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?


2 Answers 2


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 $K_a$ 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

For concentrated solutions one needs Hammett $H_0$ function and get estimates using stuff like nitrated aromatic bases, and it's still problematic to get real value. In an old paper, 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 region.


  1. Herbert H. Hyman, Martin Kilpatrick, Joseph J. Katz; J. Am. Chem. Soc. 1957, 79, 14, 3668–3671. DOI:10.1021/ja01571a016.

  2. Ronald J. Gillespie, Jack Liang; J. Am. Chem. Soc. 1988, 110, 18, 6053–6057. DOI:10.1021/ja00226a020.

  • $\begingroup$ One might argue that the dilute acid therefore is not HF. The hydrogen fluoride has reacted with water to form (H3O)F. That is the actual weak acid. $\endgroup$ Commented Apr 2, 2020 at 20:22
  • $\begingroup$ @OscarLanzi Yes, I was thinking more of equilibrium of breaking the h-bond and pair with it, but it's more reasonable to think of it as Brønsted acid. As far as "being HF", it's more of a HF then dilute HCl is HCl, I wager. $\endgroup$
    – Mithoron
    Commented Apr 4, 2020 at 14:53
  • $\begingroup$ neither hydrogen halide is really the hydrogen halide in water. The acid that's in aqueous HCl solution is (in simplified form) H3O+, which is weaker than HCl. $\endgroup$ Commented Apr 4, 2020 at 15:04
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    $\begingroup$ @OscarLanzi That was semi-jokingly, as H and F are still bound, just by h-bond :) Even "pure" H2SO4 isn't just H2SO4, at least in condensed phase. $\endgroup$
    – Mithoron
    Commented Apr 4, 2020 at 15:43

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
    Commented Aug 7, 2015 at 21:13
  • $\begingroup$ @chipbuster Of course it does I thought about asking this question just to say that ;) $\endgroup$
    – Mithoron
    Commented Aug 7, 2015 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
    Commented Aug 8, 2015 at 3:16

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