In calculating the molar solubility of silver carbonate in a solution of $\pu{0.125 M}$ sodium carbonate (knowing the $K_\mathrm{sp} = 8.46 \times 10^{-12}$) I have a question with the solution provided which is as follows:

$K_\mathrm{sp} = [\ce{Ag+}]^2 [\ce{CO3^2-}]$, the initial concentration of $\ce{CO3^2-}$ is $\pu{0.125 M}$. Let $x$ denote the change in concentration for $\ce{Ag+}$.

Therefore $K_\mathrm{sp} = x^2 \left(\frac{1}{2}x+0.125\right)$. Assuming the change for $\ce{CO3^2-}$ is negligible, that gives $K_\mathrm{sp} = x^2\cdot(0.125)$, which substituting the $K_\mathrm{sp}$ value gives $x \approx 8.227 \times 10^{-6}$, which is the molar solubility.

What I'm confused about is shouldn't the molar solubility be half that number since $\ce{Ag2CO3 <=> 2 Ag+ + CO_3^2-}$, meaning that for each mole that $\ce{Ag+}$ increases, $\ce{Ag2CO3}$ only decreases by half that amount? Could someone just explain whether my reasoning is correct or not?

  • $\begingroup$ There is justified simplification all carbonate ions come from sodium carbonate. You can verify it at the final check. $\endgroup$
    – Poutnik
    Mar 26, 2022 at 8:37
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    $\begingroup$ Hi, you might find it interesting to briefly scroll through mhchem.github.io/MathJax-mhchem -- many of the things you wrote can be done in an easier way using mhchem (the available syntax goes beyond just enclosing normal maths in \ce{...}). $\endgroup$ Mar 26, 2022 at 9:31
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    $\begingroup$ Also, yes, I'd expect the answer to be divided by a factor of 2. $\endgroup$ Mar 26, 2022 at 9:32
  • $\begingroup$ @Poutnik I don't think that is what I was asking for, I understand why the assumption was made, but I don't understand why the $x$ in the given solution actually represents the molar solubility. $\endgroup$ Mar 26, 2022 at 9:33
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    $\begingroup$ X represents c(Ag+)=2c(Ag2C3) $\endgroup$
    – Poutnik
    Mar 26, 2022 at 9:45

1 Answer 1


Suppose in equilibrium, the soubility of $\ce{Ag2CO3}$ is $s \ \pu{mol L-1}$ in an aqueous $\pu{0.125 mol L-1}$ $\ce{Na2CO3}$ solution. That means, amount of $s \ \pu{mol}$ of solid $\ce{Ag2CO3}$ would be dissolved in one leter of $\pu{0.125 mol L-1}$ $\ce{Na2CO3}$ solution and in equilibrium as:

$$\ce{Ag2CO3 (s) <=> 2Ag+ (aq) + CO3^2- (aq)}$$

Therefore, by definiion, for $\ce{Ag2CO3}$, $K_\mathrm{sp} = [\ce{Ag+}]^2[\ce{CO3^2+}]$ (all concentrations are in equilibrium).

Since, solid $\ce{Ag2CO3}$ is dissolved in an aqueous solution of $\pu{0.125 mol L-1}$ $\ce{Na2CO3}$, you may have to consider $\ce{Na2CO3}$ concentration as well. However, as Poutnik pointed out in his comment, it is justified simplification that all carbonate ions comes from sodium carbonate solution alone since it is in higher concentration compared to carbonate ion concentration from dissoled $\ce{CO2}$ in the water. Let's look at the corresponding ICE chart (ignore solid $\ce{Ag2CO3}$ in the solution):

$$\begin{array}{l|cccc} & [\ce{Ag+}] & [\ce{CO3^2-}] \\ \hline \text{Initial concentration} & 0 & 0.125 \\ \text{Change in concentration} & +2s & +s \\ \text{Equilibrium concentration} & +2s & 0.125+s \\ \hline \end{array}$$

Thus at equlibrium, $$K_\mathrm{sp} = [\ce{Ag+}]^2[\ce{CO3^2+}] = (2s)^2 \cdot (0.125+s) = (2s)^2 \cdot 0.125 = 0.5s^2$$

Since $K_\mathrm{sp} = 8.46 \times 10^{-12}$ is given,

$$0.5s^2 = K_\mathrm{sp} = 8.46 \times 10^{-12} \ \Rightarrow \ s = \left(\frac{ 8.46 \times 10^{-12}}{0.5}\right)^\frac{1}{2} \approx \pu{4.11 \times 10^{-6} mol L-1}$$

Thus, as as orthocresol pointed out in his comment, the given answer should be divided by a factor of 2.


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