Question

The generic metal hydroxide M(OH)2 has Ksp = 8.05×10−12. (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.)

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

Answer 1

a. in pure water Solubility (x) = 1.26 x 10⁻⁴M

b. in 0.202M M⁺² Solubility (x) = 9.963 x 10⁻¹²M

The large drop in solubility is consistent with the common ion effect.

Step-by-step explanation:

a. Solubility in pure water

Given: M(OH)₂ ⇄ M⁺² + 2OH⁻

I --- 0 0

C --- x 2x

E --- x 2x

Ksp = [M⁺²][OH⁻]² = (x)(2x)² = 4x³ => x = CubeRt(Ksp/4)

solubility in pure water = x = CubeRt(8.05 x 10⁻¹²/4) = 1.26 x 10⁻⁴M

b. Solubility in presence of 0.202M M⁺² as common ion.

Given: M(OH)₂ ⇄ M⁺² + 2OH⁻

I --- 0.202M 0

C --- +x +2x

E --- 0.202M + x 2x

≈ 0.202M

Ksp = [M⁺²][2x]² = (0.202)(2x)² = (0.202)(4x²) = 8.05 x 10⁻¹²

=> x = (8.05 x 10⁻¹²)/(0.202)(4) = 9.963 x 10⁻¹²M