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Ionic Equilibrium question

2020 · 6 Sep · Shift 2 · Q9
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Ionic Equilibrium question

2020 · 6 Sep · Shift 2 · Q9

JEE MainChemistryIonic EquilibriumNumerical+4 / −1
If the solubility product of AB2AB_2AB2​ is 3.20 ×\times× 10–11 M3, then the solubility of AB2AB_2AB2​ in pure water is ‾×\underline{\hspace{2cm}}\times​× 10–4 mol L–1. [Assuming that neither kind of ion reacts with water]
Numerical answer
View written solutionFree

Correct answer: 2

  1. Write the dissolution equilibrium

For the salt AB2AB_2AB2​:

AB2(s)⇌A2++2B−AB_2(s) \rightleftharpoons A^{2+} + 2B^-AB2​(s)⇌A2++2B−

If the solubility of AB2AB_2AB2​ in pure water is s mol L−1s\,\text{mol L}^{-1}smol L−1, then at equilibrium:

[A2+]=s,[B−]=2s[A^{2+}] = s, \qquad [B^-] = 2s[A2+]=s,[B−]=2s

  1. Write the expression for solubility product

Given:

Ksp=[A2+][B−]2K_{sp} = [A^{2+}][B^-]^2Ksp​=[A2+][B−]2

So,

Ksp=s(2s)2=4s3K_{sp} = s(2s)^2 = 4s^3Ksp​=s(2s)2=4s3

  1. Substitute the given value

4s3=3.20×10−114s^3 = 3.20 \times 10^{-11}4s3=3.20×10−11

s3=3.20×10−114=0.80×10−11=8.0×10−12s^3 = \frac{3.20 \times 10^{-11}}{4} = 0.80 \times 10^{-11} = 8.0 \times 10^{-12}s3=43.20×10−11​=0.80×10−11=8.0×10−12

  1. Take cube root

s=(8.0×10−12)1/3s = (8.0 \times 10^{-12})^{1/3}s=(8.0×10−12)1/3

Now,

81/3=2,(10−12)1/3=10−48^{1/3} = 2, \qquad (10^{-12})^{1/3} = 10^{-4}81/3=2,(10−12)1/3=10−4

Hence,

s=2×10−4 mol L−1s = 2 \times 10^{-4}\,\text{mol L}^{-1}s=2×10−4mol L−1

  1. Required integer

The question asks for:

‾×10−4 mol L−1\underline{\hspace{1cm}} \times 10^{-4}\,\text{mol L}^{-1}​×10−4mol L−1

So the blank is:

2\boxed{2}2​

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