Question

Consider the dissolution of AB(s):

AB(s)⇌A+(aq)+B−(aq)

Le Châtelier's principle tells us that an increase in either [A+] or [B−] will shift this equilibrium to the left, reducing the solubility of AB. In other words, AB is more soluble in pure water than in a solution that already contains A+ or B− ions. This is an example of the common-ion effect.

The generic metal hydroxide M(OH)2 has Ksp = 1.05×10⁻¹⁸. (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

A)

At equilibrium:

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

   s 2s

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

1.05*10^-18=(s)*(2s)^2

1.05*10^-18= 4(s)^3

s = 6.403*10^-7 M

Answer: 6.40*10^-7 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

1.05*10^-18=(0.202 + s)*(2s)^2

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

Above expression thus becomes:

1.05*10^-18=(0.202)*(2s)^2

1.05*10^-18= 0.202 * 4(s)^2

s = 1.14*10^-9 M

Answer: 1.14*10^-9 M