Page 184 - Modern physical chemistry

176 Equilibria in Condensed Phases

In the approximation that x is small with respect to 0.0100 M, we have

112
X = 1.749 x 10- 7 M2 ) = 4.18 x 10-4 M.
(
Then introducing this value for x on the right side of the preceding equation yields

112
(
X = 1.676 x 10- 7 M2 ) = 4.09 x 10-4 M.
Repeating the procedure gives us
x = 4.09 x 10-4 M.

Example 8.4
For the ionization of acetic acid, constant Kym is 1.754 X 10- 5 m, while for the ioniza-
tion of water, constant Kym is 1.008 X 10-- 14 at 25° C. Calculate the hydroxide ion concen-
tration in 0.100 m sodium acetate.
The acetate ion hydrolyzes following the reaction

CH 3COO- + H2 0 ~ CH 3COOH + OH-.
At appreciable acetate ion concentrations, this swamps the contribution from the ion-
ization of water and
mCH3COOH =m OH - =X.
As an approximation, we take the activity coefficients to be 1. So Qy for both equi-
librium constants is set equal to 1. From the ionization of water, we now have

The acetate ion concentration is reduced by the hydrolysis reaction so that
mCH3COO- =0.100 m-x.

For the acetic acid reaction, we obtain the relationship

14
mc coo-m + (0.100m- xX1.008x1O- m2)/x
H3 H = = 1.754 x 10- 5 m.
mCH3COOH x
We expect x to be small with respect to 0.100 m, so we replace the last equality with

(0.100 mX1.008 x 10- m2) -5
14
-'-----'-'-------'- = 1. 754 x 10 m
x 2
whence
x = 7.55 x 10-6 m.

8.3 Conditions Determining Equilibria

In a common problem, a person starts with a given amount of certain reactants
and products in a given volume or mass of solvent kept at a certain temperature and
pressure. One or more reactions proceed until equilibrium is established. The final