Consider the equilibrium:
#H-X + H_2O rightleftharpoons H_3O^+ + X^-#
Acid strength, i.e. the degree that the forward reaction goes to completion, depends on (i) the strength of the #HX# bond, and (ii) the degree that the halide is solvated.
For #HF#, there is better overlap in #H-X# because the fluoride ligand is smaller than the lower halides; this is an enthalpy effect. Moreover, the fluoride ligand, #F^-#, is smaller and more polarizing, and more effectively solvated by solvent molecules. Since fluoride is smaller and more polarizing, it causes more solvent order and the forward reaction is disfavoured.
Since both entropy and enthalpy is disfavoured with respect to fluoride, the result is that #HF# is a poor acid compared to #HCl#, #HBr#, etc. The entropy effect is probably most significant.
Note that fluoride salts give basic solutions in water:
#NaF(aq) + H_2O(l) rightleftharpoons HF(aq) + ""^(-)OH + Na^+#
How is this consistent with what we have said before?
