Consider the dissolution of AB(s): AB(s)A+(aq)+B(aq) Le Châtelier\'s principle t

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 = 6.85×1012. (NOTE: In this particular problem, because of the magnitude of the Ksp and the stoichiometry of the compound, the contribution of OHfrom 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?

Explanation / Answer

Part A:-

Let the solubility of M(OH)2 be s moles/litre

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

s 2s

Ksp = [M2+][OH-]2

= s.(2s)2

=4s3

6.85*10-12 = 4s3

s3 = 1.71*10-12

s = (1.71*10-12)1/3

   = 1.195*10-4mol/litre

Solubility in pure water = 1.195*10-4mol/litre.

Part B:-

Solubility in 0.202M soln of M(NO3)2

Common ion here is M2+

So,

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

[M2+] = s+0.202

[OH-] = 2s

Ksp = [M2+][OH-]2

   = (s+0.202)(2s)2

s+0.202 is equivalent to 0.202 since s<<0.202

Ksp = 0.202*4s2

   = 0.808s2

6.85*10-12 = 0.808s2

s2= 8.56*10-12

s = (8.56*10-12)1/2

   = 2.93*10-6mol/L

Solubility in 0.202M M(NO3)2 = 2.93*10-6mol/L

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