A solution containing a mixture of `0.05M NaCI` and `0.05M` Nal is taken. `(K_(sp) of AgCI = 10^(-10)`and `K_(sp)`of `AgI = 4 xx 10^(-16))`. When `AgNO_(3)` is added to such a solution:

Options

A. The concentartion fo `Ag^(o+)` required to precipitate `CI^(Theta)` is `2 xx 10^(-9) mol L^(-1)`.
B. The concentartion of `Ag^(o+)` required to precipitate `I^(Theta)` is `8 xx 10^(-15) mol L^(-1)`.
C. `AgCI` and `AgI` will be precipitate togther.
D. First `AgI` will be precipitated.

Answer

A, B, D

Explanation

`{:(NaCI+,NaI),(0.05M,0.005):}`
When `AgNO_(3)` is added, find the minimum concentration of `Ag^(o+)` required to start precipitation.
`[Ag^(o+)]_(AgCI) = (K_(sp(NaCI)))/([CI^(Theta)]) = (10^(-10))/(0.05) = 2 xx 10^(-9)M`
`[Ag^(o+)]_(AgI) = (K_(sp(AgI)))/([I^(Theta)]) = (4xx10^(-16))/(0.05) = 8 xx 10^(-5)M`
Since, there is a very high difference between the minimum concentrations of `Ag^(o+)`, precipitation is selective and `AgI` will precipitate first.