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

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(D) The dissociation of $PbI_2$ is given by: $PbI_2(s) \rightleftharpoons Pb^{2+}(aq) + 2I^-(aq)$.In the presence of $0.2 \ M \ Pb(NO_3)_2$,the concen

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$\sqrt{K_{sp}/0.8}$

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(D) The dissociation of $PbI_2$ is given by: $PbI_2(s) ightleftharpoons Pb^{2+}(aq) + 2I^-(aq)$.In the presence of $0.2 \ M \ Pb(NO_3)_2$,the concentration of $Pb^{2+}$ ions is dominated by the strong electrolyte $Pb(NO_3)_2$.$[Pb^{2+}] = 0.2 \ M + S \approx 0.2 \ M$ (since $S $[I^-] = 2S$.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 $S$: $S^2 = K_{sp}/0.8$.Therefore,$S = \sqrt{K_{sp}/0.8}$.

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

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