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Ammonium, $\ce{NH_4^+}$, is the conjugate acid of ammonia, $\ce{NH_3}$. I have searched the Internet and so many different answers pop up. The rule of thumb that I read was that a strong acid has a weak conjugate counterpart.

For example, a strong acid has a weak conjugate base. Is ammonium a strong or weak acid? Is ammonia a strong or weak base? Is there something to the rule of thumb that I mentioned?

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2 Answers 2

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Both ammonia is a weak base and ammonium ion is a weak acid. Many, even most, acid/base conjugate pairs are like that. We should be using comparative instead of absolute adjectives in the rule about conjugate acid-base strengths:

A weaker acid has a stronger conjugate base, not necessarily a totally strong one.

A weaker base has a stronger conjugate acid, not necessarily a totally strong one.

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  • $\begingroup$ Stronger in respect to what...the weaker acid we have? $\endgroup$ Mar 26, 2018 at 5:41
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    $\begingroup$ @Hydrous Caperilla, yes. $\endgroup$
    – Serid
    Jun 26, 2018 at 12:35
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    $\begingroup$ @HydrousCaperilla actually if I understand your question, the phrase "stronger conjugate base" does not actually mean stronger than the weaker acid. What the first statement means is: If you have acid A1 and its conjugate base B1, and you have acid A2 and its conjugate base B2, then if A1 is a weaker acid than A2, the conjugate B1 will be a stronger base than B2. $\endgroup$
    – Bennett
    May 25, 2020 at 1:19
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Consider these reactions: \begin{equation} \ce{HA + H2O <=> A- + H3O+}\tag{1} \end{equation} \begin{equation} \ce{A- + H2O <=> HA + OH-}\tag{2} \end{equation}

The equilibrium constant for reaction (1) is called "acid dissociation constant", $K_{\rm a}$, and the second is "base association constant", $K_{\rm b}$. (Note that the names are there for historical reasons, they're not correct, strictly speaking) We can factor out the water's concentration to a good approximation in dilute solutions. Consider that

\begin{equation}K_{\rm a} \times K_{\rm b} = \frac{\ce{[A- ][H3O+ ]}}{\ce{[HA]}} \times \frac{\ce{[HA][OH- ]}}{\ce{[A- ]}} = \ce{[H3O+ ][OH- ]} = K_{\rm w}\tag{3}\end{equation}

$K_{\rm w}$ is constant in constant temperature, hence $K_{\rm a}$ and $K_{\rm b}$ are inversely proportional to each other. Again, to a good approximation, we have

\begin{equation}-\log{K_{\rm a}} + (-\log{K_\rm b})=-\log{K_\rm w}\: \Longrightarrow {\rm p}K_{\rm a} + {\rm p}K_{\rm b} = 14\tag{4}\end{equation}

p$K_{\rm a}$s between -1.7 ($\ce{H3O+}$/$\ce{H2O}$) and 15.7 ($\ce{H2O}$/$\ce{OH-}$) are said to be those of weak acids in water. Acids with $K_{\rm a}$s in this range do not dissociate fully in water, and named weak acids (and weak bases, for that matter1) in water.

Now, think of it like this: $x+y=14$. If $x$ is 5, $y$ is 9, and if $x$ is -5, $y$ is 19. (4) is where your rule of thumb originates from. So

  • A weak acid like acetic acid with a p$K_{\rm a}$ of 4.76 will have a weak conjugate base with a p$K_{\rm b}$ of $14-4.76=9.24$.
  • A strong acid like $\ce{HCl}$ with a p$K_{\rm a}$ of -7 will have a conjugate base with a p$K_{\rm b}$ of $14-(-7)=21$.

Thus as Oscar's answer points out, it's inaccurate the way your guideline is phrased, but it's naturally following if we use comparatives for phrasing. I'd dare say most conjugate bases of weak acids mentioned in textbooks are weak ones.


1: As the comment points out, the weak base range is p$K_{\rm b}$>1. That slipped.

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    $\begingroup$ The definition of "strong" as having $pK$ below -1.7 may not be entirely true. In en.wikipedia.org/wiki/Base_(chemistry)#Strong_bases a "strong base" is considered one whose conjugate acid has $pK_a>13$, so the strong base needs merely $pK_b<1$ not $<-1.7$. $\endgroup$ May 15, 2016 at 22:38
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    $\begingroup$ @Oscar Noted; that somehow slipped through. $\endgroup$
    – M.A.R.
    May 16, 2016 at 6:55

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