Title : Determination of Phase Diagram for Ethanol/ Toluene/ Water System
Theory (Three-Component Systems)
Aim : To determine the phase diagram for three-component
liquid system consisting of Ethanol , Toluene
and Water .
Date : 03.11.2015
Introduction:
A phase diagram is a coordinated triangular diagram that state the
composition and relationship of three components system at constant temperature
and pressure . It shows the conditions at which thermodynamically distinct
phases can occur at equilibrium . The number of phases of a system that can
exist in equilibrium at any time depends on the conditions of temperature,
concentration and composition.
In the diagram above , the three
corners or apexes of the triangle represent a pure component each , that is 100%
by weight of one component (A, B, or C). Thus , the same apex represent 0% of
the other two components. For example, the top corner point represents 100% A
with 0% B and 0% C. Each side of the triangle represents a two component system
or binary mixture . Any line drawn
parallel to one of the sides shows constant percentage value for a component.
The three lines joining the corner points represent two-component mixture of
the three possible combinations of A, B and C. By dividing each line into 100
equal units, the location of a point along the line can be directly related to
the percentage concentration of one component in a two-component system. The
area within the triangle represents all the possible combinations of A, B and C
to give three-component system .
When a third component is added to a pair of miscible
liquid, it may affect their mutual solubility. If the third component is more
soluble in one of the liquids than in the other, then the miscibility between
that pair of liquids decreases. But, if the third component is soluble in both
components, then the mutual solubility is increase. Thus , when ethanol is
added to a mixture of benzene and water , the solubility of mixture increased
to a point where the mixture become homogenous . This shows that solubility
differs when different components are mixed together.
The knowledge on mutual solubility and phase diagram
can be applied in the case of preparing pharmaceutical formulation as it often involve
the mixing of more than one components while the resulting formulation need to be
in homogeneous form. This requirements can be made possible by knowing the
exact ratio of each component needed to be mixed and take into consideration
the temperature and pressure of the surrounding.
In this
experiment, the three components involved were Ethanol, Toluene and Water. If
water and toluene mixed together with ethanol in suitable proportion, they can
form homogeneous solution at equilibrium. The theory behind this
phenomena is that solutions are homogeneous because the ratio of solute to
solvent remains the same throughout the solution even if homogenized with
multiple sources, and stable because the solute will not settle out after any
period of time, and it cannot be removed by a filter or centrifuge. There are 3 components but only 1 phase
exists.
Apparatus: Burette , pipette , retort stand,
conical flask, measuring cylinder, test tube, conical flask stopper
Material: Toluene, ethanol, distilled water
Procedure :
1.
Mixtures of ethanol and toluene in sealed container
with a total volume of 100cm³ was prepared containing the following
percentages of ethanol (in percent): 10,25,35,50,65,75,90 and 95.
2. 20mL of each
mixture was prepared by filling a certain volume using a burette accurately.
3 3. Each mixture was
titrated with water until cloudiness is observed due to the existence of a
second phase.
5 5. The room
temperature was measured.
6. The percentage
based on the volume of each component was calculated when the second phase
starts to appear/separate.
7 7. The points were
plotted onto a triangular paper to give a triple phase diagram at the recorded
temperature.
8 8. A few more
measurements had been done if necessary. The experiment is repeated twice in
order to obtain a more accurate measurement.
Result:
%ethanol
|
% toluene
|
Volume of Water
Used (mL)
|
Average
|
|
Titration I
|
Titration II
|
|||
10
|
90
|
1.20
|
1.00
|
1.10
|
25
|
75
|
1.30
|
1.30
|
1.30
|
35
|
65
|
1.50
|
1.40
|
1.45
|
50
|
50
|
2.00
|
1.70
|
1.85
|
65
|
35
|
2.70
|
2.70
|
2.70
|
75
|
25
|
4.30
|
4.40
|
4.35
|
90
|
10
|
10.40
|
11.10
|
10.75
|
95
|
5
|
14.00
|
14.60
|
14.30
|
Calculation:
Total
volume
|
Water
|
Toluene
|
Ethanol
|
|||
Volume (mL)
|
%
|
Volume (mL)
|
%
|
Volume (mL)
|
%
|
|
21.10
|
1.10
|
5.21
|
18.0
|
85.31
|
2.0
|
9.48
|
21.30
|
1.30
|
6.10
|
15.0
|
70.42
|
5.0
|
23.47
|
21.45
|
1.45
|
6.76
|
13.0
|
60.61
|
7.0
|
32.63
|
21.85
|
1.85
|
8.47
|
10.0
|
45.77
|
10.0
|
45.77
|
22.70
|
2.70
|
11.89
|
7.0
|
30.84
|
13.0
|
57.27
|
24.35
|
4.35
|
17.86
|
5.0
|
20.53
|
15.0
|
61.60
|
30.75
|
10.75
|
34.96
|
2.0
|
6.5
|
18.0
|
58.54
|
34.30
|
14.30
|
41.70
|
1.0
|
2.92
|
19.0
|
55.40
|
Room
Temperature : 27ᵒC
Questions:
1.
