JEE Main 2025 Jan 23 Shift 1, Chemistry Q18: Solubility Equilibria Of Sparingly Soluble Salts

JEE Main2025Jan 23, Shift 1Chemistry

Which of the following happens when $N H_{4} O H$ is added gradually to the solution containing 1M $A^{2 +}$ and 1M $B^{3 +}$ ions ?

Given : $K_{s p} [A (O H)_{2}] = 9 \times 1 0^{- 10}$ and $K_{s p} [B (O H)_{3}] = 27 \times 1 0^{- 18}$ at 298 K.

A
$B (O H)_{3}$ will precipitate before $A (O H)_{2}$
B
$A (O H)_{2}$ and $B (O H)_{3}$ will precipitate together
C
$A (O H)_{2}$ will precipitate before $B (O H)_{3}$
D
Both $A (O H)_{2}$ and $B (O H)_{3}$ do not show precipitation with $N H_{4} O H$

Solution

A precipitate first appears when the ionic product just exceeds $K_{s p}$ . Whichever hydroxide needs a lower $[O H^{-}]$ to start precipitating will appear first.

For $A (O H)_{2}$ : $[A^{2 +}] [O H^{-}]^{2} = K_{s p} \Rightarrow [O H^{-}]_{min} = \frac{9 \times 1 0 ^{- 10}}{1} = 3 \times 1 0^{- 5}$ M.

For $B (O H)_{3}$ : $[B^{3 +}] [O H^{-}]^{3} = K_{s p} \Rightarrow [O H^{-}]_{min} = 3 \frac{27 \times 1 0 ^{- 18}}{1} = 3 \times 1 0^{- 6}$ M.

$B (O H)_{3}$ needs only $3 \times 1 0^{- 6}$ M $O H^{-}$ , far less than $A (O H)_{2}$ . So $B (O H)_{3}$ precipitates first.

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JEE Main 2025 Jan 23 Shift 1, Chemistry Q18: Solubility Equilibria Of Sparingly Soluble Salts

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