Which of the following happens when $\mathrm{NH}_{4} \mathrm{OH}$ is added gradually to the solution containing $1 \mathrm{M} \mathrm{A}^{2+}$ and $1 \mathrm{M} \mathrm{B}^{3+}$ ions? Given : $\mathrm{K}_{\mathrm{sp}}\left[\mathrm{A}(\mathrm{OH})_{2}\right]=9 \times 10^{-10}$ and \[ \mathrm{K}_{\mathrm{sp}}\left[\mathrm{~B}(\mathrm{OH})_{3}\right]=27 \times 10^{-18} \text { at } 298 \mathrm{~K} . \]

(A) $\mathrm{B}(\mathrm{OH})_{3}$ will precipitate before $\mathrm{A}(\mathrm{OH})_{2}$

(B) $\mathrm{A}(\mathrm{OH})_{2}$ and $\mathrm{B}(\mathrm{OH})_{3}$ will precipitate together

(C) $\mathrm{A}(\mathrm{OH})_{2}$ will precipitate before $\mathrm{B}(\mathrm{OH})_{3}$

(D) Both $\mathrm{A}(\mathrm{OH})_{2}$ and $\mathrm{B}(\mathrm{OH})_{3}$ do not show precipitation with $\mathrm{NH}_{4} \mathrm{OH}$

JEE Main · 2025 · 23 Jan 2025 · Shift 1 · Equilibrium

Which of the following happens when

NH_{4} OH

is added gradually to the solution containing

1 M A^{2 +}

and

1 M B^{3 +}

ions?

Given :

K_{sp} [A (OH)_{2}] = 9 \times 1 0^{- 10}

and

K_{sp} [B (OH)_{3}] = 27 \times 1 0^{- 18} at 298 K .

(A) $B (OH)_{3}$ will precipitate before $A (OH)_{2}$ ✓ Correct
(B) $A (OH)_{2}$ and $B (OH)_{3}$ will precipitate together
(C) $A (OH)_{2}$ will precipitate before $B (OH)_{3}$
(D)Both $A (OH)_{2}$ and $B (OH)_{3}$ do not show precipitation with $NH_{4} OH$

Answer: (A)

Precipitation begins once the ionic product exceeds

K_{s p}

. With

[A^{2 +}] = [B^{3 +}] = 1 M

:

For

A (OH)_{2}

[O H^{-}]^{2} > 9 \times 1 0^{- 10} \Rightarrow [O H^{-}] > 3 \times 1 0^{- 5} M

For

B (OH)_{3}

[O H^{-}]^{3} > 27 \times 1 0^{- 18} \Rightarrow [O H^{-}] > 3 \times 1 0^{- 6} M

Since

B (OH)_{3}

reaches saturation at a much lower hydroxide concentration, as

N H_{4} OH

is added gradually and

[O H^{-}]

rises slowly from zero,

B (OH)_{3}

precipitates before

A (OH)_{2}

JEE Main · 2025 · 23 Jan 2025 · Shift 1 · Single correct option