CH 223 Chapter Eighteen Part II Concept Guide
1. Solubility Product
Problem
The solubility of silver iodide, AgI, at 25 °C is 2.1 x 10-6 g/L. Calculate the solubility product of this salt from
its solubility.
Approach
First, write the chemical equation, letting x represent the molar solubility. Then, write the equilibrium constant
expression. Third, calculate the molar solubility from the solubility of AgI. Last, calculate the solubility product
of AgI from the molar solubility calculated earlier.
Solution
Step 1. The chemical equation is:
Step 2. The equilibrium constant expression is:
Ksp = [Ag+][I-] = (x)(x) = x2
Step 3. Calculate the molar solubility from the solubility of AgI.
Step 4. Calculate the solubility product of AgI from the molar solubility.
Ksp = x2 = (8.9 x 10-9)2 = 7.9 x 10-17
2. Solubility Product
Problem
The solubility product of silver sulfate, Ag2SO4 is 1.5 x 10-5 at 25 °C. Calculate the solubility of this salt from
its solubility product. Report the solubility in moles per liter and in grams per liter.
Approach
First, write the chemical equation, letting x represent the molar solubility. Then, write the equilibrium constant
expression. Finally, solve the Ksp expression to find the solubility of Ag2SO4.
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Solution
Step 1. The chemical equation is:
Step 2. The equilibrium constant expression is:
Ksp = [Ag+]2[SO42-] = (2x)2(x) = 4x3 = 1.5 x 10-5
Step 3. Solve the Ksp expression to find the solubility of Ag2SO4.
Ksp = [Ag+]2[SO42-] = (2x)2(x) = 4x3 = 1.5 x 10-5
x = 0.016 mol/L or 5.0 g/L
3. Solubility Product
Problem
The solubility product of AlPO4 is 5.8 x 10-19 at 25 °C. Calculate the solubility of this salt from its solubility
product in moles per liter.
Approach
First, write the chemical equation, letting x represent the molar solubility. Then, write the equilibrium constant
expression. Finally, solve the Ksp expression to find the solubility of AlPO4.
Solution
Step 1. The chemical equation is:
Step 2. The equilibrium constant expression is:
Ksp = [Al3+][PO43-] = (x)(x) = x2 = 5.8 x 10-19
Step 3. Solve the Ksp expression to find the solubility of AlPO4.
Ksp = [Ag3+][PO43-] = (x)(x) = x2 = 5.8 x 10-19
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4. Ranking Solubilities
Question
Given the following information, which has a greater solubility: CuBr or CuCl?
CuBr Ksp at 25 °C = 5.2 x 10-9
CuCl Ksp at 25 °C = 1.2 x 10-6
Solution
The stoichiometries for these two salts are the same: 1:1. Thus, we can compare their Ksp values to determine
which salt is more soluble. The larger the value of Ksp, the more soluble the salt. CuCl has, therefore, a greater
solubility than CuBr.
5. Precipitation Reactions
Question
The concentration of chromate ion in an aqueous solution is 0.012 mol/L.
(a) Will the precipitation of silver chromate take place if a concentration of silver ion of 1 x 10-6 mol/L is
introduced by adding solid silver nitrate to the solution?
(b) If the answer to (a) is "no," what concentration of silver ion would be necessary to begin precipitation? The
Ksp value of Ag2CrO4 is 2.5 x 10-12.
Ag2CrO4(s)
2 Ag+(aq) + CrO42-(aq)
Solution
(a) The ion product for the reaction of Ag2CrO4 is:
Q = [Ag+]2[CrO42-] = (1 x 10-6 mol/L)2(0.012 mol/L) = 1 x 10-14
Q < Ksp
Because Q is less than Ksp, the forward reaction is favored and no precipitate forms.
(b) Precipitation will begin when Q = Ksp. The [Ag+] at which this occurs is calculated by substituting [CrO42-] =
0.012 mol/L into the Ksp expression and solving for [Ag+].
Ksp = [Ag+]2[CrO42-] = 2.5 x 10-12
Ksp = [Ag+]2[0.012 mol/L] = 2.5 x 10-12
[Ag+] = 1.4 x 10-5 mol/L
For precipitation of silver chromate to begin, [Ag+] must be 1.4 x 10-5 mol/L.
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6. Precipitation Reactions
Question
Will a precipitate of Mg(OH)2 form if 10. mL of 0.010 M NaOH is added to 1.000 L of 0.015 mol/L MgCl2?
Assume that the volume of resulting solution is 1.015 L. The Ksp value of Mg(OH)2 is 7.1 x 10-12.
Solution
The concentrations of the ions in the resulting solution are:
The ion product is:
Q = [Mg2+][OH-]2 = (1.5 x 10-2)(9.9 x 10-5)2 = 1.5 x 10-10
Q > Ksp
Because Q is greater than Ksp, a precipitate of Mg(OH)2 will form.
7. Common Ion Effect
Problem
The solubility constant of Ag2SO4 is 1.5 x 10-5. Calculate the solubility of this salt in an aqueous solution that
contains 0.25 mol/L of Na2SO4. Is there a common ion effect seen here? Assume that none of the ions reacts
appreciably with water for form H+ or OH-.
Approach
First, write the chemical equation, letting x represent the molar solubility. Then, write the equilibrium constant
expression to calculate the solubility of the salt in pure water from the solubility product. Then, calculate the
molar solubility in Na2SO4 by solving for x. Last, comment on the calculated molar solubility in Na2SO4
compared to the solubility of the salt.
