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(D) The dissociation reaction for lead iodide is: $PbI_2(s) ⇌ Pb^{2+}(aq) + 2I^{-}(aq)$.For a salt of type $AB_2$,the solubility product $K_{sp}$ is g

Vedclass · Accepted Answer

$32 	imes 10^{-9}$

Vedclass · Answer

(D) The dissociation reaction for lead iodide is: $PbI_2(s) ⇌ Pb^{2+}(aq) + 2I^{-}(aq)$.For a salt of type $AB_2$,the solubility product $K_{sp}$ is given by $K_{sp} = [Pb^{2+}][I^{-}]^2$.If $S$ is the solubility,then $[Pb^{2+}] = S$ and $[I^{-}] = 2S$.Therefore,$K_{sp} = (S)(2S)^2 = 4S^3$.Given $S = 2 	imes 10^{-3} \ mol \ L^{-1}$.Substituting the value of $S$ into the expression: $K_{sp} = 4 	imes (2 	imes 10^{-3})^3 = 4 	imes (8 	imes 10^{-9}) = 32 	imes 10^{-9}$.

If the concentration of lead iodide in its saturated solution at $25 \, ^oC$ is $2 \times 10^{-3} \ mol \ L^{-1}$,then its solubility product is:

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