Molar Solubility of Compound

Question

What is the solubility of PbF₂ in a solution that contains 0.0900 M F⁻ ions? (Ksp of PbF₂ is 3.60 × 10⁻⁸)

Answer 1

Common Ion Effect:

PbF₂(s) <==> Pb²⁺(aq) + 2F^-(aq)

Ksp = [Pb²⁺][F^-]²

3.60x10^-8 = [Pb²⁺][0.0900]² = [Pb²⁺][0.0081]

[Pb²⁺] = 4.44x10^-6 M

This would be the solubility of PbF2 under the described conditions.