# Problem: The concentration of Mg2+ in seawater is 0.052 M. At what pH will 99% of the Mg 2+ be precipitated as the hydroxide salt? [Ksp for Mg(OH)2 = 8.9 X 10 -12.]

###### FREE Expert Solution

99% of Mg2+ is precipitated → 1% is left in the solution

[Mg2+] = 99% (0.052 M)

[Mg2+] = 5.2×10‒4 M

when 99% of Mg2+ is precipitated, 5.2×10‒4 M of Mg2+ is left in the solution

Ksp Mg(OH)2 = 8.9×10‒12 M

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

[OH] = 1.3082×10‒4 M

pOH = ‒log [OH]

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###### Problem Details

The concentration of Mg2+ in seawater is 0.052 M. At what pH will 99% of the Mg 2+ be precipitated as the hydroxide salt? [Ksp for Mg(OH)2 = 8.9 X 10 -12.]