At 25 , ^circ C,the solubility product (K_{sp | vedclass

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(C) The solubility product expression for $Mg(OH)_2$ is: $K_{sp} = [Mg^{2+}][OH^-]^2$. Given $K_{sp} = 1.0 	imes 10^{-11}$ and $[Mg^{2+}] = 0.001 \,

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$10$

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(C) The solubility product expression for $Mg(OH)_2$ is: $K_{sp} = [Mg^{2+}][OH^-]^2$. Given $K_{sp} = 1.0 	imes 10^{-11}$ and $[Mg^{2+}] = 0.001 \, M = 10^{-3} \, M$. Substituting the values: $1.0 	imes 10^{-11} = (10^{-3})[OH^-]^2$. $[OH^-]^2 = \frac{1.0 	imes 10^{-11}}{10^{-3}} = 10^{-8}$. $[OH^-] = \sqrt{10^{-8}} = 10^{-4} \, M$. Now,$pOH = -\log[OH^-] = -\log(10^{-4}) = 4$. Since $pH + pOH = 14$,we have $pH = 14 - 4 = 10$.

At $25 \, ^\circ C$,the solubility product $(K_{sp})$ of $Mg(OH)_2$ is $1.0 \times 10^{-11}$. What is the minimum $pH$ required to precipitate $Mg(OH)_2$ from a $0.001 \, M \, Mg^{2+}$ solution?

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