Page 357 - Physical Chemistry

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Chapter 11 we can approximate the activity of most pure solids and liquids as 1. This approxi-
Reaction Equilibrium mation is not valid for a substance with a large V , such as a polymer.
in Nonideal Systems m
As an example, consider the equilibrium
CaCO 1s2 ∆ CaO1s2 CO 1g2 (11.24)
3
2
Equation (11.6) gives K° a[CaO(s)]a[CO (g)]/a[CaCO (s)]. If P is not high, we can
2
3
take the activity of each solid as 1 and the activity of the gas (assumed ideal) as
P(CO )/P° [Eq. (10.96) with fugacity replaced by pressure]. Therefore
2
K° a3CO 1g24 P1CO 2>P° where P° 1 bar (11.25)
2
2
Thus, at a given T, the CO pressure above CaCO (s) is constant. Note, however, that
2
3
in the calculation of G°using G° n m°, it would be wrong to omit G for the
i
i
i
m
solids CaCO and CaO. The fact that a for each solid is nearly 1 means that m m°
3
( RT ln a) of each solid is nearly 0. However, m° for each solid is nowhere near zero
and must be included in calculating G°.
Now consider the equilibrium between a solid salt M X and a saturated aque-
n n
ous solution of the salt. The reaction is
z
M X 1s2 ∆ n M 1aq2 n X 1aq2 (11.26)
z

n n
where z and z are the charges on the ions and n and n are the numbers of posi-

tive and negative ions. Choosing the molality scale for the solute species, we have as
the equilibrium constant for (11.26)
1a 2 1a 2 n 1g m >m°2 1g m >m°2 n
n
n

K°
a3M X 1s24 a3M X 1s24
n n n n
where a , g , and m are the activity, molality-scale activity coefficient, and molality

of the ion M (aq). Provided the system is not at high pressure, we can take a 1
z
for the pure solid salt. Dropping m° from K° and using (g ) n n (g ) (g ) n
n

[Eq. (10.43)], we have as the solubility product (sp) equilibrium constant
K 1g 2 n n 1m 2 1m 2 n (11.27)*
n

sp
Equation (11.27) is valid for any salt, but its main application is to salts only slightly
soluble in water. For a highly soluble salt, the ionic strength of a saturated solution is
high, the mean ionic activity coefficient g differs substantially from 1, and its value

may not be accurately known. Moreover, ion-pair formation may be substantial in con-
centrated solutions of a salt.
EXAMPLE 11.4 Solubility-product equilibria
2
2
K for AgCl in water is 1.78 10 10 mol /kg at 25°C. Find the solubility of
sp
AgCl at 25°C in (a) pure water; (b) a 0.100 mol/kg KNO (aq) solution; (c) a
3
0.100 mol/kg KCl(aq) solution.

(a) For AgCl(s) ∆ Ag (aq) Cl (aq), we have

2
K g m1Ag 2m1Cl 2

sp
Because of the very small value of K , the ionic strength of a saturated AgCl so-
sp

lution is extremely low and g can be taken as 1. Since m(Ag ) m(Cl ) in a

2
2
solution containing only dissolved AgCl, we have 1.78 10 10 mol /kg

2
[m(Ag )] . Hence m(Ag ) 1.33 10 5 mol/kg. The solubility of AgCl in
pure water at 25°C is 1.33 10 5 mole per kilogram of solvent.
(b) The ionic strength of a 0.100 mol/kg KNO solution is 0.100 mol/kg.
3

The Davies equation (10.68) gives g 0.78. Setting m(Ag ) m(Cl ), we

2
2
2
2
have 1.78 10 10 mol /kg (0.78) [m(Ag )] . Hence m(Ag ) 1.71 10 5