Determine concentration of phosphate after copper(II) oxalate is precipitated

Solution with $\ce{Na3PO4}$ with $\pu{0.02 M}$ concentration and $\ce{Na2C2O4}$ with $\pu{0.03 M}$ concentration. I add excess $\ce{Cu(NO3)2}$ very carefully – tiny amounts. 


What will be my $\ce{PO4^3-}$ concentration at the moment when precipitation of $\ce{CuC2O4}$ will occur?


Given: $K\_\mathrm{sp}\left(\ce{Cu3(PO4)2}\right) = \pu{1.40e-37}$, $K\_\mathrm{sp}\left(\ce{CuC2O4}\right) = \pu{4.43e-10}$.


I know that the reactions will be:


$$
\begin{align}
\ce{Cu3(PO4)2 &<=> 3Cu^2+ + 2PO4^3-}\\
\ce{CuC2O4 &<=> Cu^2+ + C2O4^2-}
\end{align}
$$


Now, solving some math I find that $\ce{Cu3(PO4)2}$ precipitates first. But then I start asking my self what is happening with the $\ce{Cu}$ amount. Never did lab lessons or thought about the process behind the scene.




Clearly the phosphate will precipitate first. As you add the copper nitrate copper(II) phosphate will precipitate taking equivalent amounts of copper(II) and phosphate. Eventually there isn't enough phosphate left to take out the incremental additions of copper and its concentration begins to rise. Eventually the saturation limit with respect to oxalate is reached. So find that limiting copper concentration (at which oxalate precipitation begins) from $\ce{[Cu^{+2}] = KspCu2Ox/[Ox]}$. I get 4.92222e-07 M. Then find the saturation level for phosphate at that copper concentration from
$\ce{[PO4^{-3}]}$ = $\sqrt{K\_{sp\ce{Cu3(PO4)\_2}}/\ce{[Cu]^3}}$
I get 1.08348e-09 M. A pKsp of nearly 37 suggests that copper phosphate isn't very soluble.


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