Great question. Here's my shot at it. Fluorine is very electronegative which makes it the ideal halogen for having a very acidic hydrohalide( HX). However, fluorine has a little problem when it comes to acidity...and I do mean "little"..its size doesnt allow for stabilization of the added electron from hydrogen. After dissociation in a aqueous solution, the H+ and -F are on their own. -F is too unstable because its not successfully delocalizing that electron over its relatively small nucleus. In order for reactions to make sense, they end game is stability. Just remember stability governs reactions even LIFE.
Although this is a little different than the kind of data I'm seeking, I'll try to explain. 10 electrons is the most stable state of fluorine. If comparing HF and F(-), each has 10 electrons in each so I don't imagine any instability. Boron, carbon, nitrogen, oxygen, fluorine, neon, sodium, and magnesium are all in their most stable states with 10 electrons. Except for boron, these atoms are all less stable if they have less than 10 electrons. Adding an electron to fluorine increases its stability. Adding an electron is a redox reaction and is a different reaction with different energies than the forces holding a proton to an electron pair (my question). I further argue that for methane, ammonia, and water, the electrons are arranged in pairs in a tetrahedral arrangement. By analogy, HF and F(-) should have the same structure. If this is so, then all electron pairs may be treated equally and delocalization shouldn't change the structure of the atom or the basicity of fluoride (or any atom). Basicity is a property of an electron pair. I argue basicity should be an inverse square force, the closer, the stronger. If you compare the bond lengths, CH>NH>OH>HF. The nuclear charge of carbon is less so its electron pair extends further from its nucleus than the others, hence its electron pair should be more basic to a proton. By analogy, HF would be the least basic of this row.
If this model holds (and I think it does), then it should hold for all atoms. For the series BH4(-), CH4, NH3, H2O, and HF, they all contain 10 electrons so the difference in forces should be found in nuclear-electron pair, nucleus-proton, and electron pair-proton distances. We do not know the locus of electrons, but I argue they should still follow the laws of physics, namely inverse square forces. Anyone can do the calculations as I have and discover HF should be the weakest acid. It should be weaker than HI because of the greater charge of iodine pulls its valence electrons proportionately closer to its nucleus increasing its nucleus-proton repulsion and reducing the electron pair-proton attraction or acidity.
I know how to calculate the energy differences in reactions. What I do not know how to do is to adjust for the various changes that must be made to accommodate for the difference between gas phase and solution phase energies. For example, HF is a strong bond in the gas phase but weaker in solution. HCl, HBr, and HI are similar. The loss of an electron by sodium is endothermic in the gas phase but exothermic in solution. We should still be able to calculate the energy differences in reactions. Acidity is a solution based force and is different from gas phase forces. How do we adjust for the values to be used?