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Ionic Equilibrium question

2022 · 25 Jun · Shift 2 · Q4
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  5. /2022 · 25 Jun · Shift 2 · Q4

Ionic Equilibrium question

2022 · 25 Jun · Shift 2 · Q4

JEE MainChemistryIonic EquilibriumMCQ+4 / −1
The Ksp for bismuth sulphide (Bi2S3Bi_2S_3Bi2​S3​) is 1.08 ×\times× 10 −-− 73. The solubility of Bi2S3Bi_2S_3Bi2​S3​ in mol L −-− 1 at 298 K is :
  1. A
    1.0 ×\times× 10 −-− 15
  2. B
    2.7 ×\times× 10 −-− 12
  3. C
    3.2 ×\times× 10 −-− 10
  4. D
    4.2 ×\times× 10 −-− 8
View written solutionFree

Correct answer: A

  1. Write the dissolution equilibrium

Bi2S3(s)⇌2Bi3++3S2−Bi_2S_3(s) \rightleftharpoons 2Bi^{3+} + 3S^{2-}Bi2​S3​(s)⇌2Bi3++3S2−

If the solubility of Bi2S3Bi_2S_3Bi2​S3​ is sss mol L−1^{-1}−1, then at equilibrium:

[Bi3+]=2s,[S2−]=3s[Bi^{3+}] = 2s, \qquad [S^{2-}] = 3s[Bi3+]=2s,[S2−]=3s

  1. Write the solubility product expression

Ksp=[Bi3+]2[S2−]3K_{sp} = [Bi^{3+}]^2 [S^{2-}]^3Ksp​=[Bi3+]2[S2−]3

Substituting the concentrations:

Ksp=(2s)2(3s)3K_{sp} = (2s)^2(3s)^3Ksp​=(2s)2(3s)3

Ksp=4s2⋅27s3=108s5K_{sp} = 4s^2 \cdot 27s^3 = 108s^5Ksp​=4s2⋅27s3=108s5

  1. Substitute the given value of KspK_{sp}Ksp​

Given:

Ksp=1.08×10−73K_{sp} = 1.08 \times 10^{-73}Ksp​=1.08×10−73

So,

108s5=1.08×10−73108s^5 = 1.08 \times 10^{-73}108s5=1.08×10−73

s5=1.08×10−73108s^5 = \frac{1.08 \times 10^{-73}}{108}s5=1081.08×10−73​

Since 108=1.08×102108 = 1.08 \times 10^2108=1.08×102,

s5=10−75s^5 = 10^{-75}s5=10−75

  1. Take the fifth root

s=(10−75)1/5=10−15s = (10^{-75})^{1/5} = 10^{-15}s=(10−75)1/5=10−15

Thus,

s=1.0×10−15 mol L−1\boxed{s = 1.0 \times 10^{-15}\ \text{mol L}^{-1}}s=1.0×10−15 mol L−1​

  1. Match with the options

This corresponds to:

Option A: 1.0×10−151.0 \times 10^{-15}1.0×10−15

  1. Comparison with stored correct answer

Stored correct answer = A

Derived answer = A

So, the derived answer agrees with the stored correct answer.

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