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Ionic Equilibrium question

2003 · Shift 0 · Q7
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Ionic Equilibrium question

2003 · Shift 0 · Q7

JEE MainChemistryIonic EquilibriumMCQ+4 / −1
The solubility in water of a sparingly soluble salt AB2AB_2AB2​ is 1.0 ×\times× 10-5 mol L-1. Its solubility product number will be :
  1. A
    4 ×\times× 10-10
  2. B
    1 ×\times× 10-15
  3. C
    1 ×\times× 10-10
  4. D
    4 ×\times× 10-15
View written solutionFree

Correct answer: D

  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−

  1. Let the solubility be sss

Given solubility:

s=1.0×10−5 mol L−1s = 1.0 \times 10^{-5}\ \text{mol L}^{-1}s=1.0×10−5 mol L−1

So at equilibrium:

[A2+]=s=1.0×10−5[A^{2+}] = s = 1.0 \times 10^{-5}[A2+]=s=1.0×10−5

[B−]=2s=2.0×10−5[B^-] = 2s = 2.0 \times 10^{-5}[B−]=2s=2.0×10−5

  1. Write the solubility product expression

For AB2AB_2AB2​:

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

Substitute the concentrations:

Ksp=(1.0×10−5)(2.0×10−5)2K_{sp} = (1.0 \times 10^{-5})(2.0 \times 10^{-5})^2Ksp​=(1.0×10−5)(2.0×10−5)2

  1. Calculate

(2.0×10−5)2=4.0×10−10(2.0 \times 10^{-5})^2 = 4.0 \times 10^{-10}(2.0×10−5)2=4.0×10−10

Therefore,

Ksp=(1.0×10−5)(4.0×10−10)=4.0×10−15K_{sp} = (1.0 \times 10^{-5})(4.0 \times 10^{-10}) = 4.0 \times 10^{-15}Ksp​=(1.0×10−5)(4.0×10−10)=4.0×10−15

  1. Match with the options

Ksp=4×10−15K_{sp} = 4 \times 10^{-15}Ksp​=4×10−15

So the correct option is D.

  1. Comparison with stored answer

Stored correct answer: D

My derived answer: D

They match.

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