# Chemistry - Criterion for gravimetric analysis of silver

## Solution 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.

## Solution 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 $K_\mathrm{sp, \ce{Ag2CO3}}^{\frac 13}$