Page 203 - Modern physical chemistry

8.10 Debye-HiickeJ Equations 195

As in example 8.9, we also have
YBA = 1.00
and
mBA :;:: O.l00-x.

Substituting into the equilibrium constant expression gives

x 2
---= 1.754 x 10- 5 m
0.100-x
whence
x 2 = 1.754 x 10-{i -1.754 x 10-5 x.
Neglecting the last term yields
x = 1.32 x 10-3 m.
The corresponding ionic strength is

J.l = 1.32 x 10- 3 m.
Using the ionic diameters from table 8.5 and interpolating in table 8.6 leads to

YH+ =0.963, Y A- = 0.961.
The equilibrium expression is now

x 2 1.00 -5 -5
X ) 1.754 x 10 m = 1.895 x 10 m,
0.100-x = ( 0.963 0.961

whence
x 2 = 1.895 x 1O-{i -1.895 x 10- 5 x.
Neglecting the last term yields
x = 1.377 x 10- 3 m.
Then introducing this approximate value into the last term and solving for x gives us

x = 1.37 x 10- 3 m.
The activity coefficients calculated from the corresponding ionic strength do not differ
in three Significant figures from those used. So further iterations are not needed.

Example 8. 11

Calculate the solubility of AgCI in 0.100 m KN0 3 at 25° C.
For the equilibrium

table 8.3 tells us that

logK = -13.5085+22.9900-19.2341 = -9.7526
whence
K = 1.768 x 10- 10 m 2