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