The solubility of calcium carbonate in pure water is 6.9 × 10−5 moles per liter. What is its molar solubility in a 0.16 M CaCl2 solution? Answer in mol/L

Answer 1

This is a common ion problem. The common ion is calcium (Ca²⁺)

CaCO3(s) <==> Ca²⁺(aq) + CO₃^2-(aq)

Ksp = [Ca²⁺][CO₃^2-]

6.9x10^-5 = [0.16 M][CO₃^2-]

[CO₃^2-] = 4.3x10^-4 M = 4.3x10^-4 moles/L = solubility of CaCO₃ in 016 M CaCl₂