Calculate the solubility of PbF2 in water at 25 °C.

Question

Calculate the solubility of PbF₂ in water at 25 °C.

PbF₂ K_sp = 3.3 x 10^-8

Round your answer to 2 significant digits.

___g/L

Answer 1

PbF₂(s) <==> Pb²⁺(aq) + 2F^-(aq)

Ksp = [Pb²⁺][F^-]²

Let x = [Pb²⁺] and then [F^-] = 2x

3.3x10^-8 = (x)(2x)² = 4x³

x³ = 8.25x10^-9

x = 2.02x10^-3 M = 2.0x10^-3 M (2 sig. figs.) = solubility of PbF₂