Page 204 - Modern physical chemistry

196 Equilibria in Condensed Phases

and
8 10-10
2
mAg+m CI - = 1 1.76 x m .
r Ag+r CI-
Since the principal contribution to the ionic strength is from the KN0 3 , we have
J.-l ~ 0.100 m.
Taking the ionic diameters from table 8.5, table 8.6 now yields

r Ag+ = 0.745, r CI- = 0.755.

Since the only source of Ag+ and Cl is the AgCI, we have

So the equilibrium expression becomes

x 2 = 1 1.768 x 10- 10 m 2 = 3.143 x 10- 10 m 2 ,
(0.745)(0.755)

whence
x=1.77x10 5 m
and
w =(1.77 x 10- 5 mX143.321 g mOrl) = 2.54x 10-3 gper 1000 g H 20.

8.11 Equilibria Among Phases Revisited

In section 6.17, each constituent i was considered to move from phase to phase inde-
pendently. But when the constituents are ions, such movement is subject to the electri-
cal neutrality condition.
Suppose that the phases contain ions 1, 2, and 3 carrying Zl positive charges, Z2 negative
charges, and Z3 negative charges. Furthermore, we consider fluctuations in which only these
ions travel from phase (2) to phase (1) or vice versa. Equation (6.99) then reduces to

2
2
2
1
1
1
1
1
1
(J.-ll ( ) - J.-ll ( ) ) dnl ( ) + (J.-l2 ( ) - J.-l2 ( )) dn2 ( ) + (J.-l3 ( ) - J.-l3 ( ) ) dn3 ( ) = O. [8.98]
Electrical neutrality of the two phases is maintained if
ZI dnl (1) = Z2 dn2 (1) + Z3 dn3(I). [8.99]
Substituting this into equation (8.98) and rearranging yields
J.-ll(I) + J.-l2(I) _ J.-ll(2) _ J.-l2(2) lZ2 dn2(I) +[J.-ll(I) + J.-l3(I) _ J.-ll(2) _ J.-l3(2) lZ3 dn3(I) = O. [8.100]
[
ZI Z2 ZI Z2 ZI z3 ZI z3

Fluctuations dn 2(1) and dn3(1) are independent. So the expressions in parentheses must
vanish at equilibrium.
If the first two ions make up the strong electrolyte A" B v , then
1 2
[8.101]