If the solubility product of lead iodide (PbI | vedclass

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(A) The dissociation of lead iodide is given by: $PbI_2(s) \rightleftharpoons Pb^{2+}(aq) + 2I^-(aq)$.Let the solubility be $S \ mol/L$. Then $[Pb^{2+

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$2 	imes 10^{-3}$

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(A) The dissociation of lead iodide is given by: $PbI_2(s) ightleftharpoons Pb^{2+}(aq) + 2I^-(aq)$.Let the solubility be $S \ mol/L$. Then $[Pb^{2+}] = S$ and $[I^-] = 2S$.The solubility product expression is $K_{sp} = [Pb^{2+}][I^-]^2$.Substituting the values: $K_{sp} = (S)(2S)^2 = 4S^3$.Given $K_{sp} = 3.2 	imes 10^{-8}$,we have $4S^3 = 3.2 	imes 10^{-8}$.$S^3 = 0.8 	imes 10^{-8} = 8 	imes 10^{-9}$.Taking the cube root,$S = \sqrt[3]{8 	imes 10^{-9}} = 2 	imes 10^{-3} \ mol/L$.

If the solubility product of lead iodide $(PbI_2)$ is $3.2 \times 10^{-8}$,then its solubility in $moles/litre$ will be

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