CHEMISTRY 12
Chapter 9
Aqueous Solutions
and Solubility Equilibria
Solutions for Practice Problems
Student Textbook page 424
1. Problem
Predict whether an aqueous solution of each salt is neutral, acidic, or basic.
NaCN
LiF
Mg(NO3)2
NH4I
(a)
(b)
(c)
(d)
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 Strategy
Determine whether the cation is from a strong or weak base, and whether the anion is
from a strong or weak acid. Ions derived from weak bases or weak acids react with water
and 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 Solution
Check that you have correctly identified whether the cation is from a strong or weak
base, and whether the anion is from a strong or weak acid.
2. Problem
Is 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 that
represent the solutions that are acidic or basic.
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CHEMISTRY 12
What Is Given?
The formula of each salt is given.
Plan Your Strategy
Determine whether the cation is from a strong or weak base, and whether the anion is
from a strong or weak acid. Ions derived from weak bases or weak acids react with water
and 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 Solution
The equations that represent the reactions with water support the prediction that
NH4BrO4 and NH4Br dissolve to form an acidic solution and NaOBr dissolves to
form 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. Problem
Ka for benzoic acid, C6H5COOH, is 6.3 × 10−5 . Ka for phenol, C6H5OH, is
1.3 × 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 Strategy
For a conjugate acid-base pair, KaKb = 1.0 × 10−14 . Therefore, a small value of Ka for
an acid results in a large value of Kb for the conjugate base.
Act on Your Strategy
Because 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 Solution
Using the equation KaKb = 1.0 × 10−14 , you can calculate the Kb for each conjugate
base. 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.
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4. Problem
Sodium 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 Strategy
The 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 farthest
to completion.
Act on Your Strategy
The ion reactions with water are
HSO3−(aq) + H2O() SO32−(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 = Kw
Ka
−14
= 1.0 × 10 −2
1.4 × 10
= 7.1 × 10−13
The equilibrium constant (Ka) for the hydrogen sulfite ion acting as an acid is greater
than the equilibrium constant (Kb) for the ion acting as a base. Therefore, the solution
is acidic.
Check Your Solution
It is a common difficulty in this type of problem to identify correctly the base reaction
and the corresponding value for Kb . Kb must be calculated using the Ka value for
H2SO3(aq) , because HSO3−(aq) is the conjugate base of H2SO3(aq) .
Solutions for Practice Problems
Student Textbook page 428
5. Problem
After titrating sodium hydroxide with hydrofluoric acid, a chemist determined that the
reaction had formed an aqueous solution of 0.020 mol/L sodium fluoride. Determine
the pH of the solution.
What Is Required?
You need to calculate the pH of the solution.
What Is Given?
[NaF] = 0.020 mol/L
From Appendix E, Ka for HF(aq) = 6.3 × 10−4
Plan Your Strategy
Step 1 Decide which ion reacts with water. Write the equation that represents the
hydrolysis reaction.
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CHEMISTRY 12
Step 2
Step 3
Step 4
Step 5
Step 6
Determine the equilibrium constant for the ion involved in the
hydrolysis reaction.
Divide the ion concentration by the appropriate equilibrium constant to
determine whether the change in concentration of the ion can be ignored.
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.
Write the equilibrium expression. Substitute the equilibrium concentrations
into the expression, and solve for x.
Use the value of x to determine [H3O+]. Then calculate the pH of
the solution.
Act on Your Strategy
Sodium fluoride is the salt of a strong base (NaOH) and a weak acid (HF).
Thus, only the fluoride ion reacts with water.
HF(aq) + OH−(aq)
F−(aq) + H2O() Step 2 Kb for the fluoride ion can be calculated using Ka for HF(aq) .
Kb = Kw
Ka
−14
= 1.0 × 10 −4
6.3 × 10
= 1.6 × 10−11
Na+(aq) + F−(aq)
Step 3 NaF(aq) ∴[F−] = 0.020 mol/L
0.020
F− =
Kb
1.6 × 10−11
= 1.2 × 109
Because this value is much greater than 500, the change in concentration of
F−(aq) can be ignored compared with its initial concentration.
Step 1
Step 4
Concentration (mol/L)
F −(aq)
Initial
0.020
0
~0
−x
+x
+x
0.020 − x ≈ 0.020
x
x
Change
Equilibrium
Step 5
+
H2O()
2HF(aq)
+
OH−(aq)
−
]
Kb = [HF][OH
−
[F ]
1.6 × 10−11 = (x)(x)
0.020
x = 3.2 × 10−13
= 5.6 × 10−7 mol/L
Step 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 Solution
The 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.
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CHEMISTRY 12
6. Problem
Part 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 Strategy
Step 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 Strategy
Step 1 Amount NaOH(aq) = (0.10 mol/L) × (2.0 × 10−2 L)
= 2.0 × 10−3 mol
Amount HCl(aq) = (0.10 mol/L) × (3.0 × 10−2 L)
= 3.0 × 10−3 mol
Step 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 mol
Total volume = (2.0 × 10−2 L) + (3.0 × 10−2 L) = 5.0 × 10−2 L
−3
[HCl] = 10 × 10 −2mol
5.0 × 10 L
= 2.0 × 10−2 mol/L
Step 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 Solution
The solutions have the same molar concentration. Because the volume of hydrochloric acid is greater, the solution should be acidic and the pH less than 7.
