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 describes this nicely and cites several sources 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. It prevents from detecting the true strength of $\ce{HF}$ with methods like acid–base titration - they show dissociation of ionic pair. Similar effect is present in ion exchange resins.

Strength of this acid is revealed in more concentrated solutions, where $\ce{HF}$ molecules replace hydronium ions, creating 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 $\ce{K_a}$ 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 $\ce{K_a}$ for strong acids aren't precise). Decrease in acidity between hydrogen halides is described in answer to this question

For concentrated solutions one needs to use Hammett $H_0$ function which gets lower than -10 in pure $\ce{HF}$ (pure sulfuric acid has -12) - it means it's very strong, although not superacidic.

See also:

The Hammett Acidity Function $H_0$ for Hydrofluoric Acid Solutions


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. (credit to Wildcat)

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    $\begingroup$ Fascinatingly, this paper claims that the HF actually dissociates nearly completely in water, but the resulting hydronium ion remains bound in an $\ce{H3O+ + F-}$ complex. I'm not certain exactly how that differs from $\ce{H2O + HF}$, but an interesting idea. $\endgroup$ – chipbuster Aug 7 '15 at 21:13
  • $\begingroup$ @chipbuster Of course it does I thought about asking this question just to say that ;) $\endgroup$ – Mithoron Aug 7 '15 at 23:51
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    $\begingroup$ HF is a strange molecule. I remember that Chemguide used to have an article trying to figure out why it was a "weak acid." He tried an electronegativity argument, an enthalpy of solvation argument, and a free energy argument, all of which showed that HF shouldn't really be weaker than any other halide acid. That section seems to have been removed and replaced with the above link though. $\endgroup$ – chipbuster Aug 8 '15 at 3:16

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