JEE Main 2025 Apr 2 Shift 1, Chemistry Q9: Solubility Equilibria Of Sparingly Soluble Salts

JEE Main2025Apr 2, Shift 1Chemistry

If equal volumes of $A B_{2}$ and XY (both are salts) aqueous solutions are mixed, which of the following combination will give a precipitate of $A Y_{2}$ at 300 K?

(Given $K_{s p}$ (at 300 K) for $A Y_{2} = 5.2 \times 1 0^{- 7}$ )

A
$3.6 \times 1 0^{- 3}$ M $A B_{2}$ , $5.0 \times 1 0^{- 4}$ M XY
B
$2.0 \times 1 0^{- 4}$ M $A B_{2}$ , $0.8 \times 1 0^{- 3}$ M XY
C
$2.0 \times 1 0^{- 2}$ M $A B_{2}$ , $2.0 \times 1 0^{- 2}$ M XY
D
$1.5 \times 1 0^{- 4}$ M $A B_{2}$ , $1.5 \times 1 0^{- 3}$ M XY

Solution

Mixing equal volumes halves each species' concentration. Precipitation requires the reaction quotient to exceed the solubility product:

$Q_{s p} = [A^{2 +}] [Y^{-}]^{2} > K_{s p} = 5.2 \times 1 0^{- 7}$

After mixing, $[A^{2 +}] = \frac{[ A B _{2} ] _{0}}{2}$ and $[Y^{-}] = \frac{[ X Y ] _{0}}{2}$ .

Evaluate each option:

(1) $Q_{s p} = (1.8 \times 1 0^{- 3}) (2.5 \times 1 0^{- 4})^{2} \approx 1.1 \times 1 0^{- 10}$ $< K_{s p}$ .

(2) $Q_{s p} = (1.0 \times 1 0^{- 4}) (4.0 \times 1 0^{- 4})^{2} = 1.6 \times 1 0^{- 11}$ $< K_{s p}$ .

(3) $Q_{s p} = (1.0 \times 1 0^{- 2}) (1.0 \times 1 0^{- 2})^{2} = 1.0 \times 1 0^{- 6}$ $> K_{s p}$ $\Rightarrow$ precipitation occurs.

(4) $Q_{s p} = (7.5 \times 1 0^{- 5}) (7.5 \times 1 0^{- 4})^{2} \approx 4.2 \times 1 0^{- 11}$ $< K_{s p}$ .

$∴$ Only option (3) yields a precipitate.

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JEE Main 2025 Apr 2 Shift 1, Chemistry Q9: Solubility Equilibria Of Sparingly Soluble Salts

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