7. Problem
0.025 mol/L benzoic acid, C6H5COOH, is titrated with 0.025 mol/L sodium
hydroxide 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 Strategy
Write the balanced chemical equation for the reaction.
Calculate the concentration of the salt formed, based on the amount
(in mol) and the total volume of the solution.
Step 1
Step 2
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CHEMISTRY 12
Step 3
Step 4
Step 5
Step 6
Step 7
Step 8
Decide which ion reacts with water. Write the equation that represents
the reaction.
Determine the equilibrium constant for the ion that is involved in the
hydrolysis reaction.
Divide the concentration of the ion identified in Step 3 by the appropriate
ionization constant to determine whether the change in concentration of the
ion can be ignored.
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.
Write the equilibrium expression. Substitute the equilibrium concentrations
into the expression, and solve for x.
Use the value of x to determine [H3O+]. Then calculate the pH of
the solution.
Act on Your Strategy
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/L
2
= 0.0125 mol/L
Step 3 The salt forms Na+(aq) and C6H5COO−(aq) in solution. Na+(aq) is the cation
of a strong base, so it does not react with water. C6H5COO−(aq) is the conjugate base of a weak acid, so it does react with water. The pH of the solution
is 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
−14
= 1.0 × 10 −5
6.3 × 10
= 1.6 × 10−10
Step 1
Step 5
[C6H5COO−]
= 0.0125−10
Kb
1.6 × 10
= 7.8 × 107
This is well above 500, so the change in [C6H5COO−] can be ignored.
Step 6
Concentration (mol/L) C6H5COOH−(aq)
Initial
Step 7
C6H5COOH(aq)
+
OH−(aq)
0.0125
0
~0
−x
+x
+x
0.0125 − x ≈ 0.0125
x
x
Change
Equilibrium
+ H2O()
[C6H5COOH][OH−]
[C6H5COO−]
1.6 × 10−10 = (x)(x)
0.0125
√
x = 2.0 × 10−12
Kb =
= 1.4 × 10−6 mol/L
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CHEMISTRY 12
Step 8
x = [OH−] = 1.4 × 10−6 mol/L
pOH = −log[OH−]
= −log(1.4 × 10−6)
= 5.85
pH = 14.00 − 5.85
= 8.15
Check Your Solution
The titration forms an aqueous solution of a salt derived from a strong base and a weak
acid. The solution should be basic, which is supported by the calculation of the pH.
8. Problem
50.0 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 Strategy
Step 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 the
ion 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.
Step 8 Write the equilibrium expression. Substitute the equilibrium concentrations
into the expression, and solve for x.
Step 9 Use the value of x to determine [H3O+]. Then calculate the pH of
the solution.
Act on Your Strategy
Step 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 volume
of aqueous ammonia added must be the same as the volume of hydrobromic
acid. Therefore, the volume of NH3(aq) is 50 mL, and the total volume of the
solution is twice the volume of HBr(aq) used. Thus, the [NH4Br] must be half
the initial [HBr].
Therefore, [NH4Br] = 0.050 mol/L
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CHEMISTRY 12
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 solution will be acidic, and the pH is determined by the extent of the following
reaction.
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 relationship
KaKb = Kw.
Ka = Kw
Kb
−14
= 1.0 × 10 −5
1.8 × 10
= 5.6 × 10−10
Step 4
Step 6
0.050
[NH4+] =
Ka
5.6 × 10−10
= 8.9 × 107
This is well above 500, so the change in [NH+4 ] can be ignored.
Step 7
NH4+(aq) +
Initial
0.050
0
~0
−x
+x
+x
0.050 − x ≈ 0.050
x
x
Change
Equilibrium
Step 8
H3O+(aq)
Concentration (mol/L)
H2O()
NH3(aq)
+
[NH3][H3O+]
[NH4+]
5.6 × 10−10 = (x)(x)
0.050
√
x = 2.8 × 10−11
Ka =
= 5.3 × 10−6 mol/L
Step 9 x = [H3O+] = 5.3 × 10−6 mol/L
pH = −log[H3O+]
= −log(5.3 × 10−6)
= 5.28
Check Your Solution
The titration forms an aqueous solution of a salt derived from a weak base and a strong
acid. The solution should be acidic, which is supported by the calculation of the pH.
Solutions for Practice Problems
Student Textbook page 432
9. Problem
Write the balanced chemical equation that represents the dissociation of each compound
in water. Then write the corresponding solubility product expression.
(a) copper(I) chloride
(b) barium fluoride
(c) silver sulfate
(d) calcium phosphate
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CHEMISTRY 12
Solution
First, write a balanced equation for the equilibrium between excess solid and dissolved
ions in a saturated aqueous solution. Then use the balanced equation to write the
expression for Ksp.
Cu+(aq) + Cl−(aq)
(a) CuCl(s) Ksp = [Cu+][Cl−]
Ba2+(aq) + 2F−(aq)
(b) BaF2(s) Ksp = [Ba2+][F−]2
2Ag+(aq) + SO42−(aq)
(c) Ag2SO4(s) Ksp = [Ag+]2[SO42−]
3Ca2+(aq) + 2PO43−(aq)
(d) Ca3(PO4)2(s) 2+ 3
Ksp = [Ca ] [PO43−]2
Check Your Solution
The Ksp expressions are based on balanced equations for saturated solutions of slightly
soluble ionic compounds. The exponents in the Ksp expressions match the corresponding coefficients in the chemical equations. The coefficient 1 is not written, following
chemistry conventions.
