If the K_{sp} of Mg(OH)_2 is 1 times 10^{-12} | vedclass

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Question

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

Vedclass · Accepted Answer

$9$

Vedclass · Answer

(B) The dissociation of $Mg(OH)_2$ is given by: $Mg(OH)_2 ightleftharpoons Mg^{2+} + 2OH^-$The solubility product expression is: $K_{sp} = [Mg^{2+}][OH^-]^2$Given $K_{sp} = 1 	imes 10^{-12}$ and $[Mg^{2+}] = 0.01 \ M = 10^{-2} \ M$.Substituting the values: $1 	imes 10^{-12} = (10^{-2})[OH^-]^2$$[OH^-]^2 = \frac{1 	imes 10^{-12}}{10^{-2}} = 10^{-10}$$[OH^-] = \sqrt{10^{-10}} = 10^{-5} \ M$Now,calculate $pOH$: $pOH = -\log[OH^-] = -\log(10^{-5}) = 5$Using the relation $pH + pOH = 14$ at $25^{\circ}C$:$pH = 14 - 5 = 9$Therefore,precipitation will occur when the $pH$ is $9$.

If the $K_{sp}$ of $Mg(OH)_2$ is $1 \times 10^{-12}$,at what $pH$ will $Mg(OH)_2$ precipitate from a $0.01 \ M$ solution of $Mg^{2+}$ ions?

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