Question

Consider the dissolution of AB(s):

AB(s)⇌A+(aq)+B−(aq)

The generic metal hydroxide M(OH)2 has Ksp = 6.85×10⁻¹². (NOTE: In this particular problem, because of the magnitude of the Ksp and the stoichiometry of the compound, the contribution of OH− from water can be ignored. However, this may not always be the case.)

1. Le Châtelier's principle tells us that an increase in either [A+] or [B−] will shift this equilibrium to the left, reducing the solubility of AB. In other words, AB is more soluble in pure water than in a solution that already contains A+ or B− ions. This is an example of the common-ion effect.

What is the solubility of M(OH)2 in a 0.202 M solution of M(NO3)2?

Answer 1

0.202 M solution of M(NO₃)₂ will give 0.202 M of M²⁺ in the solution.

M(OH)₂(s) $\rightleftharpoons$ M²⁺(aq) + 2 OH^-(aq)

Initial concentration (M)                                                          0.202        0

Change in concentration (M)                                                    S            2S

Equilibrium concentration (M)                                          (0.202 + S)     2S

Given, K_sp = 6.85×10⁻¹²

Now,

K_sp = [M²⁺][OH^-]²

or, 6.85 x 10^-12 = (0.202 + S)(2S)²

or, 6.85 x 10^-12 = (0.202)(2S)²       [(0.202 + S) $\approx$ 0.202 as S<<0.202]

or, 0.808S² = 6.85 x 10^-12

or, S² = 8.48 x 10^-12

or, S = 2.91 x 10^-6 M

Hence, the solubility of M(OH)₂ in a 0.202 M solution of M(NO₃)₂ = 2.91 x 10^-6 M