Medium MCQ +4 / -1 PYQ · JEE Mains 2021

A solution is 0.1 M in Cl$-$ and 0.001 M in CrO$_4^{2 - }$. Solid AgNO3 is gradually added to it. Assuming that the addition does not change in volume and Ksp(AgCl) = 1.7 $\times$ 10$-$10 M2 and Ksp(Ag2CrO4) = 1.9 $\times$ 10$-$12 M3.

Select correct statement from the following :

  1. A AgCl precipitates first because its K<sub>sp</sub> is high.
  2. B Ag<sub>2</sub>CrO<sub>4</sub> precipitates first as its K<sub>sp</sub> is low.
  3. C Ag<sub>2</sub>CrO<sub>4</sub> precipitates first because the amount of Ag<sup>+</sup> needed is low.
  4. D AgCl will precipitate first as the amount of Ag<sup>+</sup> needed to precipitate is low. Correct answer

Solution

Conc. of Cl<sup>$-$</sup> = 0.1 M = 10<sup>$-$1</sup> M<br><br>Conc. of CrO$_4^{2 - }$ = 0.001 M = 10<sup>$-$3</sup> M<br><br>K<sub>sp</sub>(AgCl) = [Ag<sup>+</sup>][Cl<sup>$-$</sup>]<br><br>[Ag<sup>+</sup>]<sub>AgCl</sub> = ${{1.7 \times {{10}^{ - 10}}} \over {{{10}^{ - 1}}}} = 1.7 \times {10^{ - 9}}$<br><br>${K_{sp}}(A{g_2}Cr{O_4}) = {[A{g^ + }]^2}[CrO_4^{2 - }]$<br><br>$$[A{g^ + }] = \sqrt {{{1.9 \times {{10}^{ - 12}}} \over {{{10}^{ - 3}}}}} = \sqrt {19} \times {10^{ - 4}}$$<br><br>$\therefore$ AgCl will be precipitate first

About this question

Subject: Chemistry · Chapter: Equilibrium · Topic: Chemical Equilibrium and Kc, Kp

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