10. Problem
Write a balanced dissolution equation and solubility product expression for silver
carbonate, Ag2CO3 .
Solution
First, write a balanced equation for the equilibrium between excess solid and dissolved
ions 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 Solution
The Ksp expression is based on the balanced equation between excess solid and dissolved
ions. The exponents in the Ksp expression match the corresponding coefficients in the
chemical equation.
11. Problem
Write a balanced dissolution equation and solubility product expression for magnesium
ammonium phosphate, MgNH4PO4 .
Solution
First, write a balanced equation for the equilibrium between excess solid and dissolved
ions in a saturated aqueous solution. Then use the balanced equation to write the
expression for Ksp.
MgNH4PO4(s) Mg2+(aq) + NH4+(aq) + PO43−(aq)
2+
Ksp = [Mg ][NH4+][PO43−]
Check Your Solution
The Ksp expression is based on the balanced equation between excess solid and dissolved
ions. The exponents in the Ksp expression match the corresponding coefficients in the
chemical equation.
12. Problem
Iron (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?
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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 equation
to 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 hydroxide and dissolved ions in a saturated aqueous solution. Then use the balanced
equation to write the expression for Ksp. Finally, compare the two equilibrium
expressions and decide which expression is larger.
Al3+(aq) + 3OH−(aq)
Al(OH)3(s) 3+
Ksp = [Al ][OH−]3
The equilibrium expressions have the same form. In both cases, if x mol/L of salt
dissolves, Ksp is given by x 4 . Therefore, the most soluble salt will have the larger
value of Ksp. The value of Ksp for iron(III) nitrate is smaller than Ksp for aluminum
hydroxide.
Check Your Solution
Each Ksp expression is based on the balanced equation between excess solid and dissolved
ions. The exponents in the Ksp expressions match the corresponding coefficients in the
chemical equations. The equilibrium expressions are of the same form, so the least
soluble salt has the lower value for Ksp.
Solutions for Practice Problems
Student Textbook page 433
13. Problem
The 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 Strategy
Step 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 Strategy
Ag+(aq) + CN−(aq)
Step 1 AgCN(s) Step 2 Ksp = [Ag+][CN−]
Step 3 [Ag+] = [CN−] = [AgCN] = 1.5 × 10−8 mol/L
Step 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 Solution
The value of Ksp is less than the concentration of the salt, as expected. It has the correct
number of significant digits.
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CHEMISTRY 12
14. Problem
A saturated solution of copper(II) phosphate, Cu3(PO4)2 , has a concentration of
6.1 × 10−7g Cu3(PO4)2 per 1.00 × 102 mL of solution at 25˚C. What is Ksp for
Cu3(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 Strategy
Step 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 Strategy
3Cu2+(aq) + 2PO43−(aq)
Step 1 Cu3(PO4)2(s) 2+ 3
Step 2 Ksp = [Cu ] [PO43−]2
Step 3 The molar mass of Cu3(PO4)2 is 380.6 g/mol.
6.1 × 10−7g
Amount of Cu3(PO4)2 =
380.6 g/mol
= 1.6 × 10−9 mol
−9
[Cu3(PO4)2] = 1.6 × 10 mol
0.100 L
= 1.6 × 10−8 mol/L
2+
[Cu ] = 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 Solution
Check that you wrote the balanced chemical equation and the corresponding Ksp equation correctly. Pay attention to molar relationships and to the exponent of each term.
Check the concentration calculations carefully.
15. Problem
A saturated solution of CaF2 contains 1.2 × 1020 formula units of calcium fluoride per
litre 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 .
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CHEMISTRY 12
Plan Your Strategy
Step 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 Strategy
Ca2+(aq) + 2F−(aq)
Step 1 CaF2(s) Step 2 Ksp = [Ca2+][F−]2
Step 3 The amount of CaF2 is given by the following formula:
Amount = number of formula units
Avogadro’s constant
1.2 × 1020 formula units
6.0 × 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 Solution
Check that you wrote the balanced chemical equation and the corresponding Ksp equation correctly. Pay attention to molar relationships and to the exponent of each term.
Check the concentration calculations carefully.
16. Problem
The concentration of mercury(I) iodide, Hg2I2 , in a saturated solution at 25˚C is
1.5 × 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 Strategy
Use the dissolution equation to write the solubility product expression.
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.
Step 1
Step 2
Chapter 9 Aqueous Solutions and Solubility Equilibria • MHR
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CHEMISTRY 12
Act on Your Strategy
(a) Step 1 Ksp = [Hg22+][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 × 10−7g
The molar mass of Hg2I2 is 655 g/mol.
1.5 × 10−7 g
Amount of Hg2I2 =
655 g/mol
= 2.3 × 10−10 mol
[Hg2I2] = 2.3 × 10−10 mol/L
[Hg22+] = [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 to
two assumptions in this problem. Firstly, that the solution is so dilute it can be
treated 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 Solution
Check that you wrote the Ksp equation correctly. Pay attention to molar relationships
and to the exponent of each term. Check the concentration calculations carefully.
