PART A
Determination of Phase Diagram foe Ethanol / Toluene / Water system Theory Three-Component System
DATE
2
November 2015
OBJECTIVE
To determine
the phase diagram for Ethanol/ Toluene/ Water System
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 represent 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 shoes 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 decreased. However, if it is soluble in both liquids,
the mutual solubility is increased. This, 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 liquids 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 the 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 1mL
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 1mL 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.
CHEMICAL
Ethanol, toluene and
distilled water
APPARATUS
Conical flask, conical flask
stopper, retort stand and clamp, pipette, and burette.
PROCEDURE
1.
Each
determination in the experiment has been done twice.
2.
A
mixtures of ethanol and toluene were prepared in sealed containers measuring
100cm3 containing the following percentages of ethanol (in percent)
: 10, 25,35, 50,65, 75, 90 and 95.
3.
20
mL of each mixtures was prepared by filling a certain volume using a burette
(accurately).
4.
Each
of the mixtures was titrated with water until cloudiness is observed due to
existence of a second phase.
5.
Little
amount of water was added and was shake well after each addition. The room
temperature was measured.
6.
The
percentage based on the volume of each component when the second phase starts
to appear/separate was calculated.
7.
The
points were plotted on the triangular paper to give a triple phase diagrams at
the recorded temperature.
8.
A
few more measurements were done if necessary.
RESULTS
% ethanol
(v/v)
|
Total volume
(x+20)mL
|
Water
|
Toluene
|
Ethanol
|
|||
Volume used(mL)
|
%
|
Volume used(mL)
|
%
|
Volume used(mL)
|
%
|
||
10
|
20.95
|
0.95
|
4.53
|
18
|
85.92
|
2
|
9.55
|
25
|
21.30
|
1.30
|
6.10
|
15
|
70.42
|
5
|
23.47
|
35
|
22.05
|
2.05
|
9.30
|
13
|
58.96
|
7
|
31.75
|
50
|
22.65
|
2.65
|
11.70
|
10
|
44.15
|
10
|
44.15
|
65
|
23.55
|
3.55
|
15.07
|
7
|
29.78
|
13
|
55.20
|
75
|
24.15
|
4.15
|
17.18
|
5
|
20.70
|
15
|
62.11
|
90
|
31.20
|
11.2
|
35.90
|
2
|
6.41
|
18
|
57.70
|
95
|
37.10
|
17.1
|
46.09
|
1
|
2.70
|
19
|
51.21
|
DISCUSSION
Ternary phase diagram
contains three components which are water, ethanol and toluene. In this
experiment, the mixture of ethanol and toluene once sealed in measuring
container containing various percentages of ethanol and toluene. The mixture is
prepared into 20mL of each containing various percentages of ethanol such as
(in percent) 10,25,35,50,65,75,90 and 95. The mixture is titrated with
water until cloudiness appear due to existence of second phase. Addition of water to the mixture of
ethanol and toluene increases the mutual solubility of the liquid pair until at
one point the mixture become homogenous.This means two phases present in this experiment is detected and the
volume of water used until the cloudiness is appears is recorded.
A phase diagram shows the phases
existing in equilibrium at any given condition. According to the Phase Rule, a
maximum of four intensive variables (intensive properties) must be specified to
completely define the state of a three-component system. The intensive
variables that are usually chosen are pressure, temperature and
concentration. According to the phase
rule, a single phase in a three-component system may possess four degrees of
freedom.
According to phase rule:
P + F = C + 2, where
P is the number of phases in the system
C is the minimum number of chemical components required to constitute all the
phases in the system
F is the number of degrees of freedom in the system (also referred to as the
variance of the system).
The
curve of the plotted graph is termed as binomial curve. The region under the
curve shows that the presence of 2 phases that is water and toluene whereas the
region above the curve boundary shows one phase of homogenous solution. Based
on the results obtained, when there is a higher percentage of ethanol compared
to the percentage of toluene in the mixture, the volume of water needed to
titrate the mixture until cloudiness is observed is higher. The
addition of water as the third component to a pair of miscible liquids of
toluene and ethanol can change their mutual solubility. If the third
component is more soluble in one of the two different component mixtures the
mutual solubility of the liquid pair is decreased and thus two phases liquids
will be appears.
However, the
binomial curve is incomplete. From the
triangle above, the points are deviated a bit from theoretical points which are
aligned in parallel line. So, there are some errors were occurs during
conducting the experiment.
1) The glass
wares are contaminated.
2) The eyes
level of the observer was not perpendicular to the reading scales.
3) The
temperature during conducting the experiment was not consistent.
4) The tendency
of cloudiness for each mixture was not the same.
Hence, precaution steps must be concerned in the experiment.
1) The glass
wares must be rinsed before used.
2) The eyes
level of the observer must be perpendicular to the reading scale to avoid
parallax error.
3) The
temperature of the surrounding must be fixed.
4) The tendency
of cloudiness for each mixture must be consistent to avoid the deviation of the
final result.
QUESTIONS
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 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 binomial 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 binomial curve. Therefore, a clear
single liquid phase of solution is formed.
(c) Ethanol: 1
part of mixture + 4 parts of ethanol
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 experiment was done by mixing different proportion of ethanol
and toluene. Then, water was titrated into the mixture. As the mixture turns
cloudy, it indicates that two phase system were established. The reaction of
water, ethanol and toluene will appears as two phase’s system due to the decreasing
in solubility of the mixture. From the experiment, as the volume of toluene
decrease, more water needed to break the homogeneity.
References
: Alexander T Florence, David Attwood, Physicochemical
Principles of Pharmacy, Fifth Edition(2011)
Oxtoby, David W., H. P.
Gillis, and Alan Campion. Principles of Modern Chemistry. Belmont, CA: Thomson
Brooks Cole, 2008.
Robert G. Mortimer, Physical Chemistry, Third Edition(2008)
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