# Do weak acid/weak base neutralisation reactions go to completion?

I'm wondering if reactions that involve a weak acid and a weak base go to completion. For example, say we have equal amounts and equal volumes of acetic acid and ammonia being mixed together. Would this neutralisation reaction go to completion or would it reach a state of equilibrium?

$\ce{NH_3 + CH_3COOH <=> NH_4^+ + CH_3COO^-}$

Instinctively, I'm thinking that the reaction would be an equilibrium, since proton transfer should be able to occur between the products ($\ce{NH_4^+ + CH_3COO^-}$) to reform the reactants. Can anyone clarify if this is true?

• Yes, there is an equilibrium.
– MaxW
Sep 30 '17 at 4:26

Weak acids and bases do indeed remain in an equilibrium state with a certain concentration of the free acid and the free base alongside a certain concentration of the deprotonated acid anion and the protonated base cation.

In general, you can consider any acid-base reaction to be essentially the following:

$$\ce{HA <=> H+ + A-}\tag{1}$$

Which in turn means that the $\mathrm pK_\mathrm a$ value is defined as in $(2)$:

$$K_\mathrm a = \frac{[\ce{H+}][\ce{A-}]}{[\ce{HA}]}\tag{2}$$

Reactions $(1)$ and $(2)$ are usually written out in water with hydronium in place of a naked proton but the same principle applies. Of course, a base reaction can be considered essentially the reverse, meaning that the acid constant typically given is actually $\mathrm pK_\mathrm a (\ce{HB+})$ rather than anything directly corresponding to the free base. If we now take the reaction of a (weak) acid and a (weak) base, we get equation $(3)$ and the equilibrium constant as given in $(4)$:

\begin{align}\ce{HA + B &<=> A- + HB+}\tag{3}\\[0.7em] K &= \frac{[\ce{A-}][\ce{HB+}]}{[\ce{HA}][\ce{B}]}\tag{4}\end{align}

We can now perform simple mathematics with $(4)$ to arrive at the modified equation $(5)$ as below:

\begin{align}K &= \frac{[\ce{A-}][\ce{HB+}]}{[\ce{HA}][\ce{B}]}\tag{4}\\[0.7em] &= \frac{[\ce{A-}][\ce{HB+}][\ce{H+}]}{[\ce{HA}][\ce{B}][\ce{H+}]}\\[0.7em] &= \frac{[\ce{H+}][\ce{A-}]}{[\ce{HA}]} \times \frac{[\ce{HB+}]}{[\ce{B}][\ce{H+}]}\\[0.7em] K &= \frac{K_\mathrm a(\ce{A})}{K_\mathrm a (\ce{HB+})}\tag{5}\end{align}

We have thus arrived at a way to determine the equilibrium constant of the proton transfer reaction just from the acidity constants of both participants. If both acids weak and have similar $\mathrm pK_\mathrm a$ values — as ammonium and acetic acid do — the equilibrium constant will be close to $1$ and thus both sides of the equation will have similar concentrations.

Using $(5)$, you can also predict a ‘degree of completion’ for any acid-base reaction.

• I'll add another bit of terminology, the equivalence point. So given a solution of acetic acid you add an equivalent amount of ammonia, a base. However because both are weak the final pH need not be 7.00 (neutral).
– MaxW
Sep 30 '17 at 5:50
• @HernanMiraola That is indeed a typo (or a brainfart, I’m not sure).
– Jan
Sep 30 '17 at 14:23