Physical Chemistry Third Edition

7.3 Chemical Equilibrium in Solutions 323

γ± 10 −^0.^01523  0. 9655

Assume thatγ(HA)≈ 1

1. 752 × 10 −^5 

(0.9655)^2 x^2

(0. 100 −x)

1. 752 × 10 −^5

(0.9655)^2

 1. 879 × 10 −^5 

x^2

(0. 100 −x)

x^2 +(1. 879 × 10 −^5 )x− 1. 879 × 10 −^6  0

x

− 1. 879 × 10 −^5 ±

√

(1. 879 × 10 −^5 )^2 +4(1. 879 × 10 −^6 )

2

 1. 361 × 10 −^3

This corresponds to 1.361% ionization. We stop the successive approximations at this

point, since this approximation has not changed the result by very much. A further approx-

imation would make an even smaller change.

pH−log 10 (0.9655)(1. 361 × 10 −^3 ) 2. 88

Exercise 7.9

Find the percentage of acetic acid molecules that ionize and the pH in a solution prepared from

1.000 mol of acetic acid and 1.000 kg of water and maintained at 298.15 K. Use the Davies

equation to estimate activity coefficients. Neglect any H+ions contributed by the ionization

of water.

If a weak acid is quite dilute or if an acid has a very small ionization constant the

hydrogen ions that come from the water ionization cannot be neglected and we must

solve the equations for two equilibria simultaneously.

EXAMPLE7.13

Find the pH at 298.15 K of a solution made from 1. 00 × 10 −^7 mol of acetic acid and 1.000 kg

of water.

Solution

The ionization of acetic acid corresponds to

CH 3 CO 2 H
H++CH 3 CO− 2

which we abbreviate as

HA
H++A−