[Rutherford]

The solubility product of calcium oxalate is 2.1*10^-9. The pK_a values of oxalic acid are 1.23 and 4.28 (at 25°C). Calculate the solubility (in gdm^-3) of calcium oxalate monohydrate in a plant cell in which the buffer system regulates the pH to 6.5.

K_sp=[Ca²⁺][C₂O₄^2-]
As [Ca²⁺]≠[C₂O₄^2-] because of the buffer, I can't use the square root of K_sp to calculate the solubility. Therefore:
K_a1K_a2=[C₂O₄^2-][H⁺]²/[H₂C₂O₄]
I don't have [H₂C₂O₄] and [C₂O₄^2-]. What to do now?

[Hunter2]

But you have the pH

[Rutherford]

Yes, [H⁺] is known but there are 2 unknowns, as I wrote.

[Borek]

If there are several unknowns, you have to look for several equations.

[Rutherford]

Okay, I tried with charges equality:
[H⁺]+2[Ca²⁺]=[HC₂O₄^-]+2[C₂O₄^2-]
Now there is are 2 new unknowns [HC₂O₄^-] and [Ca²⁺] .
The starting concentration of H₂C₂O₄ is:
c=[HC₂O₄^-]+[C₂O₄^2-]
But it is again a new unknown.
c is related to [H₂C₂O₄]:
[H₂C₂O₄]=c-cα=c-c*sqrt(K_a1/c), but there is still one more unknown than there are equations ([Ca²⁺])? Again stuck  .

[Borek]

If you will write all existing equilibria, all mass balances and charge balance, you will have enough equations. Just be systematic. Once you have the full system ready, look for ways to simplify it.

[Rutherford]

I am not sure what do you mean by mass balance?

[Borek]

A mass balance, also called a material balance, is an application of conservation of mass to the analysis of physical systems.

So for example sum of concentrations of all oxalate forms in the solution must be equal to the amount of oxalate dissolved - which in turn means it is also equal to the sum of concentrations of all forms of calcium present in the solution.

[Rutherford]

So, I got:
1)K_sp=[Ca²⁺][C₂O₄^2-]
2)K_a1K_a2=[C₂O₄^2-][H⁺]²/[H₂C₂O₄]
3)[H⁺]+2[Ca²⁺]=[HC₂O₄^-]+2[C₂O₄^2-]
4)[Ca²⁺]=[C₂O₄^2-]+[H₂C₂O₄]+[HC₂O₄^-]
I will try to solve this now.

[AWK]

3)[H⁺]+2[Ca²⁺]=[HC₂O₄^-]+2[C₂O₄^2-]

Use expression for K_a2 to eliminate of [HC₂O₄^-]

[Rutherford]

I removed it by deducting the 3rd and the 4th equation-I will mark the new equation with 5). Then I expressed [H₂C₂O₄] from 2) and added to 5). Then, I expressed [Ca²⁺] from 5) and added to 1). I got that [C₂O₄^2-]=4.598*10^-5mol/dm³ or 4.05*10^-3g/dm³. The answer is 6.7*10^-3, so I got a mistake of around 40% which is probably not acceptable  .

[AWK]

At pH 6.5 solubility of calcium oxalate should be very close to that of pure water.
Then your first result 4.6*10^-5 is quite reliable.

[Borek]

Plug your values into equations to check if they are correct or not.

[Rutherford]

[H₂C₂O₄] is too small to compare. [Ca²⁺] that I get from equation 5) is okay with equation 1). Does this mean that the right answer isn't right?

[Borek]

No idea about what the right answer is.

Have you checked also dissociation constants? If you have a set of equations, you need to check them all. If everything is OK, your solution is OK. If not - it is not.