The `K_(sp)` of `Ag_(2)CrO_(4),AgCl,AgBr` and AgI are respectively, `1.1xx10^(-12),1.8xx10^(-10),5.0xx10^(-13),8.3xx10^(-17)`. Which one of the following salts will precipitate last if `AgNO_(3)` solution is added to the solution containing equal moles of NaCl,NaBr,NaI and `Na_(2)CrO_(4)` ?

Options

A. `Agl`
B. `AgCl`
C. `AgBr`
D. `Ag_(2)CrO_(4)`

Answer

D

Explanation

`Ag_(2)CrO_(4)hArr 2Ag^(+)+CrO_(4)^(2-)`
Solubility product
`K_(sp)=(2s)^(2)xxS=4s^(3)`
`K_(sp)=(1.1xx10^(-12))` (given)
`S= root(3)((K_(sp))/(4))=0.65xx10^(-4)`
`AgCl hArr Ag^(+)+Cl^(-)`
`K_(sp)=SxxS " " (K_(sp)=1.8xx10^(-10))`
`S=sqrt(K_(sp))=1.34xx10^(-5)`
`AgBr hArr Ag^(+)+Br^(-)`
`K_(sp)=SxxS " " (K_(sp)=5xx10^(-13))`
`S= sqrt(K_(sp))=0.71xx10^(-6)`
`AgI hArr Ag^(+)+I^(-)`
`K_(sp)=SxxS " " (K_(sp)=8.3xx10^(-17))`
`S = sqrt(K_(sp))=0.9xx10^(-8)`
`because` Solubility of `Ag_(2)CrO_(4)` is highest.
So, it will precipitate last.