Chemistry, Third edition

292 16 · ACID–BASE EQUILIBRIA

solution is that, unlike strong acids, the equilibrium concentration of hydronium ion

in the solution is not the same as the concentration of the acid originally added, i.e.

[H 3 O(aq)]CA.

The H 3 O(aq) concentration of a very weak acid with one acidic hydrogen can be

estimated by re-arranging the following equation:

Ka(T)

[H 3 O(aq)]^2

(16.5)

CA

This equation is derived in Box 16.1. The equation is a good approximation pro-

vided the percentage of acid molecules that are ionized,

percentage of acid molecules ionized 

[H 3 O(aq)]

 100

CA

does not exceed about 5%.

Example 16.3

Some white vinegar contains 0.50% by mass of ethanoic acid,

i.e. 100 g of vinegar contains 0.50 g of acid. Estimate its pH at

25 °C. What percentage of the ethanoic acid molecules are

ionized? (1.0 cm^3 of vinegar has a mass of 1.0 g.)

Answer

Consider exactly 1 dm^3 of vinegar. 1 dm^3 of vinegar contains 5.0 g CH 3 COOH. The

molar mass of ethanoic acid, M(CH 3 COOH), is 60 g mol^1.

number of moles of acid 

5.0 g

 0.083

60 g mol^1

concentration of acid 

0.083 mol

0.083 mol dm^3

1dm^3

Rearranging equation (16.5) gives

[H 3 O(aq)] (Ka(T)CA)

CA0.083 mol dm^3 andKa1.8 10 ^5 mol dm^3 (Table 16.2), so that

[H 3 O(aq)] (1.8 10 ^5 0.083)1.2 10 ^3 mol dm^3

pHlog[H 3 O(aq)]log(1.2 10 ^3 )2.92

The percentage of ethanoic acid molecules that are ionized in vinegar is calculated

as

[H 3 O(aq)]

 100 

1.2 10 ^3  100

1.4%

CA 0.083

Comment

The percentage of ionized acid molecules is less than 5%, and this confirms that

we are able to use equation (16.5) to calculate [H 3 O(aq)] for this acid.