Verified Solution • 2025–26 Session moderate

QJEEM21Ch9fc4: JEE Main 2021 Chemistry - Class 11 JEEM21Ch9fc4 Solution: A solution is 0.1 M in Cl and 0.001 M in CrO . Solid AgNO 3 is…

A solution is 0.1 M in Cl −- and 0.001 M in CrO 42−_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 :

Multiple Choice Options

  • A AgCl precipitates first because its Ksp is high.
  • B Ag2CrO4 precipitates first as its Ksp is low.
  • C Ag2CrO4 precipitates first because the amount of Ag+ needed is low.
  • D AgCl will precipitate first as the amount of Ag+ needed to precipitate is low.

Step-by-Step Verified Solution

Conc. of Cl −- = 0.1 M = 10 −- 1 M

Conc. of CrO 42−_4^{2 - } = 0.001 M = 10 −- 3 M

Ksp(AgCl) = [Ag+][Cl −- ]

[Ag+]AgCl = 1.7×10−1010−1=1.7×10−9{{1.7 \times {{10}^{ - 10}}} \over {{{10}^{ - 1}}}} = 1.7 \times {10^{ - 9}}

Ksp(Ag2CrO4)=[Ag+]2[CrO42−]{K_{sp}}(A{g_2}Cr{O_4}) = {[A{g^ + }]^2}[CrO_4^{2 - }]

[Ag+]=1.9×10−1210−3=19×10−4[A{g^ + }] = \sqrt {{{1.9 \times {{10}^{ - 12}}} \over {{{10}^{ - 3}}}}} = \sqrt {19} \times {10^{ - 4}}

∴\therefore AgCl will be precipitate first

Related Learning Resources & Exam Preparation

JEE Main PYQ Question Bank
Browse year-wise authentic previous years' questions with solutions
Authentic CBT Mock Tests
Simulate official exam environment with live timer and percentile scoring