JEE Main 2021 — Chemistry Physical Chemistry

2021mcqhard

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
$1.6\times {10}^{-6}$

Did you get this right?

Solution

Let, the solubility be $x$

Given that,

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

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

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

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

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

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

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

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

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

Your note

JEE Main Physical Chemistry in other years

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JEE Main 2021 — Chemistry Physical Chemistry

2021mcqhard

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
$1.6\times {10}^{-6}$

Did you get this right?

Solution

Let, the solubility be $x$

Given that,

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

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

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

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

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

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

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

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

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

Your note

JEE Main Physical Chemistry in other years

Keep practicing Physical Chemistry

Work through every JEE Main Physical Chemistry PYQ, year by year.

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JEE Main 2021 — Chemistry Physical Chemistry

Options

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Solution

Your note

JEE Main Physical Chemistry in other years

More JEE Main Physical Chemistry questions

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JEE Main 2021 — Chemistry Physical Chemistry

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