Solution
Step 1. The chemical equation is:
Step 2. The equilibrium constant expression for the salt in pure water is:
Ksp = [Ag+]2[SO42-] = (2x)2(x) = 4x3 = 1.5 x 10-5
Solve the Ksp expression to find the solubility of Ag2SO4 in pure water.
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Ksp = [Ag+]2[SO42-] = (2x)2(x) = 4x3 = 1.5 x 10-5
Step 3. In this example, there are two sources of SO42- in the solution: the dissolved Ag2SO4 and the initial
Na2SO4 in the solution. The equilibrium constant expression becomes:
Ksp = [Ag+]2[SO42-] = (2x)2(0.25 + x) = 1.5 x 10-5
Step 4. The Ksp is small relative to the 0.25 mol/L initial SO42-, therefore, we can assume that (0.25 + x) is
approximately 0.25. Solving for x,
The approximation is valid. The molar solubility is 0.0039 mol/L, thus the common ion effect has reduced the
solubility from 0.016 mol/L for Ag2SO4 in pure water to 0.0039 mol/L in the Na2SO4 solution.
8. Solubility and pH
Question
A 0.015 mol sample of solid Fe(OH)3 was added to 1.0 L of water, and a strong acid was added until the
Fe(OH)3 precipitate dissolved. At what pH was all of the Fe(OH)3(s) dissolved? Assume negligible volume
change due to the addition of the acid. The Ksp for Fe(OH)3 is 3 x 10-39.
Fe(OH)3(s)
Fe3+(aq) + 3 OH-(aq)
Solution
When all of the Fe(OH)3 dissolved, 0.015 mol of Fe3+ was present in 1.0 L of solution. The concentration of OHat the point when dissolution was complete may be found from the solubility product expression for Fe(OH)3.
Ksp = [Fe3+][OH-]3 = 3 x 10-39
To calculate the pH, use the formula for pOH, where pOH = - log[OH-]. Then, calculate the pH from pOH.
pOH = - log[OH-] = - log (6 x 10-13) = 12.2
pH = 14.0 - pOH = 14.0 - 12.2 = 1.8
When all of the solid Fe(OH)3 dissolved, the pH of the solution was 1.8.
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9. Solubility and pH
Question
A 0.020 mol sample of solid Fe(OH)2 was added to 1.0 L of water, and a strong acid was added until the
Fe(OH)2 precipitate dissolved. At what pH was all of the Fe(OH)2(s) dissolved? Assume negligible volume
change due to the addition of the acid. The Ksp for Fe(OH)2 is 8 x 10-16.
Fe(OH)2(s)
Fe2+(aq) + 2 OH-(aq)
Solution
When all of the Fe(OH)2 dissolved, 0.020 mol of Fe2+ was present in 1.0 L of solution. The concentration of
OH- at the point when dissolution was complete may be found from the solubility product expression for
Fe(OH)2.
Ksp = [Fe2+][OH-]2 = 8 x 10-16
To calculate the pH, use the formula for pOH, where pOH = - log[OH-]. Then, calculate the pH from pOH.
pOH = - log[OH-] = - log (2 x 10-7) = 6.7
pH = 14.0 - pOH = 14.0 - 6.7 = 7.3
When all of the solid Fe(OH)2 dissolved, the pH of the solution was 7.3.
10. Solubility and Complex Ion Formation
Question
Zn(OH)2 precipitates from an acidic aqueous solution by the addition of OH- ion to Zn2+ ion. There is the
danger, however, of redissolving the Zn(OH)2 by addition of too much OH- ion because of the formation of the
[Zn(OH)4]2- complex ion.
(a) What is the pH at which the concentration of Zn2+ falls below 1 x 10-4 mol/L due to precipitation of
Zn(OH)2?
(b) What is the pH at which the concentration of [Zn(OH)4]2- increases above 1 x 10-4 mol/L? The pH range
between these points is that in which almost all of the Zn2+ is present as a solid. The Ksp for Zn(OH)2 is 1.2 x
10-17 and the K of dissociation (Kd) for [Zn(OH)4]2- is 5 x 10-21.
Zn(OH)2(s)
Zn2+(aq) + 2 OH-(aq)
Ksp = [Zn2+] [OH-]2
Solution
(a) The formation of Zn(OH)2 becomes important as the pH increases. The OH- ion concentration in
equilibrium when the concentration of Zn2+ is 1 x 10-4 mol/L is
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[OH-] = 3 x 10-7 mol/L
To calculate the pH, use the formula for pOH, where pOH = - log[OH-]. Then, calculate the pH from pOH.
pOH = - log[OH-] = - log (3 x 10-7) = 6.5
pH = 14.0 - pOH = 14.0 - 6.5 = 7.5
The pH at which the concentration of Zn2+ falls below 1 x 10-4 mol/L due to precipitation of Zn(OH)2 is 7.5.
(b) The formation of Zn(OH)2 becomes even more important as the pH increases from neutral to alkaline
values. The solubility of Zn2+ may be described by the equilibrium between the precipitate and the complex ion.
The OH- ion concentration in equilibrium with the concentration of Zn(OH)42- being 1 x 10-4 mol/L is
[OH-] = 2 x 10-4 mol/L
To calculate the pH, use the formula for pOH, where pOH = - log[OH-]. Then, calculate the pH from pOH.
pOH = - log[OH-] = - log (2 x 10-4) = 3.7
pH = 14.0 - pOH = 14.0 - 3.7 = 10.3
The pH at which the concentration of [Zn(OH)4]2- increases above 1 x 10-4 mol/L is 10.3. Therefore, the
optimum precipitation of Zn2+ will result by controlling the pH of the solution between 7.5 and 10.3.
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