# How do you rank Bronsted acids?

Sep 5, 2016

Well, not by awarding them gold, silver, or bronze medals. We assess the extent of the following equilibrium:

$H A + {H}_{2} O r i g h t \le f t h a r p \infty n s {H}_{3} {O}^{+} + H {O}^{-}$

#### Explanation:

For strong acids, the given equilibrium lies strongly to the right. For $H X \left(X \ne F\right)$, ${H}_{2} S {O}_{4}$, $H C l {O}_{4}$, a given concentration of the acid will be stoichiometric in ${H}_{3} {O}^{+}$ (almost double in the case of sulfuric diacid) as protonolysis proceeds almost quantitatively.

For weaker acids, e.g. ${H}_{3} C - C {O}_{2} H , H F , {H}_{3} P {O}_{4}$, the protonolysis reaction does not lie completely to the right, and at equilibrium significant concentrations of the free acid are present in solution.

So the degree of ionization in aqueous solution determines the strength of the acid.