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Ionic Equilibrium question

2024 · 9 Apr · Shift 2 · Q14
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  5. /2024 · 9 Apr · Shift 2 · Q14

Ionic Equilibrium question

2024 · 9 Apr · Shift 2 · Q14

JEE MainChemistryIonic EquilibriumMCQ+4 / −1
For a sparingly soluble salt AB2\mathrm{AB}_2AB2​, the equilibrium concentrations of A2+\mathrm{A}^{2+}A2+ ions and B−B^{-}B− ions are 1.2×10−4M1.2 \times 10^{-4} \mathrm{M}1.2×10−4M and 0.24×10−3M0.24 \times 10^{-3} \mathrm{M}0.24×10−3M, respectively. The solubility product of AB2\mathrm{AB}_2AB2​ is :
  1. A
    0.069×10−120.069 \times 10^{-12}0.069×10−12
  2. B
    0.276×10−120.276 \times 10^{-12}0.276×10−12
  3. C
    6.91×10−126.91 \times 10^{-12}6.91×10−12
  4. D
    27.65×10−1227.65 \times 10^{-12}27.65×10−12
View written solutionFree

Correct answer: C

  1. Write the dissolution equilibrium

For the sparingly soluble salt AB2\mathrm{AB_2}AB2​:

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

So, the solubility product expression is:

Ksp=[A2+][B−]2K_{sp} = [\mathrm{A^{2+}}][\mathrm{B^-}]^2Ksp​=[A2+][B−]2
  1. Substitute the given equilibrium concentrations

Given:

[A2+]=1.2×10−4 M[\mathrm{A^{2+}}] = 1.2 \times 10^{-4}\,\mathrm{M}[A2+]=1.2×10−4M [B−]=0.24×10−3 M=2.4×10−4 M[\mathrm{B^-}] = 0.24 \times 10^{-3}\,\mathrm{M} = 2.4 \times 10^{-4}\,\mathrm{M}[B−]=0.24×10−3M=2.4×10−4M

Hence,

Ksp=(1.2×10−4)(2.4×10−4)2K_{sp} = (1.2 \times 10^{-4})(2.4 \times 10^{-4})^2Ksp​=(1.2×10−4)(2.4×10−4)2
  1. Calculate step-by-step

First,

(2.4×10−4)2=2.42×10−8=5.76×10−8(2.4 \times 10^{-4})^2 = 2.4^2 \times 10^{-8} = 5.76 \times 10^{-8}(2.4×10−4)2=2.42×10−8=5.76×10−8

Now multiply by 1.2×10−41.2 \times 10^{-4}1.2×10−4:

Ksp=(1.2×10−4)(5.76×10−8)K_{sp} = (1.2 \times 10^{-4})(5.76 \times 10^{-8})Ksp​=(1.2×10−4)(5.76×10−8) Ksp=6.912×10−12K_{sp} = 6.912 \times 10^{-12}Ksp​=6.912×10−12
  1. Match with the options
Ksp≈6.91×10−12K_{sp} \approx 6.91 \times 10^{-12}Ksp​≈6.91×10−12

This corresponds to Option C.

  1. Comparison with stored correct answer

Stored correct answer: C

My derived answer: C

They match.

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