298 16 · ACID–BASE EQUILIBRIA
BOX 16.2
pH of a jug of water contaminated with a drop of dilute
hydrochloric acid
Suppose we add 1 drop of ‘bench hydrochloric acid’ (concentration 2.0 mol dm^3 ) to 1 dm^3
of deionized water. What is the pH of the resulting solution?
The volume of the drop ( 0.05 cm^3 ) is negligible compared with the water, so we may take the
solution volume as 1 dm^3. The number of moles of HCl in the 0.05 cm^3 drop is
0.05
2.0 1 10 ^4 mol
1000
and the concentration of HCl in the solution is therefore
1 10 ^4 mol
1 10 ^4 mol dm^3
1dm^3
The pH of the solution is calculated as follows:
pH log[H 3 O(aq)] log[1 10 ^4 ]4.0
The pH of deionized water in the laboratory is about 6 because it contains dissolved carbon
dioxide. Therefore,
the addition of only one drop of dilute acid has reduced its pH from 6 to 4.Fig. 16.2Changes in pH when (a) 0.1 mol dm^3 HCl, (b) water and
(c) 0.1 mol dm^3 NaOH, are added to a buffer consisting of 25 cm^3 of
0.1 mol dm^3 ethanoic acid and 25 cm^3 of 0.1 mol dm^3 sodium ethanoate
(temperature 25 °C; data kindly supplied by K. Morgan).
How buffers work
Buffers either consist of a weak base and one of its salts, or a weak acid and one of its
salts. As an example, we look at a mixture containing ethanoic acid and sodium
ethanoate. The ions present in such a mixture are shown by the following equations:H 2 O
CH 3 COO,Na(s) —CH 3 COO(aq)Na(aq)CH 3 COOH(aq)H 2 O(l)\===\CH 3 COO(aq)H 3 O(aq) (16.9)The salt (since it is an ionic compound) is fully ionized, and generates a relatively
high concentration of ethanoate ions.
Now consider the ionization of ethanoic acid in the presence of sodium
ethanoate. The expression for the acidity constant of ethanoic acid is:Ka(T)[CH 3 COO(aq)][H 3 O(aq)]
(16.10)
[CH 3 COOH(aq)]If ethanoate ions are added to an aqueous solution of ethanoic acid, the equilibrium
composition will shift in order to keep the right-hand side of this expression equal to
Ka(T). As a result of the shift, [H 3 O(aq)] becomes very low, meaning that very little