The K_{sp} of Mg(OH)_2 is 1.0 times 10^{-12}. | vedclass

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(A) For the precipitation of $Mg(OH)_2$ from a $0.01 \ M$ $Mg^{2+}$ solution,the ionic product must exceed $K_{sp}$.$K_{sp} = [Mg^{2+}][OH^-]^2 = 1.0

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$pH = 10$

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(A) For the precipitation of $Mg(OH)_2$ from a $0.01 \ M$ $Mg^{2+}$ solution,the ionic product must exceed $K_{sp}$.$K_{sp} = [Mg^{2+}][OH^-]^2 = 1.0 	imes 10^{-12}$.$[OH^-]^2 = \frac{1.0 	imes 10^{-12}}{0.01} = 1.0 	imes 10^{-10}$.$[OH^-] = 1.0 	imes 10^{-5} \ M$.$pOH = -\log(1.0 	imes 10^{-5}) = 5$.$pH = 14 - pOH = 9$.For solubility in pure water: $K_{sp} = (s)(2s)^2 = 4s^3$.$4s^3 = 1.0 	imes 10^{-12} \implies s^3 = 0.25 	imes 10^{-12} = 25 	imes 10^{-14}$.$s = \sqrt[3]{25 	imes 10^{-14}} \approx 6.3 	imes 10^{-5} \ M$.

The $K_{sp}$ of $Mg(OH)_2$ is $1.0 \times 10^{-12}$. At which $pH$ does $0.01 \ M$ $Mg^{2+}$ solution begin to precipitate? Calculate the solubility of $Mg(OH)_2$ in pure water.

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