Consider the dissolution of AB(s): AB(s) A^+(aq) + B^- (aq) Le Chareller\'s tell

Question

Consider the dissolution of AB(s): AB(s) A^+(aq) + B^- (aq) Le Chareller's 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 stready contains A^+ or B^- ions. This is an example of the common-ion effect. The generic metal hydroxide M(OH)_2 has K_ap = 6.65times10^-18. What is the soubility M(OH)_2 in pure water? Express your answer with the appropriate units. What is the solubility of M(OH)_2 in a 0.202 M solution of M(NO_3)_2? Express your answer with the appropriate units.

Explanation / Answer

part A) The equilibrium will be :
...........M(OH)2 <---> M2+ + 2OH-

equili. ..x ............x ........2x

Ksp = [M2+] [ OH-]2
Let x = mole/L M(OH)2 that dissolve in water
We would get x mol/L M2+ and 2x mol/L OH-
Substitute these in the Ksp expression
6.65 x 10-18 = [x] [2x]2 = 4x3
x = 1.1846 x 10-6 M

part B) M(NO3)2 is a strong electrolyte:
.............M(NO3)2 ----------> M2+ + 2NO3-
[M2+] = 0.202 M
at equilibrium
[M2+]= x + 0.202
[OH-]= 2x
6.65 x 10-18 = [x+0.202] [2x]2 ........{NOTE : x <<<0.202 so neglect x}
x = molar solubility = 2.87 x 10-9 M

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