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

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(B) The dissociation of $Mg(OH)_2$ is given by: $Mg(OH)_2(s) \leftrightarrow Mg^{2+}(aq) + 2OH^{-}(aq)$.The solubility product expression is $K_{sp} =

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$10$

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(B) The dissociation of $Mg(OH)_2$ is given by: $Mg(OH)_2(s) \leftrightarrow Mg^{2+}(aq) + 2OH^{-}(aq)$.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: $1.0 	imes 10^{-11} = (10^{-3})[OH^{-}]^2$.$[OH^{-}]^2 = \frac{1.0 	imes 10^{-11}}{10^{-3}} = 10^{-8}$.$[OH^{-}] = \sqrt{10^{-8}} = 10^{-4}\, M$.Now,$pOH = -\log[OH^{-}] = -\log(10^{-4}) = 4$.Since $pH + pOH = 14$ at $25\,^{\circ}C$,we have $pH = 14 - 4 = 10$.

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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