JEE Main 2026 — Jan 24, Shift 1 — Chemistry solution

JEE Main2026Jan 24, Shift 1Chemistry

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

(Given :

$K_{s p} (X S) = 1 \times 1 0^{- 22}$ at $2 5^{\circ} C$ , $K_{s p} (Y S) = 4 \times 1 0^{- 16}$ at $2 5^{\circ} C$ ,

$[H_{2} S] = 0.1$ M in solution, $K_{a 1} \times K_{a 2} (H_{2} S) = 1.0 \times 1 0^{- 21}$ ,

$lo g 2 = 0.30$ , $lo g 3 = 0.48$ , $lo g 5 = 0.70$ )

Your answer

Solution

For YS to begin precipitating, $[Y^{2 +}] [S^{2 -}] \geq K_{s p} (Y S)$ .

$[S^{2 -}] \geq \frac{4 \times 1 0 ^{- 16}}{0.01} = 4 \times 1 0^{- 14}$ M.

From the two-step ionisation of $H_{2} S$ :

$[S^{2 -}] = \frac{K _{a 1} K _{a 2} [ H _{2} S ]}{[ H ^{+} ] ^{2}} = \frac{1 0 ^{- 21} \times 0.1}{[ H ^{+} ] ^{2}} = \frac{1 0 ^{- 22}}{[ H ^{+} ] ^{2}}$ .

Setting this $\geq 4 \times 1 0^{- 14}$ :

$[H^{+}]^{2} \leq \frac{1 0 ^{- 22}}{4 \times 1 0 ^{- 14}} = \frac{1}{4} \times 1 0^{- 8}$

$\Rightarrow [H^{+}] \leq \frac{1}{2} \times 1 0^{- 4}$

$\Rightarrow p H \geq 4 + lo g 2 = 4.30$ .

Nearest integer: $4$ .

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