The molar solubility of PbI_2 in 0.2 M Pb(N | vedclass

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

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$(\frac{K_{sp}}{0.8})^{1/2}$

Vedclass · Answer

(C) The dissociation of $PbI_2$ is given by: $PbI_2(s) ightleftharpoons Pb^{2+}(aq) + 2I^-(aq)$.Let the molar solubility of $PbI_2$ be $S \ mol/L$.The concentration of $Pb^{2+}$ ions from $Pb(NO_3)_2$ is $0.2 \ M$.Total concentration of $[Pb^{2+}] = (0.2 + S) \approx 0.2 \ M$ (since $S$ is very small).Total concentration of $[I^-] = 2S$.The solubility product expression is $K_{sp} = [Pb^{2+}][I^-]^2$.Substituting the values: $K_{sp} = (0.2)(2S)^2$.$K_{sp} = 0.2 	imes 4S^2 = 0.8S^2$.Solving for $S$: $S^2 = \frac{K_{sp}}{0.8}$,therefore $S = (\frac{K_{sp}}{0.8})^{1/2}$.

The molar solubility of $PbI_2$ in $0.2 \ M \ Pb(NO_3)_2$ solution in terms of $K_{sp}$ (solubility product) is

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