The concentration of \(\mathrm{Mg}^{2+}\) in seawater is \(0.052 \mathrm{M}\). At what \(\mathrm{pH}\) will \(99 \%\) of the \(\mathrm{Mg}^{2+}\) be precipitated as the hydroxide salt? \(\left[K_{\mathrm{sp}} \text { for } \mathrm{Mg}(\mathrm{OH})_{2}=8.9 \times 10^{-12} .\right]\)

We are given that \(99\%\) of the magnesium ions will be precipitated. We need to find the remaining concentration of \(\mathrm{Mg}^{2+}\) in the solution after the precipitation. Let's calculate it as follows: After precipitation, Remaining concentration of \(\mathrm{Mg}^{2+} = 0.052 \mathrm{M} \times (1 - 0.99) = 0.052 \mathrm{M} \times 0.01 = 5.2 \times 10^{-4} \mathrm{M}\).

The balanced equation for the dissolution of \(\mathrm{Mg(OH)_2}\) is: \(\mathrm{Mg(OH)_2 \rightleftharpoons Mg^{2+} + 2 OH^{-}}\) Then, the solubility product expression (\(K_{\mathrm{sp}}\)) for \(\mathrm{Mg(OH)_2}\) is: \(K_{\mathrm{sp}} = [\mathrm{Mg}^{2+}][\mathrm{OH}^{-}]^2\) We are given that \(K_{\mathrm{sp}} = 8.9 \times 10^{-12}\). Now, let's find the ion product (Q), calculated as: \(Q = [\mathrm{Mg}^{2+}][\mathrm{OH}^{-}]^2\) We don't know the concentration of \(\mathrm{OH^-}\) ions directly, but we can use pH to represent it indirectly. Since we know \([\mathrm{Mg}^{2+}] = 5.2 \times 10^{-4} \mathrm{M}\), \(Q = (5.2 \times 10^{-4})([\mathrm{OH}^{-}])^2\) Note that \(Q\) must be equal to \(K_{\mathrm{sp}}\) when the precipitation occurs.

Since we know the relationship between \([\mathrm{OH}^{-}]\) and \([\mathrm{H}^{+}]\): \(K_{\mathrm{w}} = [\mathrm{H}^{+}][\mathrm{OH}^{-}]\), where \(K_{\mathrm{w}} = 1 \times 10^{-14}\) at \(25 ^{\circ}\mathrm{C}\), and \(pH = -\log_{10}{[\mathrm{H}^{+}]}\) We can solve for \([\mathrm{OH}^{-}]\) and substitute it into the Q equation. First, express \([\mathrm{H}^{+}]\) in terms of \([\mathrm{OH}^{-}]\): \([\mathrm{H}^{+}] = \dfrac{K_w}{[\mathrm{OH}^{-}]}\) Substitute this into the pH equation: \(pH = -\log_{10}{\left(\dfrac{K_w}{[\mathrm{OH}^{-}]}\right)}\) Now, let's plug the value of Q in the equation \(Q = K_{\mathrm{sp}}\): \((5.2 \times 10^{-4})([\mathrm{OH}^{-}])^2 = 8.9 \times 10^{-12}\) Solve for \([\mathrm{OH}^{-}]\): \([\mathrm{OH}^{-}] = \sqrt{\dfrac{8.9 \times 10^{-12}}{5.2 \times 10^{-4}}} = 1.3 \times 10^{-4} \mathrm{M}\) Now plug this into the pH equation: \(pH = -\log_{10}{\left(\dfrac{1 \times 10^{-14}}{1.3 \times 10^{-4}}\right)} \approx 10.62\) Therefore, the pH when \(99\%\) of the \(\mathrm{Mg}^{2+}\) will be precipitated as hydroxide salt is approximately \(10.62\).