The solubility product of PbCl_2 at 298 K is | vedclass

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(B) The solubility equilibrium for $PbCl_2$ is given by: $PbCl_{2(s)} \rightleftharpoons Pb^{2+}_{(aq)} + 2Cl^{-}_{(aq)}$ Let the solubility be $S \ m

Vedclass · Accepted Answer

$2 	imes 10^{-2}$

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(B) The solubility equilibrium for $PbCl_2$ is given by: $PbCl_{2(s)} ightleftharpoons Pb^{2+}_{(aq)} + 2Cl^{-}_{(aq)}$ Let the solubility be $S \ mol \ dm^{-3}$. Then,$[Pb^{2+}] = S$ and $[Cl^-] = 2S$. The solubility product expression is: $K_{sp} = [Pb^{2+}][Cl^-]^2 = (S)(2S)^2 = 4S^3$ Given $K_{sp} = 3.2 	imes 10^{-5}$. $4S^3 = 3.2 	imes 10^{-5}$ $S^3 = \frac{3.2 	imes 10^{-5}}{4} = 0.8 	imes 10^{-5} = 8 	imes 10^{-6}$ $S = \sqrt[3]{8 	imes 10^{-6}} = 2 	imes 10^{-2} \ mol \ dm^{-3}$

The solubility product of $PbCl_2$ at $298 \ K$ is $3.2 \times 10^{-5}$. What is its solubility in $mol \ dm^{-3}$?

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