Recall that the three component liquid-liquid eqilibrium system at fixed temperature and pressure (always stated or implied) has, by the Phase Rule, only a single degree of freedom. The classic `triangular diagram' can in fact be regarded as a plot of three separate quantities against this single degree of freedom, which can be taken up by choosing the mol fraction of any one component in either phase.

In the diagram discussed, species 1 and 2 are partially miscible nad species 3 is completely miscible with both.

Consider the single independent variable to be the mol fraction of species 1, plotted along the horizontal axis.

Consider the left hand side of the 2-phase envelope, from its intersection
with the horizontal axis to the plait point. It is
effectively a plot of the solubility of species 1 in species 2 (which
predominates) as the amount of species 3 is varied. The intersection
with the horizontal axis, which represents binary mixtures of 1 and 2,
is the solubility of species 1 in species 2. We will call this
species 2 rich phase the *x* phase.

Correspondingly, the right hand side of the envelope represents the
solubility of species 2 in species 1 as a function of the concentration
of the third component. This will be refered to as the *y* phase.

The third piece of information on this diagram is the relationship between the
concentration of species 3 in a species 1 rich * y* phase (right hand side)
to that of species 3 in a 2 rich *x* phase (left hand side). This is
given by the tie lines.

The quantitative measure of nonideality is the * liquid activity coefficient *
.
It is a function of phase composition and temperature. For a component
in a mixture which behaves ideally, it will be unity.
In vapour liquid
equilibrium this modifies the ideal `k-value' and in the present
case of liquid-liquid equilibrium it defines the equilibrium between
two liquid phases. There will be a value of
for each component in each phase, and these will in general all be different.

If we identify the phases as the `x' and `y' phases where the respective mol fraction concentrations are given by these symbols the equilibrium condition is:

The largest absolute value of ln_{1} occurs when * x*_{2}
is one. This corresponds to a zero concentration of species one, which
is said to be at * infinite dilution. * Thus substances behave
most nonideally when they are present in only trace quantities.

The above expressions apply to the `x' phase. Two similar equations can be written for the `y' phase.
Given the solubility of species 1 in species 2, which we will
call *x*^{o}_{1} and which
on the triangular diagram is at the left hand side of the
two phase region on the horizontal axis, we know that the equilibrium
pahse will be *y*^{o}_{1} at the
other side of the two phase region. These enable us to determine
the constant *A* for the mixture.

We can write two sets of equilibrium relations, one for each
component, and we have two known equilibrium mol fractions in each.
Because the equations for the activity coefficients have only a single
adjustable parameter, all the equations can only be consistent
if the solubility of species 1 in species 2 is the same as that of
species 2 in 1, i.e. if
*y*^{o}_{2} = *x*^{o}_{1}.

Note that if the substances are very nearly insoluble
then the `y' phase is nearly all species 1, so
*y*^{o}_{1}~1 and so is
^{y}_{1}.
The `x' phase correspondingly is nearly all species 2 and so also
*x*^{o}_{2}~1.

A quick estimate of
*A* may be made:

1 x 1 =

R

- Species 1 and 2 are partially miscible and have activities given by the one parameter Margules equation.
- Species 3 forms ideal binary mixtures with both 1 and 2.

- Given solubility of species 1 in 2, or of 2 in 1, determine
*y*^{o}_{1}(=*x*^{o}_{2}) -
Determine;
*A*= (ln*y*^{o}_{1}- ln (1-*y*^{o}_{1}) ) / (*y*^{o}_{1}- (1-*y*^{o}_{1})) - Determine plait point compositions:
*y*^{p}_{3}= 1 - (2/*A*)

*y*^{p}_{1}=*y*^{p}_{2}=*x*^{p}_{1}=*x*^{p}_{2}= (1-*y*^{p}_{3})/2 - For
*y*_{1}from*y*^{p}_{1}to*y*^{o}_{1}....- Solve:

*A*(2*y*_{1}- 1 -*y*_{3}) = ln (*y*_{1}/(1-*y*_{1}-*y*_{3})

*y*_{3 }must lie between 0 and*y*^{p}_{3}.

- Solve:
- This determines the right hand boundary of the 2-phase region.
- Determine the left hand boundary from the values obtained above since;

*x*_{3}=*y*_{3}

*x*_{1}=*y*_{2}= 1 -*y*_{1}-*y*_{3}