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: PbF₂(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.

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