A solution contains `Ag^+,Ca^(2+)` and `Al^(3+)` each having a concentration of 0.1(M). `Na_3PO_4` is added to the solution. Which will be precipitated first when the concentration of `PO_4^(-3)` in the solution is the lowest ? [ Given : `K_(sp)` values of `Ag_3PO_4,Ca_3(PO_4)_2` and `AlPO_4` and `10^(-16)`, `10^(-33)` and `10^(-20)` respectively]

Explanation

`Ag_3PO_4(s) `K_(sp)=[Ag^+]^3[PO_4^(3-)]` or
`[PO_3^(3-)]=(10^(-6))/((0.1)^3)=10^(-3)(M)`
`Ca_3(PO_4)_2(s) `K_(sp)=[Ca^(2+)]^3[PO_4^(3-)]^2`
`[PO_4^(3-)]^2=(10^(-33))/((0.1)^3)=10^(-30)(M)`
`:.[PO_4^(3-)]=10^(-15)(M)`
`AlPO_4(s) `K_(sp)=[Al^(3+)][PO_4^(3-)]`
or `[PO_4^(3-)]=(10^(-20))/(0.1)=10^(-19)(M)`
So, minimum concentration of `PO_4^(3-)` ions is required for the precipitation of `AlPO_4`