What are the equilibrium concentrations of pb2 + and f - in a saturated solution of lead (ii) fluoride if the ksp of pbf2 is 3.20*10-8?

Answer is: the equilibrium concentrations fluorine anion are 0.004 M and lead cation are 0.002 M.

Chemical reaction: PbF ₂ (aq) → Pb²⁺ (aq) + 2F⁻ (aq).

Ksp = 3,2·10 ⁻⁸.

[Pb²⁺] = x.

[F⁻] = 2[Pb²⁺] = 2x

Ksp = [Pb² ⁺] · [F⁻]².

Ksp = x · 4x².

3,2·10⁻⁸ = 4x³.

x = ∛3,2·10⁻⁸ : 4.

x = [Pb²⁺] = 0,002M = 2·10⁻³ M.

[F⁻] = 2 · 0,002M = 0,004 M = 4·10⁻³ M.