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)

Also, any further links to resources about homoassociation would be appreciated. 

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.
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

\begin{aligned}
\ce{H2O + HF &amp;~&lt;=&gt; [H3O^+ {.} F^{-}]&amp;&amp;(I)}\\
\ce{[H3O^+ {.} F^{-}] &amp;~&lt;=&gt; H3O^+ + F^{-}&amp;&amp;(II)}\\
\ce{[H3O^+ {.} F^{-}] + HF &amp;~&lt;=&gt; H3O^+ + HF2^{-}&amp;&amp;(III)}\\
\ce{3HF &amp;~&lt;=&gt; H2F^+ + HF2^{-}&amp;&amp;(IV)}\\
\end{aligned}

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.