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 concentration of `K_(2)Cr_(2)O_(4) = 0.152/0.50 = 0.304 M`
Also for the following equilibrium:
`Ag_(2)CO_(3)(s) + K_(2)Cr_(2)O_(4)(aq)underset("0.340 - x") rarr Ag_(2)C_(2)O_(4)(s) + K_(2)CO_(3)underset("x")`
`K = [CO_(3)^(2-)]/[C_(2)O_(4)^(2-)] xx [Ag^(+)]^(2)/[Ag^(+)]^(2) = K_(sp)(Ag_(2)CO_(3))/K_(sp)(Ag_(2)C_(2)O_(4))`
Given, 0.304 - x = 0.0358
`implies x = 0.2682 implies K = 0.2682/0.358 = 7.5`
`K_(sp)(Ag_(2)CO_(3)) = K xx K_(sp)(Ag_(2)CO_(4)) = 7.5 xx 1.29 xx 10^(-11) = 9.675 xx 10^(-11)`