Instant Notes: Analytical Chemistry

CH 3 COOH [ CH 3 COO-+H 3 O+

and an equilibrium constant is written (in terms of concentrations)

Ka=c(Ac-)¥ c(H 3 O+)/c(HAc) = 1.75¥ 10 -^5

using the abbreviation Ac for the CH 3 COO-group. Converting to logarithmic

form, and recalling that pK =-log(K):

pH =pKa +log [c(salt)/c(acid)]

This is the Henderson-Hasselbalch equation.

If we make a mixture containing both the free acid, HAc, and its salt sodium

ethanoate, NaAc, then the equilbrium and the concentrations of acid and salt

will determine the concentration of hydrogen ions and the pH.

Example 1

For a mixture of 50 cm^3 of 0.1 M HAc with 40 cm^3 of 0.1M NaAc, giving a total

volume of 90 cm^3 ,

c(H 3 O+) =1.75¥ 10 -^5 ¥ [(50¥ 0.1/90)/(40¥ 0.1/90)]

=2.19¥ 10 -^5 M, so that

pH =4.66

Addition of acid to this buffer shifts the above equilibrium to the left and most

of the added hydrogen ion combines with the anion. Adding 10 cm^3 of 0.1 M

HCl lowers the pH only to about 4.45. If this amount of acid were added to

90 cm^3 of water, the pH would be 2.0. Similarly, when alkali is added, the

hydroxyl ions react with the acid to produce more salt. 10 cm^3 of 0.1 M NaOH

will raise the pH only to around 4.85. If this amount of alkali were added to

90 cm^3 of water, the pH would rise to 12.

Weak bases and their salts behave in much the same way. For example,

ammonia and ammonium chloride:

NH 3 +H 2 O [ NH 4 ++OH-

Kb=c(OH-)¥ c(NH 4 +)/c(NH 3 ) =1.75¥ 10 -^5

or, rewriting the Henderson-Hasselbalch equation:

pOH =pKb+log[c(salt)/c(base)]

or, since pH +pOH =14.0

pH =14.0 - pKb- log[c(salt)/c(base)]

For a mixture of equal amounts of 0.1 M ammonia and 0.1 M ammonium

chloride

pH =14.0 -4.75 =9.25

A most useful range of buffers is obtained by using salts of a dibasic (or

tribasic) acid such as phosphoric acid, H 3 PO 4 - for example, potassium

dihydrogen phosphate, KH 2 PO 4 , and disodium hydrogen phosphate, Na 2 HPO 4.

The equilibrium involved here is:

H 2 PO 4 - [ H 3 O++HPO 42 -

For this equilibrium, the second dissociation constant of phosphoric acid, Ka2, is

close to 1¥ 10 -^7 , or pKa2=7. Figure 2 shows the effect of adding acid or alkali on

76 Section C – Analytical reactions in solution