The solubility product of… - JEE Main 2021 Chemistry PYQ

JEE Main 2021 — Chemistry Physical Chemistry

2021integermedium

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

Official previous-year question

Held on 24 Feb 2021 · Verified 6 Jul 2026.

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Solution

Given: ${[{K}_{\mathrm{sp}}]}_{{\mathrm{PbI}}_{2}}=8\times {10}^{-9}$

To calculate : solubility of ${\mathrm{PbI}}_{2}$ in $0.1M$ solution of $\mathrm{Pb}{({\mathrm{NO}}_{2})}_{2}$

$(I)\mathrm{Pb}{({\mathrm{NO}}_{3})}_{2}\rightarrow {\mathrm{Pb}}_{(\mathrm{aq})}^{+2}+2{\mathrm{NO}}_{3}^{-}(\mathrm{aq})$

$\begin{matrix}0.1M & - & - \\ - & 0.1M & 0.2M\end{matrix}$

$(\mathrm{II}){\mathrm{PbI}}_{2}(s)\rightleftharpoons {\mathrm{Pb}}^{+2}(\mathrm{aq})+2{I}^{-}(\mathrm{aq})$

$\begin{matrix}s & 2s\end{matrix}$

$[{\mathrm{Pb}}^{+2}]=s+0.1$

$\simeq 0.1$

Now: ${K}_{\mathrm{sp}}=8\times {10}^{-9}=[{\mathrm{Pb}}^{+2}]{[{I}^{-}]}^{2}$

$\Rightarrow 8\times {10}^{-9}=0.1\times {(2s)}^{2}$

$\Rightarrow 8\times {10}^{-8}=4{s}^{2}\Rightarrow s=\sqrt{2}\times {10}^{-4}$

$\Rightarrow S=141\times {10}^{-6}M$

$\Rightarrow x=141$

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JEE Main 2021 — Chemistry Physical Chemistry

2021integermedium

Official previous-year question

Held on 24 Feb 2021 · Verified 6 Jul 2026.

Did you get this right?

Solution

Given: ${[{K}_{\mathrm{sp}}]}_{{\mathrm{PbI}}_{2}}=8\times {10}^{-9}$

To calculate : solubility of ${\mathrm{PbI}}_{2}$ in $0.1M$ solution of $\mathrm{Pb}{({\mathrm{NO}}_{2})}_{2}$

$(I)\mathrm{Pb}{({\mathrm{NO}}_{3})}_{2}\rightarrow {\mathrm{Pb}}_{(\mathrm{aq})}^{+2}+2{\mathrm{NO}}_{3}^{-}(\mathrm{aq})$

$\begin{matrix}0.1M & - & - \\ - & 0.1M & 0.2M\end{matrix}$

$(\mathrm{II}){\mathrm{PbI}}_{2}(s)\rightleftharpoons {\mathrm{Pb}}^{+2}(\mathrm{aq})+2{I}^{-}(\mathrm{aq})$

$\begin{matrix}s & 2s\end{matrix}$

$[{\mathrm{Pb}}^{+2}]=s+0.1$

$\simeq 0.1$

Now: ${K}_{\mathrm{sp}}=8\times {10}^{-9}=[{\mathrm{Pb}}^{+2}]{[{I}^{-}]}^{2}$

$\Rightarrow 8\times {10}^{-9}=0.1\times {(2s)}^{2}$

$\Rightarrow 8\times {10}^{-8}=4{s}^{2}\Rightarrow s=\sqrt{2}\times {10}^{-4}$

$\Rightarrow S=141\times {10}^{-6}M$

$\Rightarrow x=141$

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