JEE Main 2020 · 9 January · evening · Single correct · +4 / −1

The solubility product of Cr(OH)₃ at 298 K is 6.0 × 10 –31 . The concentration of hydroxide ions in a saturated solution of Cr(OH)₃ will be :

  1. (a) (2.22 × 10 –31 ) 1/4
  2. (b) (4.86 × 10 –29 ) 1/4
  3. (c) (18 × 10 –31 ) 1/4
  4. (d) (18 × 10 –31 ) 1/2
Answer and solution

Answer (c) (18 × 10 –31 ) 1/4

Solution

Chemistry ma’am · 3 steps

Write Kₛₚ in terms of [OH⁻] itself, and the 4th power with a factor of 3 falls right out.

  1. Set up dissolution. Cr(OH)₃ ⇌ Cr³⁺ + 3OH⁻. If the solubility is s, then [Cr³⁺] = s and [OH⁻] = 3s.
  2. Use x for hydroxide. Let [OH⁻] = x, so [Cr³⁺] = x/3. Then Kₛₚ = [Cr³⁺][OH⁻]³ = x/3 × x³ = x⁴/3.

    x⁴ = 3Kₛₚ = 3 × 6.0 × 10⁻³¹ = 18 × 10⁻³¹

  3. So. [OH⁻] = (18 × 10⁻³¹)^(1/4), which is option (c).

The idea to keep. For a salt that gives ions in a 1 : n ratio, write one ion's concentration in terms of the other before substituting. The total number of ions sets the power of the root.

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 2020.