The solubility of PbI_2 at 25 , ^circ C is 0. | vedclass

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(C) The solubility $S$ in $mol \, L^{-1}$ is calculated as: $S = \frac{0.7 \, g \, L^{-1}}{461.2 \, g \, mol^{-1}} \approx 1.5178 	imes 10^{-3} \, mo

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$14.0 	imes 10^{-9}$

Vedclass · Answer

(C) The solubility $S$ in $mol \, L^{-1}$ is calculated as: $S = \frac{0.7 \, g \, L^{-1}}{461.2 \, g \, mol^{-1}} \approx 1.5178 	imes 10^{-3} \, mol \, L^{-1}$.For the salt $PbI_2$,the dissociation is $PbI_2(s) ightleftharpoons Pb^{2+}(aq) + 2I^-(aq)$.The solubility product expression is $K_{sp} = [Pb^{2+}][I^-]^2 = (S)(2S)^2 = 4S^3$.Substituting the value of $S$: $K_{sp} = 4 	imes (1.5178 	imes 10^{-3})^3$.$K_{sp} = 4 	imes (3.50 	imes 10^{-9}) = 14.0 	imes 10^{-9}$.

The solubility of $PbI_2$ at $25 \, ^\circ C$ is $0.7 \, g \, L^{-1}$. What is the solubility product $(K_{sp})$ of $PbI_2$ at this temperature? (Molar mass of $PbI_2 = 461.2 \, g \, mol^{-1}$)

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