I performed an experiment using egg shells (calcium carbonate) to represent teeth and reacted the calcium carbonate with acids. When measuring the rate of reaction citric acid had a higher rate of reaction and from my research I saw that citric acid forms a more stable compound with calcium hence it chelates the calcium more than ethanoic acid.

In the mouth this would remove calcium ions from $$\ce{Ca10(PO4)6(OH)2(s) + 2H+ (aq) <=> 3Ca3(PO4)2(s) + Ca^2+(aq) +2H2O(l)}$$ that equation hence the position of equilibrium shifts to the right so more tooth decay.

But with egg shells: $$\ce{CaCO3(s) + 2H+ (aq) -> Ca^2+(aq) +CO2(g) +H2O(l)}$$

there is no equilibrium so if more calcium ions are chelated why does this increase the rate of reaction?

  • $\begingroup$ Well, it's also more acidic. $\endgroup$
    – Mithoron
    Commented Dec 26, 2015 at 23:50
  • $\begingroup$ thanks but in terms of chelating does the chelation not have any effect? $\endgroup$
    – Radhika
    Commented Dec 27, 2015 at 13:22

1 Answer 1


Acetic acid has a $\mathrm{p}K_\mathrm{a}$ value of $4.76$ according to Wikipedia.

Citric acid has a $\mathrm{p}K_\mathrm{a,1}$ value of $3.13$ and a $\mathrm{p}K_\mathrm{a,2}$ value of $4.76$, also according to Wikipedia. It is about ten times as acidic as acetic acid is.

The greater acidity translates to a lower $\mathrm{pH}$ of the resulting solution. The lower $\mathrm{pH}$ translates to bases being protonated more and quicker. Considering that the decomposition reaction of carbonate is actually:

$$\ce{HCO3- + H+ <=> H2CO3 <=>> H2O + CO2 ^ }$$

The $\mathrm{pH}$ effect alone should be enough to explain a faster reactivity.

However, you can also invoke the following equilibrium which will be right-side shifted with citric acid (but obviously not so much with acetic acid):

$$\ce{CaCO3 (s) + cit^3- <=> Ca^2+ + CO3^2- + cit^3- <=>> [Ca(cit)]- + CO3^2-}$$

Where $\ce{cit}$ is citrate or the anion of citric acid. Complexation of calcium will remove free calcium ions from solution, and the solubility product only considers free ions in solution. Thus by complexing the free ions, more calcium can be dissolved.

I would still perform control experiments to make sure it is actually the complexation effect of citrate and not its acidity. For example, try reaction $\ce{CaCO3}$ with acetic acid in the presence of $\ce{Na3cit}$ or $\ce{Na2Hcit}$. These should be equally good complexing agents.


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