Doesn’t the weak acid dissociate to produce the conjugate base on its own?

I understand that when a strong base is added to the buffer solution, the hydroxide ions will react with the hydrogen ions formed via the dissociation of the weak acid. The equilibrium position will shift rightwards to replenish these hydrogen ions, ensuring that few hydroxide ions remain in the solution.

Conversely, when a strong acid is added to the buffer solution, the hydrogen ions will react with the conjugate base to form the weak acid molecule, thereby ensuring that most of the additional hydrogen ions are removed from the solution, and the pH of the solution remains stable. In the literature I’ve been reading so far, it is mentioned that a salt solution containing the conjugate base ensures that there is a continuous supply of the ions, but can’t the ions simply be replenished by the dissociation of more weak acid molecules, as in the previous case?


1 Answer 1


You are correct that a weak acid will only partially dissociate in water. Taking for example the weakly acidic salt $\ce{NH4Cl}$ of which the ammonium cation is acidic, in water we get reaction (1).

$$\ce{NH4+ + H2O <=> NH3 + H3O+}\tag{1}$$

To analyse this reaction we need the equilibrium constant which in the case of acids is defined as an acidity constant $K_\mathrm a$ as shown in (2). (Wether we use $\ce{H+}$ or $\ce{H3O+}$ is irrelevant.)

$$K_\mathrm a = \frac{[\ce{NH4+}][\ce{H+}]}{[\ce{NH3}]}\tag{2}$$

Using the tabulated $\mathrm pK_\mathrm{a}$ value of 9.25, we can calculate the amount of $\ce{NH3}$ in a $\pu{1M}$ solution of $\ce{NH4Cl}$:

$$\begin{align}K_\mathrm a &= \frac{[\ce{NH4+}]^2}{[\ce{NH3}]}\\[0.3em] K_\mathrm a (c_0 - [\ce{NH4+}]) &= [\ce{NH4+}]^2\\[0.5em] [\ce{NH4+}]^2 + K_\mathrm a [\ce{NH4+}] - K_\mathrm a c_0 &= 0\\ [\ce{NH4+}] &= \frac{-K_\mathrm a \pm \sqrt{K_\mathrm a^2 + 4 K_\mathrm a c_0}}{2}\\ [\ce{NH4+}] &= \pu{2.37e-5 M}\tag{3}\end{align}$$

Whoops. We may have a good concentration of ammmonium chloride but only a very minor amount of it is still ammonium, $99.998~\%$ are conjugate base. That doesn’t look like a good buffer system. If you were to add an acid, that would be fine since the excess base can capture that. But if you add more base the pH will go up rapidly, because there are no ammonium ions in solution to buffer it.

Only if you add both a significant amount of an acid and its conjugate base will you have significant amounts of both compounds in solution which are then able to buffer by capturing excess protons or hydroxides.

  • $\begingroup$ Or you can add some strong base to $\ce{NH4Cl(aq)}$ [or some weak acid to $\ce{NH3(aq)}$] to get a better balance of the conjugate weak acid: weak base combination. This is the fastest way of making a buffer (called "adjusting a buffer's pH"). $\endgroup$
    – Karsten
    Mar 10, 2020 at 12:30

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