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270 ACIDS AND BASES

of hydrogen phosphate and dihydrogen phosphate, in the following equilibrium:

2− → − +
H 2 PO 4 HPO 4 + H
(6.48)

conjugate acid conjugate base

If this example were to proceed in the reverse direction, then the hydrogen phosphate
(on the right) would be the base, since it receives a proton, and the dihydrogen
phosphate (on the left) would be the conjugate the acid.
The equilibrium constant of the reaction in Equation (6.48) is given by

2−
+
[H ][HPO ]
K a = 4 (6.49)
−
[H 2 PO ]
4
Notice how the equilibrium constant K in Equation (6.49) is also an acidity constant,
hence the subscripted ‘a’. The value of K remains constant provided the temperature
is not altered.
Now imagine adding some acid to the solution – either by mistake or deliberately.
+
Clearly, the concentration of H will increase. To prevent the value of K a changing,
some of the hydrogen phosphate ions combine with the additional protons to form di-
hydrogen phosphate (i.e. Equation (6.48) in reverse). The position of the equilibrium
adjusts quickly and efficiently to ‘mop up’ the extra protons in the buffer solution. In
summary, the pH is prevented from changing because protons are consumed.

How do we make a ‘constant-pH solution’?

The Henderson–Hasselbach equation
We often need to prepare a solution having a constant pH. Such solutions are vital in
the cosmetics industry, as well as when making foodstuffs and in the more traditional
experiments performed by the biologist and physical chemist.
To make such a solution, we could calculate exactly how many moles of acid to add
to water, but this method is generally difficult, since even small errors in weighing
the acid can cause wide fluctuations in the pH. Furthermore, we cannot easily weigh
out one of acid oxides such as NO. Anyway, the pH of a weak acid does not clearly
follow the acid’s concentration (see p. 254).
The Henderson–Hasselbach equation, Equation (6.50), relates
In some texts, Equa- the pH of a buffer solution to the amounts of conjugate acid and
tion (6.50) is called the conjugate base it contains:
Henderson–
−
HasselbaLch equation. [A ]
pH = pK a + log (6.50)
10
[HA]
We follow the usual pattern here by making a buffer with a weak acid HA and a
−
solution of its conjugate base, such as the sodium salt of the respective anion, A .