Solutions for Practice Problems
Student Textbook page 436
17. Problem
Ksp 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 a
mass/volume percent) of AgCl at 25˚C.
What Is Given?
At 25˚C, Ksp for AgCl is 1.8 × 10−10.
Plan Your Strategy
Write the dissolution equilibrium equation.
Use the equilibrium equation to write an expression for Ksp.
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).
(a) Step 1
Step 2
Step 3
Chapter 9 Aqueous Solutions and Solubility Equilibria • MHR
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CHEMISTRY 12
Act on Your Strategy
Ag2+(aq) + Cl−(aq)
(a) Step 1 AgCl(s) Step 2 Ksp = [Ag+][Cl−]
Step 3
Concentration (mol/L)
Initial
Change
Equilibrium
AgCl(s)
Ag+(aq)
+
Cl−(aq)
0
0
+x
+x
x
x
Ksp = [Ag+][Cl−] = (x)(x)
Therefore, 1.8 × 10−10 = x 2
√
x = 1.8 × 10−10
= 1.3 × 10−5 mol/L
The 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
mass of solute (g)
× 100%
(c) Mass/volume percent =
volume of solution (mL)
Step 4
Molar mass of AgCl = 143.3 g/mol
Mass in 1 L of solution = (1.3 × 10−5 mol/L) × (143.3 g/mol)
= 1.9 × 10−3 g/L
1.9 × 10−3 g
Mass/volume percent =
× 100%
1000 mL
The solubility of AgCl at 25˚C is 1.9 × 10−4% (m/v).
Check Your Solution
Substitute the values of [Ag+] and [Cl−] into the Ksp equation. You should get the given
Ksp. Check the number of significant digits given in the question, and the number of
digits in the answers.
18. Problem
Iron(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 Strategy
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.
Chapter 9 Aqueous Solutions and Solubility Equilibria • MHR
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CHEMISTRY 12
Act on Your Strategy
Fe3+(aq) + 3OH−(aq)
Step 1 Fe(OH)3(s) 3+
Step 2 Ksp = [Fe ][OH−]3
Step 3
Concentration (mol/L)
Fe(OH)3(s)
Fe3+(aq)
Initial
Change
3OH−(aq)
0
0
+x
+3x
x
3x
Equilibrium
Step 4
+
Ksp = [Fe3+][OH−]3
= (x) × (3x)3
= 27x 4
2.8 × 10−39 = 27x 4
−39
4
x = 2.8 × 10
27
= 1.0 × 10−10 mol/L
The molar solubility of Fe(OH)3 in water is 1.0 × 10−10 mol/L.
Check Your Solution
Recall that x = [Fe3+]eq and 3x = [OH−]eq . Substitute these values into the Ksp
equation. You should get the given Ksp.
19. Problem
Ksp for zinc iodate, Zn(IO3)2 , is 3.9 × 10−6 at 25˚C. Calculate the solubility (in mol
and 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 .
Plan Your Strategy
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.
Step 5 Determine the solubility in g/L.
Act on Your Strategy
Zn(IO3)2(s) Zn2+(aq) + 2IO3−(aq)
2+
Ksp = [Zn ][IO3−]2
Step 1
Step 2
Step 3
Concentration (mol/L)
Initial
Change
Equilibrium
Step 4
Ksp =
=
=
3.9 ×
Zn(IO3)2(s)
Zn2+(aq)
+
2IO3−(aq)
0
0
+x
+2x
x
2x
[Zn2+][IO3−]2
(x) × (2x)2
4x 3
10−6 = 4x 3
Chapter 9 Aqueous Solutions and Solubility Equilibria • MHR
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CHEMISTRY 12
3.9 × 10−6
4
= 9.9 × 10−3 mol/L
The 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/mol
9.9 × 10−3 mol/L = (9.9 × 10−3mol/L) × (415.2 g/mol)
= 4.1 g/L
The solubility of Zn(IO3)2 in water is 4.1 g/L.
x =
3
Check Your Solution
Recall 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. Problem
What is the maximum number of formula units of zinc sulfide, ZnS, that can dissolve
in 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.
Plan Your Strategy
Write the dissolution equilibrium equation.
Use the equilibrium equation to write an expression for Ksp.
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.
Step 5 Determine the number of formula units dissolved in 1.0 L.
Step 1
Step 2
Step 3
Act on Your Strategy
ZnS(s) Zn2+(aq) + S2−(aq)
Ksp = [Zn2+][S2−]
Step 1
Step 2
Step 3
Concentration (mol/L)
Initial
Change
Equilibrium
ZnS(s)
Zn2+(aq)
+
S2−(aq)
0
0
+x
x
x
x
Ksp = [Zn2+][S2−] = x 2
Therefore, 2.0 × 10−22 = x 2
√
x = 2.0 × 10−22
= 1.4 × 10−11 mol/L
The 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.
Step 4
Check Your Solution
Recall that x = [Zn2+]eq = [S2−]eq . Substitute these values into the Ksp equation.
You should get the given Ksp.
