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Question

(A) Step $1$: Calculate the molarity $(S)$ of the solution.Concentration = $8.2 	imes 10^{-4} \% \ w/V = 8.2 	imes 10^{-4} \ g \ per \ 100 \ mL = 8.

Vedclass · Accepted Answer

$1.12 	imes 10^{-11}$

Vedclass · Answer

(A) Step $1$: Calculate the molarity $(S)$ of the solution.Concentration = $8.2 	imes 10^{-4} \% \ w/V = 8.2 	imes 10^{-4} \ g \ per \ 100 \ mL = 8.2 	imes 10^{-3} \ g \ L^{-1}$.$S = \frac{	ext{Concentration in } g \ L^{-1}}{	ext{Molar mass}} = \frac{8.2 	imes 10^{-3}}{58.3} \approx 1.4065 	imes 10^{-4} \ mol \ L^{-1}$.Step $2$: Calculate the solubility product $(K_{sp})$.$Mg(OH)_2 ightleftharpoons Mg^{2+} + 2OH^-$.$K_{sp} = [Mg^{2+}][OH^-]^2 = (S)(2S)^2 = 4S^3$.$K_{sp} = 4 	imes (1.4065 	imes 10^{-4})^3 \approx 4 	imes 2.78 	imes 10^{-12} \approx 1.112 	imes 10^{-11} \approx 1.12 	imes 10^{-11}$.

Metal sulfide	Solubility product
$A$. $PbS$	$I$. $4.0 \times 10^{-53}$
$B$. $HgS$	$II$. $8.0 \times 10^{-28}$
$C$. $MnS$	$III$. $1.6 \times 10^{-24}$
$D$. $ZnS$	$IV$. $2.5 \times 10^{-13}$

The concentration of a saturated solution of $Mg(OH)_2$ is $8.2 \times 10^{-4} \% \ w/V$. Calculate its solubility product. Its molecular mass is $58.3 \ g \ mol^{-1}$.

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