At 298 , K,the solubility of PbCl_2 is 2 time | vedclass

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(D) The dissociation of $PbCl_2$ is given by: $PbCl_2(s) \rightleftharpoons Pb^{2+}(aq) + 2Cl^{-}(aq)$.If $S$ is the solubility of $PbCl_2$,then $[Pb^

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$3.2 	imes 10^{-5}$

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(D) The dissociation of $PbCl_2$ is given by: $PbCl_2(s) ightleftharpoons Pb^{2+}(aq) + 2Cl^{-}(aq)$.If $S$ is the solubility of $PbCl_2$,then $[Pb^{2+}] = S$ and $[Cl^{-}] = 2S$.The solubility product constant is $K_{sp} = [Pb^{2+}][Cl^{-}]^2$.Substituting the values: $K_{sp} = (S)(2S)^2 = 4S^3$.Given $S = 2 	imes 10^{-2} \, mol/L$,we have:$K_{sp} = 4 	imes (2 	imes 10^{-2})^3 = 4 	imes (8 	imes 10^{-6}) = 3.2 	imes 10^{-5}$.

At $298 \, K$,the solubility of $PbCl_2$ is $2 \times 10^{-2} \, mol/L$,then $K_{sp} = ?$

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