The image below shows how oxalic acid dissociates and forms calcium oxalate ($\ce{CaC2O4}$ or $\ce{Ca(OOC-COO) as written below}$).

enter image description here

When the pH increases, does this result in more or less $\ce{Ca(OOC-COO)}$ forming?

I can think of two factors:

  1. As $\ce{pH}$ increases, $\ce{OH-}$ increases. Greater $\ce{OH-}$ means more of the $\ce{Ca^{2+}}$ will react with $\ce{OH-}$, forming $\ce{Ca(OH)2}$. So the reaction in the third line below will shift to the left, resuling in less $\ce{Ca(OOC-COO)}$.

  2. As $\ce{pH}$ increases, $\ce{H+}$ decreases. Less $\ce{H+}$ results in the reactions (in line 1 and 2) shifting to the right. Hence, there will be more $\ce{(-OOC-COO-)}$ and so more $\ce{Ca(OOC-COO)}$.

These give opposing results. Which one outweighs the other, and why?

The original question from by textbook:

Drinking strong antacids to relieve indigestion produces kidney fluid that is more alkaline than usual.

Are kidney stones more likely to occur under these conditions?

  1. Yes, as at a higher pH a higher concentration of oxalate ions will form.
  2. Yes, as both calcium and oxalate ions are bivalent.
  3. No, as the presence of hydrogen oxalate ion (HOOCCOO–) means that a buffer is present.
  4. No, as the higher concentration of hydroxide ion (OH–) will remove calcium ion by forming calcium hydroxide.
  • $\begingroup$ You seem to have the specifications for the problem and your work mixed together. Would you first state the problem as posed to you and then your assumptions/work? $\endgroup$ – MaxW Jan 21 '17 at 21:58
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    $\begingroup$ Do you mean just add the textbook problem? If so, I have now edited it in. Thank you. $\endgroup$ – K-Feldspar Jan 21 '17 at 22:01

In the body we'll assume that:

  • There is some low level of $\ce{Ca^{2+}}$ present in the blood or urine.
  • The pH isn't nearly high enough to precipitate out $\ce{Ca(OH)2}$.
  • Under "normal" bodily conditions calcium oxlate doesn't precipitate out either so the overall oxalate concentration is low.

Let's let $x$ be the concentration of all the oxalate species. By knowing $x$, the pH and the two pka's we can calculate the exact concentrations of all three oxalate species.

$x = \ce{[HOOC-COOH] + [HOOC-COO^-] + [^-OOC-COO^-] }$

Now the pKa values for oxalic acid are far enough apart that at most two of the three oxlate species will have an appreciable concentration. For normal bodily functions only $\ce{[HOOC-COO^-]}$ and $\ce{[^-OOC-COO^-]}$ will be the only significant species. Since only a couple of significant figures will be of interest, we can use the Henderson–Hasselbalch equation to calculate the relative concentration of the two species to reasonable accuracy. (The essence is that the third species has a negligibly small concentration.)

pH = $\text{pK}_{a2} + \text{log}(\dfrac{\ce{[^-OOC-COO^-]}}{\ce{[HOOC-COO^-]}})$

Thus as the pH increases relatively more $\ce{[^-OOC-COO^-]}$ is formed and less $\ce{Ca^{2+}}$ will be soluble.

But normal blood is alkaline already. Thus there would be a negligibly small amount of $\ce{[HOOC-COO^-]}$ in the blood to start with since the $\text{pK}_{a2} = 4.2$. So calcium oxalate will be unlikley to ppt in the blood.

However the kidneys make urine which has a pH from 4.5 to 8. So if the pH in the urine changes as much as from 4.5 to 8 then there could be a greater tendency to ppt calcium oxalate in the urine.


(4) is out since if body was that alkaline your tissue would dissolve.

(3) is out since you can see from the HH equation specified that the presence of another buffer would effect the absolute pH but not the relative oxalate species at that pH.

(2) is true, but the answer doesn't note any effect of pH. So the answer doesn't really address the problem.

Thus the answer must be (1).

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  • $\begingroup$ Thank you for the answer. Could you please explain the significance of this line "x=[HOOC−COOH]+[HOOC−COO−]+[−OOC−COO−]" It looks to be the sum of each of the reactants except Ca2+. Is that right, and if so, why are they being summed? $\endgroup$ – K-Feldspar Jan 21 '17 at 22:46
  • $\begingroup$ Ok, tried to expand that point. The simplification doesn't make any real difference for a computer program which can solve for all three species concentrations iteratively in a fraction of a second, but using the HH equation makes hand calculations a lot simpler than solving a cubic equation. Note that pka's and Ksp only have two significant figures. $\endgroup$ – MaxW Jan 21 '17 at 23:06

At a higher pH a higher concentration of oxalate ions will form (Le Chatelier's principle). Thus the formation of kidney stones is more likely.

If calcium hydroxide precipitates in your body you're most likely dead.

The solubility product of $\ce{Ca(OH)_2}$ is $5.5 \cdot 10^{-6}$. The calcium concentration in blood is about $3 \cdot 10^{-3}~\pu{mol\cdot l^{-1}}$. It needs a pH above 13 to precipitate calcium hydroxide in blood.

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  • $\begingroup$ Thank you for the answer. Would you please be able to point out why "a higher pH a higher concentration of oxalate ions will form" (which does make sense) outweighs: a higher pH = more [OH-] = Ca2+ being used up to form Ca(OH)2 = reaction in 3rd line shifting to the left = less calcium oxalate or kidney stones. $\endgroup$ – K-Feldspar Jan 21 '17 at 22:42
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    $\begingroup$ Added the pH required to precipitate $\ce{Ca(OH)_2}$ from blood. $\endgroup$ – aventurin Jan 21 '17 at 23:14

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