`0.01` mole of `AgNO_(3)` is added to 1 litre of a solution which is `0.1M` in `Na_(2)CrO_(4)` and `0.005M` in `NaIO_(3)`. Calculate the mole of precipitate formed at equilibrium and the concentrations of `Ag^(+), IO_(3)^(-)` and `CrO_(4)^(2-)`. `(K_(sP)` values of `Ag_(2)CrO_(4)` and `AgIO_(3)` are `10^(-8)` and `10^(-13)` erspectively)

Explanation

The `K_(SP)` values of `Ag_(2)CrO_(4)` and `AgIO_(3)` reveals that `CrO_(4)^(2-)` and `IO_(3)^(-)` will be precipitated on addition of `AgNO_(3)` as:
`[Ag^(+)][IO_(3)^(-)]=10^(-13)`
`[Ag^(+)]_("needed")=(10^(-13))/([0.005])=2xx10^(-11)`
`[Ag^(+)]^(2)[CrO_(4)^(2-)]=10^(-8)`
`[Ag^(+)]_("needed")=sqrt((10^(-8))/(0.1))=3.16xx10^(-4)`
Thus, `AgIO_(3)` will be precipitated first.
Now, in order to precipitate `AgIO_(3)`, one can show:
`{:(AgNO_(3)+,NaIO_(3)rarr,AgIO_(3)+,NaNO_(3)),(0.01,0.05,0,0),(0.005,0,0.005,0.005):}`
The left mole of `AgNO_(3)` are now used to precipitate `Ag_(2)CrO_(4)`.
`{:(2AgNO_(3)+,NaCrO_(4)rarr,Ag_(2)CrO_(4)+,2NaNO_(3)),(0.005,0.1,0,0),(0,0.0975,0.0025,0.005):}`
Thus, `[CrO_(4)^(2-)]` left solution `=0.0975`
Now, solution has
`{:(AgIO_(3(s))+,Ag_(2)CrO_(4(s))+,CrO_(4)^(2-)ions),(0.005,0.0025,0.0975):}`
`:. [Ag^(+)]"left" = (K_(SP)ofAg_(2)CrO_(4))/([CrO_(4)^(2-)])`
`=sqrt((10^(-8))/(0.00975))=3.2xx10^(-4)M`
`:. [IO_(3)^(-)["left"= (K_(SP)of AgIO_(3))/([Ag^(+)])`
`=(10^(-13))/(3.2xx10^(-4))=3.1xx10^(-10)M`