is consistent with a dilute aqueous solution of the salt of a strong base and a weak acid.

Concentration (mol/L)

Initial

Change

Equilibrium

F−(aq) OH−

(aq)+ +H2O(�)

0.020

−x

0.020 − x ≈ 0.020

0

+x

x

~0

+x

x

2HF(aq)


CHEMISTRY 12

6. ProblemPart way through a titration, 2.0 × 101 mL of 0.10 mol/L sodium hydroxide has been added to 3.0 × 101 mL of 0.10 mol/L hydrochloric acid. What is the pH of the solution?

What Is Required?You need to calculate the pH of the solution.

What Is Given?20 mL of 0.10 mol/L NaOH(aq) has been added to 30 mL of 0.10 mol/L HCl(aq).

Plan Your StrategyStep 1 Calculate the amount of each reagent using the following equation:

Amount (mol) = concentration (mol/L) × volume (L)Step 2 Write the chemical equation for the reaction and determine the

excess reagent.Step 3 Calculate the concentration of the excess reagent using the following equation:

Concentration (mol/L) = amount in excess (mol)total volume (L)

Step 4 Calculate the pH of the solution.

Act on Your StrategyStep 1 Amount NaOH(aq) = (0.10 mol/L) × (2.0 × 10−2 L)

= 2.0 × 10−3 molAmount HCl(aq) = (0.10 mol/L) × (3.0 × 10−2 L)

= 3.0 × 10−3 molStep 2 NaOH(aq) + HCl(aq) → NaCl(aq) + H2O(�)

Because the acid and base react in a 1:1 ratio, the HCl(aq) is in excess.Step 3 Amount of excess HCl(aq) = (3.0 × 10−3 mol) − (2.0 × 10−3 mol)

= 1.0 × 10−3 molTotal volume = (2.0 × 10−2 L) + (3.0 × 10−2 L) = 5.0 × 10−2 L

[HCl] = 10 × 10−3 mol5.0 × 10−2 L

= 2.0 × 10−2 mol/LStep 4 HCl(aq) is a strong acid. Therefore, [H3O+] = 2.0 × 10−2 mol/L .

pH = −log[H3O+]

= −log(2.0 × 10−2)= 1.70

Check Your SolutionThe solutions have the same molar concentration. Because the volume of hydro-chloric acid is greater, the solution should be acidic and the pH less than 7.

7. Problem0.025 mol/L benzoic acid, C6H5COOH, is titrated with 0.025 mol/L sodiumhydroxide solution. Calculate the pH at equivalence.

What Is Required?You need to calculate the pH at equivalence.

What Is Given?You know that 0.025 mol/L C6H5COOH(aq) reacts with 0.025 mol/L NaOH(aq) .Tables of Ka and Kb values are in Appendix E.

Plan Your StrategyStep 1 Write the balanced chemical equation for the reaction.Step 2 Calculate the concentration of the salt formed, based on the amount

(in mol) and the total volume of the solution.


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Step 3 Decide which ion reacts with water. Write the equation that represents the reaction.

Step 4 Determine the equilibrium constant for the ion that is involved in thehydrolysis reaction.

Step 5 Divide the concentration of the ion identified in Step 3 by the appropriateionization constant to determine whether the change in concentration of theion can be ignored.

Step 6 Set up an ICE table for the ion that is involved in the reaction with water.Let x represent the change in the concentration of the ion that reacts.



Act on Your StrategyStep 1 The following chemical equation represents the reaction.

C6H5COOH(aq) + NaOH(aq) → NaC6H5COO(aq) + H2O(�)

Step 2 The acid and base react in a 1:1 ratio, and the concentrations are the same.Therefore, the volume of each reagent must be the same and the total volume will be double the initial volume.Therefore,

[NaC6H5COO] = 0.025 mol/L2

= 0.0125 mol/LStep 3 The salt forms Na+

(aq) and C6H5COO−(aq) in solution. Na+

(aq) is the cationof a strong base, so it does not react with water. C6H5COO−

(aq) is the conju-gate base of a weak acid, so it does react with water. The pH of the solutionis therefore determined by the extent of the following reaction.C6H5COO−

(aq) + H2O(�) → C6H5COOH(aq) + OH−(aq)

Step 4 Therefore,Kb for C6H5COO(aq) = Kw

Ka

= 1.0 × 10−14

6.3 × 10−5

= 1.6 × 10−10

Step 5[C6H5COO−]

Kb= 0.0125

1.6 × 10−10

= 7.8 × 107

This is well above 500, so the change in [C6H5COO−] can be ignored.Step 6

Step 7 Kb = [C6H5COOH][OH−][C6H5COO−]

1.6 × 10−10 = (x)(x)0.0125

x =√

2.0 × 10−12

= 1.4 × 10−6 mol/L


Initial

Change

Equilibrium

OH−(aq)+ +H2O(�)C6H5COOH−

(aq)

0.0125

−x

0.0125 − x ≈ 0.0125

0

+x

x

~0

+x

x

C6H5COOH(aq)


CHEMISTRY 12

Step 8 x = [OH−] = 1.4 × 10−6 mol/LpOH = −log[OH−]

= −log(1.4 × 10−6)= 5.85

pH = 14.00 − 5.85= 8.15

Check Your SolutionThe titration forms an aqueous solution of a salt derived from a strong base and a weakacid. The solution should be basic, which is supported by the calculation of the pH.

8. Problem50.0 mL of 0.10 mol/L hydrobromic acid is titrated with 0.10 mol/L aqueous ammonia. Determine the pH at equivalence.

What Is Required?You need to calculate the pH at equivalence.

What Is Given?You know that 50 mL of 0.10 mol/L HBr(aq) reacts with 0.10 mol/L NH3(aq). Tables of Ka and Kb values are in Appendix E.

Plan Your StrategyStep 1 Write the balanced chemical equation for the reaction.Step 2 Find the volume of NH3(aq) added to neutralize the hydrobromic acid based

on the amount and concentration of HBr(aq) .Step 3 Calculate the concentration of the salt formed, based on the amount (in mol)

and the total volume of the solution.Step 4 Decide which ion reacts with water. Write the equation that represents

the reaction.Step 5 Determine the equilibrium constant for the ion that is involved in the

hydrolysis reaction.Step 6 Divide the concentration of the ion identified in Step 4 by the appropriate

ionization constant to determine whether the change in concentration of theion can be ignored.

Step 7 Set up an ICE table for the ion that is involved in the reaction with water. Let x represent the change in the concentration of the ion that reacts.



Act on Your StrategyStep 1 The following chemical equation represents the reaction.

NH3(aq) + HBr(aq) → NH4Br(aq)

Step 2 Amount (in mol) of NH3(aq) = 0.20 mol/L × 0.020 L= 4.0 × 10−3 mol

Step 3 NH3(aq) and HBr(aq) react in a 1:1 ratio. Because [NH3] = [HBr], the volumeof aqueous ammonia added must be the same as the volume of hydrobromicacid. Therefore, the volume of NH3(aq) is 50 mL, and the total volume of thesolution is twice the volume of HBr(aq) used. Thus, the [NH4Br] must be halfthe initial [HBr].Therefore, [NH4Br] = 0.050 mol/L


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Step 4 The titration results in an aqueous solution of ammonium bromide, NH4Br(aq).NH4

+(aq) is the conjugate acid of a weak base, so it reacts with water. Br−

(aq)

is the conjugate base of a strong acid, so it does not react with water. The sol-ution will be acidic, and the pH is determined by the extent of the followingreaction.NH4

+(aq) + H2O(�) → NH3(aq) + H3O+

(aq)

Step 5 From Appendix E, Kb for NH3(aq) is 1.8 × 10−5 .Ka for the conjugate base, NH4(aq), can be calculated using the relationshipKaKb = Kw.

Ka = KwKb

= 1.0 × 10−14

1.8 × 10−5

= 5.6 × 10−10

Step 6 [NH4+]

Ka= 0.050

5.6 × 10−10

= 8.9 × 107

This is well above 500, so the change in [NH+4] can be ignored.

