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

Answer

B

Explanation

`Ag_(2)CrO_(4)hArr2Ag+CrO_(4)^(2-), K_(sp)=(2s)^(2)xxs=4s^(3), s=((K_(sp))/(4))^(1//3)=((1.1xx10^(-12))/(4))^(1//3)`
`=0.65xx10^(-4)`
`AgCl hArr Ag^(+)+Cl^(-),K_(sp)=sxxs=s^(2)`,
`s=sqrt(K_(sp))=sqrt(1.8xx10^(-10))=1.34xx10^(-5)`
`AgBr hArr Ag^(+)+Br^(-), K_(sp)=sxxs=s^(2)`,
`s=sqrt(K_(sp))=sqrt(5.0xx10^(-13))=0.71xx10^(-6)`
`AgI hArr Ag^(+)+I, K_(sp)=sxxs=s^(2)`,
`s=sqrt(K_(sp))=sqrt(8.3xx10^(-17))=0.9xx10^(-8)`
As solubility of `Ag_(2)CrO_(4)` is highest, it will be precipitated last of all.