Calculate the molar solublity for PbF_2 given a Ksp= 0.000000347 in a 0.397 M Pb(NO3)_2 solution

Question

Calculate the molar solublity for PbF_2 given a Ksp= 0.000000347 in a 0.397 M Pb(NO3)_2 solution

Answer 1

Common Ion Effect: Common ion is Pb²⁺

PbF₂(s) <==> Pb²⁺(aq) + 2F^-(aq)

Ksp = [Pb²⁺][F-]²

Ksp = 3.47x10^-7

[Pb²⁺ ] = 0.397 M

[F-]= ?

3.47x10^-7 = (0.397)(F^-)²

[F^-] = 9.35x10^-4 M

Molar solubility of PbF₂ in 0.397 M Pb²⁺ = 1/2 x 9.35x10^-4 M = 4.68x10^-4 M