JEE Main · 2022 · Shift-IImediumIEQ-016

The Ksp for bismuth sulphide (Bi2S3) is 1.08 10-73. The solubility of Bi2S3 in mol\ L-1 at 298 K is

Ionic Equilibrium · Class 11 · JEE Main Previous Year Question

Question

The KspK_{sp} for bismuth sulphide (Bi2S3\mathrm{Bi_2S_3}) is 1.08×10−731.08 \times 10^{-73}. The solubility of Bi2S3\mathrm{Bi_2S_3} in mol L−1\mathrm{mol\ L^{-1}} at 298 K is

Options
  1. a

    1.0×10−151.0 \times 10^{-15}

    ✓
  2. b

    2.7×10−122.7 \times 10^{-12}

  3. c

    3.2×10−103.2 \times 10^{-10}

  4. d

    4.2×10−84.2 \times 10^{-8}

Correct Answera

1.0×10−151.0 \times 10^{-15}

Detailed Solution

Step 1 — Dissociation of Bi2S3\mathrm{Bi_2S_3}: Bi2S3⇌2Bi3++3S2−[Bi3+]=2s,[S2−]=3s\mathrm{Bi_2S_3 \rightleftharpoons 2Bi^{3+} + 3S^{2-}} [\mathrm{Bi^{3+}}] = 2s, \quad [\mathrm{S^{2-}}] = 3s

Step 2 — Write KspK_{sp}: Ksp=(2s)2(3s)3=4s2⋅27s3=108s5K_{sp} = (2s)^2(3s)^3 = 4s^2 \cdot 27s^3 = 108s^5

Step 3 — Solve for s: s5=1.08×10−73108=10−75s=(10−75)1/5=10−15 mol L−1s^5 = \frac{1.08 \times 10^{-73}}{108} = 10^{-75} s = (10^{-75})^{1/5} = 10^{-15}\ \mathrm{mol\ L^{-1}}

Answer: Option (1) — 1.0×10−151.0 \times 10^{-15}

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