Freezing point of a solvent is decreased due to the added solute particles in it. e.g. ions. Therefore electrolytes contribute to lowering freezing point (say for physiological conditions).

If a particular solution has a non-electrolyte solute, which is held together with hydrogen bonds (forming dimers), I want to know how do H-bonds affect the freezing point?

Pure water itself forms hydrogen bonds. Yet at 0'C it freezes theoretically. How to approximate effect of the hydrogen bonds in an acidic solution to its freezing point depression theoretical value? e.g. do H-bonds in CH3COOH halve the freezing point depression value?


1 Answer 1


The addition of one molecule (or salt) of solute produces $i$ effective solute particles. This defines the van't Hoff factor $i$, which encodes the strength of the solute with regards to colligative properties.

As simple (idealized) examples in aqueous solution, the van't Hoff factor of $\ce{NaCl}$ is 2, as $\ce{NaCl}$ dissociates into the two solute particles $\ce{Na+}$ and $\ce{Cl-}$ in solution, whereas the van't Hoff factor of glucose is 1, as glucose doesn't dissociate in solution.

These examples are idealized because they do not account for intermolecular or interionic interactions, and are therefore most valid in dilute solutions. In concentrated solutions, intermolecular and interionic effects are much more important: the presence of hydrogen bonding, for example, or ion pairs in solution. Attractive interactions "clump" the solute particles together and reduce the van't Hoff factor.

Returning to your example, if we assume that acetic acid does not dissociate in solution, and we're under conditions in which acetic acid dimer predominates, then indeed we will have a van't Hoff factor $i = 0.5$. It's important to note, however, that the hydrogen-bonding interaction in acetic acid is particularly strong, and in general intermolecular and interionic interactions will not change the van't Hoff factor so drastically.

It remains to consider how to correct for these interactions. Intuitively, it seems to me that the activity coefficient should be the desired correction factor, considering that the activity coefficient is a measure of how ideal (interaction-less) a solution is, but I do not have a more quantitative argument along these lines.

  • 1
    $\begingroup$ Appreciate this answer. It's not that I haven't looked into it. I take time to really sort things out and understand - because there are new theories I want to do some calculations. Once I do, I will reply to you again. $\endgroup$
    – bonCodigo
    Jul 27, 2017 at 9:16

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.