A solution is found to contain `[CI^(Theta)] = 1.5 xx 10^(-1)M, [Br^(Theta)] = 5.0 xx 10^(-4)M, [CrO_(4)^(2-)] = 1.9xx 10^(-2)M`.
A solution of `AgNO_(3)` (100 % dissociated) is added to the above solution drop by drop. Which silver salt will precipiate first ? Given: `K_(sp) (AgCI) = 1.5 xx 10^(-10), K_(sp) (AgBr) = 5.0 xx 10^(-13),K_(sp) (Ag_(2)CrO_(4)) =1.9 xx 10^(-12)`.

Options

A. `AgCI`
B. `AgBr`
C. `Ag_(2)CrO_(4)`
D. `AgCI` and `AgBr` togther

Answer

D

Explanation

For precipitation, `[Ag^(o+)]` ion should be minimum in the solution.
For `AgCI: [Ag^(o+)]_(min) = (K_(sp)AgCI)/([CI^(Theta)]) = (1.5 xx 10^(-10))/(1.5 xx 10^(-1)) = 10^(-9)M`
For `AgBr: [Ag^(o+)]_(min) = (K_(sp)AgBr)/([Br^(Theta)]) = (5.0xx10^(-13))/(5.0xx10^(-4)) = 10^(-9)M`
For `Ag_(2)CrO_(4): [Ag^(o+)]_(min)`
`= ((K_(sp)Ag_(2)CrO_(4))/([CrO_(4)^(2)]))^((1)/(2)) = ((1.9xx10^(-12))/(1.9xx10^(-2)))=10^(-5)M`
Therefore, `[Ag^(o+)]` min in solution is in `AgCI` and `AgBr`, so both will be precipitated.