Physical Chemistry Third Edition

6.2 Henry’s Law and Dilute Nonelectrolyte Solutions 251

pressure of exactly 1 bar, for this standard state. If the pressure on the solution is greatly

different from 1 bar, this would require a small correction term like that in Eq. (6.1-5),

which we neglect. Since the solvent obeys Raoult’s law if the solute obeys Henry’s law,

the standard state for the solvent in a dilute solution is the same as for a component of

an ideal solution. It is the pure solvent at a pressure equal toP◦.

Distribution of a Dilute Solute between Two

Solvents. Extraction

Consider the equilibration of two solutions containing the same solute but with different

solvents that are nearly insoluble in each other. For example, if I 2 is dissolved in water

most of the I 2 can be extracted from the water by equilibrating this phase with carbon

tetrachloride. For dilute solutions, it is found experimentally that the equilibrium mole

fraction of I 2 in the water phase is proportional to the mole fraction of I 2 in the carbon

tetrachloride phase. This fact is calledNernst’s distribution law. For a soluteiand two

phases denoted by A and B, this empirical law is given by

Kd

x(B)i,eq

x(A)i,eq

(Nernst’s distribution law) (6.2-4)

wherex(A)i,eqis the equilibrium mole fraction of the solute in phase A andx(B)i,eqis the

equilibrium mole fraction of the solute in phase B. The constantKdis called the

distribution constantordistribution coefficient. For a given solute, the value ofKd

depends on temperature and on the identities of the two solvents.

We now show that Nernst’s distribution law is valid for solutions that obey Henry’s

law and obtain a formula for the distribution coefficient. The chemical potential of the

solute in the two phases is given by

μ

(A)

i μ

◦(H,A)

i +RTln(x

(A)

i,eq) (6.2-5a)

and

μ

(B)

i μ

◦(H,B)

i +RTln(x

(B)

i,eq) (6.2-5b)

At equilibrium,

μ(A)i μ(B)i (6.2-5c)

We solve Eq. (6.2-5a) forxi,Aand solve Eq. (6.2-5b) forxi,B, and after using Eq. (6.2-3),

we can write

Kd

x

(B)

i,eq

x(A)i,eq

exp

[

μ◦i(H,A)−μ◦i(H,B)

RT

]



k(A)i

k(B)i

(6.2-6)

whereki,Aandki,Bare the Henry’s law constants for substanceiin phases A and B,

respectively. The mole fraction of the solute is greater in the phase in which its Henry’s

law constant is smaller. Equilibrating the two phases allows the solute to be extracted

from the other phase.