At 25 ^circ C, the solubility product of Mg( | vedclass

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(B) The dissociation equilibrium is: $Mg(OH)_2 \rightleftharpoons Mg^{2+} + 2OH^-$The solubility product expression is: $K_{sp} = [Mg^{2+}][OH^-]^2$Gi

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$10$

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(B) The dissociation equilibrium is: $Mg(OH)_2 ightleftharpoons Mg^{2+} + 2OH^-$The solubility product expression 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} 	imes [OH^-]^2$$[OH^-]^2 = \frac{1.0 	imes 10^{-11}}{10^{-3}} = 10^{-8}$$[OH^-] = \sqrt{10^{-8}} = 10^{-4} \ M$Calculating $pOH$: $pOH = -\log[OH^-] = -\log(10^{-4}) = 4$Since $pH + pOH = 14$ at $25 \ ^\circ C,$$pH = 14 - 4 = 10.$

At $25 \ ^\circ C,$ the solubility product of $Mg(OH)_2$ is $1.0 \times 10^{-11}.$ At which $pH$ will $Mg^{2+}$ ions start precipitating in the form of $Mg(OH)_2$ from a solution of $0.001 \ M \ Mg^{2+}$ ions?

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