Chapter 9 Aqueous Solutions and Solubility Equilibria • MHR
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CHEMISTRY 12
Solutions for Practice Problems
Student Textbook page 439
21. Problem
Determine 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 solution
of NaCl.
What Is Given?
From Appendix E, Ksp for AgCl = 1.77 × 10−10 . You know the concentration of the
solution of NaCl.
Plan Your Strategy
Write the dissolution equilibrium equation.
Use the equilibrium equation to write an expression for Ksp.
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) 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.
(a) Step 1
Step 2
Step 3
Act on Your Strategy
AgCl(s) Ag+(aq) + Cl−(aq)
Ksp = [Ag+][Cl−] = 1.77 × 10−10
(a) Step 1
Step 2
Step 3
Concentration (mol/L)
AgCl(s)
Initial
Change
Equilibrium
Step 4
(b) Step 1
+
Cl−(aq)
0
0
+x
+x
x
x
Ksp = [Ag+][Cl−]
1.77 × 10−10 = x 2
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.
Concentration (mol/L)
Initial
Change
Equilibrium
Step 2
Ag+(aq)
AgCl(s)
Ag+(aq)
+
Cl−(aq)
0
0.15
+x
+x
x
0.15 + x
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 is
more than 500 times greater than Ksp, as shown on the next page.
Chapter 9 Aqueous Solutions and Solubility Equilibria • MHR
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CHEMISTRY 12
0.15
0.15 =
Ksp
1.77 × 10−10
= 8.5 × 108 > 500
Therefore, 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 simplify
the 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/L
The solubility of AgCl in a solution of 0.15 mol/L NaCl is
1.18 × 10−9 mol/L.
Check Your Solution
Your approximation in Step 6 was reasonable. The solubility x is much smaller than
0.15. Le Châtelier’s principle predicts a smaller solubility in a solution containing a
common ion, and this is confirmed by the calculations.
22. Problem
Determine 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 the
solution of NaI.
Plan Your Strategy
Write the dissolution equilibrium equation.
Use the equilibrium equation to write an expression for Ksp.
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 is
contributed by NaI.
Step 4 Substitute your expressions into the expression for Ksp, and solve for x.
Step 1
Step 2
Step 3
Act on Your Strategy
Pb2+(aq) + 2I−(aq)
Step 1 PbI2(s) Step 2 Ksp = [Pb2+][I−]2
Step 3
Concentration (mol/L)
PbI2(s)
Initial
Change
Equilibrium
Step 4
Pb2+(aq)
+
2I−(aq)
0
0.050
+x
+2x
x
0.050 + 2x
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 is
more than 500 times greater than Ksp, as shown.
0.050 = 0.050
Ksp
9.8 × 10−9
= 5.1 × 106 > 500
Chapter 9 Aqueous Solutions and Solubility Equilibria • MHR
175
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 simplify
the 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/L
The solubility of PbI2 in a solution of 0.050 mol/L NaI is 3.9 × 10−6 mol/L.
Check Your Solution
Your approximation in Step 4 was reasonable. The solubility 2x is much smaller than
0.050. You can substitute your calculated values of equilibrium concentrations into the
expression you wrote for Ksp. The calculated value should equal the given value for Ksp,
within the errors introduced by mathematical rounding.
23. Problem
Calculate 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 solution
of Na2SO4.
What Is Given?
From Appendix E, Ksp for CaSO4 = 4.93 × 10−5 . You know the concentration of the
solution of Na2SO4 .
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.
Set up an ICE table. The initial concentrations are based on the solution
of Na2SO4 . Let x represent the concentration of sulfate (the common ion)
that is contributed by Na2SO4 .
Step 2 Solve for x.
(b) Step 1
Act on Your Strategy
Ca2+(aq) + SO42−(aq)
(a) Step 1 CaSO4(s) Step 2 Ksp = [Ca2+][SO42−]
= 4.93 × 10−5
Step 3
Concentration (mol/L)
CaSO4(s)
Initial
Change
Equilibrium
Step 4
Ca2+(aq)
+
SO42−(aq)
0
0
+x
+x
x
x
Ksp = [Ca+][SO42−] = x 2
4.93 × 10−5 = x 2
x = 4.93 × 10−5
x = ±7.02 × 10−3
Chapter 9 Aqueous Solutions and Solubility Equilibria • MHR
176
CHEMISTRY 12
The negative root has no physical meaning. The solubility of CaSO4 is
7.02 × 10−3 mol/L.
(b) Step 1
Concentration (mol/L)
Initial
Change
Equilibrium
Step 2
CaSO4(s)
Ca2+(aq)
+
SO42−(aq)
0
0.25
+x
+x
x
0.25 + x
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 0.25
is more than 500 times greater than Ksp, as shown below.
0.25
0.25 =
Ksp
4.93 × 10−5
= 5.1 × 103 > 500
Therefore, 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 simplify
the 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/L
The solubility of AgCl in a solution of 0.15 mol/L NaCl is
1.97 × 10−4 mol/L.
Check Your Solution
Your approximation in Step 6 was reasonable. The solubility x is a much smaller than
0.25. Le Châtelier’s principle predicts a smaller solubility in a solution containing a
common ion, and this is confirmed by the calculations.
24. Problem
Ksp 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 a
solution of CaCl2 .
What Is Given?
