JEE Main · 2021 · Shift-IImediumIEQ-017

The solubility of Ca(OH)2 in water is: [Ksp of Ca(OH)2 = 5.5 10-6]

Ionic Equilibrium · Class 11 · JEE Main Previous Year Question

Question

The solubility of Ca(OH)2\mathrm{Ca(OH)_2} in water is:

[Ksp[K_{sp} of Ca(OH)2=5.5×10−6]\mathrm{Ca(OH)_2} = 5.5 \times 10^{-6}]

Options
  1. a

    1.77×10−21.77 \times 10^{-2}

  2. b

    1.77×10−61.77 \times 10^{-6}

  3. c

    1.11×10−21.11 \times 10^{-2}

    ✓
  4. d

    1.11×10−61.11 \times 10^{-6}

Correct Answerc

1.11×10−21.11 \times 10^{-2}

Detailed Solution

Step 1 — Dissociation of Ca(OH)2\mathrm{Ca(OH)_2}: Ca(OH)2⇌Ca2++2OH−[Ca2+]=s,[OH−]=2s\mathrm{Ca(OH)_2 \rightleftharpoons Ca^{2+} + 2OH^-} [\mathrm{Ca^{2+}}] = s, \quad [\mathrm{OH^-}] = 2s

Step 2 — Write KspK_{sp}: Ksp=s(2s)2=4s3=5.5×10−6K_{sp} = s(2s)^2 = 4s^3 = 5.5 \times 10^{-6}

Step 3 — Solve for s: s3=5.5×10−64=1.375×10−6s=(1.375×10−6)1/3=(13.75×10−7)1/3≈2.4×10−71/3≈1.11×10−2 mol L−1s^3 = \frac{5.5 \times 10^{-6}}{4} = 1.375 \times 10^{-6} s = (1.375 \times 10^{-6})^{1/3} = (13.75 \times 10^{-7})^{1/3} \approx 2.4 \times 10^{-7^{1/3}} \approx 1.11 \times 10^{-2}\ \mathrm{mol\ L^{-1}}

Answer: Option (3) — 1.11×10−21.11 \times 10^{-2}

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