# Why is HF a weak acid, and not a strong acid?

Aug 8, 2016

Both entropy and enthalpy reduce the acidity of $H F$ with respect to the lower hydrogen halides.

#### Explanation:

We assess the extent of the following rxn:

$H - X \left(a q\right) + {H}_{2} O r i g h t \le f t h a r p \infty n s {H}_{3} {O}^{+} + {X}^{-}$

For the lower hydrogen halides, $X = C l , B r , I$, 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 $H - C l$, and $H - B r$. 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 solvent order. Entropy favours the reverse reaction for ${F}^{-}$.

And thus both enthalpy and entropy conspire to REDUCE the acidity of $H F$ relative to $H C l$ and $H B r$ and $H I$, 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.