If the maximum concentration of PbCl_2 in wat | vedclass

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(D) The solubility product constant $(K_{sp})$ of $PbCl_2$ is calculated from its solubility $(s)$ in water: $PbCl_2(s) \rightleftharpoons Pb^{2+}(aq)

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$4 	imes 10^{-4} \, M$

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(D) The solubility product constant $(K_{sp})$ of $PbCl_2$ is calculated from its solubility $(s)$ in water: $PbCl_2(s) ightleftharpoons Pb^{2+}(aq) + 2Cl^-(aq)$ $K_{sp} = [Pb^{2+}][Cl^-]^2 = (s)(2s)^2 = 4s^3$ Given $s = 0.01 \, M = 10^{-2} \, M$,$K_{sp} = 4 	imes (10^{-2})^3 = 4 	imes 10^{-6}$. In $0.1 \, M \, NaCl$,the concentration of $Cl^-$ is $0.1 \, M$ (common ion effect). Let the new solubility be $s'$. $K_{sp} = [Pb^{2+}][Cl^-]^2 = (s')(0.1)^2 = 4 	imes 10^{-6}$ $s' 	imes 0.01 = 4 	imes 10^{-6}$ $s' = \frac{4 	imes 10^{-6}}{10^{-2}} = 4 	imes 10^{-4} \, M$.

If the maximum concentration of $PbCl_2$ in water at $298 \, K$ is $0.01 \, M$,what is its maximum concentration in $0.1 \, M \, NaCl$ solution?

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