Page 84 - Chemical equilibria Volume 4

60 Chemical Equilibria
If, now, the equilibrium is obtained in a condensed phase, we need to
consider the particular – and very commonplace – case of highly-dilute
solutions, where the solvent is one of the reagents. This situation is
frequently encountered with dilute aqueous solutions. In such cases, the
solvent has a chemical potential which practically depends only on the
*
temperature and is given by: μ = g ; in addition, its activity is practically
0
0 0
one ( a ≅ ), so that the law of mass action is written by using relation [3.3],
1
0
in which the term relating to the solvent in the right-hand side no longer
plays a part. Note, however, that the term relating to the solvent always
comes into play on the left-hand side of the equation, i.e. in the definition of
the equilibrium constant according to the last equality in relation [3.2].
Consider a dilute solution of solutes A i in a solvent A 0. Let v denote the
0
0
molar volume molar of the pure solvent, and x 0 and x i the respective molar
fractions of the solvent and solute A i.

If we look again at relation [3.2], applied to the reference state (III) –
molar solution – we have:

Δ μ 0(III)
K (III) = exp− r [3.16]
RT

However, in the reference solution, in view of relation [A1.16] (see
Appendix 1), the chemical potentials of reference state (III) and the
infinitely-dilute reference state (II) are equal, so the equilibrium constant is
the same in both cases:

K (III) = K (II) [3.17]

If we now combine relations [A1.5] and [3.17], we obtain the
following (only the solutes are involved):

s ν
s ν
∏ (γ s (III) C s ) = ∏ (γ (II) x s ) = K (II) [3.18]
s
s s
If the reference concentration is 1 mole/L, then in a sufficiently-dilute
solution, in light of relation [A1.18], we obtain an equilibrium constant