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Chemistry · Ionic Equilibrium

JEE Advanced 2020 — Paper 2 — Question 33

An acidified solution of 0.05MZn2+0.05 \mathrm{M} \mathrm{Zn}^{2+} is saturated with 0.1MH2 S0.1 \mathrm{M} \mathrm{H}_{2} \mathrm{~S}. What is the minimum molar concentration

(M) of H+\mathrm{H}^{+}required to prevent the precipitation of ZnS ?

Use Ksp(ZnS)=1.25×10−22K_{s p}(\mathrm{ZnS})=1.25 \times 10^{-22} and overall dissociation constant of

H2 S.KNET=K1K2=1×10−21\mathrm{H}_{2} \mathrm{~S} . \mathrm{K}_{\mathrm{NET}}=K_{1} K_{2}=1 \times 10^{-21}

Answer: 0.2

Numerical answer — enter this value.

Step-by-step solution

ZnS(s)⇌Zn2+(aq)+S2(aq)\mathrm{ZnS}(\mathrm{s}) \rightleftharpoons \mathrm{Zn}^{2+}(\mathrm{aq})+\mathrm{S}^{2}(\mathrm{aq})

[S2−]=Ksp [Zn2+]=1.25×10−220.05=25×10−22H2 S⇌2H++S2−10−21=[H+]2×25×10−220.1[H+]2=10−22×102225=125[H+]=15=0.20M\begin{aligned} & {\left[\mathrm{S}^{2-}\right]=\frac{\mathrm{K}_{\text {sp }}}{\left[\mathrm{Zn}^{2+}\right]}=\frac{1.25 \times 10^{-22}}{0.05}=25 \times 10^{-22}} \\& \mathrm{H}_{2} \mathrm{~S} \rightleftharpoons 2 \mathrm{H}^{+}+\mathrm{S}^{2-} \\& 10^{-21}=\frac{\left[\mathrm{H}^{+}\right]^{2} \times 25 \times 10^{-22}}{0.1} \\& {\left[\mathrm{H}^{+}\right]^{2}=\frac{10^{-22} \times 10^{22}}{25}=\frac{1}{25}} \\& {\left[\mathrm{H}^{+}\right]=\frac{1}{5}=0.20 \mathrm{M}} \end{aligned}

Answer key and solution verified before publishing.

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Exam
JEE Advanced 2020
Paper
Paper 2
Subject
Chemistry
Chapter
Ionic Equilibrium
Topic
Sparingly Soluble Salts, Solubility Product & Precipitation Conditions

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