The solubility of PbCl_2 is | vedclass

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(C) The dissociation of $PbCl_2$ is represented as: $PbCl_2(s) \rightleftharpoons Pb^{2+}(aq) + 2Cl^-(aq)$.Let the solubility of $PbCl_2$ be $S \ mol/

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$\sqrt[3]{\frac{K_{sp}}{4}}$

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(C) The dissociation of $PbCl_2$ is represented as: $PbCl_2(s) ightleftharpoons Pb^{2+}(aq) + 2Cl^-(aq)$.Let the solubility of $PbCl_2$ be $S \ mol/L$.Then,$[Pb^{2+}] = S$ and $[Cl^-] = 2S$.The solubility product constant expression is: $K_{sp} = [Pb^{2+}][Cl^-]^2$.Substituting the values: $K_{sp} = (S)(2S)^2$.$K_{sp} = S 	imes 4S^2 = 4S^3$.Therefore,$S^3 = \frac{K_{sp}}{4}$,which gives $S = \sqrt[3]{\frac{K_{sp}}{4}}$.

The solubility of $PbCl_2$ is

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