Question

The generic metal hydroxide M(OH)2 has Ksp = 9.05×10⁻¹². (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.)

Part A: What is the solubility of M(OH)2 in pure water? Express your answer with the appropriate units.

Part B: What is the solubility of M(OH)2 in a 0.202 M solution of M(NO3)2? Express your answer with the appropriate units.

Answer 1

A)

At equilibrium:

M(OH)2 ----> M2+ + 2 OH-

   s 2s

Ksp = [M2+][OH-]^2

9.05*10^-12=(s)*(2s)^2

9.05*10^-12= 4(s)^3

s = 1.31*10^-4 M

Answer: 1.31*10^-4 M

B)

M(NO3)2 here is Strong electrolyte

It will dissociate completely to give [M2+] = 0.202 M

At equilibrium:

M(OH)2 ----> M2+ + 2 OH-

   0.202 +s 2s

Ksp = [M2+][OH-]^2

9.05*10^-12=(0.202 + s)*(2s)^2

Since Ksp is small, s can be ignored as compared to 0.202

Above expression thus becomes:

9.05*10^-12=(0.202)*(2s)^2

9.05*10^-12= 0.202 * 4(s)^2

s = 3.35*10^-6 M

Answer: 3.35*10^-6 M