The K_{sp} of Mg(OH)_2 is 1 times 10^{-12}. 0 | vedclass

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(B) The solubility product expression is $K_{sp} = [Mg^{2+}][OH^-]^2$. Given $K_{sp} = 1 	imes 10^{-12}$ and $[Mg^{2+}] = 0.01 \ M$. Substituting the

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$9$

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(B) The solubility product expression is $K_{sp} = [Mg^{2+}][OH^-]^2$. Given $K_{sp} = 1 	imes 10^{-12}$ and $[Mg^{2+}] = 0.01 \ M$. Substituting the values: $1 	imes 10^{-12} = 0.01 	imes [OH^-]^2$. $[OH^-]^2 = \frac{1 	imes 10^{-12}}{10^{-2}} = 1 	imes 10^{-10}$. $[OH^-] = 1 	imes 10^{-5} \ M$. Calculating $pOH$: $pOH = -\log[OH^-] = -\log(1 	imes 10^{-5}) = 5$. Calculating $pH$: $pH = 14 - pOH = 14 - 5 = 9$. Therefore,precipitation will occur when $pH > 9$.

The $K_{sp}$ of $Mg(OH)_2$ is $1 \times 10^{-12}$. $0.01 \ M \ MgCl_2$ will show precipitation in a solution of $pH$ greater than:-

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