We take the following quantities and mix them together:

  • $\pu{20 mL}$ solution of $\ce{AgNO3}$ $\pu{0.001 mol//dm3}$
  • $\pu{20 mL}$ solution of $\ce{H2C2O4}$ $\pu{0.200 mol//dm3}$
  • $\pu{40 mL}$ buffer solution $\mathrm{pH} = 5.00$

What is the apparent solubility of the insoluble salt formed, knowing: $K_\mathrm{a1}=6.50\cdot 10^{-2}$, $K_\mathrm{a2}=6.10\cdot 10^{-5}$ (for $\ce{H2C2O4}$) and $K_\mathrm{s}=1.10\cdot 10^{-12}$ (for $\ce{Ag2C2O4}$)?

The initial concentrations for $\ce{Ag+}$ and $\ce{H2C2O4}$ are: \begin{align} [\ce{Ag+}]&=\pu{2.50E-4 mol//dm3}\\ [\ce{H2C2O4}]&=\pu{5.00E-2 $$mol//dm3}$$ \text{(excess)} \end{align}

We're dealing with the following equilibria: \begin{align} \ce{Ag2C2O4 &<=> 2Ag+ + C2O4^{2-}}\\ \ce{H2C2O4 &<=> HC2O4- + H+}\\ \ce{HC2O4- &<=> C2O4^{2-} + H+} \end{align}

The pH of the solution is above $\mathrm{p}K_\mathrm{a2}$, therefore $\ce{H2C2O4}$ is found mostly as $\ce{C2O4^{2-}}$. By using the Henderson–Hasselbalch equation, we find that $[\ce{C2O4^{2-}}]=\pu{4.30E-2 mol//dm3}$.

Since we started with an excess of acid, we can consider that the variation of concentration for that particular species (after precipitation completes) is negligible. By using $K_\mathrm{s}$, we find that:

$$2S=[\ce{Ag+}]=\pu{5.06E-6 mol//dm3}$$

Is my approach correct? I'm confused because of their approach, where they invoke the following equation:


Which doesn't really make sense since we started with an excess of $\ce{H2C2O4}$.

In my opinion, this equation would make sense if we made a solution only with the buffer and an arbitrary quantity of insoluble salt.

  • 1
    $\begingroup$ Your work looks good to me. (I just followed your logic and didn't check the math.) I also don't understand the equation $$S=\frac{1}{2}\ce{[Ag+]}=[\ce{H2C2O4}]+[\ce{HC2O4-}]+[\ce{C2O4^{2-}}]$$ $\endgroup$ – MaxW May 1 '16 at 15:45
  • $\begingroup$ @MaxW Thanks for checking out! If we consider a certain case, the equation can be used to work out the apparent solubility: in a flask containing a buffer solution of arbitrary $pH$, we add an arbitrary quantity of $\ce{Ag2C2O4}$. We get the equilibria as seen in the post and that relation represents a simple conservation law. The equation becomes: $$\frac{1}{2}[\ce{Ag+}]=\frac{1}{2}\sqrt{\frac{K_{s}}{[\ce{C2O4^{2-}}]}}=\frac{[\ce{C2O4^{2-}}][\ce{H+}]^2}{K_{a1}K_{a2}}+\frac{[\ce{C2O4^{2-}}][\ce{H+}]}{K_{a2}}+[\ce{C2O4^{2-}}]$$ Which is solvable and gives us the apparent solubility. :D $\endgroup$ – L3ul May 1 '16 at 19:44
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    $\begingroup$ Ah.. yes since there are two silver ions, the molar solubility of the salt would be half the silver ion concentration. So $$S(\ce{Ag2C2O4})=2.53\cdot 10^{-6} \frac{\mathrm{moles}}{\mathrm{dm^3}}$$ $\endgroup$ – MaxW May 1 '16 at 19:52
  • $\begingroup$ Oups, my solution forgot to take that into account. Corrected. And as a matter of fact, considering their approach which applies to the case I presented, we get: $$S=6.84\cdot10^{-5} \frac{\mathrm{moles}}{\mathrm{dm^3}}$$ Close, but not quite. $\endgroup$ – L3ul May 1 '16 at 20:01

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