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(A) $1$. First,calculate the molar solubility $(S)$ in $mol \ L^{-1}$:$S = \frac{	ext{Concentration in } g \ L^{-1}}{	ext{Molar mass of } Mg(OH)_2}

Vedclass · Accepted Answer

$1.121 	imes 10^{-14}$

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(A) $1$. First,calculate the molar solubility $(S)$ in $mol \ L^{-1}$:$S = \frac{	ext{Concentration in } g \ L^{-1}}{	ext{Molar mass of } Mg(OH)_2} = \frac{8.2 	imes 10^{-4} \ g \ L^{-1}}{58 \ g \ mol^{-1}} \approx 1.414 	imes 10^{-5} \ mol \ L^{-1}$.$2$. The dissociation of $Mg(OH)_2$ is: $Mg(OH)_2(s) ightleftharpoons Mg^{2+}(aq) + 2OH^-(aq)$.$3$. The solubility product expression is: $K_{sp} = [Mg^{2+}][OH^-]^2 = (S)(2S)^2 = 4S^3$.$4$. Substitute the value of $S$: $K_{sp} = 4 	imes (1.414 	imes 10^{-5})^3$.$5$. $K_{sp} = 4 	imes (2.828 	imes 10^{-15}) = 1.131 	imes 10^{-14} \ (mol \ L^{-1})^3$.

The concentration of a saturated solution of $Mg(OH)_2$ is $8.2 \times 10^{-4} \ g \ L^{-1}$ at $298 \ K$. Calculate its solubility product $(K_{sp})$. (Molar mass of $Mg(OH)_2 = 58 \ g \ mol^{-1}$)

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