Solubility of PbI_2 is 0.005 M. Then,the sol | vedclass

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(D) The dissociation of $PbI_2$ is given by: $PbI_2(s) ightleftharpoons Pb^{2+}(aq) + 2I^-(aq)$.Let the solubility be $s = 0.005 \ M = 5 	imes 10^{

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None of these

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(D) The dissociation of $PbI_2$ is given by: $PbI_2(s) ightleftharpoons Pb^{2+}(aq) + 2I^-(aq)$.Let the solubility be $s = 0.005 \ M = 5 	imes 10^{-3} \ M$.The solubility product expression is $K_{sp} = [Pb^{2+}][I^-]^2$.Substituting the concentrations in terms of solubility: $K_{sp} = (s)(2s)^2 = 4s^3$.$K_{sp} = 4 	imes (5 	imes 10^{-3})^3$.$K_{sp} = 4 	imes 125 	imes 10^{-9} = 500 	imes 10^{-9} = 5 	imes 10^{-7}$.Since $5 	imes 10^{-7}$ is not among the given options,the correct answer is $D$.

Solubility of $PbI_2$ is $0.005 \ M$. Then,the solubility product of $PbI_2$ is

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