JEE Main · 2020 · Shift-IImediumIEQ-064

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…

Ionic Equilibrium · Class 11 · JEE Main Previous Year Question

Question

The solubility product of Cr(OH)3\mathrm{Cr(OH)_3} at 298 K is 6.0×10−316.0 \times 10^{-31}. The concentration of hydroxide ions in a saturated solution of Cr(OH)3\mathrm{Cr(OH)_3} will be

Options
  1. a

    (2.22×10−31)1/4(2.22 \times 10^{-31})^{1/4}

  2. b

    (18×10−31)1/4(18 \times 10^{-31})^{1/4}

    ✓
  3. c

    (18×10−31)1/2(18 \times 10^{-31})^{1/2}

  4. d

    (4.86×10−29)1/4(4.86 \times 10^{-29})^{1/4}

Correct Answerb

(18×10−31)1/4(18 \times 10^{-31})^{1/4}

Detailed Solution

Step 1 — Dissociation of Cr(OH)3\mathrm{Cr(OH)_3}: Cr(OH)3⇌Cr3++3OH−[Cr3+]=s,[OH−]=3s\mathrm{Cr(OH)_3 \rightleftharpoons Cr^{3+} + 3OH^-} [\mathrm{Cr^{3+}}] = s, \quad [\mathrm{OH^-}] = 3s

Step 2 — Write KspK_{sp}: Ksp=s(3s)3=27s4=6.0×10−31s4=6.0×10−3127=627×10−31=29×10−31K_{sp} = s(3s)^3 = 27s^4 = 6.0 \times 10^{-31} s^4 = \frac{6.0 \times 10^{-31}}{27} = \frac{6}{27} \times 10^{-31} = \frac{2}{9} \times 10^{-31}

Step 3 — Find [OH−][\mathrm{OH^-}]: [OH−]=3s(3s)4=81s4=81×29×10−31=18×10−31[OH−]=(18×10−31)1/4[\mathrm{OH^-}] = 3s (3s)^4 = 81s^4 = 81 \times \frac{2}{9} \times 10^{-31} = 18 \times 10^{-31} [\mathrm{OH^-}] = (18 \times 10^{-31})^{1/4}

Answer: Option (2) — (18×10−31)1/4(18 \times 10^{-31})^{1/4}

Practice this question with progress tracking

Want timed practice with adaptive difficulty? Solve this question (and hundreds more from Ionic Equilibrium) inside The Crucible, our adaptive practice platform.

Practice in The Crucible →Browse Ionic Equilibrium chapter →

More JEE Main Ionic Equilibrium PYQs

2025hard

If equal volumes of AB2 and XY (both are salts) aqueous solutions are mixed, which of the following combination will give a precipitate of AY2 at 300 K? (Given Ksp for AY2 = 5.2 10-7)

2025medium

A weak acid HA has degree of dissociation x. Which option gives the correct expression of pH - pKa?

2024easy

For a sparingly soluble salt AB2, the equilibrium concentrations of A2+ ions and B- ions are 1.2 10-4 M and 0.24 10-3 M, respectively. The solubility product of AB2 is:

2024medium

Given below are two statements: Statement I: On passing HCl gas through a saturated solution of BaCl2, at room temperature white turbidity appears. Statement II: When HCl gas is passed through a…

2024hard

Solubility of calcium phosphate (molecular mass M) in water is W g per 100 mL at 25 °C. Its solubility product at 25 °C will be approximately:

View all 49 Ionic Equilibrium PYQs →
The solubility product of Cr(OH)3 at 298 K is 6.0 10-31. The concentration of hydroxide ions in a… (JEE Main 2020) | Canvas Classes