Does the mixture containing 70% ethanol, 20% water and 10% toluene ( volume ) appear clear or does it form two
phases?
At
these concentrations, the mixture will remain clear and form one liquid
phase.
2.
What will happen if you dilute 1 part of the mixture with 4 parts of (a) water;
(b) toluene; (c) ethanol?
1
part mixture x 20% water = 1 x 20/100 = 0.2 part of water
1
part mixture x 10% toluene = 1 x 10/100 = 0.1 part of toluene
1
part mixture x 70% ethanol = 1 x 70/100 = 0.7 part of ethanol
Therefore, there are 0.7 part of ethanol;
0.2 part of water; 0.1 part of toluene in the mixture.
(a) 1 part of
mixture + 4 parts of water:
Water =
(0.2+4 / 1+4) x 100% = 84%
Toluene
= (0.1 /1+4) x 100% =2%
Ethanol
= (0.7/1+4) x 100% =14%
From the phase diagram,
this mixture is under the area of the binomial curve. Therefore, a 2 phase is
formed.
(b) 1 part of mixture +
4 parts of toluene:
Water =
(0.2 / 1+4) x 100% = 4%
Toluene
= (0.1+4 / 1+4) x 100% =82%
Ethanol
= (0.7 / 1+4) x 100% =14%
From the phase diagram, this mixture is
outside the area of the binomial curve. Therefore, a clear single liquid phase
of solution is formed.
(c) 1 part of mixture + 4 parts
of ethanol:
Water =
(0.2/ 1+4) x 100% = 4%
Toluene
= (0.1 / 1+4) x 100% =2%
Ethanol
= (0.7+4 / 1+4) x 100% =94%
From the phase diagram,
this mixture is outside the area of the binomial curve. Therefore, a clear
single liquid phase of solution is formed.
Discussion:
Phase
diagram is a diagram representing the limits of stability of the various phases
in a chemical system at equilibrium, with respect to variables of parameters
such as concentration and temperature. Therefore, to define the state of each
phase, knowledge of several variables is needed and essential. Thus, regarding
to this issue, phase rule is formulated by J. Willard Gibbs.
Phase rule is used to relate the effect of the minimal number of independent
variables upon the various phases that can exist in an equilibrium system that
containing a given number of components. The phase rule is expressed as
follows:
F = C – P + 2
where,
F = Number of degree of freedom P = Number
of phases
C= Number of components
By using phase rule equation in this
experiment, the number of degree of freedom, F can be determined. In this
experiment, we use three component as each components are composed of different
elements. At the beginning of the experiment, we mixed toluene and ethanol
which eventually resulting only one liquid phase (homogenous solution). Thus
the F obtain in this experiment is 4 (as F = 3-1+2) for a non-condensed system.
The degree of freedom at the beginning of the experiment is comprised of
temperature, pressure and the each of the concentration of the components. As
water is added to the system, F will be reduced to 2 (as F= 3-3+2) for a
condensed system as the number of phases are increased from one liquid phase to
two liquid phase and addition of one vapour phase. As the number of components
increase, the number of degree of freedom also will be increased. However, as F
is increased, the system will become more complex, it becomes necessary to fix
more variable to define the system. Ironically, the greater the number of
phases in equilibrium, the fewer the degrees of freedom.
