`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 `AgCrO_(4)` and `AgIO_(3)` reveals that `CrO_(4)^(2-)` and `IO_(3)^(Theta)` will be precipitated on addition of `AgNO_(3)` as:
`[Ag^(o+)] [IO_(3)^(Theta)] = 10^(-13)`
`[Ag^(o+)]_("needed") = (10^(-13))/([0.005]) = 2 xx 10^(-11)`
`[Ag^(o+)]^(2)[CrO_(4)^(2-)] = 10^(-8)`
`[Ag^(o+)]_("needed") = sqrt((10^(-8))/(0.1)) = 3.16 xx 10^(-4)`
Thus, `AgIO_(#)` will be preicipitated first.
Now, in order to precipitate `AgIO_(3)`, we get
`{:(AgNO_(3)+,NaIO_(3)rarr,AgIO_(3)+,NaNO_(3),),(0.01,0.005,0,0,),(0.005,0,0.005,0.005,):}`
The left moles of `AgNO_(3)` are now used to precipitate `Ag_(2)CrO_(4)`
`{:(2AgNO_(3)+,Na_(2)CrO_(4)rarr,Ag_(2)CrO_(4)+,NaNO_(3),),(0.01,0.005,0,0,),(0.005,0,0.005,0.005,):}`
Thus, `[CrCO_(4)^(2-)]` left in solution `= 0.0975`
Now, solutions has `{:(AgIO_(3)(s)+,AgCrO_(4)(s)+,CrO_(4)^(2-),,),(0.005,0.0025,0.0975,,):}`
[`Ag^(o+)]_("left") = (K_(sp)Ag_(2)CrO_(4))/([CrO_(4)^(2-)]) = sqrt((10^(-8))/(0.0975))`
`= 3.2 xx 10^(-4)M`
`[IO_(3)^(Theta)]_("left") = (K_(sp)AgIO_(3))/([Ag^(o+)]) = (10^(-13))/(3.2xx10^(-4)) = 3.1 xx 10^(-10)M`.