The solubility of \(\mathrm{Pb}\left(\mathrm{IO}_{3}\right)_{2}(s)\) in a \(0.10-M \mathrm{KIO}_{3}\) solution is \(2.6 \times 10^{-11} \mathrm{~mol} / \mathrm{L}\). Calculate \(K_{\mathrm{sp}}\) for \(\mathrm{Pb}\left(\mathrm{IO}_{3}\right)_{2}(s) .\)

First, let's write down the balanced chemical equation for the dissociation of \(\mathrm{Pb}\left(\mathrm{IO}_{3}\right)_{2}\) when it dissolves in water: \[\mathrm{Pb}\left(\mathrm{IO}_{3}\right)_{2}(s) \rightleftharpoons \mathrm{Pb}^{2+}(aq) + 2\mathrm{IO}_{3}^{-}(aq)\] Now, let's write the \(K_{sp}\) expression: \[K_{sp} = [\mathrm{Pb}^{2+}][\mathrm{IO}_{3}^{-}]^2\]

The given solubility of \(\mathrm{Pb}\left(\mathrm{IO}_{3}\right)_{2}\) is \(2.6 \times 10^{-11}\,\mathrm{mol}/\mathrm{L}\). From the balanced chemical equation, we can see that for every mole of \(\mathrm{Pb}\left(\mathrm{IO}_{3}\right)_{2}\) that dissolves in solution, it produces 1 mole of \(\mathrm{Pb}^{2+}\) and 2 moles of \(\mathrm{IO}_3^-\) ions. Denote the solubility of \(\mathrm{Pb}\left(\mathrm{IO}_{3}\right)_{2}\) as \(s\), then at equilibrium, the concentrations of the dissociated ions are: \[[\mathrm{Pb}^{2+}] = s = 2.6 \times 10^{-11}\,\mathrm{M}\] \[[\mathrm{IO}_{3}^{-}] = 2s + [\mathrm{KIO}_{3}]\] Since the initial concentration of \(\mathrm{KIO}_{3}\) is given as \(0.10\,\mathrm{M}\), we have: \[[\mathrm{IO}_{3}^{-}] = 2(2.6 \times 10^{-11}) + 0.10 = 0.100000052\,\mathrm{M}\]

Now that we have the equilibrium concentrations of \(\mathrm{Pb}^{2+}\) and \(\mathrm{IO}_{3}^{-}\), we can substitute these values into the \(K_{sp}\) expression: \[K_{sp} = [\mathrm{Pb}^{2+}][\mathrm{IO}_{3}^{-}]^2 = (2.6 \times 10^{-11})(0.100000052)^2\] Calculate the value of \(K_{sp}\): \[K_{sp} \approx 6.76 \times 10^{-23}\]

The solubility product constant, \(K_{sp}\), for \(\mathrm{Pb}\left(\mathrm{IO}_{3}\right)_{2}(s)\) is approximately \(6.76 \times 10^{-23}\).