The solubility product of Mg(OH)_2 is 1 times | vedclass

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(B) The solubility product expression is: $K_{sp} = [Mg^{2+}][OH^{-}]^2$ Rearrange the above expression to solve for $[OH^-]$: $[OH^{-}] = \sqrt{\frac

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$9$

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(B) The solubility product expression is: $K_{sp} = [Mg^{2+}][OH^{-}]^2$ Rearrange the above expression to solve for $[OH^-]$: $[OH^{-}] = \sqrt{\frac{K_{sp}}{[Mg^{2+}]}}$ Substitute the given values ($K_{sp} = 1 	imes 10^{-12}$ and $[Mg^{2+}] = 0.01 \ M$): $[OH^{-}] = \sqrt{\frac{1 	imes 10^{-12}}{0.01}} = \sqrt{1 	imes 10^{-10}} = 1 	imes 10^{-5} \ M$ Calculate $pOH$ from the hydroxide ion concentration: $pOH = -\log [OH^{-}] = -\log (1 	imes 10^{-5}) = 5$ Calculate $pH$ from $pOH$: $pH = 14 - pOH = 14 - 5 = 9$ Thus,precipitation will occur at $pH = 9$.

The solubility product of $Mg(OH)_2$ is $1 \times 10^{-12}$. At what $pH$ will precipitation occur in a $0.01 \ M$ $MgCl_2$ solution?

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