Step 7

Step 8 Ka = [NH3][H3O+][NH4

+]

5.6 × 10−10 = (x)(x)0.050

x =√

2.8 × 10−11

= 5.3 × 10−6 mol/LStep 9 x = [H3O+] = 5.3 × 10−6 mol/L

pH = −log[H3O+]

= −log(5.3 × 10−6)= 5.28

Check Your SolutionThe titration forms an aqueous solution of a salt derived from a weak base and a strongacid. The solution should be acidic, which is supported by the calculation of the pH.


9. ProblemWrite the balanced chemical equation that represents the dissociation of each compoundin water. Then write the corresponding solubility product expression.(a) copper(I) chloride(b) barium fluoride(c) silver sulfate(d) calcium phosphate


Initial

Change

Equilibrium

+ +H2O(�) H3O+(aq)NH4

+(aq)

0.050

−x

0.050 − x ≈ 0.050

0

+x

x

~0

+x

x

NH3(aq)


CHEMISTRY 12

SolutionFirst, write a balanced equation for the equilibrium between excess solid and dissolvedions in a saturated aqueous solution. Then use the balanced equation to write the expression for Ksp.(a) CuCl(s) ⇀↽ Cu+

(aq) + Cl−(aq)

Ksp = [Cu+][Cl−](b) BaF2(s) ⇀↽ Ba2+

(aq) + 2F−(aq)

Ksp = [Ba2+][F−]2

(c) Ag2SO4(s) ⇀↽ 2Ag+(aq) + SO4

2−(aq)

Ksp = [Ag+]2[SO42−]

(d) Ca3(PO4)2(s) ⇀↽ 3Ca2+(aq) + 2PO4

3−(aq)

Ksp = [Ca2+]3[PO43−]2

Check Your SolutionThe Ksp expressions are based on balanced equations for saturated solutions of slightlysoluble ionic compounds. The exponents in the Ksp expressions match the correspond-ing coefficients in the chemical equations. The coefficient 1 is not written, followingchemistry conventions.

10. ProblemWrite a balanced dissolution equation and solubility product expression for silver carbonate, Ag2CO3.

SolutionFirst, write a balanced equation for the equilibrium between excess solid and dissolvedions in a saturated aqueous solution. Then use the balanced equation to write the expression for Ksp.Ag2CO3(s) ⇀↽ 2Ag+

(aq) + CO3−

(aq)

Ksp = [Ag+]2[CO3−]

Check Your SolutionThe Ksp expression is based on the balanced equation between excess solid and dissolvedions. The exponents in the Ksp expression match the corresponding coefficients in thechemical equation.

11. ProblemWrite a balanced dissolution equation and solubility product expression for magnesiumammonium phosphate, MgNH4PO4.

SolutionFirst, write a balanced equation for the equilibrium between excess solid and dissolvedions in a saturated aqueous solution. Then use the balanced equation to write the expression for Ksp.MgNH4PO4(s) ⇀↽ Mg2+

(aq) + NH4+

(aq) + PO43−

(aq)

Ksp = [Mg2+][NH4+][PO4

3−]

Check Your SolutionThe Ksp expression is based on the balanced equation between excess solid and dissolvedions. The exponents in the Ksp expression match the corresponding coefficients in thechemical equation.

12. ProblemIron (III) nitrate has very low solubility.(a) Write the solubility product expression for iron(III) nitrate.(b) Do you expect the value of Ksp of iron(III) nitrate to be larger or smaller than the

Ksp for aluminum hydroxide, which has a slightly higher solubility?


CHEMISTRY 12

Solution(a) Write a balanced equation for the equilibrium between excess solid iron(III) nitrate

and dissolved ions in a saturated aqueous solution. Then use the balanced equationto write the expression for Ksp.Fe(NO3)3(s) ⇀↽ Fe3+

(aq) + 3NO3−

(aq)

Ksp = [Fe3+][NO3−]3

(b) Write a balanced equation for the equilibrium between excess solid aluminum hy-droxide and dissolved ions in a saturated aqueous solution. Then use the balancedequation to write the expression for Ksp. Finally, compare the two equilibrium expressions and decide which expression is larger.Al(OH)3(s) ⇀↽ Al3+

(aq) + 3OH−(aq)

Ksp = [Al3+][OH−]3

The equilibrium expressions have the same form. In both cases, if x mol/L of saltdissolves, Ksp is given by x4 . Therefore, the most soluble salt will have the largervalue of Ksp. The value of Ksp for iron(III) nitrate is smaller than Ksp for aluminumhydroxide.

Check Your SolutionEach Ksp expression is based on the balanced equation between excess solid and dissolvedions. The exponents in the Ksp expressions match the corresponding coefficients in thechemical equations. The equilibrium expressions are of the same form, so the least soluble salt has the lower value for Ksp.


13. ProblemThe maximum solubility of silver cyanide, AgCN, is 1.5 × 10−8 mol/L at 25˚C.Calculate Ksp for silver cyanide.

What Is Required?You need to find the value of Ksp for AgCN at 25˚C.

What Is Given?You know the solubility of AgCN at 25˚C.

Plan Your StrategyStep 1 Write an equation for the dissolution of AgCN.Step 2 Use the equation to write the solubility product expression.Step 3 Find the concentration (in mol/L) of each ion.Step 4 Substitute the concentrations into the solubility product expression,

and calculate Ksp.

Act on Your StrategyStep 1 AgCN(s) ⇀↽ Ag+

(aq) + CN−(aq)

Step 2 Ksp = [Ag+][CN−]Step 3 [Ag+] = [CN−] = [AgCN] = 1.5 × 10−8 mol/LStep 4 Ksp = [Ag+][CN−]

= (1.5 × 10−8)(1.5 × 10−8)= 2.2 × 10−16

Ksp for silver cyanide is 2.2 × 10−16 at 25˚C.

Check Your SolutionThe value of Ksp is less than the concentration of the salt, as expected. It has the correctnumber of significant digits.


CHEMISTRY 12

14. ProblemA saturated solution of copper(II) phosphate, Cu3(PO4)2, has a concentration of6.1 × 10−7g Cu3(PO4)2 per 1.00 × 102 mL of solution at 25˚C. What is Ksp forCu3(PO4)2 at 25˚C ?

What Is Required?You need to find the value of Ksp for Cu3(PO4)2 at 25˚C.

What Is Given?You know 6.1 × 10−7 g of Cu3(PO4)2 is dissolved in 1.00 × 102 mL of solution at 25˚C.

Plan Your StrategyStep 1 Write an equation for the dissolution of Cu3(PO4)2.Step 2 Use the equation to write the solubility product expression.Step 3 Find the concentration (in mol/L) of copper(II) phosphate.

Next, find the concentration (in mol/L) of each ion.Step 4 Substitute the concentrations into the solubility product expression,

and calculate Ksp.

Act on Your StrategyStep 1 Cu3(PO4)2(s) ⇀↽ 3Cu2+

(aq) + 2PO43−

(aq)

Step 2 Ksp = [Cu2+]3[PO43−]2

Step 3 The molar mass of Cu3(PO4)2 is 380.6 g/mol.

Amount of Cu3(PO4)2 = 6.1 × 10−7g380.6 g/mol

= 1.6 × 10−9 mol

[Cu3(PO4)2] = 1.6 × 10−9 mol0.100 L

= 1.6 × 10−8 mol/L[Cu2+] = 3 × [Cu3(PO4)2]

= 3 × (1.6 × 10−8 mol/L)

= 4.8 × 10−8 mol/L

[PO43−] = 2 × [Cu3(PO4)2] = 3.2 × 10−8 mol/L

Step 4 Ksp = [Cu2+]3[PO43−]2

= (4.8 × 10−8)3(3.2 × 10−8)2

= 1.1 × 10−37

Ksp for copper(II) phosphate is 1.1 × 10−37 at 25˚C.

