The solubility product of magnesium hydroxide | vedclass

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Question

(A) For $Mg(OH)_2 \rightleftharpoons Mg^{2+} + 2OH^-$,the solubility product $K_{sp} = [Mg^{2+}][OH^-]^2 = (s)(2s)^2 = 4s^3$. In pure water: $4s^3 = 1

Vedclass · Accepted Answer

$1.44 	imes 10^{-4} \ M, 4.8 	imes 10^{-9} \ M$

Vedclass · Answer

(A) For $Mg(OH)_2 ightleftharpoons Mg^{2+} + 2OH^-$,the solubility product $K_{sp} = [Mg^{2+}][OH^-]^2 = (s)(2s)^2 = 4s^3$. In pure water: $4s^3 = 1.2 	imes 10^{-11} \implies s^3 = 0.3 	imes 10^{-11} = 3.0 	imes 10^{-12} \implies s = \sqrt[3]{3.0} 	imes 10^{-4} \approx 1.44 	imes 10^{-4} \ M$. In $0.05 \ M$ $NaOH$,$[OH^-] \approx 0.05 \ M$. $K_{sp} = [Mg^{2+}][OH^-]^2 \implies 1.2 	imes 10^{-11} = s(0.05)^2 \implies s = \frac{1.2 	imes 10^{-11}}{2.5 	imes 10^{-3}} = 4.8 	imes 10^{-9} \ M$.

The solubility product of magnesium hydroxide $Mg(OH)_2$ is $1.2 \times 10^{-11}$. Calculate its solubility in pure water and $0.05 \ M$ $NaOH$.

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