You know that Ksp for PbCl2 is 1.6 × 10−5 , and you know the concentration of the
solution of CaCl2 .
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) 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.
Chapter 9 Aqueous Solutions and Solubility Equilibria • MHR
177
CHEMISTRY 12
Act on Your Strategy
Pb2+(aq) + 2Cl−(aq)
(a) Step 1 PbCl2(s) Step 2 Ksp = [Pb2+][Cl−]2
= 1.6 × 10−5
Step 3
Concentration (mol/L)
PbCl2(s)
Initial
Change
Equilibrium
Pb2+(aq)
+
2Cl−(aq)
0
0
+x
+2x
x
2x
Ksp = [Pb2+][Cl−]2
1.6 × 10−5 = x(2x)2
1.6 × 10−5 = 4x 3
√
3
x = 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 4
Concentration (mol/L)
Initial
Change
Equilibrium
Step 2
PbCl2(s)
Pb2+(aq)
+
2Cl−(aq)
0
0.20
+x
+2x
x
0.20 + 2x
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 is
more than 500 times greater than Ksp, as shown below.
0.20
0.20 =
Ksp
1.6 × 10−5
= 1.2 × 104 > 500
Therefore, 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 simplify
the 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/L
The solubility of PbCl2 in a solution of 0.10 mol/L CaCl2 is
4.0 × 10−4 mol/L.
Check Your Solution
Your approximation in Step 6 was reasonable. The solubility 2x is a much smaller than
0.20. Le Châtelier’s principle predicts a smaller solubility in a solution containing a
common ion, and this is confirmed by the calculations.
Chapter 9 Aqueous Solutions and Solubility Equilibria • MHR
178
CHEMISTRY 12
Solutions for Practice Problems
Student Textbook Page 441
25. Problem
A buffer solution is made by mixing 250 mL of 0.200 mol/L aqueous ammonia and
400 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 ammonium
chloride are mixed. From Appendix E, Kb for NH3(aq) is 1.8 × 10−5 .
Plan Your Strategy
Step 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 the
value of x.
Step 5 Use the following equations:
pOH = −log[OH−]
pH = 14.00 = pOH
Act on Your Strategy
Step 1 When the solutions are mixed, the total volume is:
(250 mL + 400 mL) = 650 mL
[NH3] = 0.250 L × 0.200 mol/L
0.650 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
Concentration (mol/L)
Initial
NH3(aq)
7.69 × 10−2
NH4+(aq) +
H2O()
−2
7.69 × 10 − x
OH−(aq)
9.23 × 10−2
~0
+x
+x
9.23 × 10 + x
x
−x
Change
Equilibrium
+
−2
+
−
Kb = [NH4 ][OH ]
[NH3]
−2
= (9.23 × 10 −2+ x)(x)
(7.69 × 10 − x)
Step 4 Assume that (7.69 × 10−2 − x) ≈ 7.69 × 10−2 and
(9.23 × 10−2 + x) ≈ 9.23 × 10−2 .
−2
)(x) = 1.8 × 10−5
Kb = (9.23 × 10 −2
(7.69 × 10 )
Therefore, x = 1.5 × 10−5 mol/L = [OH−]
Step 3
Chapter 9 Aqueous Solutions and Solubility Equilibria • MHR
179
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 Solution
The value of x (1.5 × 10−5) is negligible compared with the initial concentration of
each component. A buffer that is made using a weak base and its conjugate acid
should have a pH that is greater than 7.
26. Problem
A buffer solution is made by mixing 200 mL of 0.200 mol/L aqueous ammonia and
450 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?
200 mL of 0.200 mol/L aqueous ammonia and 450 mL of 0.150 mol/L ammonium
chloride are mixed. From Appendix E, Kb for NH3(aq) is 1.8 × 10−5 .
Plan Your Strategy
Step 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 the
value of x.
Step 5 Use the following equations:
pOH = −log[OH−]
pH = 14.00 − pOH
Act on Your Strategy
Step 1 When the solutions are mixed, the total volume is (250 + 400) = 650 mL .
[NH3] = 0.200 L × 0.0200 mol/L
0.650 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
Concentration (mol/L)
Initial
NH3(aq)
6.15 × 10−2
Step 3
H2O()
NH4+(aq) +
−2
6.15 × 10 − x
OH−(aq)
1.04 × 10−1
~0
+x
+x
1.04 × 10 + x
x
−x
Change
Equilibrium
+
−1
+
−
Kb = [NH4 ][OH ]
[NH3]
−1
= (1.04 × 10 −2+ x)(x)
(6.15 × 10 − x)
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Assume that (6.15 × 10−2 − x) ≈ 6.15 × 10−2 and
(1.04 × 10−1 + x) ≈ 1.04 × 10−1 .
−1
Kb = (1.04 × 10 +−2x)(x) = 1.8 × 10−5
(6.15 × 10 )
Therefore, x = 1.1 × 10−5 mol/L = [OH−]
Step 5 pOH = −log(1.1 × 10−5)
= 4.96
pH = 14.00 − 4.96
= 9.04
The pH of the buffer solution is 9.04.
Step 4
Check Your Solution
The 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 acid
should have a pH that is greater than 7.
