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CH 223 Chapter Fifteen 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) = x^2
Step 3. Calculate the molar solubility from the solubility of AgI.
Step 4. Calculate the solubility product of AgI from the molar solubility.
Ksp = x^2 = (8.9 x 10-9)^2 = 7.9 x 10-
2. Solubility Product
Problem
The solubility product of silver sulfate, Ag 2 SO 4 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 Ag 2 SO 4.
Solution
Step 1. The chemical equation is:
Step 2. The equilibrium constant expression is:
Ksp = [Ag+]^2 [SO 4 2-] = (2x)^2 (x) = 4x^3 = 1.5 x 10-
Step 3. Solve the Ksp expression to find the solubility of Ag 2 SO 4.
Ksp = [Ag+]^2 [SO 4 2-] = (2x)^2 (x) = 4x^3 = 1.5 x 10-
x = 0.016 mol/L or 5.0 g/L
3. Solubility Product
Problem
The solubility product of AlPO 4 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 AlPO 4.
Solution
Step 1. The chemical equation is:
Step 2. The equilibrium constant expression is:
Ksp = [Al3+][PO 4 3-] = (x)(x) = x^2 = 5.8 x 10-
Step 3. Solve the Ksp expression to find the solubility of AlPO 4.
Ksp = [Ag3+][PO 4 3-] = (x)(x) = x^2 = 5.8 x 10-
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 MgCl 2? 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-
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 Ag 2 SO 4 is 1.5 x 10-5. Calculate the solubility of this salt in an aqueous solution that contains 0.25 mol/L of Na 2 SO 4. 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 Na 2 SO 4 by solving for x. Last, comment on the calculated molar solubility in Na 2 SO 4 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 [SO 4 2-] = (2x)^2 (x) = 4x^3 = 1.5 x 10-
Solve the Ksp expression to find the solubility of Ag 2 SO 4 in pure water.
Ksp = [Ag+]^2 [SO 4 2-] = (2x)^2 (x) = 4x^3 = 1.5 x 10-
Step 3. In this example, there are two sources of SO 4 2-^ in the solution: the dissolved Ag 2 SO 4 and the initial Na 2 SO 4 in the solution. The equilibrium constant expression becomes:
Ksp = [Ag+]^2 [SO 4 2-] = (2x)^2 (0.25 + x) = 1.5 x 10-
Step 4. The Ksp is small relative to the 0.25 mol/L initial SO 4 2-, 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 Ag 2 SO 4 in pure water to 0.0039 mol/L in the Na 2 SO 4 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 OH- at 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-
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.
pH = 14.0 - pOH = 14.0 - 12.2 = 1.
When all of the solid Fe(OH) 3 dissolved, the pH of the solution was 1.8.
[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.
pH = 14.0 - pOH = 14.0 - 6.5 = 7.
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) 4 2-^ being 1 x 10-4^ mol/L is
Zn(OH) 4
2 −
1 x 10
− 4
[OH
− 1
]
2
= = = 4 x 10
− 8
KspKf (1.2 x 10
− 17
)(2 x 10
20
[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.
pH = 14.0 - pOH = 14.0 - 3.7 = 10.
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.
