Page 91 - Basic physical chemistry for the atmospheric sciences

Solution chemistry and aqueous equilibria 77

ions of the saturated solution (called a common ion) reduces the sol u­
bility of the solute and causes precipitation of the excess solid.
Exercise 4.5. What is the molar solubility of Ag2Cr0is) in a 0.05
M solution of K 2 Cr04(aq)? The solubility product for Ag2Cr0is) is
2.4 x 1 0 - 12• (Assume that K 2 Cr04(aq) remains subsaturated as the
Ag 2 Cr04(s) is added. )
Solution. K2Cr04(aq) acts a s a source o f Cro�- ( aq) ions i n the
n
solutio . Therefore, this is the common ion problem discussed qualita­
tively in Exercise 4 . 4 . Let C be the molar solubility of Ag 2Cr04(s).
Then, from Reaction (4. 1 3 ) , we have from the dissolving of the
=
Ag 2 Cr0is ) , [Ag+ (aq)] = 2C and [Cro�- (aq)] C . But there is a further
0.05 M of Cro�- (aq) from the Ag2Cr04( ) . We can summarize the
s
situation as follows

From Ag 2 Cr04(s): 2C c mol L - 1
From 0.05 M . of K 2 Criaq) : 0.05 mol L - 1
Total concentrations: 2C C + 0.05 mol L - 1

The usual K,P relation must be satisfied. Therefore,
2
2
[Ag+ ( aq)] [Cro�- ( aq)] + (2C) (C + 0.05) = 2 . 4 x 1 0 - 1 2 (4. 16)
An exact solution for C would involve solving a cubic equation. How­
ever, we can make a simplification. We know from Exercise 4.4(b)
that C will be less than it was in Exercise 4 . 4(a), that is, less than
4
0.84 x 1 0 - M . Hence, C + 0.05 = 0.05. Therefore, Eq. (4. 1 6) becomes
4C2(0. 0 5) = 2.4 x 1 0 - 1 2

or,
C = 4 x 1 0 - 6 M

Hence, the equilibrium molar solubility of Ag 2 Cr0is) in a 0.05 M
solution of K2Cr0iaq) is 4 x 10- 6 M . This i s 2 1 times less than the
solubility of Ag 2 Cr0is) in pure water which was calculated in Exer­
cise 4.4(a).
Certain solutes produce a greater effect on colligative properties
than expected from the relations given in Section 4.4. This can be
allowed for empirically by introducing the van't Hoff a ctor (i), which
f
i s defined as