27. Problem
A buffer solution contains 0.200 mol/L nitrous acid, HNO2(aq), and 0.140 mol/L
potassium nitrite, KNO2(aq) . What is the pH of the buffer solution?
What Is Required?
You need to determine 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 Strategy
Step 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
concentrations. Check the validity of this assumption when you find the
value of x.
Step 4 Use the equation pH = −log[H3O+].
Act on Your Strategy
HNO2 + H2O() H3O+(aq) + NO2−(aq)
Step 1
NO2−(aq) + H3O+(aq)
Concentration (mol/L)
HNO2(aq) +
Initial
0.200
0.140
~0
−x
+x
+x
0.200 − x
0.140 + x
x
Change
Equilibrium
H2O()
[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) = 5.6 × 10−4
(0.200)
Step 2
Ka =
∴x = 8.0 × 10−4 mol/L
Step 4 pH = −log(8.00 × 10−4)
= 3.10
The pH of the buffer solution is 3.10.
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CHEMISTRY 12
Check Your Solution
The value of x (8.0 × 10−4) is negligible compared with the initial concentration of
each component. A buffer that is made using a weak acid and its conjugate base
should have a pH that is less than 7.
28. Problem
A buffer solution is prepared by dissolving 1.80 g of benzoic acid, C6H5COOH, and
1.95 g of sodium benzoate, NaC6H5COO, in 800 mL of water. Calculate the pH of
the buffer solution.
What Is Required?
You need to determine the pH of the 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 Strategy
Step 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
concentrations. Check the validity of this assumption when you find the
value of x.
Step 5 Use the following equation:
pH = −log[H3O+]
Act on Your Strategy
Step 1 The molar mass of C6H5COOH is 122.1 g/mol.
1.80 g
[C6H5COOH] =
122.1 g/mol
= 1.47 × 10−2 mol
The molar mass of NaC6H5COO is 144.1 g/mol.
1.95 g
[NaC6H5COO] =
144.1 g/mol
= 1.35 × 10−2 mol
Step 2
Concentration (mol/L) C6H5COOH(aq) +
Initial
Change
Equilibrium
H2O()
C6H5COOH−(aq) +
H3O+(aq)
1.47 × 10−2
1.35 × 10−2
~0
−x
+x
+x
1.47 × 10−2 − x
1.35 × 10−2 + x
x
[C6H5COO−][H3O3]
[C6H5COOH]
−2
= (1.35 × 10 −2+ x)(x)
(1.47 × 10 − x)
Step 4 Assume that (1.35 × 10−2 + x) ≈ 1.35 × 10−2 and
(1.47 × 10−2 − x) ≈ 1.47 × 10−2 .
−2
Ka = (1.35 × 10 +−2x)(x) = 6.3 × 10−5
(1.47 × 10 )
∴x = 6.9 × 10−5mol/L
Step 3
Ka =
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CHEMISTRY 12
Step 5
pH = −log(6.9 × 10−5)
= 4.16
The pH of the buffer solution is 4.16.
Check Your Solution
The value of x (6.9 × 10−5) is negligible compared with the initial concentration of
each component. A buffer that is made using a weak acid and its conjugate base
should have a pH that is less than 7.
Solutions for Practice Problems
Student Textbook page 446
29. Problem
A 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 .
Solution
Write 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 Solution
It seems reasonable that a precipitate formed, since Ksp for lead(II) chloride is small
compared with the concentration of chloride ions and lead ions.
30. Problem
One 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 Strategy
Step 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.
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CHEMISTRY 12
Act on Your Strategy
Step 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 other
words, assume that the final volume of the reaction mixture is 0.250 L.
AgNO3(aq) → Ag+(aq) + NO3−(aq)
∴[Ag+] = [AgNO3] = 0.10 mol/L
K2CrO4(aq) → 2K+(aq) + CrO42−(aq)
[CrO42−] = [K2CrO4] = c × Vinitial
Vfinal
−3
= (1.5 mol/L)(0.050 mL × 1.0 × 10 L/mL)
0.250 L
−4
= 3.0 × 10 mol/L
Step 2 The possible precipitate is silver chromate.
2Ag+(aq) + CrO42− → 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 Solution
The units are correct in the calculation of the concentration of silver ions. It seems reasonable that a precipitate formed, since Ksp for silver chromate is very small compared
with the concentration of chromate ions and silver ions.
31. Problem
A 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 chemical
equation 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 Strategy
Step 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.
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CHEMISTRY 12
Act on Your Strategy
Step 1 Na2CO3(aq) → 2Na+(aq) + CO32−(aq)
∴[CO32−] = [Na2CO3] = 0.0015 mol/L
The molar mass of CaCl2 is 111.0 g/mol. Therefore,
0.010 g
Amount CaCl2 =
111.0 g/mol
= 9.0 × 10−5 mol
−5
[CaCl2] = 9.0 × 10 mol
0.50 L
= 1.8 × 10−4 mol/L
CaCl2(aq) → Ca2+(aq) + 2Cl−(aq)
Therefore, [Ca2+] = [CaCl2] = 1.8 × 10−4 mol/L
Step 2 The possible precipitate is calcium carbonate.
Ca2+(aq) + CO32− → 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 Solution
The units are correct in the calculation of the concentration of silver ions. It seems
reasonable that a precipitate formed, since Ksp for calcium carbonate is small
compared with the concentration of calcium ions and carbonate ions.
