What is the solubility of PbF₂ in water? (Ksp of PbF₂ is 3.6 × 10⁻⁸)

Answer 1

PbF2(s) <==> Pb2+(aq) + 2F-(aq)

Ksp = [Pb2+][F-]²

3.6x10^-8 = (x)(2x)²

3.6x10^-8 = 4x³

x³ = 9x10^-9

x = 3x10^-3 M = molar solubility of PbF₂