This question already has an answer here:

I'm learning about salt hydrolysis in school and this just jumped out at me. How can $\ce{NH_4^+}$ be both a weak acid and a strong conjugate acid? Since $\ce{NH_4^+}$ is a weak acid it has a strong conjugate base $\ce{NH_3}$. However, as part of a dissolved salt, since $\ce{NH_4OH}$ is a weak base, then $\ce{NH_4^+}$ is considered a "strong" conjugate acid. How can we reconcile these facts? Are the definitions slightly different for a salt ion and a normal acid?


marked as duplicate by Jan, Todd Minehardt, Klaus-Dieter Warzecha, Wildcat, getafix Nov 27 '16 at 10:31

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.


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:

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

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.


Not the answer you're looking for? Browse other questions tagged or ask your own question.