Sample Problems and Answers - Lecture 11
1.
Identify the one correct statement concerning the solubility of Mg(OH)2.
(a)
(b)
(c)
(d)
(e)
pH has no effect on the solubility of Mg(OH)2.
Mg(OH)2 is less soluble at pH 10 than at pH 7.
Mg(OH)2 is more soluble in 0.1 M MgCl2 solution than in water.
Mg(OH)2 is less soluble at pH 4 than at pH 7.
The solubility product constant for Mg(OH)2 is greatest at pH 7.
The solubility equilibrium is:
Mg(OH)2(s)
Mg2+(aq) + 2OH-(aq)
The solubility product is:
Ksp = [Mg2+][OH-]2
As OH-(aq) is a product of the reaction, the equilibrium (and hence the
solubility) is very pH sensitive – answer (a) is incorrect.
Increasing the pH leads to an increase in [OH-]. The equilibrium shifts to the left
and the solubility decreases. ([Mg+][OH-]2 must be constant so if [OH-] is
increased, [Mg2+] must decrease). Answer (b) is correct.
If [Mg2+] is increased by adding MgCl2, the equilibrium shifts to the left,
precipitating Mg(OH)2. Answer (c) is incorrect.
Lowering the pH leads to a decrease in [OH-). The equilibrium shifts to the right
and the solubility increases. ([Mg+][OH-]2 must be constant so if [OH-] is
decreased, [Mg2+] must increase). Answers (d) and (e) are incorrect.
2.
The Ksp for Sc(OH)3 is 2 × 10–31. What is the solubility of Sc(OH)3 (in mol L-1) of a
solution buffered at 6.7?
The dissolution reaction is:
Sc(OH)3(s)
Sc3+(aq) + 3OH–(aq)
Hence, the solubility product is
Ksp = [Sc3+(aq)][OH-(aq)]3
As pH + pOH = 14.0,
pOH = 14.0 – 6.7 = 7.3
As pOH = -log[OH-(aq)], [OH-(aq)] = 10-7.3 M.
From above,
[Sc3+(aq)] = Ksp / [OH-(aq)]3 = (2 × 10-31) / (10-7.3)3 = 1.6 × 10-9.
As dissolution of 1 mol of Sc(OH)3 gives 1 mol of Sc3+(aq), this is also the
solubility:
solubility = 1.6 × 10-9 M