Page 391 - Physical Chemistry

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Chapter 12 a a a b
Multicomponent Phase Equilibrium n x n 0.0213.0 mol2 0.06 mol, n 0.4811.9 mol2 0.94 mol
B
4
B
B
6
a
a
a
n n n 3.0 mol 0.06 mol 2.9 mol
4
8
W
B
a
b
n n n 14.0 2.9 2 mol 1.0 mol
W
W
W
2
8
Alternatively, the lever rule can be used.
Exercise
Solve this problem using the lever rule.
Exercise
Repeat this example for 3.0 mol of W and 1.0 mol of B shaken together at 90°C.
a
b
a
(Answer: n 1.3 mol, n 0.02 mol, n b 1.7 mol, n 0.98 mol.)
W B 6 W B
Partition Coefficients
Suppose solvents A and B are partly miscible at temperature T and form the phases a
(a dilute solution of B in solvent A) and b (a dilute solution of A in B) when shaken
at T. If we add solute i to the system, it will distribute itself between the phases a and
b
a
b so as to satisfy m m . Using the concentration scale, we have [Eq. (10.31)]
i
i
b
,b
a
a
,a
b
m° RT ln 1g c >c°2 m° RT ln 1g c >c°2
c,i i
c,i i
c,i
c,i
a
a
,a
,b
b
b
ln 1g c >g c 2 1m° m° 2>RT
c,i i
c,i i
c,i
c,i
c a i g b c,i
,a
,b
K AB,i a exp3 1m° m° 2>RT4 (12.45)
c,i
c,i
c b i g c,i
b
a
The quantity K AB,i c /c is the partition coefficient (or distribution coefficient)
i
i
for solute i in solvents A and B. (Recall separatory-funnel extractions done in organic
chem lab). K AB,i is not accurately equal to the ratio of solubilities of i in A and B be-
cause phases a and b are not pure A and pure B. The exponential in (12.45) is a func-
tion of T and weakly a function of P. The equation preceding (12.45) is the relation
G° RT ln K° for the “reaction” i(b) → i(a).
As the amounts of i in phases a and b change, the activity-coefficient ratio in
(12.45) changes, and the concentrations of B in phase a and A in phase b also change
(see Sec. 12.12). Therefore K AB,i depends on how much i was added to the system and
is not a true constant at fixed T and P, unless a and b are ideally dilute solutions. The
K AB,i value tabulated in the literature is the value corresponding to very dilute solu-
tions of i in a and b, where the activity coefficients are very close to 1 and the com-
positions of phases a and b are very close to what they would be in the absence of
solute i.
The octanol/water partition coefficient K ow of a solute between the phases
oct
wat
formed by 1-octanol and water is c /c , where oct denotes the octanol-rich phase.
K ow is widely used in drug and environmental studies as a measure of how a solute dis-
tributes itself between an organic phase and an aqueous phase. The octanol–water
liquid–liquid phase diagram resembles Fig. 12.18; at 25°C, the phases in equilibrium
have x a octanol 0.793 and x b H 2 O 0.993.
A drug with too high a K ow will tend to accumulate in fatty tissue of the body and
might not reach its intended target. A drug with too low a K ow will not readily go
through cell membranes (which are lipidlike).
Fish swimming in polluted water may have concentrations of a pollutant such as
DDT that are thousands of times the concentration in the water, due to the high
solubility of the pollutant in the fish’s fatty tissue. The bioconcentration factor BCF is