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:
$$S=\frac{1}{2}\ce{[Ag+]}=[\ce{H2C2O4}]+[\ce{HC2O4-}]+[\ce{C2O4^{2-}}]$$
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.