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?


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.


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

  • 1
    $\begingroup$ That is a nice short cut for comparing solubilities. $\endgroup$
    – M. Farooq
    Jan 8 '21 at 15:15

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