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

$\ce{NH3}$ is a weak base so I would have expected $\ce{NH4+}$ to be a strong acid. I can't find a good explanation anywhere and am very confused. Since only a small proportion of $\ce{NH3}$ molecules turn into $\ce{NH4+}$ molecules, I would have expected a large amount of $\ce{NH4+}$ molecules to become $\ce{NH3}$ molecules.

• I'd rather avoid using the vague terms like "weak", "strong", "big", "small" without any numerical characteristic. They don't really mean anything. There are layers upon layers of different kinds of "small". See, you are very small compared to an elephant, and elephant is a good deal smaller than the Empire State Building, which in turn is so small it is barely visible from space. Ditto for "weak". – Ivan Neretin Dec 28 '16 at 19:03
• @IvanNeretin Strong and weak in terms of acids and bases is defined as $\mathrm{p}K_x > 0$ (weak) or $\mathrm{p}K_x < 0$ (strong) — at least that’s the definition that followed me through school and uni. – Jan Dec 28 '16 at 22:04

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.

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

• Regarding the opposite direction it might be worth noting that it works with the negations. $$\mathrm{p}K_\mathrm{a} (\ce{HA}) \geq 0 \Longleftarrow \mathrm{p}K_\mathrm{b} (\ce{A-}) \leq 0\tag{1b}$$ This is known as Modus Tollens. – Anonymous Coward Dec 29 '16 at 10:55
• @JoseAntonioDuraOlmos Yes, the reverse of the negation is always true afaik. So the conjugate acid of a strong base is a weak acid. – Jan Dec 29 '16 at 14:20

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 $\ce{HCl + H2O -> Cl- + H3O+}$.

• One warning: if you have both acid and base in sufficient quantities, make sure to use the Henderson-Hasselbalch Equation, in addition to the above. – Ryan Dec 28 '16 at 21:34
• For example: pH = -log10(kA of NH4+) + log10([NH3]/[NH4+]) – Ryan Dec 28 '16 at 21:35
• Why is there a need to use H-H under "different conditions"? This is not true because the H-H equation is simply a different restatement of the definition of the Ka (take log10 on both sides). As such it is entirely equivalent to the Ka equation you gave above and likewise the scope of use is exactly the same. – orthocresol Dec 29 '16 at 4:41
• I only meant that the requester needs to be careful if he has other solutions in the mixture (e.g. this solution being a buffer for a strong acid or base). Apologies for the confusion. – Ryan Jan 5 '17 at 0:45

In case a less abstract answer will help:

A base, like $\ce{NH_3}$, is a base because it has a significant chance of picking up protons in water. You can almost think of it as a competition between the $\ce{NH_3}$'s and the $\ce{H_2O}$'s to pick up the free protons.

It is a weak base because it is not a certainty that all the $\ce{NH_3}$'s in a given sample will pick up a proton and hold onto it. In any given sample at any point in time, a certain portion of the $\ce{NH_4^+}$'s ($\ce{NH_3}$'s that won the proton) will give their protons up and a certain number of $\ce{NH_3}$'s will pick up new protons. Eventually, the system reaches a steady state (quantifiable with $K_b$).

Our definition of an "acid" is just something that donates protons, as the $\ce{NH_4^+}$ does above.

It's not that they're really different things: the $\ce{NH_3}$'s a base when its picking up protons and its an acid when it lets them go.