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 = 1.05×1018. (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.)

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

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

[M2+]= S , [OH-] = 2S ,          where S= solubility or amount fo M(OH)2 dissociated

Ksp = [M2+][OH-]^2 = ( S) ( 2S)^2 = 4S^3 = 1.05 x 10^-18

S = solubility of M(OH)2 = 6.403 x 10^-7 M

B) when we have 0.202 M(NO3)2   we have 0.202 M2+ since M(NO3)2 is strong electrolyte and dissociates completly to give M2+ and 2NO3-

now Ksp = [M2+][OH-]^2              now M2+ are from two sources oen is M(OH)2 and other is M(NO3)2

1.05 x 10^-18 = ( 0.202+S ) x ( 2S)^2                   ( now 0.202+S =0.202 since we get S is negligible value)

1.05 x 10^-18 = ( 0.202) ( 4S^2)

S = 1.13 x 10^-9 M is solubility

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