Physical Chemistry Third Edition

6.2 Henry’s Law and Dilute Nonelectrolyte Solutions 255

Solution

For a dilute solution,

ci

ni

V

≈

ni

n 1 Vm,1

≈

xi

Vm,1

xi≈ciVm,1

Henry’s law is

Pikixi≈kiVm,1cik(c)i ci

This equation becomes more nearly exact as the concentration becomes smaller.

For a dilute solution, the chemical potential can be expressed in terms of the molar

concentration. We begin with the relation in terms of the mole fraction:

μiμ◦i(H)+RTln(xi)

Expressing the mole fraction in terms of the molar concentration,

μiμi◦(H)+RTln(ciVm,1)μ◦i(H)+RTln(c◦iVm,1)+RTln(ci/c◦)

wherec◦1 mol L−^1 or 1 mol m−^3 (exactly). We let

μ◦i(c)μ◦i(H)+RTln(ci◦Vm,1)

and obtain the result

μiμ◦i(c)+RTln(ci/c◦) (dilute solution) (6.2-19)

In addition to the mole fraction, molality, and concentration, the composition of a

solution can be represented by the mass percentage, by parts per million by mass, or by

volume percentage. For dilute solutions, Henry’s law can be expressed in terms of any

of these composition measures, since all of them are proportional to the mole fraction

in a dilute solution.

Exercise 6.12

Find the expressions for the Henry’s law constants using parts per million, percentage by volume,

and percentage by mass.

The Solubility of a Gas in a Liquid

At equilibrium the amount of a gas that dissolves in a liquid is very nearly proportional

to the partial pressure of the gas, thus obeying Henry’s law. Instead of thinking of a

solution producing a vapor phase at equilibrium with the solution, we think of a gas

dissolving to produce a dilute liquid solution at equilibrium with the gas.