Question

The molar solubility of `Ag_2CO_3` is `1.3xx10^-4*mol*L^-1`. What is `K_"sp"` for this salt?

Answer 1

`K_"sp"=8.80xx10^-12`..........

Explanation:

We interrogate the equilibrium:

`Ag_2CO_3(s) rightleftharpoons2Ag^(+) + CO_3^(2-)`

And thus `K_"sp"=[Ag^+]^2[CO_3^(2-)]`

But from the stoichiometry, `[CO_3^(2-)]=S_(Ag_2CO_3)`, and `[Ag^+]=2S_(Ag_2CO_3)`........and so

`K_"sp"=Sxx(2S)^2=4S^3=4xx(1.3xx10^-4)^3=8.80xx10^-12.....`

This site reports that `K_"sp"`, `Ag_2CO_3`, is `8.46xx10^-12`, so our estimate is kosher.