Find the solubility of PbI_2 if its solubilit | vedclass

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(A) For $PbI_2$,the dissociation equilibrium is: $PbI_{2(s)} \rightleftharpoons Pb^{2+}_{(aq)} + 2I^{-}_{(aq)}$ Let the solubility be $S$. Then $[Pb^{

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$1.21 	imes 10^{-3} \ mol \ L^{-1}$

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(A) For $PbI_2$,the dissociation equilibrium is: $PbI_{2(s)} ightleftharpoons Pb^{2+}_{(aq)} + 2I^{-}_{(aq)}$ Let the solubility be $S$. Then $[Pb^{2+}] = S$ and $[I^-] = 2S$. The solubility product expression is: $K_{sp} = [Pb^{2+}][I^-]^2 = (S)(2S)^2 = 4S^3$ Given $K_{sp} = 7.0 	imes 10^{-9}$. $4S^3 = 7.0 	imes 10^{-9}$ $S^3 = \frac{7.0 	imes 10^{-9}}{4} = 1.75 	imes 10^{-9}$ $S = \sqrt[3]{1.75 	imes 10^{-9}} = 1.205 	imes 10^{-3} \ mol \ L^{-1} \approx 1.21 	imes 10^{-3} \ mol \ L^{-1}$.

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

Find the solubility of $PbI_2$ if its solubility product is $7.0 \times 10^{-9}$.

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