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A solution is 0.1 M in Cl^(-) and 0.001...

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_(SP) (AgCl) = 1.7 xx 10^(-10) M^(2) and K_(SP)(Ag_(2)CrO_(4)) = 1.9 xx 10^(-12) M^(3)`. Select correct statement from the following:

A

AgCl will precipitate first as the amount of `Ag^(+)` needed to precipitate is low

B

`Ag_(2)CrO_(4)` precipitates first because its `K_(sp)` is high

C

AgCl precipitates first because its `K_(sp)` is high

D

`Ag_(2)CrO_(4)` precipitates first as its `K_(sp)` is low

Text Solution

Verified by Experts

The correct Answer is:
A
Conc.of `Cl^(-)=0.1M=10^(-1)M`
Conc.of `CrO_(4)^(2-)=0.001M=10^(-3)M`
`K_(sp)(AgCl)=[Ag^(+)][Cl^(-)]`
`[Ag^(+)]_(AgCl)=(1.7xx10^(-10))/(10^(-1))=1.7xx10^(-9)`
`K_(sp)(Ag_(2)CrO_(4))=[Ag^(+)]^(2)[CrO_(4)^(2-)]`
`[Ag^(+)]=sqrt((1.9xx10^(-12))/(10^(-3)))=sqrt(19)xx10^(-4)`
`:.AgCl` will be precipitated first.
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