The solubility of AgCN in a… - JEE Main 2021 Chemistry PYQ

JEE Main 2021 — Chemistry Physical Chemistry

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The solubility of $\mathrm{AgCN}$ in a buffer solution of $\mathrm{pH}=3$ is $x$ . The value of $x$ is: [Assume : No cyano complex is formed ; ${K}_{\mathrm{sp}}(\mathrm{AgCN})=2.2\times {10}^{-16}$ and ${K}_{a}(\mathrm{HCN})=6.2\times {10}^{-10}$ ]

Official previous-year question

Held on 25 Feb 2021 · Verified 6 Jul 2026.

Options

A
$1.9\times {10}^{-5}$
B
$2.2\times {10}^{-16}$
C
$0.625\times {10}^{-6}$
D
$1.6\times {10}^{-6}$

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Solution

Let, the solubility be $x$

Given that,

${K}_{\mathrm{sp}}(\mathrm{AgCN})=2.2\times {10}^{-16}$ and ${K}_{a}(\mathrm{HCN})=6.2\times {10}^{-10}$

$\mathrm{AgCN}\rightleftharpoons \underset{x}{{\mathrm{Ag}}^{+}}+\underset{x}{{\mathrm{CN}}^{-}}{K}_{\mathrm{sp}}=2.2\times {10}^{-16}$

${H}^{+}+{\mathrm{CN}}^{-}\rightleftharpoons \mathrm{HCN}K=\frac{1}{{K}_{a}}=\frac{1}{6.2\times {10}^{-10}}$

${K}_{\mathrm{sp}}\times \frac{1}{{K}_{a}}=[{\mathrm{Ag}}^{+}][{\mathrm{CN}}^{-}]\times \frac{[\mathrm{HCN}]}{[{H}^{+}][{\mathrm{CN}}^{-}]}$

$2.2\times {10}^{-16}\times \frac{1}{6.2\times {10}^{-10}}=\frac{[S][S]}{{10}^{-3}}$

${S}^{2}=\frac{2.2}{6.2}\times {10}^{-9}$

${S}^{2}=3.55\times {10}^{-10}$

$S=\sqrt{3.55\times {10}^{-10}}$

$S=1.88\times {10}^{-5}\Rightarrow 1.9\times {10}^{-5}$

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Solution

Let, the solubility be $x$

Given that,

${K}_{\mathrm{sp}}(\mathrm{AgCN})=2.2\times {10}^{-16}$ and ${K}_{a}(\mathrm{HCN})=6.2\times {10}^{-10}$

$\mathrm{AgCN}\rightleftharpoons \underset{x}{{\mathrm{Ag}}^{+}}+\underset{x}{{\mathrm{CN}}^{-}}{K}_{\mathrm{sp}}=2.2\times {10}^{-16}$

${H}^{+}+{\mathrm{CN}}^{-}\rightleftharpoons \mathrm{HCN}K=\frac{1}{{K}_{a}}=\frac{1}{6.2\times {10}^{-10}}$

${K}_{\mathrm{sp}}\times \frac{1}{{K}_{a}}=[{\mathrm{Ag}}^{+}][{\mathrm{CN}}^{-}]\times \frac{[\mathrm{HCN}]}{[{H}^{+}][{\mathrm{CN}}^{-}]}$

$2.2\times {10}^{-16}\times \frac{1}{6.2\times {10}^{-10}}=\frac{[S][S]}{{10}^{-3}}$

${S}^{2}=\frac{2.2}{6.2}\times {10}^{-9}$

${S}^{2}=3.55\times {10}^{-10}$

$S=\sqrt{3.55\times {10}^{-10}}$

$S=1.88\times {10}^{-5}\Rightarrow 1.9\times {10}^{-5}$

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Options

Did you get this right?

Solution

Your note

JEE Main Physical Chemistry in other years

More JEE Main Physical Chemistry questions

Other Chemistry topics for JEE Main

Organic Chemistry

Inorganic Chemistry

Keep practicing Physical Chemistry