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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.

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    $\begingroup$ The numeric value denoted by the cryptic letters $\rm Ksp$ is going to be useless to you unless you know what it is. $\endgroup$ – Ivan Neretin Jan 22 '18 at 18:07
  • $\begingroup$ @Ivan I know that the very small value of Cu(PO4)2 tells me exactly that he is less prone to precipitate. As higher the value of our constant so the more likely he will 1st to precipitate. But in calculus i find the opposite, because the amount of Cu in Cu(PO4)2 is so small - he will be the first to precipitate. Little bit confused her. Hence, I cant understand what remains with other amount of Cu that reacts with C2O4? And , how at all the precipitation will look like? the reaction will occur separetely different timing? or at the same time both mixed. $\endgroup$ – Mabadai Jan 22 '18 at 18:28
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    $\begingroup$ What you know is wrong, and calculus is right. Also, check the formula of that salt, otherwise it will turn out terribly wrong anyway. What is the charge of copper ion? What is the charge of phosphate ion? What is the formula, then? $\endgroup$ – Ivan Neretin Jan 22 '18 at 18:30
  • $\begingroup$ The stoichiometry will be in 1 --> 1 : 2 ratio $\endgroup$ – Mabadai Jan 22 '18 at 18:33
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    $\begingroup$ dah... check your chemical formulas for the ppts. Copper has a +2 charge, phosphate a -3 charge, and oxalate has a -2 charge. $\endgroup$ – MaxW Jan 22 '18 at 18:36
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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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    $\begingroup$ Without the correct formula for copper phosphate and copper oxalate (and the formula given in the question is clearly incorrect) correct calculation is impossible. Side note: use proper formatting. MathJax isn't that hard to learn. $\endgroup$ – Weijun Zhou Jan 22 '18 at 22:08
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    $\begingroup$ It should be $\ce{Cu3(PO4)2}$. I don't expect redox reaction to take place here. And by the way, I am not the downvoter. $\endgroup$ – Weijun Zhou Jan 22 '18 at 22:43
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    $\begingroup$ @A.J.deLange We try very hard to maintain a professional atmosphere here on the site. As such, please choose your language carefully when interacting with others (see more information at the help center). $\endgroup$ – jonsca Jan 22 '18 at 23:48
  • $\begingroup$ Agree, there is no place her for ego competition, i just need a little clarification. Yes, it can be called a "homework" question, but my point is not to do it and forget, but to understand deeply the process that stands behind all the theory and math of chemistry world. $\endgroup$ – Mabadai Jan 22 '18 at 23:52
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    $\begingroup$ Please visit this page, this page and this one on how to format your future posts better with MathJax and Markdown. $\endgroup$ – andselisk Jan 23 '18 at 0:24

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