Question

Given a set of acids, will the weakest acid have the strongest conjugate base?

Answer 1

My word....

Explanation:

Given a solvent, acidity is measured on the basis of the ability of the studied acid to donate a proton to a solvent....i.e. we measure the equilibrium...

`HX(aq) + H_2O(l) rightleftharpoons H_3O^+ + X^-`

For strong acids, `X=Cl, Br, I`, the equilibrium lies almost entirely to the right as we face the page. In other words, the halide, `X^-` completes poorly for the proton, and is thus, by definition, A WEAK BASE...

On the other hand, for `X=F`...the equilibrium lies to the left as we face the page....

`HF(aq) + H_2O(l) rightleftharpoons H_3O^+ + F^-`

In other words, fluoride anion, because of its size, and charge density, and also because it is disfavoured entropically, competes STRONGLY for the proton....and thus fluoride anion is a moderately strong conjugate base...whereas the rest of the series are weak bases..

Most of the time, we limit our discussion of acid-base behaviour to the water solvent. But other solvents exist....and for a more basic regime...we would turn to another water-like solvent, liquid ammonia....which exhibits another acid-base equilibrium reaction....

`2NH_3(l) rightleftharpoonsNH_4^+ + NH_2^(-)`

Here, the amide ion, the conjugate base, is a mighty basic species, so much so that it does not exist in water...

And likewise we could go to a more acidic regime than water, acetic acid or liquid `HF`...

`2HF(l)rightleftharpoonsH_2F^+ + F^-`

For a quantitative treatment in water, see here. And also see here.