We address the equilibrium....
#HX(aq) + H_2O(l) rightleftharpoonsH_3O^+ + X^(-)#
For strong acids, this equilibrium lies strongly to the right; for weak acids the equilibrium lies to the left, and at equilibrium significant quantities of the free acid, i.e. HX(aq), remain in solution. The #"hydronium ion"#, #H_3O^+#, is a REPRESENTATION of the acid in the solution, we could likewise represent the equilibrium as...
#HX(aq) + stackrel(H_2O)rightleftharpoonsH^+ + X^(-)#
The actual acidium species is PROBABLY a cluster of 3-4 water molecules with an EXTRA #H^+#, i.e. #H_7O_3^+# or #H_9O_4^+#...the protium is mobile and tunnels if you like in solution to give a substantial ionic mobility.
So, as chemists, as physical scientists, let us look at the data. This site quotes #pK_a=-9.3, -8.7, -6.3# for #HI#, #HBr#, and #HCl# respectively (the LOWER the #pK_a# the farther to the right the equilibrium lies...i.e. #pK_a=-log_10[H_3O^+]#)
On the other hand, for #HF#, #pK_a=+3.17#...so hydrogen fluoride is a weaker acid by NINE orders of magnitude in comparison to #HCl#. Two factors influence this:
#i.# #"The strength of the H-X bond, i.e. enthalpy."# As a smaller atom than the lower halogens, fluorine more effectively overlaps the hydrogen orbitals than does chlorine and bromine, resulting in a stronger bond, and weaker acidities for hydrogen fluoride with respect to hydrogen chloride and hydrogen bromide.
#ii.# #"The charge density of the halide ion, i.e. entropy."# As a smaller atom than the lower halogens, fluoride ion is more effectively solvated by the water molecules, and this results in UNFAVOURABLE solvent ORDER.
Both factors conspire to reduce the acidity of #HF# as compared to the acidities of #HCl# and #HBr#...
This is probably a second year inorganic treatment of the problem. I don't know at which level you study, so pick out the appropriate portions of the answer.