Check Your SolutionCheck that you wrote the balanced chemical equation and the corresponding Ksp equa-tion correctly. Pay attention to molar relationships and to the exponent of each term.Check the concentration calculations carefully.

15. ProblemA saturated solution of CaF2 contains 1.2 × 1020 formula units of calcium fluoride perlitre of solution. Calculate Ksp for CaF2.

What Is Required?You need to find the value of Ksp for CaF2.

What Is Given?You know one litre of solution contains 1.2 × 1020 formula units of CaF2.


CHEMISTRY 12

Plan Your StrategyStep 1 Write an equation for the dissolution of CaF2.Step 2 Use the equation to write the solubility product expression.Step 3 Find the concentration (in mol/L) of calcium fluoride. Next, find the

concentration (in mol/L) of each ion.Step 4 Substitute the concentrations into the solubility product expression,

and calculate Ksp.

Act on Your StrategyStep 1 CaF2(s) ⇀↽ Ca2+

(aq) + 2F−(aq)

Step 2 Ksp = [Ca2+][F−]2

Step 3 The amount of CaF2 is given by the following formula:

Amount = number of formula unitsAvogadro’s constant

= 1.2 × 1020 formula units6.0 × 1023 formula units/mol

= 2.0 × 10−4 mol[CaF2] = 2.0 × 10−4 mol/L[Ca2+] = [CaF2] = 2.0 × 10−4 mol/L[F−] = 2 × [CaF2]

= 2 × (2.0 × 10−4 mol/L)= 4.0 × 10−4 mol/L

Step 4 Ksp = [Ca2+][F−]2

= (2.0 × 10−4)(4.0 × 10−4)2

= 3.2 × 10−11

Ksp for calcium fluoride is 3.2 × 10−11.

Check Your SolutionCheck that you wrote the balanced chemical equation and the corresponding Ksp equa-tion correctly. Pay attention to molar relationships and to the exponent of each term.Check the concentration calculations carefully.

16. ProblemThe concentration of mercury(I) iodide, Hg2I2, in a saturated solution at 25˚C is1.5 × 10−4 ppm.(a) Calculate Ksp for Hg2I2. The solubility equilibrium is written as follows:

Hg2I2 ⇀↽ Hg22+ + 2I−

(b) State any assumptions that you made when you converted ppm to mol/L.

What Is Required?You need to find the value of Ksp for Hg2I2 at 25˚C.

What Is Given?You know the concentration of Hg2I2 in a saturated solution at 25˚C is 1.5 × 10−4 ppm.You are given the dissolution equation.

Plan Your StrategyStep 1 Use the dissolution equation to write the solubility product expression.Step 2 Find the concentration (in mol/L) of mercury(I) iodide. Then, find the

concentration (in mol/L) of each ion.Step 3 Substitute the concentrations into the solubility product expression, and

calculate Ksp.


CHEMISTRY 12

Act on Your Strategy(a) Step 1 Ksp = [Hg2

2+][I−]2

Step 2 1 ppm is equivalent to 1 mg of solute dissolved in 1 kg of solution. 1 kg of water at 4˚C has a volume of 1 L. Therefore, 1.5 × 10−4 mg of Hg2I2

is dissolved in 1.0 L of solution.1.5 × 10−4 mg = 1.5 × 10−7gThe molar mass of Hg2I2 is 655 g/mol.

Amount of Hg2I2 = 1.5 × 10−7 g655 g/mol

= 2.3 × 10−10 mol[Hg2I2] = 2.3 × 10−10 mol/L[Hg2

2+] = [Hg2I2] = 2.3 × 10−10 mol/L[I−] = 2 × [Hg2I2]

= 2 × (2.3 × 10−10 mol/L)= 4.6 × 10−10 mol/L

Step 3 Ksp = [Hg22+][I−]2

= (2.3 × 10−10)(4.6 × 10−10)2

= 4.9 × 10−29

Ksp for mercury(I) iodide is 4.9 × 10−29 at 25˚C.

(b) 1 ppm is equivalent to 1 mg of solute dissolved in 1 kg of solution. This leads totwo assumptions in this problem. Firstly, that the solution is so dilute it can betreated as 1 kg of water. Secondly, you assume that 1 kg of water has a volume of 1 L. This is only strictly true at 4˚C.

Check Your SolutionCheck that you wrote the Ksp equation correctly. Pay attention to molar relationshipsand to the exponent of each term. Check the concentration calculations carefully.


17. ProblemKsp for silver chloride, AgCl, is 1.8 × 10−10 at 25˚C.(a) Calculate the molar solubility of AgCl in a saturated solution at 25˚C.(b) How many formula units of AgCl are dissolved in 1.0 L of saturated silver

chloride solution?(c) What is the percent (m/v) of AgCl in a saturated solution at 25˚C?

What Is Required?You need to determine the solubility (in mol/L, in formula units, and expressed as amass/volume percent) of AgCl at 25˚C.

What Is Given?At 25˚C, Ksp for AgCl is 1.8 × 10−10.

Plan Your Strategy(a) Step 1 Write the dissolution equilibrium equation.

Step 2 Use the equilibrium equation to write an expression for Ksp.Step 3 Set up an ICE table. Let x represent molar solubility. Use the stoichiometry

of the equilibrium equation to write expressions for the equilibrium concentrations of the ions.

Step 4 Substitute your expressions into the expression for Ksp, and solve for x.(b) Convert the molar solubility to formula units per litre.(c) Convert the molar solubility to a percent (m/v).


CHEMISTRY 12

Act on Your Strategy(a) Step 1 AgCl(s) ⇀↽ Ag2+

(aq) + Cl−(aq)

Step 2 Ksp = [Ag+][Cl−]

Step 3

Step 4 Ksp = [Ag+][Cl−] = (x)(x)Therefore, 1.8 × 10−10 = x2

x =√

1.8 × 10−10

= 1.3 × 10−5 mol/LThe molar solubility of AgCl in water is 1.3 × 10−5 mol/L.

(b) 1.3 × 10−5 mol = (1.3 × 10−5) × (6.0 × 1023 formula units/mol)= 7.8 × 1018 formula units

The solubility of AgCl in water is 7.8 × 1018 formula units/L

(c) Mass/volume percent = mass of solute (g)volume of solution (mL)

× 100%

Molar mass of AgCl = 143.3 g/molMass in 1 L of solution = (1.3 × 10−5 mol/L) × (143.3 g/mol)

= 1.9 × 10−3 g/L

Mass/volume percent = 1.9 × 10−3 g1000 mL

× 100%

The solubility of AgCl at 25˚C is 1.9 × 10−4% (m/v).

Check Your SolutionSubstitute the values of [Ag+] and [Cl−] into the Ksp equation. You should get the givenKsp. Check the number of significant digits given in the question, and the number ofdigits in the answers.

18. ProblemIron(III) hydroxide, Fe(OH)3, is an extremely insoluble compound. Ksp for Fe(OH)3

is 2.8 × 10−39 at 25˚C. Calculate the molar solubility of Fe(OH)3 at 25˚C.

What Is Required?You need to determine the solubility (in mol/L) of Fe(OH)3 at 25˚C.

What Is Given?At 25˚C, Ksp for Fe(OH)3 is 2.8 × 10−39.

Plan Your StrategyStep 1 Write the dissolution equilibrium equation.Step 2 Use the equilibrium equation to write an expression for Ksp.Step 3 Set up an ICE table. Let x represent molar solubility. Use the stoichiometry


Step 4 Substitute your expressions into the expression for Ksp, and solve for x.


Initial

Change

Equilibrium

+Ag+(aq) Cl−

(aq)AgCl(s)

0

+x

x

0

+x

x


CHEMISTRY 12

Act on Your StrategyStep 1 Fe(OH)3(s) ⇀↽ Fe3+

(aq) + 3OH−(aq)

Step 2 Ksp = [Fe3+][OH−]3

Step 3

Step 4 Ksp = [Fe3+][OH−]3

= (x) × (3x)3

= 27x4

2.8 × 10−39 = 27x4

x = 4

√2.8 × 10−39

27

= 1.0 × 10−10 mol/LThe molar solubility of Fe(OH)3 in water is 1.0 × 10−10 mol/L.

