The solubility product of Mg(OH)_2 is 1.8 tim | vedclass

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Question

(A) The dissociation of $Mg(OH)_2$ is given by: $Mg(OH)_{2(s)} \rightleftharpoons Mg^{2+}_{(aq)} + 2OH^{-}_{(aq)}$ Let the solubility be $S \ mol \ dm

Vedclass · Accepted Answer

$1.650 	imes 10^{-4}$

Vedclass · Answer

(A) The dissociation of $Mg(OH)_2$ is given by: $Mg(OH)_{2(s)} ightleftharpoons Mg^{2+}_{(aq)} + 2OH^{-}_{(aq)}$ Let the solubility be $S \ mol \ dm^{-3}$. Then,$[Mg^{2+}] = S$ and $[OH^-] = 2S$. The solubility product expression is: $K_{sp} = [Mg^{2+}][OH^-]^2 = (S)(2S)^2 = 4S^3$. Given $K_{sp} = 1.8 	imes 10^{-11}$. $4S^3 = 1.8 	imes 10^{-11}$ $S^3 = \frac{1.8 	imes 10^{-11}}{4} = 0.45 	imes 10^{-11} = 4.5 	imes 10^{-12}$. $S = \sqrt[3]{4.5 	imes 10^{-12}} = 1.650 	imes 10^{-4} \ mol \ dm^{-3}$.

Formula	Solubility product $(K_{sp})$
$AgBrO_3$	$2.5 \times 10^{-10}$
$MgSO_4$	$2.7 \times 10^{-27}$
$CaCO_3$	$2.7 \times 10^{-18}$
$FeC_2O_4$	$2.4 \times 10^{-15}$

The solubility product of $Mg(OH)_2$ is $1.8 \times 10^{-11}$ at $298 \ K$. What is its solubility in $mol \ dm^{-3}$?

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