Physical Chemistry Third Edition

6.2 Henry’s Law and Dilute Nonelectrolyte Solutions 253

For a dilute solutionn 1 is much larger than the other amounts, and the molality is nearly

proportional to the mole fraction:

xi≈

niM 1

w 1

miM 1 (dilute solution) (6.2-10)

For a dilute solution Henry’s law can be expressed in terms of the molality:

PikimiM 1 ki(m)mi (dilute solution) (6.2-11)

wherek(m)i kiM 1 is called themolality Henry’s law constantfor substancei. For a

sufficiently dilute solution it is independent of the molality but depends on the identities

of all substances present and on the temperature.

EXAMPLE 6.6

a.From the value ofk 2 for ethanol (substance 2) in Example 6.4, find the value ofk(m) 2.

b.Find the vapor pressure of a 0.0500 mol kg−^1 solution of ethanol in benzene, assuming

the molality version of Henry’s law to hold.

Solution

a.k(m) 2 (1. 51 × 103 torr)(0.07812 kg mol−^1 )118 torr(mol kg−^1 )−^1

b.P 2 

[

118 torr(mol kg−^1 )−^1

](

0 .0500 mol kg−^1

)

 5 .90 torr

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

the molality in an equation similar to Eqs. (6.2-2) and (6.1-8). Using Eqs. (6.2-2) and

(6.2-10),

μiμ◦i(H)+RTln(miM 1 ) (dilute solution)

μiμ◦i(m)+RTln(mi/m◦) (dilute solution) (6.2-12)

where

μ◦i(m)μ◦i(H)+RTln(M 1 m◦) (6.2-13)

and wherem◦is defined to equal 1 mol kg−^1 (exactly).

The quantityμ◦i(m)is the chemical potential of substanceiin itsmolality standard

state. This standard state is componentiin a hypothetical solution withmiequal to

m◦(exactly 1 mol kg−^1 ) and with Henry’s law in the form of Eq. (6.2-11) valid at this

molality. Again we specify a pressure of exactly 1 bar for this standard state. Since the

standard state is a hypothetical solution, the actual 1-molal solution is not required to

obey Henry’s law.

Exercise 6.11

Show thatμ◦i(m)is equal to the chemical potential of substanceiin the vapor phase at equilibrium

with a 1.000 mol kg−^1 solution if Eq. (6.2-11) is valid at this molality.