# Homoassociation and Acid Strength

I understand that the strength of HF differs significantly depending on the concentration. I understand the basic theory behind this - homoassociation - and stabilization of the highly negative charge dense fluoride anion through hydrogen "bonds"/intermolecular attractions with undissociated solute $\ce{HF}$ molecules.

However, I am unclear on the Wikipedia explanation of homoassociation: http://en.wikipedia.org/wiki/Homoassociation

Regarding the equations - I cannot tell if the authors meant to refer to pure, liquid $\ce{HF}$ or a solution of HF. I'm thinking that the authors are referring to liquid $\ce{HF}$ because auto-ionization of HF is mentioned. I'm thinking that auto-ionization would only be significant in liquid $\ce{HF}$. Or in a concentrated solution.

Also, regarding the second equation - the authors make it look like the fluoride anion is forming an actual, intramolecular, bond with the $\ce{HF}$ molecule. I think the authors are implying the formation of an intermolecular hydrogen bond, correct?

On the other hand the bifluoride anion seems to be a real molecular entity (http://en.wikipedia.org/wiki/Bifluoride).

This only muddies the water further. I.e. which of the following best represents how the fluoride anion is stabilized, causing concentrated solutions of $\ce{HF}$ to ionize further than one would expect from $\ce{K_a}$ values?

$\ce{HF}$ ... $\ce{F^-}$ (hydrogen bonding)

or

$\ce{HF_2^-}$ (formation of bifluoride anion)

ETA: Also, can't the fluoride anion form hydrogen bonds with water? The fact that concentrated HF is stronger than dilute HF than what mere $\ce{Ka}$ values suggest implies that H-bonding with water isn't as effective as H-bonding with HF molecules. Why? Is water less polar? On the other hand water can form a maximum of two hydrogen bonds with the fluoride anion, while a HF molecule can only form a maximum of one hydrogen bond with a fluoride anion.

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I think this Wikipedia article on hydrogen fluoride will answer your questions. It discusses the acidity of $\ce{HF}$ from dilute to concentrated aqueous solution, and then pure $\ce{HF}$. They present four equilibria to describe these situations
In dilute solution equilibrium (I) lies far to the right and the second equibrium (II) far to the left. So even though the $\ce{HF}$ is extensively dissociated, the resultant hydrogen bonded ion pair is very stable and reluctant to undergo significant dissociation. Hence the solution is weakly acidic. As the concentration of $\ce{HF}$ is increased homodissociation comes into play [equilibrium (III)] producing the hydrogen difluoride ion as a discrete entity which is stabilized through intermolecular hydrogen bonding. The increase in $\ce{H3O+}$ makes the resultant solution very acidic. In pure $\ce{HF}$, autoionization occurs producing an exteremly acidic solution. Page 4 of this link shows a graph of HF acidity from near zero to 50% concentration.
I'm not sure of how you are using the phrase "comprised of hydrogen bonds". Methane is comprised of hydrogen bonds (to carbon). $\ce{H2F^{-}}$ is an ion with real $\ce{H-F}$ bonds. The hydrogens in these ions then form intermolecular hydrogen bonds with other other $\ce{H2F^{-}}$ molecules. –  ron Jun 21 at 19:16