Ammonium is a weak acid but isn't it considered a strong conjugate acid as a salt's ion?
Just because something is a weak base does not mean its conjugate acid has to be a strong acid. Likewise, just because something is a weak acid does not mean that its conjugate base is a strong base.

The only connection between acidity of an acid and basicity of a conjugate base, henceforth labelled $\ce{HA}$ and $\ce{A-}$, is that they are related by the ion product of water, $K_\mathrm{w}$ by the following equations:

$$\begin{align}\ce{HA &amp;&lt;=&gt; H+ + A-}&amp;\quad K_\mathrm{a} &amp;= \frac{[\ce{H+}][\ce{A-}]}{[\ce{HA}]}\tag{1}\\[0.8em]
\ce{A- + H2O &amp;&lt;=&gt; HA + OH-}&amp;\quad K_\mathrm{b} &amp;= \frac{[\ce{OH-}][\ce{HA}]}{[\ce{A-}]}\tag{2}\\[0.8em]
K_\mathrm{a} \times K_\mathrm{b} &amp;= \frac{[\ce{H+}][\ce{A-}]}{[\ce{HA}]} \times \frac{[\ce{OH-}][\ce{HA}]}{[\ce{A-}]}\\[0.6em]
&amp;= \frac{[\ce{H+}][\ce{A-}][\ce{OH-}][\ce{HA}]}{[\ce{HA}][\ce{A-}]}\\[0.6em]
&amp;= [\ce{H+}][\ce{OH-}] = K_\mathrm{w}\tag{3}\\[1.2em]
\Longrightarrow \mathrm{p}K_\mathrm{a} + \mathrm{p}K_\mathrm{b} &amp;= 14\tag{4}\end{align}$$

Therefore, the sum of the acid and base constants must equal $14$. But if $\mathrm{p}K_\mathrm{a} = 6$, then $\mathrm{p}K_\mathrm{b} = 8$ — meaning the conjugate base of a weak acid is a weak base. This basically mirrors ammonia’s case.