Page 190 - Modern physical chemistry

182 Equilibria in Condensed Phases

If the acetic acid were completely ionized and the contribution from ionization of
water were negligible, we would have

m H + = 1.00 x 10-6 m.

Put this into the first equation of the set and complete the round:

= 1.008 x 10- 14 = 1 008 10-8
m OH - • x m,
1.00 x 10-6
m A - = 1.00 x 10-6 - 0.01 x 10-6 = 0.99 x 10- 6 m,

(1.00 x 10-6 X 0.99 x 10-6 ) -7
mBA = =0.56xl0 m,
1.754 x 10--5
mw = 1.00 x 10- 6 - 0.056 x 10-6 + 0.010 x 10-6 = 0.954 x 10-6 m.

Repeating the steps with this number gives us (in the final step)

m + = 1.00 x 10-6 - 0.051 x 10-6 + 0.011 x 10-6 = 0.960 x 10-6 m.
H
The average of these two values is
mw = 0.957 x 10-6 m.

To illustrate use of equation (8.44), we replace the final equation in the set with
Y = 1.00 x 10-6 -mBA -m A -.

Again, if the acetic acid were completely ionized but the contribution from ionization of
water were negligible, we would have

Xl = (mw )1 = 1.00 x 10-6 m.

If only the water ionized, we would have
7
X2 = (mH+ )2 = 1.004 x 10- m.

With Xl the first three equations are the same as in the first round before. So
m OH - = 1.008 x 10-8 m,

m - = 0.990 x 10-6 m,
A
mBA = 0.56 x 10- 7 m,
and
Y1 = 1.00 x 10-6 - 0.056 x 10-6 - 0.990 x 10-6 = -0.046 x 10-6 m.
With X2, we find that
m _ = 1.008 X 10- 14 = 1.004 x 10-7 m,
OH 1.004 x 10-7