In
this experiment, we are using three different component, which is toluene,
ethanol, and water. It is important to know the properties of toluene and
ethanol solubility in water. Both toluene and ethanol have different properties
in water. Ethanol are soluble in water while toluene are not soluble in water.
At the beginning of the experiment, we mixed an amount of toluene and ethanol
together. Both are forming a clear mutual miscible solution and forming
homogenous solution (one liquid phase). But after adding an amount of water,
the solubility of the clear mutual miscible solution are disrupted. The clear
solution turns cloudy which indicates the solution does not longer miscible to
each other and starts to form two layer (two liquid phase) as resulting from
the different degrees of solubility of toluene and ethanol in water.
The addition of water (third component) to a
miscible liquids, which is toluene and ethanol can change their mutual
solubility. As water is more soluble in ethanol compared to toluene, the mutual
solubility of toluene and ethanol is decreased. However, if water is soluble in
both of the liquids, the mutual solubility will be increased. Thus, when water
is added to the mixture of toluene and ethanol, the mutual solubility increased
until it reached a point where the mixture becomes homogenous solution. This is
proved in the experiment, as more amount of water is needed to turn the clear
miscible solution into cloudy immiscible solution. In addition, all of the
component in this experiment can be a miscible solution when they are mixed
together if and only if correct proportions of each component is determined.
Figure 3.1
Triangular diagram
Based
on this experiment, we are discussing about the phase diagram for ternary
systems as we are dealing with three component systems. Thus, it is more
convenient for us to use triangular coordinate graph paper. In this ternary
systems, only liquid phases are involved either it will form one phase or two
phase. By using a triangular coordinate graph paper, there are certain rules
that relate to the use if triangular coordinates. The concentrations of the two
component used in this experiment must be expressed on weight-weight basis.
This is because in pharmaceutical interest, it will be an efficient method of
preparing dispersions and allows the concentration to be expressed as mole
fraction or molality. Based on the Figure 3.1, each of three corners or apexes
of the triangle represent 100% by weight (in this experiment we use volume of
the component) of one component (A: toluene, B: ethanol or C: water). As a
result, the same apex will represent 0 % of the other two components. For
instance, the top corner point in Figure 3.1 represents 100% A. The three lines
joining of the corner points (referred to point AB, BC and CA) represent
two-component mixtures of three possible combinations of A, B and C. Thus,
lines AB, BC and CA are used for two-component mixtures of A and B, B and C and
C and A, respectively.
By
dividing each line into 100 equal units, location of point along the line
(referred to line AB, BC and CA) can directly related to the percentage
concentration of one component in two-component system as the third component
of the system is equal to 0. On the hand, the area within the triangle
represents all the possible combinations of A, B and C to give three-component
systems. For example, based on Figure 3.1, the line parallel to CB that cuts
point D is equivalent to 85.31% A , resulting the system to contains 85.31% of
A and 14.69% of B and C together. Applying similar arguments to the other two
components in the system, we can say that along the line AB, C= 0. As we
proceed from line AC towards B, the concentration of B is elevated until the
apex, B=100%. The point D lies on the line parallel to line AC is equivalent to
5.21% of B. Thus, to find the concentration of C, we use 100% - (A+B) % = 100%
- 90.25% = 9.48%. This is confirmed in the Figure 3.1 as the line AB towards
apex C, the point D lies on the line is equivalent to 9.48% of C. This is
applied to all points that lies within the line in the triangle.
Conclusions:
In conducting
the experiment of phase equilibria in three-component systems, it is important
to define the number of degree of freedom to find out the least number of
variables that needs to be fixed in order to describe the system completely.
This is important as it will affected the data obtained to plot the complete
binomial curve. Besides, in this experiment also, all the three component can
achieved a homogenous solution (one liquid phase) if the correct proportions of
each component is mixed together. The mixture of water, toluene and ethanol
become more homogenous when decreasing amount of toluene and increasing amount
of ethanol is used in the system. If the proportions of the three components is
not correct in order to achieve a clear miscible solution, two liquid phase
will be form which is indicates by the appearing of the cloudiness of the mixed
solution.
Reference:
1.
Patrick
J. Sinko, Yashveer Singh. 2011. Martins Physical Pharmacy and Pharmaceutical
Pharmacy Sciences. Ed. ke6. China: Lippincott Williams & Wilkins.
