The pH at which Mg(OH)_2 [K_{sp} = 1 times 10 | vedclass

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(B) Precipitation begins when the ionic product $Q_{sp}$ equals the solubility product $K_{sp}$.For the reaction $Mg(OH)_2(s) \rightleftharpoons Mg^{2

Vedclass · Accepted Answer

$09$

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(B) Precipitation begins when the ionic product $Q_{sp}$ equals the solubility product $K_{sp}$.For the reaction $Mg(OH)_2(s) ightleftharpoons Mg^{2+}(aq) + 2OH^-(aq)$,the expression is $Q_{sp} = [Mg^{2+}][OH^-]^2$.Given $[Mg^{2+}] = 0.10 \ M$ and $K_{sp} = 1 	imes 10^{-11}$.Setting $Q_{sp} = K_{sp}$:$0.10 	imes [OH^-]^2 = 1 	imes 10^{-11}$$[OH^-]^2 = 10^{-10}$$[OH^-] = 10^{-5} \ M$.Now,calculate $pOH$:$pOH = -\log[OH^-] = -\log(10^{-5}) = 5$.Finally,calculate $pH$:$pH + pOH = 14$$pH = 14 - 5 = 9$.

The $pH$ at which $Mg(OH)_2$ $[K_{sp} = 1 \times 10^{-11}]$ begins to precipitate from a solution containing $0.10 \ M$ $Mg^{2+}$ ions is . . . . . .

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