[Ahmed Abdullah]

The hydration enthalpy of Fluoride ion is very high.
We know that HF react violently with Na-OH to form fluoride salt and water. The enthalpy of this reaction is greater than Neutralization enthalpy (H+ + OH-=H2O) of other acid base. That proves that the hydration enthalpy is sufficient enough to supply required energy for H-F bond breaking, beside producing extra heat.
So I don't understand why HF don't ionise in water, where hydration enthalpy is large enough to promote ionization.

[ATMyller]

HF is water soluble, but it doesn't ionizate because unlike all other hydrogen halogens it doesn't exist in monomolecular form. Instead it forms polymeric structures, chains that are kept together by hydrogen bonds between hydrogen and adjacent molecule's fluoride.

[vhpk]

No, I think that in water HF ionizates:
HF + H₂O --> H₃O⁺ + F^-
but after that
HF + F^- --> HF^2-
Almost acids ionizate in water to produce H³O⁺

[wilson]

I think ATMyller is right. HF is misible in water. And yes, it is a weak acid unlike other hydrogen halides because it does not dissociate (ionize) to an extent as large compared to HCl, HBr, HI.
Now here's your question, why is this so?

Actually, this is a more complicated process than simply hydration enthalpy of F-(g) to F-(aq).
Visualize this: HF is a very polar molecule with a very polar bond with high dipole moments due to fluorine as a very electronegative atom. The 3 lone pairs on the small fluorine atom also amplifies this effect. Thus, HF exist as polymeric chains in water due to strong hydrogen bonding with itself and water molecules. So note that HF does NOT exist as one molecule but a large chain with many water molecules.

Here,
You forgot about (1)the energy needed to separate the water molecules and the HF polymeric structures from hydrogen bonding to isolate just one molecule, (2)HF bond enthalpy to break the H-F bond, and (3)electron affinity of F, and ionisation energy of H to separate charges.
These are 3 other energy changes on top of hydration enthalpy which occurs after all the above.

Then on an overall, HF + H2O --> H3O+ + F- [LOW equilibrium constant]

So although hydration enthalpy is very favourable, you cannot draw a conclusion without considering other energy changes!

[wilson]

HF + H2O --> H3O+ + F-
To complete my answer (i forgot about your entire question), why the violent reaction and high heat of neutralization when HF is a weak acid?

If moles of HF in solution is high enough (concentrated),
a disproportionation reaction takes place.

2HF ? H+ + FHF?

This is interesting, as we get more H+, and yet maintaining stability of the FHF- anion by independent hydrogen bonding of the F-H-F itself.

So
This interaction does not interfere with ionization of HF as the fluoride ion would-- I am not too sure how the entropy falls into place here, but I do know that the SMALL F atom will impose orderly arrangement of the water molecules by hydrogen bonding and increase entropy which is not favourable.