An acidified solution of \(0.05 \mathrm{M} \mathrm{Zn}^{2+}\) is saturated with \(0.1 \mathrm{M} \mathrm{H}_{2} \mathrm{~S}\). What is the minimum molar concentration (M) of \(\mathrm{H}^{+}\)required to prevent the precipitation of ZnS ? Use \(K_{\mathrm{sp}}(\mathrm{ZnS})=1.25 \times 10^{-22}\) and Overall dissociation constant of \(\mathrm{H}_{2} \mathrm{~S}, K_{\mathrm{NET}}=K_{1} K_{2}=1 \times 10^{-21}\)

JEE Advanced · 2020 · 27 Sep 2020 · Paper 2 · Equilibrium

An acidified solution of

0.05 M Zn^{2 +}

is saturated with

0.1 M H_{2} S

. What is the minimum molar concentration (M) of

H^{+}

required to prevent the precipitation of ZnS ?
Use

K_{sp} (ZnS) = 1.25 \times 1 0^{- 22}

and
Overall dissociation constant of

H_{2} S, K_{NET} = K_{1} K_{2} = 1 \times 1 0^{- 21}

Answer: 0.2

To prevent precipitation of ZnS, we need:

[Zn^{2 +}] [S^{2 -}] < K_{s p}

At the threshold of precipitation:

[S^{2 -}] = \frac{K _{s p}}{[ Zn ^{2 +} ]} = \frac{1.25 \times 1 0 ^{- 22}}{0.05} = 2.5 \times 1 0^{- 21} M

For

H_{2} S

dissociation:

H_{2} S ⇌ 2 H^{+} + S^{2 -}

K_{NET} = \frac{[ H ^{+} ] ^{2} [ S ^{2 -} ]}{[ H _{2} S ]}

Substituting values:

1 \times 1 0^{- 21} = \frac{[ H ^{+} ] ^{2} \times 2.5 \times 1 0 ^{- 21}}{0.1}

[H^{+}]^{2} = \frac{1 \times 1 0 ^{- 21} \times 0.1}{2.5 \times 1 0 ^{- 21}} = \frac{0.1}{2.5} = 0.04

[H^{+}] = 0.2 M

Minimum

[H^{+}]

= 0.2 M

JEE Advanced · 2020 · 27 Sep 2020 · Paper 2 · Numerical answer