32. Problem
0.10 mg of magnesium chloride, MgCl2 , is added to 2.5 × 102 mL of 0.0010 mol/L
NaOH. Does a precipitate of magnesium hydroxide form? Include a balanced chemical
equation 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 magnesium
hydroxide is 5.61 × 10−12 .
Plan Your Strategy
Step 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.
Chapter 9 Aqueous Solutions and Solubility Equilibria • MHR
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CHEMISTRY 12
Act on Your Strategy
Step 1 NaOH(aq) → Na+(aq) + OH−(aq)
Therefore, [OH−] = [NaOH] = 0.0010 mol/L
The molar mass of MgCl2 is 95.2 g/mol. Therefore,
0.10 mg × 10−3 g/mg
Amount MgCl2 =
95.2 g/mol
= 1.1 × 10−6 mol
−6
[MgCl2] = 1.1 × 10 mol
0.25 L
= 4.2 × 10−6 mol/L
MgCl2(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 Solution
The units are correct in the calculation of the concentration of silver ions. It seems
reasonable that no precipitate formed, since the concentration of magnesium ion
is small.
Solutions for Practice Problems
Student Textbook page 447
33. Problem
1.0 × 102 mL of 1.0 × 10−3 mol/L Pb(NO3)2 is added to 40 mL of 0.040 mol/L
NaCl. 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 sodium
chloride 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 Strategy
Step 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.
Chapter 9 Aqueous Solutions and Solubility Equilibria • MHR
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CHEMISTRY 12
Act on Your Strategy
Step 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 is
not, according to guideline 2.
PbCl2(s) + NaNO3(aq)
Step 2 Pb(NO3)2(aq) + NaCl(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 × Vinitial
Vfinal
−3
= (1.0 × 10 mol/L)(100 mL)
(100 mL + 40 mL)
= 7.1 × 10−4 mol/L
[Cl−] = [NaCl] = c × Vinitial
Vfinal
(0.040
mol/L)(40
mL)
=
(100 mL + 40 mL)
= 1.1 × 10−2 mol
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.
Step 4
Check Your Solution
The units in the calculation of the concentrations of the ions are correct. It seems reasonable that no precipitate forms, since the concentrations of ions are relatively small.
34. Problem
2.3 × 102 mL of 0.0015 mol/L AgNO3 is added to 1.3 × 102 mL of 0.010 mol/L
calcium acetate, Ca(CH3COO)2 . Does a precipitate form? Include a balanced
chemical equation for the formation of the possible precipitate. Ksp for AgCH3COO
is 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 calcium
acetate solutions. For AgNO3 , c = 0.0015 mol/L and V = 230 mL. For
Ca(CH3COO)2 , c = 0.010 mol/L and V = 130 mL. As well, you have the solubility
guidelines in Table 9.3.
Plan Your Strategy
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.
Step 1
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CHEMISTRY 12
Act on Your Strategy
2AgCH3COO(s) + Ca(NO3)2(aq)
Step 1 2AgNO3(aq) + Ca(CH3COO)2(aq) Step 2 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.
[Ag+] = [AgNO3] = c × Vinitial
Vfinal
(0.0015
mol/L)(230
mL)
=
(230 mL + 130 mL)
= 9.6 × 10−4 mol/L
[CH3COO−] = 2 × [Ca(CH3COO)2] = 2 × c × Vinitial
Vfinal
(0.010
mol/L)(130
mL)
=2×
(130 mL + 230 mL)
= 7.2 × 10−3 mol/L
Qsp = [Ag+][CH3COO−]
= (9.6 × 10−4)(7.2 × 10−3)
= 6.9 × 10−6
Step 4 Since Qsp < Ksp , no precipitate is formed.
Step 3
Check Your Solution
The 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 large
value compared with the concentrations of the silver ions and acetate ions.
35. Problem
25 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 for
the formation of the possible precipitate.
What Is Given?
You know the initial concentration and volume of the sodium hydroxide and
cobalt(II) chloride solutions. For NaOH, c = 0.10 mol/L and V = 25 mL. For
CoCl2 , c = 0.00010 mol/L and V = 500 mL. As well, you have the solubility
guidelines in Table 9.3.
Plan Your Strategy
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
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.
Step 1
Act on Your Strategy
Step 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 .
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CHEMISTRY 12
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.
[Co2+] = [CoCl2] = c × Vinitial
Vfinal
= (0.00010 mol/L)(500 mL)
(500 mL + 25 mL)
= 9.5 × 10−5 mol/L
[OH−] = [NaOH] = c × Vinitial
Vfinal
= (0.10 mol/L)(25 mL)
(25 mL + 500 mL)
= 4.8 × 10−3 mol/L
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 .
Step 4
Check Your Solution
The 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. Problem
250 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 barium
chloride 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 in
Table 9.3.
Plan Your Strategy
Step 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
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.
Act on Your Strategy
Step 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 .
Chapter 9 Aqueous Solutions and Solubility Equilibria • MHR
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CHEMISTRY 12
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.
[Ba2+] = [BaCl2] = c × Vinitial
Vfinal
= (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
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 .
Step 4
Check Your Solution
The 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.
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