PHYSICAL CHEMISTRY IN BRIEF

CHAP. 11: ELECTROCHEMISTRY [CONTENTS] 383

Solution

At 0◦C the ionic product of water isKw= 1. 25 × 10 −^15. Substituting into (11.85) yields

pH=−

1

2

log(1. 25 × 10 −^15 ) = 7. 45.

At 25◦C and the standard pressure the ionic product of water isKw= 1. 00 × 10 −^14 and

pH=−

1

2

log(1. 00 × 10 −^14 ) = 7. 00.

At 100◦C the ionic product of water isKw= 9. 81 × 10 −^13 and

pH=−

1

2

log(9. 81 × 10 −^13 ) = 6. 00.

Note:In the past, water and aqueous solutions were studied almost exclusively at room

temperature and atmospheric pressure. This is the reason for the deeply rooted conviction

that the pH of pure water equals 7. The results of the above example make it clear that

water has its pH = 7 only at a temperature of 25◦C and a pressure of 101.325 kPa.

11.6.3 pH of a neutral solution

A neutral solution is such an aqueous solution of a salt that does not contain any other ions
of H+and OH−but those formed in consequence of the dissociation of water. For the pH of
a neutral solution, the same relation applies as that for the pH of pure water, i.e. equation
(11.85).

Example

Calculate the pH of a solution of NaCl at a temperature of 25◦C and a pressure of 101.325 kPa.

Solution

The presence of sodium chloride does not affectKw, the equilibrium constant of reaction (11.47)

because NaCl is an inert component in this reaction, see8.5.4. The pH is thus the same as that

of pure water at the given temperature and pressure, i.e. 7.00.