Calculate the solubility of CaCO3 (a) in pure water and (b) in a solution in which [(NH4)2CO3] = 0.216 M.

Question

Calculate the solubility of CaCO₃ (a) in pure water and (b) in a solution in which [(NH₄)₂CO₃] = 0.216 M.

Solubility in pure water =	M
Solubility in 0.216 M  (NH4)2CO3 =	M

Answer 1

Again, we need to know the Ksp for CaCO₃. I find a value of 4.7x10^-9

CaCO3(s) ==> Ca2+(aq) + CO32-(aq)

Ksp = [Ca²⁺][CO₃^2-]

4.7x10^-9 = (x)(x) = x²

x = 6.86x10^-5 M = solubility in pure water

For solubility in 0.216 M (NH₄)₂CO₃, we have a common ion effect (the common ion being CO₃^2-). So, solubility will always be less than in pure water.

4.7x10^-9 = [Ca²⁺][[0.216]

[Ca²⁺] = 2.18x10^-8 M = solubility in 0.216 M (NH₄)₂CO₃