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(D) The solubility product of $PbI_{2}$ is given by $K_{sp} = [Pb^{2+}][I^{-}]^{2} = 8.0 	imes 10^{-9}$.In $0.1 \ M$ $Pb(NO_{3})_{2}$ solution,$Pb(NO

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$141$

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(D) The solubility product of $PbI_{2}$ is given by $K_{sp} = [Pb^{2+}][I^{-}]^{2} = 8.0 	imes 10^{-9}$.In $0.1 \ M$ $Pb(NO_{3})_{2}$ solution,$Pb(NO_{3})_{2}$ dissociates completely as $Pb(NO_{3})_{2} ightarrow Pb^{2+} + 2NO_{3}^{-}$.Thus,the concentration of $Pb^{2+}$ ions from $Pb(NO_{3})_{2}$ is $0.1 \ M$.Let the solubility of $PbI_{2}$ be $s \ mol/L$. The dissociation of $PbI_{2}$ is $PbI_{2}(s) ightleftharpoons Pb^{2+}(aq) + 2I^{-}(aq)$.The total concentration of $Pb^{2+}$ is $(0.1 + s) \approx 0.1 \ M$ (since $s$ is very small) and $[I^{-}] = 2s$.Substituting these into the $K_{sp}$ expression: $8.0 	imes 10^{-9} = (0.1)(2s)^{2}$.$8.0 	imes 10^{-9} = 0.1 	imes 4s^{2} = 0.4s^{2}$.$s^{2} = \frac{8.0 	imes 10^{-9}}{0.4} = 20 	imes 10^{-9} = 2.0 	imes 10^{-8}$.$s = \sqrt{2.0 	imes 10^{-8}} = \sqrt{2} 	imes 10^{-4} = 1.41 	imes 10^{-4} = 141 	imes 10^{-6} \ mol/L$.Comparing this with $x 	imes 10^{-6} \ mol/L$,we get $x = 141$.

The solubility product of $PbI_{2}$ is $8.0 \times 10^{-9}$. The solubility of lead iodide in $0.1 \ M$ solution of lead nitrate is $x \times 10^{-6} \ mol/L$. The value of $x$ is ....... .(Rounded off to the nearest integer)
$[$Given: $\sqrt{2}=1.41]$

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The solubility product of $PbI_{2}$ is $8.0 \times 10^{-9}$. The solubility of lead iodide in $0.1 \ M$ solution of lead nitrate is $x \times 10^{-6} \ mol/L$. The value of $x$ is ....... .(Rounded off to the nearest integer)$[$Given: $\sqrt{2}=1.41]$