JEE PYQ: Ionic Equilibrium Question 6
Question 6 - 2021 (24 Feb Shift 2)
Type: Numerical
The solubility product of $\text{PbI}_2$ is $8.0 \times 10^{-9}$. The solubility of lead iodide in 0.1 molar 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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Answer: 141
Solution
$K_{sp}(\text{PbI}_2) = 8 \times 10^{-9}$
$\text{PbI}_2(s) \rightleftharpoons \text{Pb}^{2+}(aq) + 2\text{I}^-(aq)$
In 0.1 M $\text{Pb(NO}_3)_2$: $[\text{Pb}^{2+}] = S + 0.1 \approx 0.1$
$K_{sp} = (S + 0.1)(2S)^2 \approx 0.1 \times 4S^2$
$8 \times 10^{-9} = 0.1 \times 4S^2$
$S^2 = 2 \times 10^{-8}$
$S = 1.414 \times 10^{-4}$ mol/L $= 141.4 \times 10^{-6}$ mol/L
$x \approx 141$
BETA
