JEE Main · 2024 · Shift-IIeasyIEQ-048

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…

Ionic Equilibrium · Class 11 · JEE Main Previous Year Question

Question

For a sparingly soluble salt AB2\mathrm{AB_2}, the equilibrium concentrations of A2+\mathrm{A^{2+}} ions and B−\mathrm{B^-} ions are 1.2×10−41.2 \times 10^{-4} M and 0.24×10−30.24 \times 10^{-3} M, respectively. The solubility product of AB2\mathrm{AB_2} is:

Options
  1. a

    6.91×10−126.91 \times 10^{-12}

    ✓
  2. b

    0.276×10−120.276 \times 10^{-12}

  3. c

    27.65×10−1227.65 \times 10^{-12}

  4. d

    0.069×10−120.069 \times 10^{-12}

Correct Answera

6.91×10−126.91 \times 10^{-12}

Detailed Solution

Step 1 — Dissociation: AB2⇌A2++2B−Ksp=[A2+][B−]2\mathrm{AB_2 \rightleftharpoons A^{2+} + 2B^-} K_{sp} = [\mathrm{A^{2+}}][\mathrm{B^-}]^2

Step 2 — Substitute values: [A2+]=1.2×10−4 M[B−]=0.24×10−3=2.4×10−4 M[\mathrm{A^{2+}}] = 1.2 \times 10^{-4}\ \mathrm{M} [\mathrm{B^-}] = 0.24 \times 10^{-3} = 2.4 \times 10^{-4}\ \mathrm{M}

Step 3 — Calculate: Ksp=(1.2×10−4)(2.4×10−4)2=(1.2×10−4)(5.76×10−8)=6.912×10−12≈6.91×10−12K_{sp} = (1.2 \times 10^{-4})(2.4 \times 10^{-4})^2 = (1.2 \times 10^{-4})(5.76 \times 10^{-8}) = 6.912 \times 10^{-12} \approx 6.91 \times 10^{-12}

Answer: Option (1) — 6.91×10−126.91 \times 10^{-12}

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