Question

The generic metal hydroxide M(OH)2 has Ksp = 4.45×10⁻¹². (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.)

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

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

Answer 1

1)

At equilibrium:

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

   s 2s

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

4.45*10^-12=(s)*(2s)^2

4.45*10^-12= 4(s)^3

s = 1.036*10^-4 M

Answer: 1.04*10^-4 M

2)

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

4.45*10^-12=(0.202 + s)*(2s)^2

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

Above expression thus becomes:

4.45*10^-12=(0.202)*(2s)^2

4.45*10^-12= 0.202 * 4(s)^2

s = 2.347*10^-6 M

Answer: 2.35*10^-6 M