Check Your SolutionRecall that x = [Fe3+]eq and 3x = [OH−]eq. Substitute these values into the Ksp

equation. You should get the given Ksp.

19. ProblemKsp for zinc iodate, Zn(IO3)2 , is 3.9 × 10−6 at 25˚C. Calculate the solubility (in moland in g/L) of Zn(IO3)2 in a saturated solution.

What Is Required?You need to determine the solubility (in mol/L and in g/L) of Zn(IO3)2 at 25˚C.

What Is Given?At 25˚C, Ksp for Zn(IO3)2 is 3.9 × 10−6 .



Step 4 Substitute your expressions into the expression for Ksp, and solve for x.Step 5 Determine the solubility in g/L.

Act on Your StrategyStep 1 Zn(IO3)2(s) ⇀↽ Zn2+

(aq) + 2IO3−

(aq)

Step 2 Ksp = [Zn2+][IO3−]2

Step 3

Step 4 Ksp = [Zn2+][IO3−]2

= (x) × (2x)2

= 4x3

3.9 × 10−6 = 4x3


Initial

Change

Equilibrium

+Zn2+(aq) 2IO3

−(aq)Zn(IO3)2(s)

0

+x

x

0

+2x

2x


Initial

Change

Equilibrium

+Fe3+(aq) 3OH−

(aq)Fe(OH)3(s)

0

+x

x

0

+3x

3x


CHEMISTRY 12

x = 3

√3.9 × 10−6

4

= 9.9 × 10−3 mol/LThe molar solubility of Zn(IO3)2 in water is 9.9 × 10−3 mol/L.

Step 5 Molar mass of Zn(IO3)2 = 415.2 g/mol9.9 × 10−3 mol/L = (9.9 × 10−3mol/L) × (415.2 g/mol)

= 4.1 g/LThe solubility of Zn(IO3)2 in water is 4.1 g/L.

Check Your SolutionRecall that x = [Zn2+]eq and 2x = [IO3

−]eq. Substitute these values into the Ksp

equation. You should get the given Ksp. The answers have the correct number of significant digits.

20. ProblemWhat is the maximum number of formula units of zinc sulfide, ZnS, that can dissolvein 1.0 L of solution at 25˚C? Ksp for ZnS is 2.0 × 10−22.

What Is Required?You need to determine the solubility (in formula units) of ZnS at 25˚C.

What Is Given?At 25˚C, Ksp for ZnS is 2.0 × 10−22.


of the equilibrium equation to write expressions for the equilibrium concen-trations of the ions.

Step 4 Substitute your expressions into the expression for Ksp, and solve for x.Step 5 Determine the number of formula units dissolved in 1.0 L.

Act on Your StrategyStep 1 ZnS(s) ⇀↽ Zn2+

(aq) + S2−(aq)

Step 2 Ksp = [Zn2+][S2−]

Step 3

Step 4 Ksp = [Zn2+][S2−] = x2

Therefore, 2.0 × 10−22 = x2

x =√

2.0 × 10−22

= 1.4 × 10−11 mol/LThe molar solubility of ZnS in water is 1.4 × 10−11 mol/L.

Step 5 1.4 × 10−11 mol = (1.4 × 10−11 mol) × (6.0 × 1023 formula units/mol)= 8.4 × 1012 formula units

The solubility of ZnS in water is 8.4 × 1012 formula units/L.

Check Your SolutionRecall that x = [Zn2+]eq = [S2−]eq . Substitute these values into the Ksp equation. You should get the given Ksp.


Initial

Change

Equilibrium

+Zn2+(aq) S2−

(aq)ZnS(s)

0

+x

x

0

x

x


CHEMISTRY 12


21. ProblemDetermine the molar solubility of AgCl(a) in pure water(b) in 0.15 mol/L NaCl

What Is Required?You need determine the solubility of AgCl (in mol/L) in water, and then in a solutionof NaCl.

What Is Given?From Appendix E, Ksp for AgCl = 1.77 × 10−10. You know the concentration of thesolution of NaCl.




Step 4 Substitute your expressions into the expression for Ksp, and solve for x.(b) Step 1 Set up an ICE table. The initial concentrations are based on the solution

of NaCl. Let x represent the concentration of chloride (the common ion)that is contributed by AgCl.

Step 2 Solve for x.

Act on Your Strategy(a) Step 1 AgCl(s) ⇀↽ Ag+

(aq) + Cl−(aq)

Step 2 Ksp = [Ag+][Cl−] = 1.77 × 10−10

Step 3

Step 4 Ksp = [Ag+][Cl−]1.77 × 10−10 = x2

x =√

1.77 × 10−10

x = ±1.33 × 10−5

The negative root has no physical meaning. The solubility of AgCl is 1.33 × 10−5 mol/L.

(b) Step 1

Step 2 Since Ksp is very small, you can assume that x is much smaller than 0.15.To check the validity of this assumption, determine whether or not 0.15 ismore than 500 times greater than Ksp, as shown on the next page.


Initial

Change

Equilibrium

+Ag+(aq) Cl−

(aq)AgCl(s)

0

+x

x

0.15

+x

0.15 + x


Initial

Change

Equilibrium

+Ag+(aq) Cl−

(aq)AgCl(s)

0

+x

x

0

+x

x


CHEMISTRY 12

0.15Ksp

= 0.151.77 × 10−10

= 8.5 × 108 > 500Therefore, in the (0.15 + x) term, x can be ignored. In other words,(0.15 + x) is approximately equal to 0.15. As a result, you can simplifythe equation as follows:

Ksp = [Ag+][Cl−]1.77 × 10−10 = (x)(0.15 + x) ≈ (x)(0.15)Therefore, 0.15x ≈ 1.77 × 10−10

x ≈ 1.18 × 10−9 mol/LThe solubility of AgCl in a solution of 0.15 mol/L NaCl is 1.18 × 10−9 mol/L.

Check Your SolutionYour approximation in Step 6 was reasonable. The solubility x is much smaller than0.15. Le Châtelier’s principle predicts a smaller solubility in a solution containing acommon ion, and this is confirmed by the calculations.

22. ProblemDetermine the molar solubility of lead(II) iodide, PbI2, in 0.050 mol/L NaI.

What Is Required?You need to determine the solubility of PbI2 (in mol/L) in a solution of NaI.

What Is Given?From Appendix E, Ksp for PbI2 = 9.8 × 10−9. You know the concentration of thesolution of NaI.

Plan Your StrategyStep 1 Write the dissolution equilibrium equation.Step 2 Use the equilibrium equation to write an expression for Ksp.Step 3 Set up an ICE table. The initial concentrations are based on the solution of

NaI. Let x represent the concentration of iodide (the common ion) that iscontributed by NaI.


Act on Your StrategyStep 1 PbI2(s) ⇀↽ Pb2+

(aq) + 2I−(aq)

Step 2 Ksp = [Pb2+][I−]2

Step 3

Step 4 Since Ksp is very small, you can assume that 2x is much smaller than 0.050.To check the validity of this assumption, determine whether or not 0.050 ismore than 500 times greater than Ksp, as shown.0.050

Ksp= 0.050

9.8 × 10−9

= 5.1 × 106 > 500


Initial

Change

Equilibrium

+Pb2+(aq) 2I−

(aq)PbI2(s)

0

+x

x

0.050

+2x

0.050 + 2x


CHEMISTRY 12

Therefore, in the (0.050 + 2x) term, 2x can be ignored. In other words,(0.050 + 2x) is approximately equal to 0.050. As a result, you can simplifythe equation as follows:

Ksp = [Pb2+][I−]2

9.8 × 10−9 = (x)(0.050 + 2x)2 ≈ (x)(0.050)2

Therefore, (2.5 × 10−3)(x) ≈ 9.8 × 10−9

x ≈ 3.9 × 10−6 mol/LThe solubility of PbI2 in a solution of 0.050 mol/L NaI is 3.9 × 10−6 mol/L.

