The solubility product of PbI_2 is 1.08 times | vedclass

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(D) The dissociation of $PbI_2$ is given by: $PbI_2(s) \rightleftharpoons Pb^{2+}(aq) + 2I^-(aq)$.Let the solubility of $PbI_2$ be $s \ mol \ dm^{-3}$

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$3.0 	imes 10^{-3}$

Vedclass · Answer

(D) The dissociation of $PbI_2$ is given by: $PbI_2(s) ightleftharpoons Pb^{2+}(aq) + 2I^-(aq)$.Let the solubility of $PbI_2$ be $s \ mol \ dm^{-3}$.Then,$[Pb^{2+}] = s$ and $[I^-] = 2s$.The solubility product expression is $K_{sp} = [Pb^{2+}][I^-]^2$.Substituting the values: $1.08 	imes 10^{-7} = (s)(2s)^2 = 4s^3$.Solving for $s$: $s^3 = \frac{1.08 	imes 10^{-7}}{4} = 0.27 	imes 10^{-7} = 27 	imes 10^{-9}$.Taking the cube root: $s = \sqrt[3]{27 	imes 10^{-9}} = 3 	imes 10^{-3} \ mol \ dm^{-3}$.

The solubility product of $PbI_2$ is $1.08 \times 10^{-7}$. Calculate its solubility in $mol \ dm^{-3}$ at $298 \ K$?

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