If the K_{sp} of a saturated solution of Mg(O | vedclass

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(B) The dissociation of $Mg(OH)_2$ is given by: $Mg(OH)_2 \rightleftharpoons Mg^{2+} + 2OH^-$Let the solubility be $s$. Then $[Mg^{2+}] = s$ and $[OH^

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$10.3$

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(B) The dissociation of $Mg(OH)_2$ is given by: $Mg(OH)_2 ightleftharpoons Mg^{2+} + 2OH^-$Let the solubility be $s$. Then $[Mg^{2+}] = s$ and $[OH^-] = 2s$.The solubility product expression is: $K_{sp} = [Mg^{2+}][OH^-]^2 = (s)(2s)^2 = 4s^3$.Given $K_{sp} = 4 	imes 10^{-12}$,we have $4s^3 = 4 	imes 10^{-12}$,which implies $s^3 = 10^{-12}$,so $s = 10^{-4} \ M$.Now,$[OH^-] = 2s = 2 	imes 10^{-4} \ M$.$pOH = -\log[OH^-] = -\log(2 	imes 10^{-4}) = -(\log 2 + \log 10^{-4}) = -(0.3010 - 4) = 3.699 \approx 3.7$.$pH = 14 - pOH = 14 - 3.7 = 10.3$.

If the $K_{sp}$ of a saturated solution of $Mg(OH)_2$ is $4 \times 10^{-12}$,then the $pH$ is equal to:

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