# How do we assess the strength of an acid?

Mar 16, 2017

How else but by experiment.........?

#### Explanation:

Of course we know that $H B r$ is a POTENT $\text{Bronsted acid}$. And all $\text{Bronsted acids}$ are by definition potent $\text{Lewis acids}$, electron pair acceptors.

Solutions of hydrogen bromide in water are stoichiometric in $\text{hydronium ion}$:

$H B r \left(g\right) + {H}_{2} O \left(l\right) \rightarrow {H}_{3} {O}^{+} + B {r}^{-}$.

With the notable exception of $H F$, ALL of the hydrogen halides, $H X$, are strong $\text{Bronsted acids}$, i.e. the defining equilibrium lies strongly to the RIGHT:

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

Because the fluoride ion is smaller, and more charge dense, and the $H - F$ bond is strong, $H F$ is a relatively weak Bronsted acid, and fluoride salts give rise to (slightly) basic solutions:

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

This is well-known to be an entropy NOT an enthalpy phenomenon.