The solubility of `Ag_(2)C_(2)O_(4)` at `25^(@)C` is `1.20 xx 10^(-11)`. A solution of `K_(2)C_(2)O_(4)` containing `0.15mol` in `500mL` water is mixed with excess of `Ag_(2)CO_(3)` till the following equilibrium is established:
`Ag_(2)CO_(3) + K_(2)C_(2)O_(4) hArr Ag_(2)C_(2)O_(4) + K_(2)CO_(3)`
At equilibrium, the solution constains `0.03 mol` of `K_(2)CO_(3)`. Assuming that the degree of dissociation of `K_(2)C_(2)O_(4)` and `K_(2)CO_(3)` to be equal, calculate the solubility product of `Ag_(2)CO_(3)`. [Take `100%` ionisation of `K_(2)C_(2)O_(4)` and `K_(2)CO_(3)]`

Explanation

Initial `[K_(2)C_(2)O_(4)] = (0.15)/(0.5) = 0.30 M = P`
and Final `[K_(2)CO_(3)] = (0.03)/(0.5) = 0.006 M = x`
`{:(Ag_(2)CO_(3)+,K_(2)C_(2)O_(4)hArr,K_(2)CO_(3)+AgC_(2)O_(4),,),(,P,-,,),(,P-x,x,,):}`
`1mol K_(2) CO_(3) -= 1mol K_(2)C_(2)O_(4)` and since both are completely ionised.
Now find `[Ag^(o+)]` at equilibrium as:
`[Ag^(o+)] = sqrt((K_(sp)of Ag_(2)C_(2)O_(4))/([C_(2)O_(4)^(2-)]))`
`= sqrt((1.20xx10^(-11))/(P-x)) = sqrt((1.20 xx 10^(-11))/(0.3-0.06)) = 7.-07 xx 10^(-6)M`
`[CO_(3)^(2-)]_("final") = x = 0.06M`
Now, `K_(sp) of Ag_(2)CO_(3) = [Ag^(o+)]^(2)[CO_(3)^(2-)]`
`rArr K_(sp)` of `Ag_(2)CO_(3) = (7.07 xx 10^(-6))^(2) xx 0.06= 3xx10^(-12)`