Question

The generic metal hydroxide M(OH)2 has Ksp = 8.85×10⁻¹². (NOTE: In this particular problem, because of the magnitude of the Ksp and the stoichiometry of the compound, the contribution ofOH− 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?

Express your answer with the appropriate units.

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Part B

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

Express your answer with the appropriate units.

Answer 1

A)

At equilibrium:
M(OH)2 <----> M2+ + 2 OH-

s 2s   

Ksp = [M2+][OH-]^2
8.85*10^-12=(s)*(2s)^2
8.85*10^-12= 4(s)^3
s = 1.30*10^-4 M
Answer: 1.30*10^-4 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
8.85*10^-12=(0.202 + s)*(2s)^2
Since Ksp is small, s can be ignored as compared to 0.202
Above expression thus becomes:
8.85*10^-12=(0.202)*(2s)^2
8.85*10^-12= 0.202 * 4(s)^2
s = 3.31*10^-6 M

Answer: 3.31*10^-6 M