At 298 K,the solubility product of PbCl_2 is | vedclass

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Question

(A) The dissociation of $PbCl_2$ is given by: $PbCl_{2(s)} \rightleftharpoons Pb^{2+}_{(aq)} + 2Cl^{-}_{(aq)}$Let the solubility be $S \ mol \ L^{-1}$

Vedclass · Accepted Answer

$6.3 	imes 10^{-3}$

Vedclass · Answer

(A) The dissociation of $PbCl_2$ is given by: $PbCl_{2(s)} ightleftharpoons Pb^{2+}_{(aq)} + 2Cl^{-}_{(aq)}$Let the solubility be $S \ mol \ L^{-1}$. Then $[Pb^{2+}] = S$ and $[Cl^-] = 2S$.The solubility product expression is: $K_{sp} = [Pb^{2+}][Cl^-]^2 = (S)(2S)^2 = 4S^3$.Given $K_{sp} = 1.0 	imes 10^{-6}$,we have:$4S^3 = 1.0 	imes 10^{-6}$$S^3 = 0.25 	imes 10^{-6} = 250 	imes 10^{-9}$$S = (250)^{1/3} 	imes 10^{-3} \approx 6.3 	imes 10^{-3} \ mol \ L^{-1}$.

At $298 \ K$,the solubility product of $PbCl_2$ is $1.0 \times 10^{-6}$. What will be the solubility of $PbCl_2$ in $mol \ L^{-1}$?

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