What concentration of the lead ion, pb2+, must be exceeded to precipitate pbf2 from a solution that is 1.00*10-2 m in the fluoride ion, f-? ksp for lead (ii) fluoride is 3.3*10-8?

Answer is: concentration of Pb²⁺ must be exceeded is 3.3·10⁻⁴ M.

Chemical reaction : Pb²⁺ (aq) + 2F⁻ (aq) → PbF₂ (s).

Ksp (PbF ₂) = 3.3·10⁻⁸.

c (F ⁻) = 0.01 M.

Ksp (PbF ₂) = c (Pb²⁺) · c (F⁻) ².

c (Pb² ⁺) = Ksp (PbF₂) : c (Cl⁻) ².

c (Pb² ⁺) = 3.3·10⁻⁸ : (0.01 M) ².

c (Pb² ⁺) = 0.000000033 M³ : 0.0001 M².

c (Pb² ⁺) = 0.00033 M = 3.3·10⁻⁴ M.