Title:Phase Diagram for Etanol/Toluene/Water System Theory
Date:4/11/2014
OBJECTIVE
To
determine the phase diagram ethanol/ toluene/ water system theory.
INTRODUCTION
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 the 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.
Examples of three-component systems that has been studied include castor oil/
alcohol/ water; peppermint oil/ propylene glycol/ water; peppermint oil/
polyethylene glycol/ water.
The benefits of
preparing an oily substance as homogenous water in liquid are already clear.
However, what will happen to a system like this when it is diluted should also
be known and this can be explained through the understanding of the triangular
phase diagram. Figure 1 is also for the system containing components peppermint
oilpolysorbate 20-water. A concentration of 7.5% oil, 42.5% polysorbate 20 and
50% water (point A in diagram) can be diluted for 10 times with water giving a
solution that is still clear (now containing 0.75% of oil, 4.25% polysorbate 20
and 95% water). However, when 1 ml of water is added to 10ml of clear solution
B (49% oil, 5% polysorbate 20, 1% water) the solution becomes cloudy, point B'
(44.55% oil, 45.45% polysorbate 20 and 10% water). If I ml of water is further
added, the solution becomes clear, point B" (40.5% oil, 41.3% polysorbate
20, 18.2% water) but if the original solution is diluted three times (16 1/3%
water, 16 2/3% polysorbate 20, 67% water) the solution becomes cloudy.
APPARATUS
-Ethanol
-Toluene
-Burette
-Water
-Beaker
-Conical flask
PROCEDURE
1.Mixture of ethanol
and toluene was prepared in sealed container (100 cm3) according to
the percentage of ethanol (10,25,35,50,65,75,90,95).
2.20 mL of mixture was
prepared by filling into conical flask before it was titrated.
3.The mixture was
titrated with water until cloudiness was observed.It was due to the existence
of a second phase.
4.A little water was
added into the conical flask and shaked well after each addition.
5.The room temperature
was measured.
6.The volume of water
used for titration was recorded.
7.Step 1-6 was repeated
for second experiment.
8.The percentage based
on the volume of each component when the second phase starte to appear was
calculated.
9.The points were
plotte onto a triangular paper to give a triple phase diagram at the recorde
temperature.
10.A few measurement
was done.
RESULT
% of ethanol
(v/v)
|
Volume of Water Used (mL)
|
Average
|
|
Titration I
|
Titration II
|
||
10
|
0.5
|
0.7
|
0.60
|
25
|
0.9
|
1.0
|
0.95
|
35
|
1.9
|
1.7
|
1.80
|
50
|
2.0
|
2.1
|
2.05
|
65
|
2.3
|
2.7
|
2.50
|
75
|
3.3
|
3.9
|
3.60
|
90
|
9.7
|
10.5
|
10.1
|
95
|
13.7
|
13.9
|
13.8
|
Total volume
|
Water
|
Toluene
|
Ethanol
|
|||
(x + 20mL)
|
Volume (mL)
|
%
|
Volume (mL)
|
%
|
Volume (mL)
|
%
|
20.6
|
0.60
|
2.91
|
18.0
|
87.38
|
2.0
|
9.71
|
20.95
|
0.95
|
4.53
|
15.0
|
71.60
|
5.0
|
23.87
|
21.80
|
1.80
|
8.26
|
13.0
|
59.63
|
7.0
|
32.11
|
22.05
|
2.05
|
9.30
|
10.0
|
45.35
|
10.0
|
45.35
|
22.50
|
2.50
|
11.11
|
7.0
|
31.11
|
13.0
|
57.78
|
23.60
|
3.60
|
15.25
|
5.0
|
21.19
|
15.0
|
63.56
|
30.1
|
10.1
|
33.55
|
2.0
|
6.64
|
18.0
|
59.80
|
33.8
|
13.8
|
40.83
|
1.0
|
2.96
|
19.0
|
56.21
|
Discussion
Ethanol/toluene/water system
is a three-component system. In this system which containing three components
but one phase, F= 3-1+2= 4. The four degree of freedoms is temperature,
pressure, and the concentrations of two of the three components. Only two
concentrations term are required
Water and toluene are only
slightly miscible. Two phase system will be formed if they mixed together.
However, ethanol is highly miscible to both water and toluene. Hence, as these
three components were mixed until certain proportion, all three components
would be completely miscible and only single phase system is formed in which
all the three components are miscible to each other and is homogenous. This
situation is illustrated by a ternary system.
In order to display this system, triangular coordinate graph paper
called triangular diagram is used. In the practical, the ternary system
involves a pair of partially miscible liquids-water and toluene. In this
experiment, water is added through the burette to the mixture of toluene and
ethanol which is miscible to each other.
