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?
0.202 M solution of M(NO_{3})_{2} will give 0.202 M of M^{2+} in the solution.
M(OH)_{2}(s) M^{2+}(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} = [M^{2+}][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) 0.202 as S<<0.202]
or, 0.808S^{2} = 6.85 x 10^{-12}
or, S^{2} = 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(NO_{3})_{2} = 2.91 x 10^{-6} M
Consider the dissolution of AB(s): AB(s)⇌A+(aq)+B−(aq) The generic metal hydroxide M(OH)2 has Ksp...