PHYSICAL CHEMISTRY IN BRIEF

CHAP. 11: ELECTROCHEMISTRY [CONTENTS] 389

If the conditions are not as above, we have to take into account dissociation in the second
stage and solve, in the case of a weak dibasic acid, the set of equations (11.60) and (11.61). We
calculate the concentration of hydrogen ions and substitute into relation (11.84). The procedure
is similar for weak diacidic bases.
In weak polybasic acids and polyacidic bases, dissociation in the third stage and higher
usually does not have any effect on the pH value.

11.6.10pH of the salt of a weak acid and a strong base

In consequence of hydrolysis, see section11.5.8, solutions of the salts of strong bases and weak
acids exhibit alkaline reactions. When calculating the pH it is usually necessary to consider
not only the hydrolytic equilibrium but also the dissociation of water.
The set of equations (11.70) and (11.49) does not have an analytical solution. When the
degree of hydrolysis is low, however, we obtain an analytical relation for the pH

pH =−

1

2

log

KwK cst

c

, (11.97)

whereKwis the ionic product of water,Kis the dissociation constant of the acid, andcis the
initial concentration of the salt.

Example

Calculate the pH of an aqueous solution of sodium formate of the concentration c =

0. 001 mol dm−^3 at a temperature of 298 K and the standard pressure. Under these conditions

the dissociation constant of formic acid isK= 1. 77 × 10 −^4 , and the ionic product of water is

Kw= 10−^14.

Solution

Substituting into (11.97) yields

pH=−

1

2

log

(

10 −^14 × 1. 77 × 10 −^4

0. 001

)

= 7. 376.