Question

The generic metal hydroxide M(OH)2 has Ksp = 8.25×10−18. (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.)

Part A: What is the solubility of M(OH)2 in pure water?

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

Answer 1

Given: Ksp = 8.25 x 10^-18

Part A: M(OH)₂ M²⁺ + 2 OH^-

Ksp = [M²⁺] [OH^-]²

Ksp = (x) (2x)²

Ksp = 4x³ = 8.25 x 10^-18

(x = Solubility)

x = (8.25 x 10^-18 / 4)^1/3

x = 1.27 x 10^-6 M

Part B: In the presence of 0.202 M solution of M(NO3)2

As M(NO₃)2 is a strong electrolyte, it will dissociate fully

M(NO₃)₂ M²⁺ + 2 NO₃^-

Now [M²⁺] = x + 0.202 and OH- will remain same.

Ksp = (x + 0.202) (2x)² = 8.25 x 10^-18

We can neglect x in comparison to 0.202 as we know value of x is very small.

Ksp = 0.202 (4x²) = 0.808x² = 8.25 x 10^-18

x = ( 8.25 x 10^-18 / 0.808)^1/2

x = 3.19 x 10^-9 M