Hence, in this experiment, the volume of water added will be recorded
once the mixture appears cloudy.
The
experiment is carried out at a constant temperature and pressure. The
temperature is at room temperature which is 27◦C and is in atmospheric
pressure. During the experiment, there is solution containing 10% ethanol and
90% toluene. An average 0.6% of water is added to the solution in order for the
second phase to appear. Next, in the solution containing 25% ethanol and 75%
toluene, an average of 0.95mL of water is need in order for the second phase to
appear. When the solution contains 35% ethanol and 65% toluene, average 1.80mL
of water is needed for the water to appear cloudy. In the solution containing
50% of ethanol and 50% of toluene, average 2.05mL of water is needed for the
appearance of second phase in the solution. While in the solution containing
65% ethanol and 35% toluene, 2.50mL of water is needed in order for the
solution to turn cloudy or for the second phase to occur. Then, in the solution
containing 75% ethanol and 25% toluene, an average 3.60mL of water is added for
the second phase to occur. When the mixture contain 90% of ethanol and 10% of
toluene, 10.1mL of water is needed and when the mixture contain 95% of ethanol
and 5% of toluene, 13.8mL of water is needed for second phase to occur.
The
graph is plotted in a triangular coordinate graph paper called triangular
diagram. When the graph is plotted, a curve shape is obtained. It is in term of binodal curve in which the
condition at which two distinct phases coexists in one graph. The graph is
marked as region A and region B. Region A is the region bounded by the curve.
The mixture with composition contain in region A appears cloudy. This is due to
the occurrence of second phase. This indicates the amount of ethanol is not
enough for homogenous mixture or mixture with single phase to form. The mixture
which is not bounded by the curve is marked as region B. The mixture with
composition at region B appears clear and single phase. This is because the
amount of ethanol is enough to make all the three components, ethanol, water
and toluene to completely miscible to each other. The point at both end of the
curve is the limit of the solubility of toluene in water and water in toluene.
Homogenous mixture can be formed if the first point is not exceeded and the
second point is exceeded.
There
are some errors occur during the experiment. Firstly, the conical flask used is
not dry carefully after rinsed with distilled water. This caused the mixture
already in cloudy form before water is added using the burette. Hence, the
experiment needed to redo again to get a more accurate result. Secondly, when
the reading is taken from the scale of the burette or measuring cylinder, the
line of vision of the eyes are not perpendicular to the scale of reading due to
human error and cause the inaccuracy in the result obtained. There are several
precautions needed to be taking note in this experiment. First, when the
conical flask and measuring cylinder are rinsed with distilled water, make sure
the laboratory apparatus is rinsed carefully so that no droplets of water left
inside. Next, the line of vision of the eyes must always perpendicular to the
scale of reading of the laboratory apparatus to increase the accuracy of the
result obtained. During the titration, the mixture inside the conical flask
must be shaken well to ensure the mixture is well mixed.
Question
:
1.Does
the mixture containing 70% ethanol, 20% water and 10% toluene (volume) appear
clear or does it form two layers?
The mixture
will remain clear and it will 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 70% ethanol = 1 x 70/100 = 0.7 part of 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
Therefore, there are 0.7 part of
ethanol; 0.2 part of water; 0.1 part of toluene in the mixture.
(a) Water: 1 part of mixture + 4
parts of water:
Ethanol = 0.7/5 x 100% =14%
Water = (0.2+4)/5 x 100% = 84%
Toluene = 0.1/5 x 100% =2%
Therefore, 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) Toluene: 1 part of mixture + 4
parts of toluene
Ethanol = 0.7/5 x 100% =14%
Water = 0.2/5 x 100% = 4%
Toluene = 0.5/5 x 100% =82%
Therefore, 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 = 4.7/5 x 100% =94%
Water = 0.2/5 x 100% = 4%
Toluene = 0.1/5 x 100% =2%
Therefore, 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 = 4.7/5 x 100% =94%
Water = 0.2/5 x 100% = 4%
Toluene = 0.1/5 x 100% =2%
Therefore, from the phase diagram, this mixture is outside the area of the binodal curve. Therefore, a clear single liquid phase of solution is formed.
CONCLUSION
The phase diagram for Ethanol,
Toluene and Water System (Ternary System) is determined. Determination of the
binomial curve is done but it is incomplete due to some errors occur in the
experiment.
REFERENCES
1. Physical
Pharmacy: Physical Chemistry Principles in Pharmaceutical Sciences, by Martin,
A.N.
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