Chemistry, Third edition

16 · ACID–BASE EQUILIBRIA

Buffer solutions may also be prepared by mixing a weak base and one of its salts

(such as ammonia solution and ammonium chloride). The [OH(aq)] of such a

mixture may be calculated using the expression

[OH(aq)]

CBKb(T)

Cs

whereCBis the concentration of base, Csis the concentration of salt, and Kb(T)is the

basicity constant.

Data for the sodium ethanoate–ethanoic acid buffer

Let us use equation (16.11) to calculate the pH of an ethanoate–ethanoic acid buffer

at 25 °C made by mixing 25 cm^3 of 0.100 mol dm^3 sodium ethanoate solution with

25 cm^3 of 0.100 mol dm^3 ethanoic acid solution. Since the volume doubles, the

concentrations are halved and CACs0.050 mol dm^3 .Kafor ethanoic acid at

25 °C 1.8 10 ^5 mol dm^3. Substituting into equation (16.11) gives

[H 3 O(aq)]

0.0501.8 10 ^5

1.8 10 ^5 mol dm^3

0.050

The pH of the buffer mixture is

pHlog[1.8 10 ^5 ]4.74

(This is approximately the initial pH of the buffer in Fig. 16.2.) By using different

acids (different Kavalues) and their salts, buffers operating at different pH values

may be obtained. To adjust the pH of a buffer more exactly, the ratio of the initial

concentrations of acid and salt in the mixture are carefully controlled.

Buffer capacity

The amount of acid or alkali that needs to be added before the pH of a buffer changes

is called the buffer capacityof the buffer. The buffer capacity of a buffer containing a

relatively high number of moles of acid and salt is greater than the capacity of a

buffer with a lower number of moles of acid and salt. As we have already noted, the

buffer capacity of the buffer in Fig. 16.2 is equivalently to roughly 3 cm^3 of

0.1 mol dm^3 HCl or 3 cm^3 of 0.1 mol dm^3 NaOH.

300

Calculating the pH of buffer solutions

Investigate the effect of using different volumes of 0.100 mol dm^3 ethanoic acid and

0.100 mol dm^3 sodium ethanoate salt, by repeating the pH calculation starting with

(i) 25 cm^3 of acid and 50 cm^3 of salt solution, and (ii) 50 cm^3 of acid and 25 cm^3 of salt

solution.

Exercise 16K

Alternative form of equation (16.11) – the

Henderson–Hasselbalch equation

Equation (16.11) is sometimes used in a different form. Taking logs on both sides gives

log [H 3 O(aq)]logKa(T)log

CA

Cs