For this problem, we´re being asked to calculate the concentration of the lead ion Pb^{2+} that must be exceeded to precipitate PbF_{2} from a solution of fluoride ion F^{-}.

In water, the dissociation of PbF_{2} is the following:

**PbF _{2(s)} **

The K_{sp} expression for PbF_{2} is:

Where solids are ignored (reactants), so:

${\mathbf{K}}_{\mathbf{sp}}\mathbf{=}\mathit{p}\mathit{r}\mathit{o}\mathit{d}\mathit{u}\mathit{c}\mathit{t}\mathit{s}\mathbf{=}\left[{\mathbf{Pb}}^{\mathbf{2}\mathbf{+}}\right]{\left[{\mathbf{F}}^{\mathbf{-}}\right]}^{\mathbf{2}}$

Note that each concentration is raised by the stoichiometric coefficient: [Pb^{2+}] is raised to 1 and [F^{-}] is raised to 2.

What concentration of the lead ion, Pb ^{2+}, must be exceeded to precipitate PbF_{2} from a solution that is 1.00x10^{-2} *M* in the fluoride ion, F- = *K*sp for lead(II) fluoride is 3.3x10^{-8}.

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