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76 Basic physical chemistry
define it in terms of the activities of the substances (see note 5 m
Chapter ) .
1
Whenever the product of the concentrations of any two ions (raised
to the appropriate powers) in a solution is less than the corresponding
Ksp value, the solution is subsaturated.
In Exercise . 3 the ions in the saturated solution derive only from
4
the solid solute. However, as illustrated in the next two problems,
solubility products can be used to calculate the solubility of a sub
stance in a saturated solution even if some of the ions derive from
another source.
Exercise 4.4. (a) Calculate the molar solubility of silver chromate
[Ag2Cr0is)] in water, given that the solubility product is 2 . 4 x 1 0 - 1 2 .
(b) Discuss qualitatively the effects of adding some CrO� - (aq) ions
from another source to a saturated solution of silver chromate in
water.
Solution. (a) The reaction is
Ag 2 Cr0is) µ 2 Ag+ (aq) + Cro� - (aq) (4. 1 3 )
The solubility product is
2 2
[Ag + (aq)] [Cr0� - (aq)] = 2 .4 x 1 0 - 1 (4. 1 4 )
We see from Reaction (4. 1 3 ) that when 1 mole of Ag 2 Cr0is) dissolves
in water, 2 moles of Ag+(aq) and 1 mole of Cro�- (aq) appear in the
n
saturated solutio . Therefore,
[Ag+(aq)] = 2[Cro�- ( aq)] (4. 1 5)
From Eqs. (4 . 1 4 ) and (4. 1 5 )
2
{2[Cro� - (aq)J} [Cr0(aq)] = 2 . 4 x 0 -1 2
1
Therefore, [C�- (aq)] = 0 .84 x 1 0 - 4 M. S i nce, from Reaction (4. 1 3 ) , l
mole of silver chromate dissolves for every l mole of Cro�- (aq) that
forms, the solubility of silver chromate in water is 0.84 x 1 0 - 4 M .
(b) I f CrOi- ( aq) ions are added o the equilibrium situation repre
t
sented by Reaction (4. 1 3 ) , LeChatelier's principle tells us that the
reverse reaction in (4. 1 3 ) will be favored. This means that some
Ag2Cr04(s) will precipitate out from the saturated solution, thereby
reducing the concentration of Ag+ (aq).
We see from the last problem that the addition to a saturated solu
tion of a small quantity of another solution that contains one of the