Peter Sykes attributes the difference in their acidity to differential solvation. enter image description here

From what I understand, there is not much difference in how well the ions are solvated, since the Delta H is comparable, however, the change in entropy is much greater for ethanoic acid but I don't understand why?

  • $\begingroup$ On contrary it clearly says that difference is caused by solvation but between between formate and acetate $\endgroup$
    – Mithoron
    Commented May 6, 2017 at 19:11

1 Answer 1


What an interesting way of looking at acidity! How I would look at it would be the more standard way which is to compare the stability of their conjugate bases...

The ethanoate ion has an methyl group, which is electron-donating, donating electron density towards the oxygens, concentrating the negative charge. This makes it a more unstable conjugate base.

In stark contrast, the methanoate ion, does not have the electron donating group and thus, does not have such a high concentration of negative charge. This makes it the more stable conjugate base.

The explanation proposed can also explain the differential solvation of the two carboxylic acids...

With a higher charge density, ethanoic acid would form a more "ordered" hydration shell as the electrostatic attraction between the water molecules and the negative charge is greater.

With a lower charge density, methanoic acid would form a less "ordered" hydration shell as the electrostatic attraction between the water molecules and the negative charge is not as great.

Consequently, the decrease in entropy for the dissociation of ethanoic acid is much greater than that for the dissociation of methanoic acid.

Therefore, the dissociation of ethanoic acid would be less favoured than the dissociation of methanoic acid, based on both considerations of entropy and conjugate base stability, which are actually linked.

  • $\begingroup$ But by the same argument, we could say that since the negative charge density is greater on ethanoate ion, it should be solvated to a greater extent, but as seen in the Delta H, they're both solvated to roughly the same extent. Why is the change is entropy so different for both? $\endgroup$
    – xasthor
    Commented May 7, 2017 at 17:19
  • $\begingroup$ Hmmm... Why is there necessarily a relationship between enthalpy change and solvation? In this case, when the water molecules form a more ordered hydration shell, only entropy is concerned. The decrease in solvent-solvent interactions and the increase in solute-solvent interactions is more or less the same. Thus, the enthalpy change would also be more or less equivalent for both. This is just my speculation. $\endgroup$ Commented May 9, 2017 at 7:56
  • $\begingroup$ I would think there would be a relation in enthalpy since the greater negative charge density would attract and form more hydrogen bonds with the solvent. $\endgroup$
    – xasthor
    Commented May 9, 2017 at 13:17

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