molar solubility

Question

What is the molar solubility of Ca(OH)2 if Ksp=5.02×10−6 in 0.25 M CaCl2 solution.Clearly state any assumptions that were made and justify the assumption was a good assumption.

(HINT: If the x/[initial Ca+2] x 100% is within 5%, then the assumption is valid.

Answer 1

Common ion effect:

The common ion in this problem is calcium (Ca²⁺)

Ca(OH)₂(s) ===> Ca²⁺(aq) + 2OH^-(aq)

Ksp = [Ca²⁺][OH^-]²

Since the solution also contains 0.25 M CaCl₂, we can assume that the 0.25 M Ca²⁺ represents the majority of the Ca²⁺ ion in solution, i.e. the Ca²⁺ from the Ca(OH)₂ is negligible.

5.02x10^-6 = (0.25)(x)²

x = 4.5x10^-3 M= [OH^-]

Thus, [Ca²⁺] coming from Ca(OH)₂ would be only 2.2x10^-3 M which is very small (<5%) of the 0.25 M. So the assumption was valid.