Is HF the least acidic hydrogen halide?

Fluorine is the most electronegative halogen and therefore, there is larger difference in electronegativity between the atoms of $\ce{HF}$ than any other hydrogen halide, which means the positive charge on hydrogen atom is the greatest in this compound and hence a comparatively small negative charge is needed to attract it. If so, why is it classified as least acidic hydrogen halide if the $\ce{H+}$ is easy to remove?




First of all, as chipbuster says, $\ce{HF}$ in diluted solutions in water is nearly completely dissociated and therefore **shouldn't be called weak**. [Wikipedia](https://en.wikipedia.org/wiki/Hydrofluoric_acid#Properties) describes this nicely and cites [several](https://dx.doi.org/10.1021%2Fja00537a008) [sources](https://dx.doi.org/10.1021%2Fja077846o) for this claim.


It was rather difficult to prove (spectroscopic methods were used), because hydronium ions created in dissociation are mostly bound to fluorine anions with hydrogen bonds, in what is called [tight ion pairs](https://en.wikipedia.org/wiki/Tight_ion_pair). It prevents from detecting the true strength of $\ce{HF}$ with methods like [acid–base titration](https://en.wikipedia.org/wiki/Acid%E2%80%93base_titration) - they show dissociation of ionic pair. Similar effect is present in [ion exchange resins](https://en.wikipedia.org/wiki/Ion-exchange_resin).


Strength of this acid is revealed in more concentrated solutions, where $\ce{HF}$ molecules replace hydronium ions, creating [bifluoride](https://en.wikipedia.org/w/index.php?title=Bifluoride) anions, freeing them - it's homoassotiation mentioned by Wildcat.


More to the point - does it mean that $\ce{HF}$ isn't weaker acid than other hydrogen halides? The answer isn't simple, and acidity itself depends on many factors.


In terms of $K\_a$ in diluted aqueous solutions $\ce{HF}$ is probably still weaker than $\ce{HCl}$ and the rest but getting exact value is greater problem than normally (generally values of $K\_a$ for strong acids aren't precise). Decrease in acidity between hydrogen halides is described in answer to [this question](https://chemistry.stackexchange.com/questions/13609/hydrogen-halide-acidity)


For concentrated solutions one needs [Hammett $H\_0$ function](https://en.wikipedia.org/wiki/Hammett_acidity_function) and get estimates using stuff like nitrated aromatic bases, and it's still problematic to get real value. In [an old paper](http://pubs.acs.org/doi/abs/10.1021/ja01571a016), I found an estimate barely lower than -10 for pure $\ce{HF}$, while pure sulfuric acid has -12. A newer paper, on the other hand, gives value 15.1 - almost five orders of magnitude more acidic ($H\_0$ is logarithmic like pH). The new value puts pure $\ce{HF}$ firmly in the [superacidic](https://en.wikipedia.org/wiki/Superacid) region.


### References


1. Herbert H. Hyman, Martin Kilpatrick, Joseph J. Katz; J. Am. Chem. Soc. 1957, 79, 14, 3668–3671. [DOI:10.1021/ja01571a016](https://doi.org/10.1021/ja01571a016).
2. Ronald J. Gillespie, Jack Liang; J. Am. Chem. Soc. 1988, 110, 18, 6053–6057. [DOI:10.1021/ja00226a020](https://doi.org/10.1021/ja00226a020).


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Hydrofluoric acid is the least acidic hydrogen halide because of fluorine's electronegativity. Because of the fluoride ion's small size, it cannot disperse the negative charge over a larger space and will have an extremely high affinity for an electrophile (like $\ce{H+}$), and because of this it will remain mostly as $\ce{HF}$.


At high concentrations, through hydrogen bonding with its conjugate base, hydrofluoric acid is actually a significantly stronger acid due a process called [homoassociation](https://en.wikipedia.org/wiki/Homoassociation). (credit to Wildcat)


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