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JEE Main 202225 JuneEvening Shift
Exam
JEE Main
Year
2022
Date
25 June
Shift
Evening Shift
Subject
Chemistry
Chapter
Ionic Equilibrium

JEE Main 2022 · 25 June · Evening Shift · Chemistry · Ionic Equilibrium

The K _ sp for bismuth sulphide ( Bi _ 2 S _ 3 ) is 1 . 08 × 10 ^ -...

The Ksp{K}_{sp} for bismuth sulphide (Bi2S3)\left({Bi}_{2}{S}_{3}\right) is 1.08×10−731 . 08 \times {10}^{- 73}. The solubility of Bi2S3inmolL−1{Bi}_{2}{S}_{3} in {molL}^{- 1} at 298K298K is

Options

  1. A1.0×10−151 . 0 \times {10}^{- 15}
  2. B2.7×10−122 . 7 \times {10}^{- 12}
  3. C3.2×10−103 . 2 \times {10}^{- 10}
  4. D4.2×10−84 . 2 \times {10}^{- 8}

Correct answer

A. 1.0×10−151 . 0 \times {10}^{- 15}

Step-by-step solution

Bi2S3{Bi}_{2}{S}_{3} will get dissociated as:

Bi2S3⇌2Bi+3+3S−22s3s{Bi}_{2}{S}_{3} ⇌ 2{Bi}^{+ 3} + 3{S}^{- 2}2s3s

Here,

Ksp=(2s)2(3s)3=108s5s=(Ksp108)15s=(1.08×10−73108)15s=10−15molL−1{K}_{sp} = {\left(2s\right)}^{2}{\left(3s\right)}^{3} = 108{s}^{5}s = \left(\dfrac{{K}_{sp}}{108}\right)\dfrac{1}{5}s = \left(\dfrac{1 . 08 \times {10}^{- 73}}{108}\right)\dfrac{1}{5}s = {10}^{- 15}mol{L}^{- 1}

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