Book seems to say that we can have weak acid and weak conjugate base

[zenterix]

Homework Statement

In the book I am reading there is a snippet that seems to indicate

Relevant Equations

that it is possible to have a weak acid with a weak conjugate base.

Here is the relevant section from the book

Consider the equilibrium

$$\mathrm{HA(aq)+H_2O(l)\rightleftharpoons H_3O^+(aq)+A^-(aq)}\ \ \ \ \ K_a=\mathrm{\frac{[H_3O^+][A^-]}{[HA]}}$$

The values of $\mathrm{[HA]}$ and $\mathrm{[A^-]}$ that appear in $K_a$ are the equilibrium concentrations of acid and base in the solution, not the concentrations added initially.

However, a weak acid $\mathrm{HA}$ typically loses only a tiny fraction of its protons, and so $\mathrm{HA}$ is negligibly different from the concentration of the acid used to prepare the buffer, $\mathrm{[HA]_{initial}}$.

This is all fine, but then there is the following

Likewise, only a tiny fraction of the weakly basic anions $\mathrm{A^-}$ accept protons, so $\mathrm{[A^-]}$ is negligibly different from the concentration of the base used to prepare the buffer, $\mathrm{[A^-]_{initial}}$.

I am under the impression that if an acid is weak (ie, only a small fraction of the molecules donate protons) then the conjugate base is strong (a large fraction of the molecules accept protons).

But the snippet above seems to say that that both the acid and the conjugate base are weak.

How can this be?

 

[zenterix]

For a little more context, the topic is what happens when we create a buffer with the same initial concentrations of weak acid and conjugate base.

In the equilibrium in the OP, we would have $\mathrm{[HA]=[A^-]}$ and so

$$\mathrm{K_a=[H_3O^+]}$$

$$\implies \mathrm{pH=-\log{(K_a)}=pK_a}$$