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

`{:(,Ag_(2)CO_(3)+K_(2)C_(2)O_(4),hArrAg_(2)C_(2)O_(4)+,K_(2)CO_(3)),("Mole before reaction", "excess",0.1520,),("Mole at", (0.1520-0.0358),0.0358,0.0358),("equilibrium", ,=0.1162, ):}`
at equilibrium,
`:. [C_(2)O_(4)^(2-)]=(0.1162)/(0.5)=0.2324M`
`[CO_(3)^(2-)]=(0.0358)/(0.5)=0.0716M`
`:' [Ag^(+)]^(2)[C_(2)O_(4)^(2-)]= [2xx0.0358)/(0.5)]^(2)[(0.1520)/(0.5)]`
`=6.23xx10^(-3)`
Thus `[Ag^(+)]^(2)[C_(2)O_(4)^(2-)]gtK_(SP)of Ag_(2)C_(2)O_(4)`
`(1.29xx10^(-11))`
`:. Ag_(2)C_(2)O_(4)` will precipitate out.
`:. Ag_(2)CO_(3)` is solid asnd `Ag_(2)C_(2)O_(4)` is almost precipitated out. For `Ag_(2)C_(2)O_(4)` precipitation,
`[Ag^(+)]^(2)[C_(2)O_(4)^(2-)]=K_(SP_(Ag_(2)C_(2)O_(4))`
`[Ag^(+)]^(2)[0.2324]= 1.29xx10^(-11)`
`:. [Ag^(+)]^(2)=(1.29xx10^(-11))/(0.2324)=5.55xx10^(-11)`
Now for `Ag_(2)CO_(3),K_(SP)=[Ag^(+)]^(2)[CO_(3)^(2-)]`
`=[5.55xx10^(-11)][0.0716]`
`=3.97xx10^(-12)mol^(3)litre^(-3)`