What is the concentration of lead(II) ions in a saturated solution of lead(II) fluoride?

Question

What is the concentration of lead(II) ions in a saturated solution of lead(II) fluoride?

Ksp(PbF2) = (3.991x10^-8)

Answer 1

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Lead(II) fluoride = PbF₂

PbF₂(s) <==> Pb²⁺(aq) + 2F-(aq)

Ksp = [Pb²⁺][F-]²

3.991x10^-8 = (x)(2x)²

3.991x10^-8 = 4x³

x³ = 9.978x10^-9

x = 2.15x10^-3 M = [Pb²⁺]