99% of Mg^{2+} is precipitated → 1% is left in the solution

[Mg^{2+}] = 99% (0.052 M)

**[Mg ^{2+}] = 5.2×10^{‒4} M**

when 99% of Mg^{2+} is precipitated, 5.2×10^{‒4} M of Mg^{2+} is left in the solution

K_{sp} Mg(OH)_{2} = 8.9×10^{‒12} M

${\mathbf{K}}_{\mathbf{sp}}\mathbf{=}\mathbf{\left[}{\mathbf{Mg}}^{\mathbf{2}\mathbf{+}}\mathbf{\right]}{\mathbf{\left[}{\mathbf{OH}}^{\mathbf{-}}\mathbf{\right]}}^{\mathbf{2}}\phantom{\rule{0ex}{0ex}}\frac{\mathbf{8}\mathbf{.}\mathbf{9}\mathbf{\times}{\mathbf{10}}^{\mathbf{-}\mathbf{12}}}{\mathbf{(}\mathbf{5}\mathbf{.}\mathbf{2}\mathbf{\times}{\mathbf{10}}^{\mathbf{-}\mathbf{4}}\mathbf{)}}\mathbf{=}\frac{\overline{)(5.2\times {10}^{-4})}{\mathbf{\left[}{\mathbf{OH}}^{\mathbf{-}}\mathbf{\right]}}^{\mathbf{2}}}{\overline{)\mathbf{(}\mathbf{5}\mathbf{.}\mathbf{2}\mathbf{\times}{\mathbf{10}}^{\mathbf{-}\mathbf{4}}\mathbf{)}}}\phantom{\rule{0ex}{0ex}}\sqrt{\mathbf{1}\mathbf{.}\mathbf{7115}\mathbf{\times}{\mathbf{10}}^{\mathbf{-}\mathbf{8}}}\mathbf{=}\sqrt{{\mathbf{\left[}{\mathbf{OH}}^{\mathbf{-}}\mathbf{\right]}}^{\mathbf{2}}}$

**[OH^{‒}] = 1.3082×10^{‒4} M**

**pOH = ‒log [OH ^{‒}]**

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

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