Which of the following happens when NH₄ OH is added gradually to the solution containing 1 M A²⁺ and 1 M B³⁺ ions?
Given : Kₛₚ[A(OH)₂]=9 × 10⁻¹⁰ and Kₛₚ[B(OH)₃]=27 × 10⁻¹⁸ at 298 K.

Solution

Chemistry ma’am · 4 steps

Find the OH⁻ each ion needs to start precipitating; the smaller need comes out first.

1. Precipitation condition. A hydroxide starts to precipitate when its ionic product just reaches Kₛₚ. Adding NH₄OH slowly raises [OH⁻], so the ion whose threshold [OH⁻] is lower precipitates first.
2. For A2+. [A²⁺][OH⁻]² = 9 × 10⁻¹⁰ with [A²⁺] = 1 M gives [OH⁻]² = 9 × 10⁻¹⁰, so [OH⁻] = 3 × 10⁻⁵ M.
3. For B3+. [B³⁺][OH⁻]³ = 27 × 10⁻¹⁸ with [B³⁺] = 1 M gives [OH⁻]³ = 27 × 10⁻¹⁸, so [OH⁻] = 3 × 10⁻⁶ M.
4. So. B(OH)₃ needs only 3 × 10⁻⁶ M of OH⁻, ten times less than A(OH)₂ needs. It reaches its threshold first, so B(OH)₃ precipitates before A(OH)₂, which is option (d).

The idea to keep. Do not compare Kₛₚ values of salts with different formulae directly. Work out the precipitating-ion concentration each one needs; the lowest threshold precipitates first.

Answer from the NTA final answer key. Solution by PrepWiser.