PRACTICAL 1: PHASE DIAGRAM
Title:
Determination of Phase Diagram for Ethanol/ Toluene/ Water System Theory
Objective:
To determine the phase diagram for three component systems of ethanol/ toluene/ water system.
Apparatus:
Burette, conical flask, retord stand, measuring cylinder and test tubes.
Materials:
Ethanol, toluene and distilled water.
Introduction
Three component system
For three-component systems at constant temperature and
pressure, the compositions may be stated in the form of coordinates for a
triangular diagram.
In
the diagram above, each corner of the triangular diagram represents a pure
component which is 100% A, 100 % B and 100% C. Meanwhile, each side represents
two-component mixtures and within the triangular diagram itself represents
ternary components. Any line parallel to a side of the triangular diagram shows
constant percentage value for a component. For example, DE shows 20% of A with
varying amounts of B and C. So does line FG, showing all mixtures containing
50% of B. These lines intercept with each other at K, which definitely contains
20% A, 50%B as well as 30%C. Measurements can be made this way because in a
triangular diagram, the sum of all distances from K which is drawn parallel to
the three sides of the diagram is same and equals to the length of any one side
of the triangular diagram.
The
addition of a third component to a pair of miscible liquids can change their
mutual solubility. If this third component is more soluble in one of two
different components the mutual solubility of the liquid pair is decreased.
However, if it is soluble in both of the liquids, the mutual solubility is
increased. Thus, when ethanol is added to a mixture of benzene and water, the
mutual solubility of the liquid pair increased until it reached a point whereby
the mixture becomes homogenous. This approach is used in the formulation of
solutions.
Procedures
1. An amount of eight, 20 ml solution of toluene and ethanol were prepared in eight different 100cm3 conical flasks.
2. Each flash were filled so that it contain 10%, 25%, 35%, 50%, 65%, 75%, 90% and 95% of ethanol with the rest was toluene. The conical flasks were labelled A, B, C, D, E, F, G and H respectively.
3. A burette was filled with distilled water.
4. The mixtures were titrated with water, accompanied by vigorous shaking of the conical flask.
5. Titration was stopped when a cloudy mixture was formed.
6. The volume of the water used was recorded.
7. Steps 1-6 were repeated to do a second titration. The volume of water required for complete titration of each mixture was recorded.
8. Average volume of water used was calculated.
9. % volume of each component of the ternary system for when a second phase became separated was calculated.
10. These values were plotted on a graph paper with triangular axes to produce a triple phase diagram.
1. An amount of eight, 20 ml solution of toluene and ethanol were prepared in eight different 100cm3 conical flasks.
2. Each flash were filled so that it contain 10%, 25%, 35%, 50%, 65%, 75%, 90% and 95% of ethanol with the rest was toluene. The conical flasks were labelled A, B, C, D, E, F, G and H respectively.
3. A burette was filled with distilled water.
4. The mixtures were titrated with water, accompanied by vigorous shaking of the conical flask.
5. Titration was stopped when a cloudy mixture was formed.
6. The volume of the water used was recorded.
7. Steps 1-6 were repeated to do a second titration. The volume of water required for complete titration of each mixture was recorded.
8. Average volume of water used was calculated.
9. % volume of each component of the ternary system for when a second phase became separated was calculated.
10. These values were plotted on a graph paper with triangular axes to produce a triple phase diagram.
Results
Test
tube
|
% of
ethanol (v/v %)
|
Vol. of
ethanol (ml)
|
Vol of
Toluene
(ml)
|
Vol of water (ml)
|
Average
vol of water (ml)
|
|
I
|
II
|
|||||
A
|
10
|
2
|
18
|
2.9
|
1.2
|
2.05
|
B
|
25
|
5
|
15
|
2.8
|
0.9
|
1.85
|
C
|
35
|
7
|
13
|
2.0
|
1.2
|
1.60
|
D
|
50
|
10
|
10
|
1.7
|
2.2
|
1.95
|
E
|
65
|
13
|
7
|
3.2
|
2.8
|
3.00
|
F
|
75
|
15
|
5
|
4.6
|
5.2
|
4.90
|
G
|
90
|
18
|
2
|
11.0
|
10.0
|
10.5
|
H
|
95
|
19
|
1
|
15.0
|
15.0
|
15.0
|
Calculations
The
% of ethanol = (2/ 2+18+2.09) x 100%
= 16.54 %
The % of toluene = (18/ 2+18+2.09) x 100%
= 81.48 %
The
% of water = 100% - 16.54% -81.48 %
= 1.98 %
For mixture containing 25% ethanol, the components when the second phase starts to appear,
The % of ethanol = (5/5+15+1.85) x 100%
= 22.88 %
The % of toluene = (15/ 5+15+1.85) x 100%
= 69.5%
The % of water = (100 - 22.88- 69.5) %
= 7.62%
For mixture containing 25% ethanol, the components when the second phase starts to appear,
The % of ethanol = (5/5+15+1.85) x 100%
= 22.88 %
The % of toluene = (15/ 5+15+1.85) x 100%
= 69.5%
The % of water = (100 - 22.88- 69.5) %
= 7.62%
For mixture containing 35%
ethanol, the components when the second phase starts to appear,
The % of ethanol = (7/ 7+13+ 1.6)x 100%
= 32.41 %
The % of toluene = (13/ 7+13+1.6) x 100%
= 60.19 %
The % of water = 100% - 32.41% - 60.19%
= 7.4 %
For mixture containing 50%
ethanol, the components when the second phase starts to appear,
The % of ethanol = (10/ 10+10+1.5) x 100%
= 45.56 %
The % of toluene = (10/ 10+10+1.5) x 100%
= 45.56 %
The
% of water = 100% - 45.56% - 45.56 %
= 8.88 %
For mixture containing 65%
ethanol, the components when the second phase starts to appear,
The % of ethanol = (13/ 13+7+3) x 100%
= 56.52 %
The % of toluene = (7/ 13+7+3) x 100%
= 30.43 %
The
% of water = 100% - 56.52% - 30.43%
= 13.05 %
For mixture containing 75%
ethanol, the components when the second phase starts to appear,
The % of ethanol = )15/ 15+5+4.9) x 100%
= 60.24 %
The
% of toluene = (5/ 15+5+4.9) x 100%
= 20.08 %
The
% of water = 100% - 60.24 % - 20.08 %
= 19.68 %
For mixture containing 90%
ethanol, the components when the second phase starts to appear,
The % of ethanol = (18/ 18+2+10.5) x 100%
= 59.01 %
The % of toluene = (2/ 18+2+10.5) x 100%
= 6.56 %
The
% of water = 100% - 59.01% - 6.56 %
= 34.43 %
For mixture containing 95%
ethanol, the components when the second phase starts to appear,
The % of ethanol = (19/ 19+1+15) x 100%
= 54.29 %
The
% of toluene = (1/ 19+1+15) x 100%
= 2.86 %
= 2.86 %
The % of water = 100% - 54.29 % - 2.86 %
= 42.85 %
Discussion
In this
experiment, we want to
determine the of phase diagram for three component system of ethanol/ toluene/
water system. The result is shows by using a ternary phase diagram. This type of
diagram is three-dimensional but is illustrated in two-dimensions for ease of
drawing and reading. Instead of being a rectangular plot, it is a triangle. It
is called triangular diagram.
