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

JEE Main · 2018 · 08 Apr 2018 · Offline · Equilibrium

An aqueous solution contains

0.10 M H_{2} S

and 0.20 M HCl . If the equilibrium constants for the formation of

HS^{-}

from

H_{2} S

1.0 \times 1 0^{- 7}

and that of

S^{2 -}

from

HS^{-}

ions is

1.2 \times 1 0^{- 13}

then the concentration of

S^{2 -}

ions in aqueous solution is :

Answer: (C)

Multiplying the two dissociation steps of

H X_{2} S

gives the overall equilibrium

H X_{2} S 2 H X^{+} + S X^{2 -}

with constant

K = K_{a 1} K_{a 2} = (1.0 \times 1 0^{- 7}) (1.2 \times 1 0^{- 13}) = 1.2 \times 1 0^{- 20}

. The strong acid fixes

[H X^{+}] = 0.20

M and suppresses the sulfide dissociation, so with

[H X_{2} S] = 0.10

K = \frac{[ H X ^{+} ] ^{2} [ S X ^{2 -} ]}{[ H X _{2} S ]}

rearranges to

[S X^{2 -}] = \frac{1.2 \times 1 0 ^{- 20} \times 0.10}{( 0.20 ) ^{2}} = 3 \times 1 0^{- 20}

JEE Main · 2018 · 08 Apr 2018 · Offline · Single correct option