Is ammonium a strong or weak conjugate acid?

Question

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?

Answer 1

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.

Answer 2

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.

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

Answer 3

Understanding from an HSC chemistry (2019) point of view:

Acids

• the conjugate acid of a strong base will be weak, therefore never reacting water
• the conjugate acid of a moderately strong base will be moderately weak, therefore somewhat reacting with water
• the conjugate acid of a weak base will be strong, therefore completely reacting with water

Bases

• the conjugate base of a strong acid will be weak, therefore never reacting with water
• the conjugate base of a moderately strong acid will be moderately weak, therefore somewhat reacting with water
• the conjugate base of a weak acid will be strong, therefore completely reacting with water

When a salt ionises in solution, it's ions should be considered as the conjugate acid and conjugate base of their respective acid/base

e.g. CH3COONa --> CH3COO- + Na+

The Na+ is a weak conjugate acid of the strong base NaOH and therefore will never react with water

The CH3COO- is a moderately weak conjugate base of the moderately weak acid CH3COOH, and it therefore reacts with water somewhat to produce OH- ions:

CH3COO- + H2O --> CH3COOH + OH-

This makes the resultant solution basic, giving a pH of around 8-9