Check Your SolutionYour approximation in Step 4 was reasonable. The solubility 2x is much smaller than0.050. You can substitute your calculated values of equilibrium concentrations into theexpression you wrote for Ksp. The calculated value should equal the given value for Ksp,within the errors introduced by mathematical rounding.

23. ProblemCalculate the solubility of calcium sulfate, CaSO4,(a) in pure water(b) in 0.25 mol/L Na2SO4

What Is Required?You need determine the solubility of CaSO4 (in mol/L) in water, and then in a solutionof Na2SO4.

What Is Given?From Appendix E, Ksp for CaSO4 = 4.93 × 10−5. You know the concentration of thesolution of Na2SO4.





(b) Step 1 Set up an ICE table. The initial concentrations are based on the solutionof Na2SO4. Let x represent the concentration of sulfate (the common ion)that is contributed by Na2SO4.

Step 2 Solve for x.

Act on Your Strategy(a) Step 1 CaSO4(s) ⇀↽ Ca2+

(aq) + SO42−

(aq)

Step 2 Ksp = [Ca2+][SO42−]

= 4.93 × 10−5

Step 3

Step 4 Ksp = [Ca+][SO42−] = x2

4.93 × 10−5 = x2

x =√

4.93 × 10−5

x = ±7.02 × 10−3


Initial

Change

Equilibrium

+Ca2+(aq) SO4

2−(aq)CaSO4(s)

0

+x

x

0

+x

x


CHEMISTRY 12

The negative root has no physical meaning. The solubility of CaSO4 is7.02 × 10−3 mol/L.

(b) Step 1

Step 2 Since Ksp is very small, you can assume that x is much smaller than 0.25.To check the validity of this assumption, determine whether or not 0.25is more than 500 times greater than Ksp, as shown below.0.25Ksp

= 0.254.93 × 10−5

= 5.1 × 103 > 500Therefore, in the (0.25 + x) term, x can be ignored. In other words,(0.25 + x) is approximately equal to 0.25. As a result, you can simplifythe equation as follows:

Ksp = [Ca2+][SO42−]

4.93 × 10−5 = (x)(0.25 + x) ≈ (x)(0.25)Therefore, 0.25x ≈ 4.93 × 10−5

x ≈ 1.97 × 10−4 mol/LThe solubility of AgCl in a solution of 0.15 mol/L NaCl is1.97 × 10−4 mol/L.

Check Your SolutionYour approximation in Step 6 was reasonable. The solubility x is a much smaller than0.25. Le Châtelier’s principle predicts a smaller solubility in a solution containing acommon ion, and this is confirmed by the calculations.

24. ProblemKsp for lead(II) chloride, PbCl2, is 1.6 × 10−5 . Calculate the molar solubility of PbCl2.(a) in pure water(b) in 0.10 mol/L CaCl2

What Is Required?You need to determine the solubility of PbCl2 (in mol/L) in water, and then in asolution of CaCl2.

What Is Given?You know that Ksp for PbCl2 is 1.6 × 10−5 , and you know the concentration of thesolution of CaCl2.




Step 4 Substitute your expressions into the expression for Ksp, and solve for x.(b) Step 1 Set up an ICE table. The initial concentrations are based on the solution

of CaCl2. Let x represent the concentration of chloride (the common ion)that is contributed by CaCl2.

Step 2 Solve for x.


Initial

Change

Equilibrium

+Ca2+(aq) SO4

2−(aq)CaSO4(s)

0

+x

x

0.25

+x

0.25 + x


CHEMISTRY 12

Act on Your Strategy(a) Step 1 PbCl2(s) ⇀↽ Pb2+

(aq) + 2Cl−(aq)

Step 2 Ksp = [Pb2+][Cl−]2

= 1.6 × 10−5

Step 3

Step 4 Ksp = [Pb2+][Cl−]2

1.6 × 10−5 = x(2x)2

1.6 × 10−5 = 4x3

x = 3√

4.0 × 10−6

x = 1.6 × 10−2

The solubility of PbCl2 is 1.6 × 10−2 mol/L.(b) Step 1 Each formula unit of calcium chloride dissociates to form two chloride ions.

Therefore, 0.10 mol/L CaCl2 dissociates to give [Cl−] = 0.20 mol/L .

Step 2 Since Ksp is very small, you can assume that x is much smaller than 0.20.To check the validity of this assumption, determine whether or not 0.20 ismore than 500 times greater than Ksp, as shown below.0.20Ksp

= 0.201.6 × 10−5

= 1.2 × 104 > 500Therefore, in the (0.20 + 2x) term, 2x can be ignored. In other words,(0.20 + 2x) is approximately equal to 0.20. As a result, you can simplifythe equation as follows:

Ksp = [Pb2+][Cl−]2

1.6 × 10−5 = (x)(0.20 + 2x)2 ≈ (x)(0.20)2

Therefore, (4.0 × 10−2)(x) ≈ 1.6 × 10−5

x ≈ 4.0 × 10−4 mol/LThe solubility of PbCl2 in a solution of 0.10 mol/L CaCl2 is4.0 × 10−4 mol/L.

Check Your SolutionYour approximation in Step 6 was reasonable. The solubility 2x is a much smaller than0.20. Le Châtelier’s principle predicts a smaller solubility in a solution containing acommon ion, and this is confirmed by the calculations.


Initial

Change

Equilibrium

+Pb2+(aq) 2Cl−

(aq)PbCl2(s)

0

+x

x

0.20

+2x

0.20 + 2x


Initial

Change

Equilibrium

+Pb2+(aq) 2Cl−

(aq)PbCl2(s)

0

+x

x

0

+2x

2x


CHEMISTRY 12

Solutions for Practice ProblemsStudent Textbook Page 441

25. ProblemA buffer solution is made by mixing 250 mL of 0.200 mol/L aqueous ammonia and400 mL of 0.150 mol/L ammonium chloride. Calculate the pH of the buffer solution.

What Is Required?You need to determine the pH of the buffer solution.

What Is Given?250 mL of 0.200 mol/L aqueous ammonia and 400 mL of 0.150 mol/L ammoniumchloride are mixed. From Appendix E, Kb for NH3(aq) is 1.8 × 10−5 .

Plan Your StrategyStep 1 Determine the initial concentrations of NH3(aq) and NH4

+(aq) in the

buffer solution.Step 2 Write the reaction for the dissociation of NH3(aq). Set up an ICE table,

including the initial concentration of NH4+

(aq) in the buffer.Step 3 Write the equation for Kb, and substitute equilibrium terms into the equation.Step 4 Solve the equation for x. Assume that x is small compared with the initial

concentrations. Check the validity of this assumption when you find thevalue of x.

Step 5 Use the following equations:pOH = −log[OH−]pH = 14.00 = pOH

Act on Your StrategyStep 1 When the solutions are mixed, the total volume is:

(250 mL + 400 mL) = 650 mL

[NH3] = 0.250 L × 0.200 mol/L0.650 L

= 7.69 × 10−2 mol/L

[NH4+] = 0.400 L × 0.150 mol/L

0.650 L= 9.23 × 10−2 mol/L

Step 2

Step 3 Kb = [NH4+][OH−]

[NH3]

= (9.23 × 10−2 + x)(x)(7.69 × 10−2 − x)

Step 4 Assume that (7.69 × 10−2 − x) ≈ 7.69 × 10−2 and(9.23 × 10−2 + x) ≈ 9.23 × 10−2 .

