JEE Main 2026 — Chemistry Physical Chemistry

2026integerhard

Consider two Group IV metal ions $\mathrm{X}^{2+}$ and $\mathrm{Y}^{2+}$ . A solution containing $0.01 \mathrm{M} \mathrm{X}^{2+}$ and $0.01 \mathrm{M} \mathrm{Y}^{2+}$ is saturated with $\mathrm{H}_{2} \mathrm{~S}$ . The pH at which the metal sulphide YS will form as a precipitate is $\_\_\_\_$ . (Nearest integer)

(Given: $\mathrm{K}_{\mathrm{sp}}(\mathrm{XS})=1 \times 10^{-22}$ at $25^{\circ} \mathrm{C}, \mathrm{K}_{\mathrm{sp}}(\mathrm{YS})=4 \times 10^{-16}$ at $25^{\circ} \mathrm{C}$ , $\left[\mathrm{H}_{2} \mathrm{~S}\right]=0.1 \mathrm{M}$ in solution, $\mathrm{K}_{a 1} \times \mathrm{K}_{a 2}\left(\mathrm{H}_{2} \mathrm{~S}\right)=1.0 \times 10^{-21}, \log 2=0.30$ , $\log 3=0.48, \log 5=0.70$ )

Official previous-year question

Held on 24 Jan 2026 · Verified 6 Jul 2026.

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Solution

$XS(s) \rightleftharpoons X^{+2}(aq.) + S^{2-}(aq.)$

For precipitation of $XS(s)$

$[X^{+2}][S^{2-}] \geq K_{sp}(XS)$

$[S^{2-}] \geq \dfrac{1 \times 10^{-22}}{0.01} = 10^{-20}$

$YS(s) \rightleftharpoons Y^{+2}(aq) + S^{2-}(aq)$

For precipitation of $YS(s)$

$[Y^{+2}][S^{2-}] \geq K_{sp}(YS)$

$[S^{2-}] \geq \dfrac{4 \times 10^{-16}}{10^{-2}} = 4 \times 10^{-14}$

Now, $H_2S(aq) \rightleftharpoons 2H^+(aq) + S^{2-}(aq)$

$\dfrac{[S^{2-}][H^+]^2}{H_2S} = K_{a_1} \times K_{a_2} = 1 \times 10^{-21}$

$[S^{2-}] = \dfrac{1 \times 10^{-21} \times [H_2S]}{[H^+]^2} \geq 4 \times 10^{-14}$

$[H^+]^2 \leq \dfrac{1}{4} \times 10^{-7} \times 10^{-1}$

$[H^+] \leq \dfrac{1}{2} \times 10^{-4} \Rightarrow pH \geq 4.3$

Nearest integer pH = 4.

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Solution

$XS(s) \rightleftharpoons X^{+2}(aq.) + S^{2-}(aq.)$

For precipitation of $XS(s)$

$[X^{+2}][S^{2-}] \geq K_{sp}(XS)$

$[S^{2-}] \geq \dfrac{1 \times 10^{-22}}{0.01} = 10^{-20}$

$YS(s) \rightleftharpoons Y^{+2}(aq) + S^{2-}(aq)$

For precipitation of $YS(s)$

$[Y^{+2}][S^{2-}] \geq K_{sp}(YS)$

$[S^{2-}] \geq \dfrac{4 \times 10^{-16}}{10^{-2}} = 4 \times 10^{-14}$

Now, $H_2S(aq) \rightleftharpoons 2H^+(aq) + S^{2-}(aq)$

$\dfrac{[S^{2-}][H^+]^2}{H_2S} = K_{a_1} \times K_{a_2} = 1 \times 10^{-21}$

$[S^{2-}] = \dfrac{1 \times 10^{-21} \times [H_2S]}{[H^+]^2} \geq 4 \times 10^{-14}$

$[H^+]^2 \leq \dfrac{1}{4} \times 10^{-7} \times 10^{-1}$

$[H^+] \leq \dfrac{1}{2} \times 10^{-4} \Rightarrow pH \geq 4.3$

Nearest integer pH = 4.

JEE Main 2026 — Chemistry Physical Chemistry

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JEE Main 2026 — Chemistry Physical Chemistry

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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