Question

What is the solubility of Ca(OH)2(s) in an aqueous solution that initially contains 0.001 M Ca(NO3)2?

Answer 1

Ksp of Ca(OH)2 is 5.02*10^-6

Ca(NO3)2 here is Strong electrolyte

It will dissociate completely to give [Ca2+] = 0.001 M

At equilibrium:

Ca(OH)2 <----> Ca2+ + 2 OH-

   1*10^-3 +s 2s

Ksp = [Ca2+][OH-]^2

5.02*10^-6=(1*10^-3 + s)*(2s)^2

Since Ksp is small, s can be ignored as compared to 1*10^-3

Above expression thus becomes:

5.02*10^-6=(1*10^-3)*(2s)^2

s = 3.54*10^-2 M

Answer: 3.54*10^-2 M