`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

Let x and y be the molar concentration of `Ag^+` required to precipitate of `Ag_(2)CrO_(4) and AgIO_(3)` respectively.
`x=sqrt(K_(sp) (Ag_(2)CrO_(4))/([CrO_(4)^(2-)]))=sqrt((10^(-8))/(0.1))=3.16 xx 10^(-4)`
`y=K_(sp) (AgIO_(3))/([IO_(3)^(-)])=sqrt(10^(-13))/(0.005)=2.0 xx 10^(-11)`
As `y lt lt x, AgIO_(3)` will precipitate first
0.01 mole `" "` 0.005 mole
`underset(0.01-0.05=0.005"mole")AgNO_(3)+underset(0)(NaIO_(3))=underset("0.005 mole")(AgIO_(3)(s)+NaNO_(3))`
And then, for remaining 0.005 mole of `AgNO_(3)`,
`overset("0.005 mole")(2AgNO_(3))+overset("0.01 mole")(Na_(2)CrO_(4)) to underset("0.0025 mole")(Ag_(2)CrO_(4)(s)+2NaNO_(3))`
=0.0975 mole
mole of `AgIO(3)` ppt. =0.005
and mole of `Ag_(2)CrO_(4)" ppt. =0.0025"`
Now, for the requilibrium
`Ag_(2)CrO_(4) Leftrightarrow 2Ag^(+)+CrO_(4)^(2-)`
`K_(sp)=[Ag^(+)]^(2) [CrO_(4)^(2-)]`
`10^(-8)=[Ag^(+)] (0.0975) therefore [CrO_(4)^(2-)]=0.0975 M`
`[Ag^(+)]=sqrt((10^(-6))/(0.0975))=3.2 xx 10^(-4)"mole/L"`
Further for the equilibrium
`AgIO_(3) Leftrightarrow Ag^+ +IO_(3)^(-)`
`10^(-13)=(3.2 xx 10^(-4)) [IO_(3)^(-)]`
`[IO_(3)^(-)]=3.125 xx 10^(-10)"mole/L"`