McKay, Donald. "Front matter" Multimedia Environmental Models ...

chemical M mol into this hypothetical environment, then we can argue that M must
be the sum of the concentration-volume products as follows:
M = C 1V 1 + C 2V 2 + C 3V 3 + C 4V 4
= C 1V 1 + (K 21C 1)V 2 + (K 31C 1)V 3 + (K 41C 1)V 4
= C 1[V 1 + K 21V 2 + K 31V 3 + K 41V 4]
Therefore,
and
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C 1 = M/[V 1 + K 21V 2 + K 31V 3 + K 41V 4]
C 2 = K 21C 1, C 3 = K 31C 1 and C 4 = K 41C 1
It is thus possible to calculate the concentrations in each phase, the amounts in each
phase, and the percentages, and obtain a tentative picture of the behavior of this
chemical at equilibrium in an evaluative environment. This is best illustrated by an
example.
Worked Example 5.5
Benzene partitions in a hypothetical environment between air, water, sediment,
and fish (subscripted A, W, S, and F). The volumes of each phase are given below.
The dimensionless partition coefficients are also given below. Calculate the concentrations,
amounts, and percentages in each phase, assuming that a total of 10 moles
of benzene is introduced into this system.
V A = 1000 V W = 20 V S = 10 V F = 0.05 m 3
K AW = 0.2 K SW = 15 K FW = 20
Using the equation developed above,
C W = 10/(20 + 0.2 ¥ 1000 + 15 ¥ 10 + 20 ¥ 0.05) = 10/371 = 0.027 mol/m 3
Therefore,
C A = 0.0054, C S = 0.405, C F = 0.54
The amounts in each phase are the products CV, namely,
air = 5.39, water = 0.539, sediment = 4.04, fish = 0.03
from which the percentages are, respectively, 53.9, 5.4, 40.4, and 0.3.
It is clear that, in this system, benzene partitions primarily into air, mainly
because of the large volume of air. The concentration in the air is lower than in any
other phase; thus, we must discriminate between phases where the amount is large,
which depends on the product CV, and where the concentration is large. There is a