Kb = (9.23 × 10−2)(x)(7.69 × 10−2)

= 1.8 × 10−5

Therefore, x = 1.5 × 10−5 mol/L = [OH−]


Initial

Change

Equilibrium

++ OH−(aq)NH3(aq) NH4

+(aq)H2O(�)

9.23 × 10−2

+x

9.23 × 10−2 + x

7.69 × 10−2

−x

7.69 × 10−2 − x

~0

+x

x


CHEMISTRY 12

Step 5 pOH = −log(1.5 × 10−5)= 4.82

pH = 14.00 − 4.82= 9.18

The pH of the buffer solution is 9.18.

Check Your SolutionThe value of x (1.5 × 10−5) is negligible compared with the initial concentration ofeach component. A buffer that is made using a weak base and its conjugate acidshould have a pH that is greater than 7.

26. ProblemA buffer solution is made by mixing 200 mL of 0.200 mol/L aqueous ammonia and450 mL of 0.150 mol/L ammonium chloride. Calculate the pH of the buffer solution.


What Is Given?200 mL of 0.200 mol/L aqueous ammonia and 450 mL of 0.150 mol/L ammoniumchloride are mixed. From Appendix E, Kb for NH3(aq) is 1.8 × 10−5 .

Plan Your StrategyStep 1 Determine the initial concentrations of NH3(aq) and NH4

+(aq) in the

buffer solution.Step 2 Write the reaction for the dissociation of NH3(aq). Set up an ICE table,

including the initial concentration of NH4+

(aq) in the buffer.Step 3 Write the equation for Kb, and substitute equilibrium terms into the equation.Step 4 Solve the equation for x. Assume that x is small compared with the initial


Step 5 Use the following equations:pOH = −log[OH−]pH = 14.00 − pOH

Act on Your StrategyStep 1 When the solutions are mixed, the total volume is (250 + 400) = 650 mL.

[NH3] = 0.200 L × 0.0200 mol/L0.650 L

= 6.15 × 10−2 mol/L

[NH4+] = 0.450 L × 0.150 mol/L

0.650 L= 1.04 × 10−1 mol/L

Step 2

Step 3 Kb = [NH4+][OH−]

[NH3]

= (1.04 × 10−1 + x)(x)(6.15 × 10−2 − x)


Initial

Change

Equilibrium

++ OH−(aq)NH3(aq) NH4

+(aq)H2O(�)

1.04 × 10−1

+x

1.04 × 10−1 + x

6.15 × 10−2

−x

6.15 × 10−2 − x

~0

+x

x


CHEMISTRY 12

Step 4 Assume that (6.15 × 10−2 − x) ≈ 6.15 × 10−2 and (1.04 × 10−1 + x) ≈ 1.04 × 10−1 .

Kb = (1.04 × 10−1 + x)(x)(6.15 × 10−2)

= 1.8 × 10−5

Therefore, x = 1.1 × 10−5 mol/L = [OH−]Step 5 pOH = −log(1.1 × 10−5)

= 4.96pH = 14.00 − 4.96

= 9.04The pH of the buffer solution is 9.04.

Check Your SolutionThe value of x (1.1 × 10−5) is negligible compared with the initial concentration of each component. A buffer that is made using a weak base and its conjugate acidshould have a pH that is greater than 7.

27. ProblemA buffer solution contains 0.200 mol/L nitrous acid, HNO2(aq), and 0.140 mol/Lpotassium nitrite, KNO2(aq). What is the pH of the buffer solution?


What Is Given?0.200 mol/L HNO2(aq) is mixed with 0.140 mol/L potassium nitrite, KNO2(aq).From Appendix E, Ka for HNO2 is 5.6 × 10−4 .

Plan Your StrategyStep 1 Write the reaction for the dissociation of HNO2. Set up an ICE table,

including the initial concentration of NO2− in the buffer.

Step 2 Write the equation for Ka, and substitute equilibrium terms into the equation.Step 3 Solve the equation for x. Assume that x is small compared with the initial


Step 4 Use the equation pH = −log[H3O+].

Act on Your StrategyStep 1 HNO2 + H2O(�) ⇀↽ H3O+

(aq) + NO2−

(aq)

Step 2 Ka = [NO−2][H3O+]

[HNO2]

= (0.140 + x)(x)(0.200 − x)

Step 3 Assume that (0.140 + x) ≈ 0.140 and (0.200 − x) ≈ 0.200.

Ka = (1.40)(x)(0.200)

= 5.6 × 10−4

∴x = 8.0 × 10−4 mol/LStep 4 pH = −log(8.00 × 10−4)

= 3.10The pH of the buffer solution is 3.10.


Initial

Change

Equilibrium

++ H3O+(aq)HNO2(aq) NO2

−(aq)H2O(�)

0.140

+x

0.140 + x

0.200

−x

0.200 − x

~0

+x

x


CHEMISTRY 12

Check Your SolutionThe value of x (8.0 × 10−4) is negligible compared with the initial concentration ofeach component. A buffer that is made using a weak acid and its conjugate baseshould have a pH that is less than 7.

28. ProblemA buffer solution is prepared by dissolving 1.80 g of benzoic acid, C6H5COOH, and1.95 g of sodium benzoate, NaC6H5COO, in 800 mL of water. Calculate the pH ofthe buffer solution.


What Is Given?1.8 g of C6H5COOH and 1.95 g of NaC6H5COO are dissolved in 800 mL of water.From Appendix E, Ka for C6H5COOH is 6.3 × 10−5 .

Plan Your StrategyStep 1 Determine the initial concentrations of C6H5COOH and C6H5COO− in

the buffer solution.Step 2 Write the reaction for the dissociation of C6H5COOH. Set up an ICE table,

including the initial concentration of C6H5COO− in the buffer.Step 3 Write the equation for Ka, and substitute equilibrium terms into the equation.Step 4 Solve the equation for x. Assume that x is small compared with the initial


Step 5 Use the following equation:pH = −log[H3O+]

Act on Your StrategyStep 1 The molar mass of C6H5COOH is 122.1 g/mol.

[C6H5COOH] = 1.80 g122.1 g/mol

= 1.47 × 10−2 molThe molar mass of NaC6H5COO is 144.1 g/mol.

[NaC6H5COO] = 1.95 g144.1 g/mol

= 1.35 × 10−2 molStep 2

Step 3 Ka = [C6H5COO−][H3O3][C6H5COOH]

= (1.35 × 10−2 + x)(x)(1.47 × 10−2 − x)

Step 4 Assume that (1.35 × 10−2 + x) ≈ 1.35 × 10−2 and(1.47 × 10−2 − x) ≈ 1.47 × 10−2 .

Ka = (1.35 × 10−2 + x)(x)(1.47 × 10−2)

= 6.3 × 10−5

∴x = 6.9 × 10−5mol/L


Initial

Change

Equilibrium

++ H3O+(aq)C6H5COOH(aq) C6H5COOH−

(aq)H2O(�)

1.35 × 10−2

+x

1.35 × 10−2 + x

1.47 × 10−2

−x

1.47 × 10−2 − x

~0

+x

x


CHEMISTRY 12

Step 5 pH = −log(6.9 × 10−5)= 4.16

The pH of the buffer solution is 4.16.

Check Your SolutionThe value of x (6.9 × 10−5) is negligible compared with the initial concentration ofeach component. A buffer that is made using a weak acid and its conjugate baseshould have a pH that is less than 7.


29. ProblemA solution contains 0.15 mol/L of NaCl and 0.0034 mol/L Pb(NO3)2. Does a precipitate form? Include a balanced chemical equation for the formation of the possible precipitate. Ksp for PbCl2 is 1.7 × 10−5 .

SolutionWrite the ion product expression for lead(II) chloride. Then substitute the ion concentrations into the expression to determine Qsp . Compare Qsp with Ksp, and predict whether or not a precipitate will form.PbCl2(s) ⇀↽ Pb2+

(aq) + 2Cl−(aq)

Ion product expression = [Pb2+][Cl−]2

Qsp = (0.0034) × (0.15)2

= 7.6 × 10−5

Because Qsp is greater than Ksp, a precipitate will form.

