What is the solubility of PbF₂ in a solution that contains 0.0550 M F⁻ ions? (Ksp of PbF₂ is 3.60 × 10⁻⁸)

Question

What is the solubility of PbF₂ in a solution that contains 0.0550 M F⁻ ions? (Ksp of PbF₂ is 3.60 × 10⁻⁸)

Answer 1

Assuming 1 liter of solution: [F^-]₀ = .0550 M Using an ICE diagram or just thinking through the problem... For every x moles of PbF₂ dissolved, you get x M Pb²⁺ and 2x M F^-

The K expression for the equilibrium after x moles have dissolved is

3.6 x 10^-8 = (x) (.055 + 2x)²

You could solve the cubic for x, but it is reasonable to treat 2x << .055 and just solve 3.6 x 10^-8 = .055x

If you need grams or grams per 100 ml (for solubility) you will need to convert from x moles/liter.

Answer 2

Write the Ksp equilibrium:

PbF₂(s)<-->Pb²⁺(aq)+2F^-(aq)

Write the K expession:

Ksp=[Pb2+][F-]²

Substitute in the K value and the [F-]

3.60(10^-8)=[x][0.0550]²

solve for x

x=(3.6(10^-8))/(0.0550²)

x=0.00001190

x=molar solubility=1.190×10^-5 M