JEE Main · 2019 · Shift-ImediumIEQ-065

What is the molar solubility of Al(OH)3 in 0.2 M NaOH solution? [Ksp(Al(OH)3) = 2.4 10-24]

Ionic Equilibrium · Class 11 · JEE Main Previous Year Question

Question

What is the molar solubility of Al(OH)3\mathrm{Al(OH)_3} in 0.2 M NaOH solution?

[Ksp(Al(OH)3)=2.4×10−24][K_{sp}(\mathrm{Al(OH)_3}) = 2.4 \times 10^{-24}]

Options
  1. a

    3×10−193 \times 10^{-19}

  2. b

    12×10−2112 \times 10^{-21}

  3. c

    12×10−2312 \times 10^{-23}

  4. d

    3×10−223 \times 10^{-22}

    ✓
Correct Answerd

3×10−223 \times 10^{-22}

Detailed Solution

Step 1 — Common ion effect: NaOH provides [OH−]=0.2[\mathrm{OH^-}] = 0.2 M

Step 2 — Dissociation of Al(OH)3\mathrm{Al(OH)_3}: Al(OH)3⇌Al3++3OH−\mathrm{Al(OH)_3 \rightleftharpoons Al^{3+} + 3OH^-} Let solubility = ss mol/L (assume s≪0.2s \ll 0.2) Ksp=[Al3+][OH−]3=s(0.2)3=s×8×10−3K_{sp} = [\mathrm{Al^{3+}}][\mathrm{OH^-}]^3 = s(0.2)^3 = s \times 8 \times 10^{-3}

Step 3 — Solve for s: s=Ksp(0.2)3=2.4×10−248×10−3=3×10−22 mol L−1s = \frac{K_{sp}}{(0.2)^3} = \frac{2.4 \times 10^{-24}}{8 \times 10^{-3}} = 3 \times 10^{-22}\ \mathrm{mol\ L^{-1}}

Answer: Option (4) — 3×10−223 \times 10^{-22}

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