Why is EDTA ($\ce{C10H16N2O8}$) poorly soluble in water?

I am interested in analyzing the hardness of water in a sample due to the presence of $\ce{Ca^2+}$ cations. I attempted to do this by titrating with ethylene diamine tetraacetic acid ($\ce{C10H16N2O8}$) however these attempts have proven to be futile.

  • $\begingroup$ There are all sorts of standard procedures for doing a total water hardness. You can also do Ca only. Did you use a standard procedure (if so which one?), or did you try to make something up? $\endgroup$
    – MaxW
    Commented Apr 17, 2016 at 2:02
  • $\begingroup$ I've done EDTA titration of hard water. We did it in the class I used to teach. What procedure are you using? Are you using EBT? To answer your question, EDTA isn't typically considered poorly soluble in water. I believe more is going on here. $\endgroup$ Commented Apr 17, 2016 at 2:14

1 Answer 1


I ran into the same problem myself when I bought (foolishly, it turned out) $\ce{H4EDTA}$ instead of $\ce{Na2H2EDTA}$ or $\ce{Na4EDTA}$ for some of the experiments for my doctorate. It really confused me, because carboxylic acids are generally quite soluble and, with four of the things, you'd think it'd dissolve right away. But, no. There are even amines in there, too, but they apparently don't help much either.

Based on the abstract of a 1973 article by Ladd and Povey1, my guess is the apparent poor solubility of $\ce{H4EDTA}$ is actually a kinetic effect, due to a network of intra- and inter-molecular hydrogen bonds present in the solid (emphasis added):

The [EDTA] molecule exists in a cis conformation and possesses two-fold symmetry, the diad axis passing through the central $\ce{C-C}$ bond. The nitrogen atom is protonated, and takes part in a bifurcated intramolecular hydrogen bond. A short $\mathbf{\left(2.46~Å\right)}$ hydrogen bond links the oxygen atoms of adjacent molecules.

I don't have the necessary thermodynamic data at hand to argue with certainty, but my guess is that the activation energy for solvent penetration into this hydrogen-bond network is rather high, and thus the rate of solvation is slow even though the overall process is (presumably) exergonic $\left(\Delta G < 0\right)$.

Reference: Ladd, M. F. C.; Povey, D. C. Crystallographic and spectroscopic studies on ethylenediaminetetraacetic acid (edta). J. Crystal Molec. Struct. 1973, 3 (1), 15–23. DOI: 10.1007/BF01270899.

  • 2
    $\begingroup$ An option would be to dissolve EDTA through $\mathrm{pH}$ with an alkaline hydroxide, the cation of which will not form complexes, in effect re-creating $\ce{Na2H2EDTA}$ or $\ce{Na4EDTA}$ in vitro. The kinetics aren't great but much better than for the dissolution of the tetravalent acid if I remember correctly. $\endgroup$
    – user41033
    Commented Jul 20, 2018 at 8:57

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