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JEE Main 202624 JanuaryMorning Shift
Exam
JEE Main
Year
2026
Date
24 January
Shift
Morning Shift
Subject
Chemistry
Chapter
Ionic Equilibrium

JEE Main 2026 · 24 January · Morning Shift · Chemistry · Ionic Equilibrium

Consider two Group IV metal ions X^ 2+ and Y^ 2+ . A solution...

Consider two Group IV metal ions X2+and\mathrm{X}^{2+} and\mathrm{Y}^{2+}.Asolutioncontaining. A solution containing0.01 \mathrm{M} \mathrm{X}^{2+} and 0.01MY2+issaturatedwith0.01 \mathrm{M} \mathrm{Y}^{2+} is saturated with\mathrm{H}_{2} \mathrm{~S}.ThepHatwhichthemetalsulphideYSwillformasaprecipitateis. The pH at which the metal sulphide YS will form as a precipitate is\_\_\_\_.(Nearestinteger)<br>(Given:. (Nearest integer) <br> (Given:\mathrm{K}_{\mathrm{sp}}(\mathrm{XS})=1 \times 10^{-22}atat25^{\circ} \mathrm{C}, \mathrm{K}_{\mathrm{sp}}(\mathrm{YS})=4 \times 10^{-16}atat25^{\circ} \mathrm{C},[H2 S]=0.1M, \left[\mathrm{H}_{2} \mathrm{~S}\right]=0.1 \mathrm{M} in solution, Ka1×Ka2(H2 S)=1.0×10−21,log⁡2=0.30\mathrm{K}_{a 1} \times \mathrm{K}_{a 2}\left(\mathrm{H}_{2} \mathrm{~S}\right)=1.0 \times 10^{-21}, \log 2=0.30, log⁡3=0.48,log⁡5=0.70\log 3=0.48, \log 5=0.70)

Correct answer

0

Step-by-step solution

XS(s)⇌X+2(aq.)+S2−(aq.)XS(s) \rightleftharpoons X^{+2}(aq.) + S^{2-}(aq.)
For precipitation of XS(s)XS(s)
[X+2][S2−]≥Ksp(XS)[X^{+2}][S^{2-}] \geq K_{sp}(XS)
[S2−]≥1×10−220.01=10−20[S^{2-}] \geq \dfrac{1 \times 10^{-22}}{0.01} = 10^{-20}
YS(s)⇌Y+2(aq)+S2−(aq)YS(s) \rightleftharpoons Y^{+2}(aq) + S^{2-}(aq)
For precipitation of YS(s)YS(s)
[Y+2][S2−]≥Ksp(YS)[Y^{+2}][S^{2-}] \geq K_{sp}(YS)
[S2−]≥4×10−1610−2=4×10−14[S^{2-}] \geq \dfrac{4 \times 10^{-16}}{10^{-2}} = 4 \times 10^{-14}
Now, H2S(aq)⇌2H+(aq)+S2−(aq)H_2S(aq) \rightleftharpoons 2H^+(aq) + S^{2-}(aq)
[S2−][H+]2H2S=Ka1×Ka2=1×10−21\dfrac{[S^{2-}][H^+]^2}{H_2S} = K_{a_1} \times K_{a_2} = 1 \times 10^{-21}
[S2−]=1×10−21×[H2S][H+]2≥4×10−14[S^{2-}] = \dfrac{1 \times 10^{-21} \times [H_2S]}{[H^+]^2} \geq 4 \times 10^{-14}
[H+]2≤14×10−7×10−1[H^+]^2 \leq \dfrac{1}{4} \times 10^{-7} \times 10^{-1}
[H+]≤12×10−4⇒pH≥4.3[H^+] \leq \dfrac{1}{2} \times 10^{-4} \Rightarrow pH \geq 4.3
Nearest integer pH = 4.
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