HW, Chapt 3 answers
(a) Dissolution of pure toluene in water: let phase 1 represent
aqueous phase, phase 2 organic phase.
(i) Concentration of toluene in aqueous phase in mol/L and in
mole fraction units: Convert from mg/L to mol/L by converting
mg to g and dividing by molecular weight:
mg
1 mol
-3 g
515
 10
L
mg

92.13 g
 solubility of pure toluene = 5.59 x 10-3 mol/L
To convert from mol/L units to mol fraction units, multiply by
molar volume of mixture: Thus, x1 = (5.59 x 10-3 mol/L)(0.018
L/mol)
 x1 = 1.01 x 10-4 mole fraction (ii) Aqueous
activity coefficient and activity of toluene in aqueous phase. At
equilibrium, the chemical potential will be the same in each
phas:
1
1 =
x1
;
1
3
1 =
=
9.94
x
10
1.01 x 10 - 4
The activity of toluene in the aqueous phase is defined as:
activity = 1x1 from which it follows that
activity of toluene in aqueous phase = 1
(iii) Equilibrium constants for the dissolution process:
For the dissolution reaction:
toluenepure
tolueneaq
x1
K' 12 =
x2
Since in the pure toluene phase, x2 = 1, K'12 = x1 or
K'12 = 1.01 x 10-4
Alternatively, we can express a concentration based equilibrium
constant K12 as:
C1
K12 =
C2
To use this, we need to calculate the concentration of toluene in
the organic phase in mol/L units:
C2 = 0.8669
g
mL
1 mol
 103

= 9.41 mol / L
mL
L
92.13 g
5.59  10 -3 mol / L
K12 =
= 5.94  10 -4
9.41 mol / L
Note that K'12  K12.
(iv) Partial molar excess free energy of solution of toluene:
The partial molar excess free energy of solution is defined as
(SGI, p. 51)
partial molar excess free energy of solution = RT ln i
In the aqueous phase, i = 9.94 x 103, thus:
partial molar excess free energy of solution in aqueous phase =
(8.314)(298)[ln(9.94 x 103)] = 22.8 kJ/mol
In the organic phase, i = 1 (ln i = 0), and thus
partial molar excess free energy of solution in organic phase
=0
(b) Aqueous solubility of toluene if organic phase is 50:50
(vol:vol) toluene:benzene
Using the same reasoning as before, by equating the chemical
potential of the toluene in the aqueous phase with that in the
organic phase, we can show
1x1 = 2x2 (1 = aqueous phase)
We now need to determine x2, the mol fraction of the toluene in
the organic phase.
If we had 1 mL of toluene and 1 mL of benzene present, then the
total number of moles of organic liquid would be:
(
# moles = (1 mL )
0.8669 g
mL
)(
1 mol
92.13 g
)
(
+ (1 mL )
0.8876 g
mL
)(
= 2.077 x 10-2 moles in 2 mL, of which 9.410 x 10-3
moles consist of toluene and 1.136 x 10-2 moles consist of
benzene.
Thus, the mole fraction of toluene in the mixture is:
9.410 x 10 - 3
x2 =
= 0.453
2
2.077 x 10
x1 =
x2   2
(0.453)(1)
=
= 4.56 x 10 - 5 mole fraction
3
1
(9.94 x 10 )
To convert to mol/L units, divide by the molar volume of water:
4.56 x 10 - 5 moles / mol T OT
conc. in aq. phase =
= 2.53 x 10 - 3 mol / L
0.018 L / mol
1 mo
78.11