Check Your SolutionIt seems reasonable that a precipitate formed, since Ksp for lead(II) chloride is smallcompared with the concentration of chloride ions and lead ions.

30. ProblemOne drop (0.050 mL) of 1.5 mol/L potassium chromate, K2CrO4, is added to 250 mL of 0.10 mol/L AgNO3. Does a precipitate form? Include a balanced chemical equation for the formation of the possible precipitate. Ksp for Ag2CrO4

is 2.6 × 10−12.

What Is Required?Will silver chromate precipitate under the given conditions?

What Is Given?You know the concentration and volume of the potassium chromate and silver nitrate solutions. For K2CrO4, c = 1.5 mol/L and V = 0.050 mL. For AgNO3, c = 0.10 mol/L and V = 250 mL. You also know Ksp for silver chromate.

Plan Your StrategyStep 1 Determine the concentrations of silver ions and chromate ions in the

reaction mixture.Step 2 Substitute the ion concentrations into the ion product expression for silver

chromate to determine Qsp .Step 3 Compare Qsp with Ksp, and predict whether or not a precipitate will form.


CHEMISTRY 12

Act on Your StrategyStep 1 Because the volume of the potassium chromate solution is much smaller

than the volume of the silver nitrate solution, you can ignore it. In otherwords, assume that the final volume of the reaction mixture is 0.250 L.AgNO3(aq) → Ag+

(aq) + NO3−

(aq)

∴[Ag+] = [AgNO3] = 0.10 mol/LK2CrO4(aq) → 2K+

(aq) + CrO42−

(aq)

[CrO42−] = [K2CrO4] = c × Vinitial

Vfinal

= (1.5 mol/L)(0.050 mL × 1.0 × 10−3 L/mL)0.250 L

= 3.0 × 10−4 mol/LStep 2 The possible precipitate is silver chromate.

2Ag+(aq) + CrO4

2− → Ag2CrO4(s)

Qsp = [Ag+]2[CrO42−]

= (0.10)2 × (3.0 × 10−4)= 3.0 × 10−6

Step 3 Since Qsp > Ksp , Ag2CrO4 will precipitate until Qsp = 2.6 × 10−12 .

Check Your SolutionThe units are correct in the calculation of the concentration of silver ions. It seems rea-sonable that a precipitate formed, since Ksp for silver chromate is very small comparedwith the concentration of chromate ions and silver ions.

31. ProblemA chemist adds 0.010 g of CaCl2 to 5.0 × 102 mL of 0.0015 mol/L sodium carbonate,Na2CO3. Does a precipitate of calcium carbonate form? Include a balanced chemicalequation for the formation of the possible precipitate.

What Is Required?Will calcium carbonate precipitate under the given conditions?

What Is Given?You know the concentration and volume of the sodium carbonate solution. For Na2CO3, c = 0.0015 mol/L and V = 5.0 × 102 mL. For CaCl2, 0.010 g is added to the solution. From Appendix E, you also know that Ksp for calcium carbonate is 3.36 × 10−9.

Plan Your StrategyStep 1 Determine the concentrations of calcium ions and carbonate ions in the

reaction mixture.Step 2 Substitute the ion concentrations into the ion product expression for calcium

carbonate to determine Qsp .Step 3 Compare Qsp with Ksp, and predict whether or not a precipitate will form.


CHEMISTRY 12

Act on Your StrategyStep 1 Na2CO3(aq) → 2Na+

(aq) + CO32−

(aq)

∴[CO32−] = [Na2CO3] = 0.0015 mol/L

The molar mass of CaCl2 is 111.0 g/mol. Therefore,

Amount CaCl2 = 0.010 g111.0 g/mol

= 9.0 × 10−5 mol

[CaCl2] = 9.0 × 10−5 mol0.50 L

= 1.8 × 10−4 mol/LCaCl2(aq) → Ca2+

(aq) + 2Cl−(aq)

Therefore, [Ca2+] = [CaCl2] = 1.8 × 10−4 mol/LStep 2 The possible precipitate is calcium carbonate.

Ca2+(aq) + CO3

2− → CaCO3(s)

Qsp = [Ca2+][CO32−]

= (1.8 × 10−4) × (0.0015)= 2.7 × 10−7

Step 3 Since Qsp > Ksp , CaCO3 will precipitate until Qsp = 3.36 × 10−9.

Check Your SolutionThe units are correct in the calculation of the concentration of silver ions. It seemsreasonable that a precipitate formed, since Ksp for calcium carbonate is small compared with the concentration of calcium ions and carbonate ions.

32. Problem0.10 mg of magnesium chloride, MgCl2, is added to 2.5 × 102 mL of 0.0010 mol/LNaOH. Does a precipitate of magnesium hydroxide form? Include a balanced chemicalequation for the formation of the possible precipitate.

What Is Required?Will magnesium hydroxide precipitate under the given conditions?

What Is Given?You know the concentration and volume of the sodium hydroxide solution. For NaOH, c = 0.0010 mol/L and V = 2.5 × 102 mL. For MgCl2, 0.10 mg is added to the solution. From Appendix E, you also know that Ksp for magnesiumhydroxide is 5.61 × 10−12.

Plan Your StrategyStep 1 Determine the concentrations of magnesium ions and hydroxide ions in the

reaction mixture.Step 2 Substitute the ion concentrations into the ion product expression for

magnesium hydroxide to determine Qsp .Step 3 Compare Qsp with Ksp, and predict whether or not a precipitate will form.


CHEMISTRY 12

Act on Your StrategyStep 1 NaOH(aq) → Na+

(aq) + OH−(aq)

Therefore, [OH−] = [NaOH] = 0.0010 mol/LThe molar mass of MgCl2 is 95.2 g/mol. Therefore,

Amount MgCl2 = 0.10 mg × 10−3 g/mg95.2 g/mol

= 1.1 × 10−6 mol

[MgCl2] = 1.1 × 10−6 mol0.25 L

= 4.2 × 10−6 mol/LMgCl2(aq) → Mg2+

(aq) + 2Cl−(aq)

Therefore, [Mg2+] = [MgCl2] = 4.2 × 10−6 mol/L.Step 2 The possible precipitate is magnesium hydroxide.

Mg2+(aq) + 2OH−

(aq) → Mg(OH)2(s)

Qsp = [Mg2+][OH−]2

= (4.2 × 10−6) × (0.0010)2

= 4.2 × 10−12

Step 3 Since Qsp < Ksp , no precipitate forms.

Check Your SolutionThe units are correct in the calculation of the concentration of silver ions. It seemsreasonable that no precipitate formed, since the concentration of magnesium ion is small.


33. Problem1.0 × 102 mL of 1.0 × 10−3 mol/L Pb(NO3)2 is added to 40 mL of 0.040 mol/LNaCl. Does a precipitate form? Include a balanced chemical equation for the formation of the possible precipitate.

What Is Required?Decide whether a precipitate will form under the given circumstances. Write a balanced chemical equation for the formation of the possible precipitate.

What Is Given?You know the initial concentration and volume of the lead(II) nitrate and sodiumchloride solutions. For Pb(NO3)2, c = 1.0 × 10−3 mol/L and V = 1.0 × 102 mL.For NaCl, c = 0.040 mol/L and V = 40 mL. As well, you have the solubility guidelines in Table 9.3.

Plan Your StrategyStep 1 Decide whether a compound with low solubility forms when lead(II) nitrate

and sodium chloride are mixed, using the solubility guidelines.Step 2 If an insoluble compound forms, write an equation that represents the

reaction. Look up Ksp for the compound in Appendix E.Step 3 Determine the concentrations of the ions that make up the compound.Step 4 Substitute the concentrations of the ions into the ion product expression for

the compound to determine Qsp .Step 5 Compare Qsp with Ksp, and predict whether or not a precipitate forms.