In the case of toluene, ethanol, and water which water and toluene are usually
form a two-phase system because they are only slightly miscible. The heavier of
the two phases consists of water saturated with toluene, while the lighter
phase is toluene saturated with water. However, ethanol is completely miscible
with both toluene and water. Thus, the addition of sufficient amount of ethanol
to the toluene-water system would produce a single liquid phase (upper region
in the diagram) in which all the three components are miscible and the mixture
is homogenous. This is shown in the triple phase diagram that has been plotted
on the triangular diagram.
The curve of the plotted graph is termed a binodal curve or binodal. The region
bounded by this curve represents the two-phase region (labelled in the diagram).
Mixture with composition contained within the lower region of the curve is
cloudy in appearance due to the phase separation. In other words, the amount of
ethanol is not sufficient for a homogenous mixture to be produced. The region
of the graph that is not bounded by the binodal curve represents the one–phase
region. It is labelled on the diagram. Mixture with composition that falls into
this region is clear and they are homogenous. For these mixtures, the amount of
ethanol is sufficient to produce a single liquid phase.
The points that are at both ends of the curve are the limits of solubility of
toluene in water and water in toluene. Along the toluene-water line, which
represents a binary mixture of toluene and water, the liquids are able to form
a homogenous mixture as long as the first point is not exceeded. However, the
second point must be exceeded for a homogenous mixture to be formed. The length
of line between the two points represents the mixture of toluene and water with
such composition that they cannot form a homogenous mixture. This may be due to
insolubility of toluene in water or water in toluene.
In this experiment, some
errors may be happened and influence the accuracy of the result formed. Our
eyes must be parallel to the meniscus position when taking reading on burette
or pipette. It can make sure the volume taken and recorded is accurate. Besides,
the conical flask must be shaked well after each addition of water. Lastly, the
judgement of the cloudy solution formed depends on personal judgement. So,
different person in a group may have different judgement about the cloudiness
of the mixtures.
Questions
1. Will a mixture containing 70% ethanol, 20% water and 10% toluene remain clear or form two phases?
From the
graph plotted, at these concentrations, the mixture appears as single liquid
phase. Thus, the solution appears clear.
2. What
will happen if you dilute 1 part of the mixture with 4 parts of (a) water;
(b) toluene; (c) ethanol?
1 part x 70% ethanol = 1 part x 70/100 = 0.7 part
of ethanol
1 part x 20% water = 1 part x
20/100 = 0.2 part of water
1 part x 10% toluene = 1 part x 10/100 = 0.1 part
of toluene
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:
Ethanol = (0.7/ 1+4) x 100% =14%
Water = (0.2+4/ 1+4) x 100% = 84%
Toluene = (0.1/ 1+4) x 100% =2%
From the phase diagram, this mixture is outside the area of the binodal curve. Therefore, a clear single liquid phase of solution is formed.
(b) 1 part of
mixture + 4 parts of toluene
Ethanol = (0.7/ 1+4) x 100% =14%
Water = (0.2/ 1+4) x 100% = 4%
Toluene = (0.1+4/ 1+4) x 100% =82%
From the phase diagram, this mixture is within the area of the binodal curve. Therefore, a two liquid phase will form and the mixture is cloudy.
From the phase diagram, this mixture is within the area of the binodal curve. Therefore, a two liquid phase will form and the mixture is cloudy.
(c) 1 part of
mixture + 4 parts of ethanol
Ethanol = x 100% =94%
Water = x 100% = 4%
From the
phase diagram, this mixture is outside the area of the binodal curve. Therefore,
a clear single liquid phase of solution is formed.
Ethanol, toluene and water system is a ternary
system with one pair of partially miscible liquid (toluene and water). The
addition of sufficient amount of
ethanol to the toluene-water system would
produce a single liquid phase in which all the three components are miscible
and the mixture is homogenous.
References
1. Physicochemical
Principles of Pharmacy , 3rd edition (1998) . A.T. Florence and
D.Attwood. Macmillan Press Ltd.
2. Physical
Pharmacy: Physical Chemistry Principles in Pharmaceutical Sciences, by Martin,
A.N.
Appendices
Appendices
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