A solution is 0.1 {M} in {Cl}^{-}and 0.001 {M} in {CrO}_{4}^{2-}. Solid {AgNO}_{3} is gradually added to it Assuming that the addition does not change in volume and {K}_{\text{sp }}({AgCl})=1.7 \\times10^{-10} {M}^{2} and{K}_{{sp}}({Ag}_{2} {CrO}_{4})=1.9 \\times10^{-12} {M}^{3} Select correct statement from the following: [20 Jul 2021 Shift 2]

Correct Answer is : AgCl will precipitate first as the amount of Ag+ needed to precipitate is low.

{AgCl} precipitates first because its {K}_{{sp}} is high.

{Ag}_{2} {CrO}_{4} precipitates first as its {K}_{{sp}} is low.

{Ag}_{2} {CrO}_{4} precipitates first because the amount of {Ag}^{+}needed is low.

AgCl will precipitate first as the amount of {Ag}^{+} needed to precipitate is low.

Ionic Equilibrium Part 1

Section: Chemistry

Question : 33 of 100

Marks: +1, -0

A solution is 0.1M in Cl−and 0.001M in CrO42−.
Solid AgNO3 is gradually added to it Assuming that the addition does not change in volume and Ksp (AgCl)=1.7×10−10M2 andKsp(Ag2CrO4)=1.9×10−12M3
Select correct statement from the following:

[20 Jul 2021 Shift 2]

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

Validate

Go to Question:

Prev Question

Next Question