# Chemistry - Why is ammonium a weak acid if ammonia is a weak base?

## Solution 1:

First, let’s get the definition of weak and strong acids or bases out of the way. The way I learnt it (and the way everybody seems to be using it) is:

• $\displaystyle \mathrm{p}K_\mathrm{a} < 0$ for a strong acid
$\displaystyle \mathrm{p}K_\mathrm{b} < 0$ for a strong base

• $\displaystyle \mathrm{p}K_\mathrm{a} > 0$ for a weak acid
$\displaystyle \mathrm{p}K_\mathrm{b} > 0$ for a weak base

Thus strong acid and weak base are not arbitrary labels but clear definitions based on an arbitrary measurable physical value — which becomes a lot less arbitrary if you remember that this conincides with acids stronger than $\ce{H3O+}$ or acids weaker than $\ce{H3O+}$.

Your point of confusion seems to be a statement that is commonly taught and unquestionably physically correct, which, however, students have a knack of misusing:

The conjugate base of a strong acid is a weak base.

Maybe we should write that in a more mathematical way:

If an acid is strong, its conjugate base is a weak base.

Or in mathematical symbolism:

$$\mathrm{p}K_\mathrm{a} (\ce{HA}) < 0 \Longrightarrow \mathrm{p}K_\mathrm{b} (\ce{A-}) > 0\tag{1}$$

Note that I used a one-sided arrow. These two expressions are not equivalent. One is the consequence of another. This is in line with another statement that we can write pseudomathematically:

If it is raining heavily the street will be wet.

$$n(\text{raindrops}) \gg 0 \Longrightarrow \text{state}(\text{street}) = \text{wet}\tag{2}$$

I think we immediately all agree that this is true. And we should also all agree that the reverse is not necessarily true: if I empty a bucket of water on the street, then te street will be wet but it is not raining. Thus:

$$\text{state}(\text{street}) = \text{wet} \rlap{\hspace{0.7em}/}\Longrightarrow n(\text{raindrops}) \gg 0\tag{2'}$$

This should serve to show that sometimes, consequences are only true in one direction. Spoiler: this is also the case for the strength of conjugate acids and bases.

Why is the clause above on strength and weakness only true in one direction? Well, remember the way how $\mathrm{p}K_\mathrm{a}$ values are defined:

\begin{align}\ce{HA + H2O &<=> H3O+ + A-} && K_\mathrm{a} (\ce{HA}) = \frac{[\ce{A-}][\ce{H3O+}]}{[\ce{HA}]}\tag{3}\\[0.6em] \ce{A- + H2O &<=> HA + OH-} && K_\mathrm{b} (\ce{A-}) = \frac{[\ce{HA}][\ce{OH-}]}{[\ce{A-}]}\tag{4}\end{align}

Mathematically and physically, we can add equations $(3)$ and $(4)$ together giving us $(5)$:

\begin{align}\ce{HA + H2O + A- + H2O &<=> A- + H3O+ + HA + OH-}&& K = K_\mathrm{a}\times K_\mathrm{b}\tag{5.1}\\[0.6em] \ce{2 H2O &<=> H3O+ + OH-}&&K = K_\mathrm{w}\tag{5.2}\end{align}

We see that everything connected to the acid $\ce{HA}$ cancels out in equation $(5)$ (see $(\text{5.2})$) and thus that the equilibrium constant of that reaction is the autodissociation constant of water $K_\mathrm{w}$. From that, equations $(6)$ and $(7)$ show us how to arrive at a well-known and important formula:

\begin{align}K_\mathrm{w} &= K_\mathrm{a} \times K_\mathrm{b}\tag{6}\\[0.6em] 10^{-14} &= K_\mathrm{a} \times K_\mathrm{b}\\[0.6em] 14 &= \mathrm{p}K_\mathrm{a} (\ce{HA}) + \mathrm{p}K_\mathrm{b} (\ce{A-})\tag{7}\end{align}

Now let us assume the acid in question is strong, e.g. $\mathrm{p}K_\mathrm{a} (\ce{HA}) = -1$. Then, by definition the conjugate base must be (very) weak: $$\mathrm{p}K_\mathrm{b}(\ce{A-}) = 14- \mathrm{p}K_\mathrm{a}(\ce{HA}) = 14-(-1) = 15\tag{8}$$

Hence, our forward direction of statement $(1)$ holds true. However, the same is not true if we add an arbitrary weak acid to the equation; say $\mathrm{p}K_\mathrm{a} (\ce{HB}) = 5$. Then we get:

$$\mathrm{p}K_\mathrm{b} (\ce{B-}) = 14-\mathrm{p}K_\mathrm{a}(\ce{HB}) = 14-5 = 9\tag{9}$$

A base with a $\mathrm{p}K_\mathrm{b} = 9$ is a weak base. Thus, the conjugate base of the weak acid $\ce{HB}$ is a weak base.

We realise that we can generate a weak base in two ways: by plugging a strong acid into equation $(7)$ or by plugging a certain weak base. Since the sum of $\mathrm{p}K_\mathrm{a} + \mathrm{p}K_\mathrm{b}$ must equal $14$, it is easy to see that both cannot be strong. However, it is very possible that both the base and the acid are weak.

Thus, the reverse statement of $(1)$ is not true.

$$\mathrm{p}K_\mathrm{a}(\ce{HA}) < 0 \rlap{\hspace{1em}/}\Longleftarrow \mathrm{p}K_\mathrm{b} (\ce{A-}) > 0\tag{1'}$$

## Solution 2:

First, let's understand the perspective of weak acid and weak base. This is in relation to pure water — like in most general chemistry courses.

Pure Water has a $\mathrm{p}K_\mathrm{a}$ of 14.

$\ce{NH3/NH4+}$ has a $\mathrm{p}K_\mathrm{a}$ of 9.25.

We know: $$\ce{NH4+ + H2O <--> NH3 + H3O+}$$

Thus, we solve for the $K_\mathrm{a}$ (which depends on the concentration of each of your chemicals):

$$K_\mathrm{a} = \frac{[\ce{NH3}][\ce{H3O+}]}{[\ce{NH4+}]}$$

Similarly, we can solve for $K_\mathrm{b}$:

We know the ionization equation for a base is: $$\ce{B + H2O <--> HB+ + OH-}$$

Which means: $$\ce{NH3 + H2O <--> NH4+ + OH-}$$

So in order to solve for the $K_\mathrm{b}$, plug the concentration of your chemicals in:

$$K_\mathrm{b} = \frac{[\ce{NH4+}][\ce{OH-}]}{[\ce{NH3}]}$$

tl;dr: As the math shows, you can think of these as a "weak but not meaningless" acid/base. Don't let your assumptions throw you off — a weak acid generates a weak base, and vice versa. See the Henderson-Hasselbalch Equation.

$\ce{NH3}$ is not in the same class of weak bases as say, $\ce{Cl-}$. The acid to base curve isn't extreme like $\ce{HCl + H2O -> Cl- + H3O+}$.