# Does a strong acid's conjugate base act as Brønsted-Lowry base?

Why is the conjugate base of a strong acid not considered a Brønsted-Lowry base? Is it because the base is so weak that it essentially has no ability to accept a proton?

(Quote from my lecture notes: "$\ce{Br-}$ is the conjugate base of a strong acid ($\ce{HBr}$), so it will not act as a Brønsted-Lowry base").

In theory, a Brønsted-Lowry base is a chemical substance which can accept a proton.

But in the case of a strong acid, the acid will lose a proton and form a weak conjugate base. Say there is a strong acid HA.

$$\ce{HA + H2O <=> A- +H3O+}$$

So if we take a certain quantity of HA, most of it will undergo the forward reaction since HA is a strong acid. A- is a weak base and will not want to accept the proton and thus only a small quantity of the conjugate base will do undergo the backward reaction. Hence practically, you could say that the conjugate base of a strong acid is not a Brønsted-Lowry base because of its reluctancy to accept the proton.

In the case of halide ions (setting fluoride aside as its conjugate acid is weak), they can act as Lewis bases. They will not, of course, react with solvated protons in aqueous solution but will form complexes with Lewis-acidic transition metal ions.

All compounds with an accessable free lone pair can act as a Brønsted-Lowry base. That explicitly includes conjugate bases of strong acids.

The only question (and difference) remains how basic these bases are, i.e. what their $\mathrm{p}K_\mathrm{b}$ is. Since it is directly related to the $\mathrm{p}K_\mathrm{a}$ by the $\mathrm{p}K_\mathrm{ion}$ of the solvent ($14$ for water), a strong acid will give a weak conjugate base.

However, no matter how weak the base it just takes a strong enough acid to protonate it. Using superacids, even methane $\ce{CH4}$ — a compound that one really would not accuse of having any free lone pairs — could be protonated to give a carbonium ion which quickly decomposed. Its existence was confirmed by the decomposition products the formation of which required the existence of $\ce{CH5+}$.

From my understanding, the Brønsted-Lowry (BL) acid/base theory is in relation to the idea of conjugate acids and bases.

A conjugate base does not always occur when there is interaction between a Lewis acid and a Lewis base, but a conjugate base always occurs when there is a BL acid-base reaction. This is because BL acids-bases are defined by the exchange of protons (H+ ions).

When a BL acid donates a proton to a BL base, the result is a conjugate base (originally the acid) and a conjugate acid (originally the base). And, you must notice that this reaction is in equilibrium! Thus, the reverse reaction is possible. This means that the conjugate base will act like a BL base.

Now, it may be that the initial BL acid that was used was considered a very strong acid, so the conjugate base will be a very weak BL base. Thus, the equilibrium may be favoring the products (K>>>1). As a result, in a closed system, very little conjugate base will undergo the reverse reaction. Nevertheless, at least some should undergo the reverse reaction and it will then act as a BL base!