CHEMISTRY 12

Act on Your StrategyStep 1 When you mix the solutions, lead(II) chloride, PbCl2, and sodium nitrate,

NaNO3, can also form. Sodium nitrate is soluble, but lead(II) chloride isnot, according to guideline 2.

Step 2 Pb(NO3)2(aq) + NaCl(aq) ⇀↽ PbCl2(s) + NaNO3(aq)

From Table 9.2, Ksp for PbCl2 is 1.7 × 10−5 .Step 3 When calculating the volume of the reaction mixture, assume that it is equal

to the sum of the volumes of the solutions. This is a reasonable assumption,because both solutions are dilute.

[Pb2+] = [Pb(NO3)2] = c × VinitialVfinal

= (1.0 × 10−3 mol/L)(100 mL)(100 mL + 40 mL)

= 7.1 × 10−4 mol/L

[Cl−] = [NaCl] = c × VinitialVfinal

= (0.040 mol/L)(40 mL)(100 mL + 40 mL)

= 1.1 × 10−2 mol

Step 4 Qsp = [Pb2+][Cl−]2

= (7.1 × 10−4)(1.1 × 10−2)2

= 8.6 × 10−8

Step 5 Since Qsp < Ksp , no precipitate forms.

Check Your SolutionThe units in the calculation of the concentrations of the ions are correct. It seems rea-sonable that no precipitate forms, since the concentrations of ions are relatively small.

34. Problem2.3 × 102 mL of 0.0015 mol/L AgNO3 is added to 1.3 × 102 mL of 0.010 mol/Lcalcium acetate, Ca(CH3COO)2. Does a precipitate form? Include a balanced chemical equation for the formation of the possible precipitate. Ksp for AgCH3COOis 2.0 × 10−3 .

What Is Required?Decide whether a precipitate will form under the given circumstances. Write a balanced chemical equation for the formation of the possible precipitate.

What Is Given?You know the initial concentration and volume of the silver nitrate and calciumacetate solutions. For AgNO3, c = 0.0015 mol/L and V = 230 mL. ForCa(CH3COO)2, c = 0.010 mol/L and V = 130 mL. As well, you have the solubilityguidelines in Table 9.3.

Plan Your StrategyStep 1 The Ksp value for silver acetate indicates that it is not very soluble in water.

Write an equation that represents the reaction to form silver acetate.Step 2 Determine the concentrations of the ions that make up the compound.Step 3 Substitute the concentrations of the ions into the ion product expression for

the compound to determine Qsp .Step 4 Compare Ksp with Qsp , and predict whether or not a precipitate forms.


CHEMISTRY 12

Act on Your StrategyStep 1 2AgNO3(aq) + Ca(CH3COO)2(aq) ⇀↽ 2AgCH3COO(s) + Ca(NO3)2(aq)

Step 2 When calculating the volume of the reaction mixture, assume that it is equalto the sum of the volumes of the solutions. This is a reasonable assumption,because both solutions are dilute.

[Ag+] = [AgNO3] = c × VinitialVfinal

= (0.0015 mol/L)(230 mL)(230 mL + 130 mL)

= 9.6 × 10−4 mol/L

[CH3COO−] = 2 × [Ca(CH3COO)2] = 2 × c × VinitialVfinal

= 2 × (0.010 mol/L)(130 mL)(130 mL + 230 mL)

= 7.2 × 10−3 mol/L

Step 3 Qsp = [Ag+][CH3COO−]= (9.6 × 10−4)(7.2 × 10−3)= 6.9 × 10−6

Step 4 Since Qsp < Ksp , no precipitate is formed.

Check Your SolutionThe units in the calculation of the concentrations of the ions are correct. It seems reasonable that no precipitate formed, since Ksp for silver acetate is a relatively largevalue compared with the concentrations of the silver ions and acetate ions.

35. Problem25 mL of 0.10 mol/L NaOH is added to 5.0 × 102 mL of 0.00010 mol/L cobalt(II)chloride, CoCl2. Does a precipitate form? Include a balanced chemical equation forthe formation of the possible precipitate.

What Is Given?You know the initial concentration and volume of the sodium hydroxide andcobalt(II) chloride solutions. For NaOH, c = 0.10 mol/L and V = 25 mL. ForCoCl2, c = 0.00010 mol/L and V = 500 mL. As well, you have the solubility guidelines in Table 9.3.

Plan Your StrategyStep 1 Decide whether a compound with low solubility forms when sodium

hydroxide and cobalt(II) chloride are mixed, using the solubility guidelines.Step 2 If an insoluble compound forms, write an equation that represents the



Act on Your StrategyStep 1 When you mix the sodium hydroxide and cobalt(II) chloride solutions,

sodium chloride, NaCl, and cobalt(II) hydroxide, Co(OH)2, can also form.Sodium chloride is soluble, but cobalt(II) hydroxide is probably not, according to guideline 2.

Step 2 NaOH(aq) + CoCl2(aq) → NaCl(aq) + Co(OH)2(s)

From Appendix E, Ksp for Co(OH)2 is 5.92 × 10−15.


CHEMISTRY 12


[Co2+] = [CoCl2] = c × VinitialVfinal

= (0.00010 mol/L)(500 mL)(500 mL + 25 mL)

= 9.5 × 10−5 mol/L

[OH−] = [NaOH] = c × VinitialVfinal

= (0.10 mol/L)(25 mL)(25 mL + 500 mL)

= 4.8 × 10−3 mol/L

Step 4 Qsp = [Co2+][Cl−]2

= (9.5 × 10−5)(4.8 × 10−3)2

= 2.2 × 10−9

Step 5 Since Qsp > Ksp , Co(OH)2 precipitates until Qsp = 5.92 × 10−15.

Check Your SolutionThe units in the calculation of the concentrations of the ions are correct. It seems reasonable that a precipitate formed, since Ksp for cobalt hydroxide is very small compared with the concentrations of the cobalt ions and hydroxide ions.

36. Problem250 mL of 0.0011 mol/L Al2(SO4)3 is added to 50 mL of 0.022 mol/L BaCl2. Does a precipitate form? Include a balanced chemical equation for the formation of the possible precipitate.

What Is Given?You know the initial concentration and volume of the aluminum sulfate and bariumchloride solutions. For Al2(SO4)3, c = 0.0011 mol/L and V = 250 mL. For BaCl2,c = 0.022 mol/L and V = 50 mL. As well, you have the solubility guidelines inTable 9.3.

Plan Your StrategyStep 1 Decide whether a compound with low solubility forms when calcium nitrate

and sodium fluoride are mixed, using the solubility guidelines.Step 2 If an insoluble compound forms, write an equation that represents the



Act on Your StrategyStep 1 When you mix the aluminum sulfate and barium chloride solutions,

aluminum chloride, AlCl3, and barium sulfate, BaSO4, can also form.Aluminum chloride is soluble, but barium sulfate is not, according to guideline 4.

Step 2 Al2(SO4)3(aq) + 3BaCl2(aq) → 2AlCl3(aq) + 3BaSO4(s)

From Appendix E, Ksp for BaSO4 is 1.08 × 10−10.


CHEMISTRY 12


[Ba2+] = [BaCl2] = c × VinitialVfinal

= (0.022 mol/L)(50 mL)(50 mL + 250 mL)

= 3.7 × 10−3 mol/L

[SO42−] = 3 × [Al2(SO4)3] = 3 × c × Vinitial

Vfinal

= 3 × (0.0011 mol/L)(250 mL)(250 mL + 50 mL)

= 2.8 × 10−3 mol/L

Step 4 Qsp = [Ba2+][SO42−]

= (3.7 × 10−3)(2.8 × 10−3)= 1.0 × 10−5

Step 5 Since Qsp > Ksp , BaSO4 precipitates until Qsp = 1.08 × 10−10.

Check Your SolutionThe units in the calculation of the concentrations of the ions are correct. It seems reasonable that a precipitate formed, since Ksp for barium sulfate is very small compared with the concentrations of the barium ions and sulfate ions.


CHEMISTRY 12

chapter 9

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