Applications of Equilibrium Thermodynamics - Elements of Environmental Engineering: Thermodynamics and Kinetics - page 132

Many inorganic gases (e.g., carbon dioxide, sulfur dioxide, and ammonia) that
dissolve in water undergo chemical reactions depending on the pH of the solution.
For example, SO 2 dissolving in rainwater gives rise to sulfuric acid, which is the cause
of acid rain . Many organic species also undergo reactions with water depending on
the pH. For example, phenol is converted into phenolate ion under alkaline conditions.
The ionization of water gives rise to hydrogen and hydroxide ions:

H + + OH − . H 2 O (4.12)

At equilibrium, in pure water , the concentrations of both ions are equal and have a
value 1 10 − 7 M. The equilibrium constant for the reaction is K w =[ H + ][ OH − ]= ×

10 − 14 M 2 . The pH of pure water is given by H + ]= 1 7.0. Pure rainwater

is slightly acidic with a pH of 5.6. The dissolution of gases and organic compounds
that ionize in solution can change the pH of water. This process can also change the
equilibrium constant for dissolution of gases in water given by the Henry's law.
Let us consider SO 2 dissolution in water. The transfer of SO 2 from gas to water is
given by the following equilibrium reaction:

× − log [

SO 2 ( g ) + H 2 O(l) SO 2 · H 2 O ( l ) . (4.13)

The reaction equilibrium is discussed in Chapter 5 (see Section 5.6.2). For the above
reaction, the equilibrium constant is given by

K eq = [

SO 2 ·

] l

H 2 O

] l .

(4.14)

P SO 2 [ H 2 O

Since the concentration of water is a constant, it can be incorporated into the equilib-
rium constant. If we express the [SO 2 · H 2 O] l in molar concentration, we have a new

equilibrium constant,

[SO 2 × H 2 O] l
P SO 2

K eq =

(4.15)

expressed in units of mol/L/atm. Note that this is the inverse of the conventional
definition of the air-water partition constant.

P SO 2
[SO 2 ×

1
K eq

K aw =

H 2 O] l =

(4.16)

The complex species SO 2 · H 2 O(l) can undergo subsequent ionization to produce

bisulfite, HSO 3 and sulfite, SO 2 −

ions according to the following reactions:

3

HSO 3 ( l ) H + ( l )

SO 2 · H 2 O(l) + (4.17)

and

HSO 3 ( l ) SO 2 −

3

H + ( l ) . ( l ) + (4.18)