# Apparent solubility of Ag2C2O4 in a buffer solution

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.

• Your work looks good to me. (I just followed your logic and didn't check the math.) I also don't understand the equation $$S=\frac{1}{2}\ce{[Ag+]}=[\ce{H2C2O4}]+[\ce{HC2O4-}]+[\ce{C2O4^{2-}}]$$ – MaxW May 1 '16 at 15:45
• @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 – L3ul May 1 '16 at 19:44
• Ah.. yes since there are two silver ions, the molar solubility of the salt would be half the silver ion concentration. So $$S(\ce{Ag2C2O4})=2.53\cdot 10^{-6} \frac{\mathrm{moles}}{\mathrm{dm^3}}$$ – MaxW May 1 '16 at 19:52
• 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. – L3ul May 1 '16 at 20:01