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 = 3.45×1018. (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.)

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

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

Explanation / Answer

1.

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

   y y 2y

Ksp = 3.45 x 10^-18

3.45 x 10^-18 = [M2+][OH-]^2

3.45 x 10^-18 = (y)(2y)^2

y = 9.51 x 10^-7 M

Molar solubility = 9.51 x 10^-7 M

2.

Due to addition of M(NO3)2 there will be common ion effect and [M^2+] ion increases

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

   y                y + 0.202       2y

3.45 x 10^-18 = ( y + 0.202)(2y)^2

3.45 x 10^-18 = 4y^3 + 0.808y^2

4y^3 + 0.808y^2 - 3.45 x 10^-18 = 0

y = 2.6 x 10^-9 M

Solubility = 2.6 x 10^-9 M

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