Problem: 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 .

For this problem, we’re being asked to calculate the concentration of Pb²⁺ needed in order to precipitate PbF₂ from a 1.00 x 10^-2 M F^- solution.

Dissociation reaction: PbF₂(s) ⇌ Pb²⁺(aq) + 2F^–(aq)

Step 1: ICE Chart

Step 2: K_sp expression

$\boxed{{\mathbf{K}}_{\mathbf{sp}}\mathbf{=}\frac{\mathbf{products}}{\cancel{\mathbf{reactants}}}\mathbf{=}\mathbf{\left[}{\mathbf{Pb}}^{\mathbf{2}\mathbf{+}}\mathbf{\right]}{\mathbf{\left[}{\mathbf{F}}^{\mathbf{-}}\mathbf{\right]}}^{\mathbf{2}}}$

Note that each concentration is raised by the stoichiometric coefficient