JEE Main 2021 · 25 February · evening · Single correct · +4 / −1

The solubility of Ca(OH)₂ in water is :
[Given : The solubility product of Ca(OH)₂ in water = 5.5 × 10⁻⁶]

  1. (a) 1.11 × 10⁻⁶
  2. (b) 1.11 × 10⁻²
  3. (c) 1.77 × 10⁻⁶
  4. (d) 1.77 × 10⁻²
Answer and solution

Answer (b) 1.11 × 10⁻²

Solution

Chemistry ma’am · 3 steps

Each Ca(OH)₂ gives two OH⁻ ions, so set Kₛₚ = 4s³ and take a cube root.

  1. Write the dissociation. Ca(OH)₂ ⇌ Ca²⁺ + 2OH⁻. If s mol of the solid dissolves per litre, then [Ca²⁺] = s and [OH⁻] = 2s.

    Kₛₚ = [Ca²⁺][OH⁻]² = s(2s)² = 4s³

  2. Put in numbers. s³ = (5.5 × 10⁻⁶)/4 = 1.375 × 10⁻⁶. The cube root of 10⁻⁶ is 10⁻², and the cube root of 1.375 is about 1.11.
  3. So. s = 1.11 × 10⁻² mol L⁻¹, which is option (b).

The idea to keep. For a salt AB₂ or A₂B, Kₛₚ = 4s³. Always square the ion that appears twice, including its factor of 2.

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

Study Ionic Equilibrium

Asked about 100 times in JEE Main since 2020. This question is from 2021.