Both and reduce the acidity of ##HF## with respect to the lower hydrogen halides

Both and reduce the acidity of ##HF## with respect to the lower hydrogen halides.

We assess the extent of the following rxn:

##H-X(aq) + H_2O rightleftharpoons H_3O^+ + X^-##

For the lower hydrogen halides, ##X=Cl, Br, I##, the equilibrium lies strongly to the right. For ##X=F##, the equilibrium lies to the left (you will have to get your own quantitative data). So why?

i. It is a fact that the ##H-F## bond is stronger than ##H-Cl##, and ##H-Br##. Enthalpy favours the reverse reaction for ##X=F##.

ii. It is also a fact that the ##F^-## is smaller and more polarizing, and thus more likely to induce order. Entropy favours the reverse reaction for ##F^-##.

And thus both enthalpy and entropy conspire to REDUCE the acidity of ##HF## relative to ##HCl## and ##HBr## and ##HI##, which three are all strong Bronsted acids. The entropy effect is probably the most significant. I am happy to entertain further questions if you have doubts.