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 6.0×10316.0 \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

    (2.22×1031)1/4(2.22 \times 10^{-31})^{1/4}

  2. b

    (18×1031)1/4(18 \times 10^{-31})^{1/4}

  3. c

    (18×1031)1/2(18 \times 10^{-31})^{1/2}

  4. d

    (4.86×1029)1/4(4.86 \times 10^{-29})^{1/4}

Correct Answerb

(18×1031)1/4(18 \times 10^{-31})^{1/4}

Detailed Solution

Step 1 — Dissociation of Cr(OH)3\mathrm{Cr(OH)_3}: Cr(OH)3Cr3++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=6.0×1031s4=6.0×103127=627×1031=29×1031K_{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=81×29×1031=18×1031[OH]=(18×1031)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) — (18×1031)1/4(18 \times 10^{-31})^{1/4}

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