What is the solubility?

Question

What is the solubility of PbF₂ in a solution that contains 0.0850 M Pb²⁺ ions? (Ksp of PbF₂ is 3.60 × 10⁻⁸)

Answer 1

PbF₂₍s) <==> Pb²⁺(aq) + 2F^-(aq)

Ksp = [Pb²⁺][F^-]²

Let x = [Pb²⁺] and then 2x = [F^-]

Ksp = 3.60x10^-8 = (x)(2x)²

3.60x10^-8 = 4x³

x³ = 9.00x10^-9

x = 2.08x10^-3

Solubility of PbF₂ = 2.08x10^-3 M