Criterion for gravimetric analysis of silver

Question

If the only criterion for gravimetric analysis of silver is solubility of precipitate, which one of the following two compound are we supposed to choose? $K_\mathrm{s}(\ce{AgCl}) = \pu{1.78E-10}$ and $K_\mathrm{s}(\ce{Ag2CO3}) = \pu{8.13E-12}.$

a) $\ce{AgCl}$
b) $\ce{Ag2CO3}$
c) both of them are suitable
d) none of them

As I learned from textbook, one of the criteria for choosing the right one is: $K_\mathrm{s} < 10^{-7}$—thus I eliminate option d.

I also read that it is meaningless to compare the solubilities of two salts having different formulas on the basis of their $K_\mathrm{s}$ values. Therefore, I would choose option c—both have $K_\mathrm{s} < 10^{-7}.$ But from experience I found out that $\ce{AgCl}$ is used way more often—but that is probability because of other criteria—which I do not have to bear in mind in this question.

Is my reasoning correct?

Answer 1

The solubility of $\ce{AgCl}$ is equal to $\sqrt{K_\mathrm{s}} = \pu{1.3E-5 M}.$ The solubility $s$ of $\ce{Ag2CO3}$ is such that $K_\mathrm{s} = 4s^3.$ So that its solubility $s$ is equal to $s = \pu{1.2E-4 M}.$ This is ten times more than the solubility of $\ce{AgCl}.$

For gravimetric purposes, $\ce{AgCl}$ is a better choice.

Answer 2

Be aware that in the solubility comparison context, solubility product constants can be directly compared for compounds with the same number of ions created from the formula, where a greater solubility product means a greater solubility.

For compounds with different ion counts, one has to compare ( molar ) solubilities in $\pu{[mol/L]}$ , calculated from solubility products as $K_\mathrm{sp}^{\frac 1n},$ where $n$ is number of ions. $n=2$ for $\ce{AgCl}$, $n=3$ for $\ce{Ag2CO3}$.

So $K_\mathrm{sp,\ce{AgCl}}^{\frac 12}$ versus $K_\mathrm{sp, \ce{Ag2CO3}}^{\frac 13}$