Question

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.)

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)

At equilibrium:

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

   s 2s

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

6.85*10^-12=(s)*(2s)^2

6.85*10^-12= 4(s)^3

s = 1.196*10^-4 M

Answer: 1.20*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

6.85*10^-12=(0.202 + s)*(2s)^2

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

Above expression thus becomes:

6.85*10^-12=(0.202)*(2s)^2

6.85*10^-12= 0.202 * 4(s)^2

s = 2.912*10^-6 M

Answer: 2.91*10^-6 M