Page 202 - Modern physical chemistry

194 Equilibria in Condensed Phases

is 1.754 X 10-0 m when HA is acetic acid. So we have

YH+Y A- mH+mA- = 1.754 x 10-5 m
YHA mHA
whence
YHA 1.754 x 10- 5 m.
mHA YH+Y A-
Since HA is a molecule at low concentration, we take
YHA = 1.00.
The ionic strength involved in reducing the ion activity coefficients is practically all due
to the sodium chloride; we have

)1 = ~( mNa+z Na + 2 +mC1-z C1 -2)= ~(0.100 x 12 +0.100 x 12)= 0.100 m.

In table 8.5, one finds the ionic diameters

Then from the 0.100 m column of table 8.6, one obtains

Y H + = 0.825,
Since the principal source of hydrogen ions and the only source of acetate ions is the
ionization of the acetic acid, we have
mHA ~0.100-x.

Substituting into the equilibrium expression gives

x' ~( I; )1.754xIO-<im~2.743xW-5m,
0.100-x 0.825 0.775

whence
x 2 = 2.743 x 10-6 -2.743 x 1O- 5 x.
Neglecting the last term yields
x = 1.656 x 10- 3 m.
Introducing this approximate value into the last term and again solving for x leads to

x = 1.64 x 10-3 m.

Example 8.10
Calculate the hydrogen ion concentration in 0.100 m acetic acid at 25° C.
The ionic strength now is determined by the concentration of ionized acetic acid. But
initially this is not known. So we follow an iteration procedure, first taking

Yw =1.000, Y A- =1.000.