Fluorine is the most electronegative halogen and therefore, there is larger difference in electronegativity between the atoms of 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 HX+ is easy to remove?
Answer
First of all as @chipbuster says 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 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 HF molecules replace hydronium ions, creating bifluoride anions, freeing them - it's homoassotiation mentioned by Wildcat.
More to the point - does it mean that HF isn't weaker acid than other hydrogen halides? The answer isn't simple, and acidity itself depends on many factors.
In terms of KXa in diluted aqueous solutions HF is probably still weaker than HCl and the rest but getting exact value is greater problem than normally (generally values of KXa 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 H0 function which gets lower than -10 in pure HF (pure sulfuric acid has -12) - it means it's very strong, although not superacidic.
See also:
The Hammett Acidity Function H0 for Hydrofluoric Acid Solutions
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