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Ionic Equilibrium question

2021 · 20 Jul · Shift 2 · Q4
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  5. /2021 · 20 Jul · Shift 2 · Q4

Ionic Equilibrium question

2021 · 20 Jul · Shift 2 · Q4

JEE MainChemistryIonic EquilibriumMCQ+4 / −1
A solution is 0.1 M in Cl −-− and 0.001 M in CrO 42−_4^{2 - }42−​. Solid AgNO3AgNO_3AgNO3​ is gradually added to it. Assuming that the addition does not change in volume and Ksp(AgClAgClAgCl) = 1.7 ×\times× 10 −-− 10 M2 and Ksp(Ag2CrO4Ag_2CrO_4Ag2​CrO4​) = 1.9 ×\times× 10 −-− 12 M3. Select correct statement from the following :
  1. A
    AgCl precipitates first because its Ksp is high.
  2. B
    Ag2CrO4Ag_2CrO_4Ag2​CrO4​ precipitates first as its Ksp is low.
  3. C
    Ag2CrO4Ag_2CrO_4Ag2​CrO4​ precipitates first because the amount of Ag+Ag^+Ag+ needed is low.
  4. D
    AgClAgClAgCl will precipitate first as the amount of Ag+Ag^+Ag+ needed to precipitate is low.
View written solutionFree

Correct answer: D

  1. Condition for precipitation

A salt starts precipitating when the ionic product just reaches its solubility product.

  • For AgClAgClAgCl: Ksp(AgCl)=[Ag+][Cl−]=1.7×10−10K_{sp}(AgCl) = [Ag^+][Cl^-] = 1.7 \times 10^{-10}Ksp​(AgCl)=[Ag+][Cl−]=1.7×10−10

  • For Ag2CrO4Ag_2CrO_4Ag2​CrO4​: Ksp(Ag2CrO4)=[Ag+]2[CrO42−]=1.9×10−12K_{sp}(Ag_2CrO_4) = [Ag^+]^2[CrO_4^{2-}] = 1.9 \times 10^{-12}Ksp​(Ag2​CrO4​)=[Ag+]2[CrO42−​]=1.9×10−12

We calculate the minimum [Ag+][Ag^+][Ag+] required for each precipitation to begin.

  1. For AgClAgClAgCl

Given: [Cl−]=0.1 M[Cl^-] = 0.1\,M[Cl−]=0.1M

At the point of first precipitation: [Ag+][Cl−]=1.7×10−10[Ag^+][Cl^-] = 1.7 \times 10^{-10}[Ag+][Cl−]=1.7×10−10

So, [Ag+]=1.7×10−100.1=1.7×10−9 M[Ag^+] = \frac{1.7 \times 10^{-10}}{0.1} = 1.7 \times 10^{-9}\,M[Ag+]=0.11.7×10−10​=1.7×10−9M

  1. For Ag2CrO4Ag_2CrO_4Ag2​CrO4​

Given: [CrO42−]=0.001 M=10−3M[CrO_4^{2-}] = 0.001\,M = 10^{-3} M[CrO42−​]=0.001M=10−3M

At the point of first precipitation: [Ag+]2[CrO42−]=1.9×10−12[Ag^+]^2[CrO_4^{2-}] = 1.9 \times 10^{-12}[Ag+]2[CrO42−​]=1.9×10−12

Thus, [Ag+]2=1.9×10−1210−3=1.9×10−9[Ag^+]^2 = \frac{1.9 \times 10^{-12}}{10^{-3}} = 1.9 \times 10^{-9}[Ag+]2=10−31.9×10−12​=1.9×10−9

Hence, [Ag+]=1.9×10−9[Ag^+] = \sqrt{1.9 \times 10^{-9}}[Ag+]=1.9×10−9​

[Ag+]≈4.36×10−5 M[Ag^+] \approx 4.36 \times 10^{-5}\,M[Ag+]≈4.36×10−5M

  1. Compare the required [Ag+][Ag^+][Ag+] values
  • For AgClAgClAgCl: 1.7×10−9M1.7 \times 10^{-9} M1.7×10−9M
  • For Ag2CrO4Ag_2CrO_4Ag2​CrO4​: 4.36×10−5M4.36 \times 10^{-5} M4.36×10−5M

Since a much smaller concentration of Ag+Ag^+Ag+ is needed for AgClAgClAgCl, it will start precipitating first.

  1. Check the options
  • A: Incorrect. Higher KspK_{sp}Ksp​ does not imply first precipitation here; precipitation depends on both KspK_{sp}Ksp​ and ion concentrations.
  • B: Incorrect. Although Ag2CrO4Ag_2CrO_4Ag2​CrO4​ has lower KspK_{sp}Ksp​, it still needs much higher [Ag+][Ag^+][Ag+] because of the expression [Ag+]2[CrO42−][Ag^+]^2[CrO_4^{2-}][Ag+]2[CrO42−​] and given concentrations.
  • C: Incorrect. Ag2CrO4Ag_2CrO_4Ag2​CrO4​ does not need lower [Ag+][Ag^+][Ag+].
  • D: Correct. AgClAgClAgCl precipitates first because the required [Ag+][Ag^+][Ag+] is lower.
  1. Final answer

D\boxed{\text{D}}D​

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