JEE Main 2018 — Chemistry Physical Chemistry

2018mcqmedium

An aqueous solution contains $0.10M {H}_{2}S$ and $0.20M\mathrm{HCl}$ . If the equilibrium constant for the formation of ${\mathrm{HS}}^{-}$ from ${H}_{2}S$ is $1.0\times {10}^{-7}$ and that of ${S}^{2-}$ from ${\mathrm{HS}}^{-}$ ions is $1.2\times {10}^{-13}$ , then, the concentration of ${S}^{2-}$ ions in the aqueous solution is:

Official previous-year question

Held on 8 Apr 2018 · Verified 6 Jul 2026.

Options

A
$5\times {10}^{-19}$
B
$5\times {10}^{-8}$
C
$3\times {10}^{-20}$
D
$6\times {10}^{-21}$

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Solution

$\begin{matrix}\mathrm{HCl} & \rightarrow & {H}^{+} & + & {\mathrm{Cl}}^{-} \\ & & (0.2+x+y) & & 0.2\end{matrix}$

$\begin{matrix}{H}_{2}S & \rightleftharpoons & {H}^{+} & + & {\mathrm{HS}}^{-} \\ (0.1-x) & & (0.2+x+y) & & (x-y)\end{matrix};{K}_{{a}_{1}}={10}^{-7}$

$\begin{matrix}{\mathrm{HS}}^{-} & \rightarrow & {H}^{+} & + & {S}^{2-} \\ (x-y) & & (0.2+x+y) & & (y)\end{matrix};{K}_{{a}_{2}}=1.2\times {10}^{-13}$

$[{H}^{+}]=(0.2+x+y)M=0.2M$

${K}_{{a}_{1}}=\frac{[{H}^{+}][{\mathrm{HS}}^{-}]}{[{H}_{2}S]}=\frac{0.2\times [{\mathrm{HS}}^{-}]}{0.1-x}=\frac{0.2[{\mathrm{HS}}^{-}]}{0.1}$ ( $x=$ negligible)

$[{\mathrm{HS}}^{-}]=\frac{{K}_{{a}_{1}}\times 0.1}{0.2}=\frac{1}{2}\times {10}^{-7}$

${K}_{{a}_{2}}=\frac{[{H}^{+}][{S}^{2-}]}{[{\mathrm{HS}}^{-}]}=\frac{[0.2M][{S}^{2-}]}{\frac{1}{2}\times {10}^{-7}}$

$[{S}^{2-}]=\frac{{K}_{{a}_{2}}\times \frac{1}{2}\times {10}^{-7}}{0.2}=\frac{1.2\times {10}^{-13}\times \frac{1}{2}\times {10}^{-7}}{0.2}=3\times {10}^{-20}M$

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Solution

$\begin{matrix}\mathrm{HCl} & \rightarrow & {H}^{+} & + & {\mathrm{Cl}}^{-} \\ & & (0.2+x+y) & & 0.2\end{matrix}$

$\begin{matrix}{H}_{2}S & \rightleftharpoons & {H}^{+} & + & {\mathrm{HS}}^{-} \\ (0.1-x) & & (0.2+x+y) & & (x-y)\end{matrix};{K}_{{a}_{1}}={10}^{-7}$

$\begin{matrix}{\mathrm{HS}}^{-} & \rightarrow & {H}^{+} & + & {S}^{2-} \\ (x-y) & & (0.2+x+y) & & (y)\end{matrix};{K}_{{a}_{2}}=1.2\times {10}^{-13}$

$[{H}^{+}]=(0.2+x+y)M=0.2M$

${K}_{{a}_{1}}=\frac{[{H}^{+}][{\mathrm{HS}}^{-}]}{[{H}_{2}S]}=\frac{0.2\times [{\mathrm{HS}}^{-}]}{0.1-x}=\frac{0.2[{\mathrm{HS}}^{-}]}{0.1}$ ( $x=$ negligible)

$[{\mathrm{HS}}^{-}]=\frac{{K}_{{a}_{1}}\times 0.1}{0.2}=\frac{1}{2}\times {10}^{-7}$

${K}_{{a}_{2}}=\frac{[{H}^{+}][{S}^{2-}]}{[{\mathrm{HS}}^{-}]}=\frac{[0.2M][{S}^{2-}]}{\frac{1}{2}\times {10}^{-7}}$

$[{S}^{2-}]=\frac{{K}_{{a}_{2}}\times \frac{1}{2}\times {10}^{-7}}{0.2}=\frac{1.2\times {10}^{-13}\times \frac{1}{2}\times {10}^{-7}}{0.2}=3\times {10}^{-20}M$

JEE Main 2018 — Chemistry Physical Chemistry

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JEE Main 2018 — Chemistry Physical Chemistry

Options

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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