The solubility of PbI_2 at 25,^{circ}C is 0.7 | vedclass

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Question

(D) The dissociation of $PbI_2$ is given by: $PbI_2(s) \rightleftharpoons Pb^{2+}(aq) + 2I^{-}(aq)$.First,calculate the molar solubility $(s)$ in $mol

Vedclass · Accepted Answer

$14.0 	imes 10^{-9}$

Vedclass · Answer

(D) The dissociation of $PbI_2$ is given by: $PbI_2(s) ightleftharpoons Pb^{2+}(aq) + 2I^{-}(aq)$.First,calculate the molar solubility $(s)$ in $mol\, L^{-1}$:$s = \frac{	ext{solubility in } g\, L^{-1}}{	ext{molar mass}} = \frac{0.7}{461.2} \approx 1.5178 	imes 10^{-3} \, mol\, L^{-1}$.The solubility product expression is $K_{sp} = [Pb^{2+}][I^{-}]^2 = (s)(2s)^2 = 4s^3$.Substituting the value of $s$:$K_{sp} = 4 	imes (1.5178 	imes 10^{-3})^3 \approx 4 	imes 3.5 	imes 10^{-9} = 14.0 	imes 10^{-9}$.

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

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