TITLE:
EXPERIMENT 3 (PART A) – DETERMINATION OF THE PHASE DIAGRAM FOR
ETHANOL/TOLUENE/WATER SYSTEM THEORY (3 SYSTEM THEORY)
DATE OF
EXPERIMENT: 10TH NOVEMBER 2016
OBJECTIVES
1.
To determine the
solubility limit in a ternary system of water and two other liquids (ethanol
and toluene), one of which is completely miscible (ethanol) and the is partly
miscible with water (toluene)
2.
To construct a
solubility curve on the triangular diagram or ternary diagram based on the
three-system mixture that is being studied.
INTRODUCTION
A ternary plot or known as the triangle plot, simplex
plot is a graph based on three variables which equals to a constant. Graphically,
it depicts the ratios of the three variables as positioned in the equilateral
triangle where at the vertices of each triangle represents the pure composition
of the component practically in a system composed of three species or
components. In the ternary plot, the proportions of the three variables a, b, and c (the vertices
of the triangle) must sum to some constant, K. Usually, this constant is represented as 1.0 or 100%. Because
a + b + c = K for all
substances being graphed, any one variable is not independent of the others, so
only two variables must be known to find a sample's point on the graph: for
instance, c must be equal to K − a − b.
Because the three proportions
cannot vary independently: there were only two degree of freedom and it is
possible to graph the intersection of all three variables in only two
dimensions. The advantage of using
a ternary plot for depicting compositions is that three variables can be
conveniently plotted in a two-dimensional graph as shown in this experiment and
can be referred in the data presentation part. Ternary plot is used in this
experiment since it can also be used to create phase diagrams by outlining the
composition regions on the plot where different phases exist. The area bounded
by the graph in this experiment technically represents the three-system mixture
can exist in two-phase or physically when the mixture turned cloudy during the
addition of distilled water.
MATERIALS AND
APPARATUS
APPARATUS:
1.
Volumetric
flasks
2.
Retort stand
3.
Burette
4.
Dropper
5.
Measuring
cylinder
6.
Beaker
CHEMICALS:
1.
Ethanol
2.
Toluene
3.
Distilled water
METHODOLOGY
1.
8 conical flasks
were labelled as A, B, C, D, E, F, G and H.
2.
2mL of Ethanol
is added into conical flask A, and 18mL of Toluene is added into the same
volumetric flask for a total volume of the mixture is 20mL in the conical
flask.
3.
A burette is
filled with distilled water and clamped to the retort stand
4.
The initial
reading of the burette is recorded.
5.
The content in
the conical flask was titrated with the distilled water in the burette and the
conical flask was swirled gently until the mixture turns cloudy and the final
reading of the burette is recorded.
6.
Step 2-5 is
repeated with volumetric flask B, C, D, E, F, G and H with different percentage
composition of Toluene and Ethanol in each volumetric flask respectively with
the reference to the table below:
ETHANOL
|
TOLUENE
|
||
VOLUMETRIC
FLASK
|
PERCENTAGE
(%)
|
VOLUME
(mL)
|
VOLUME
(mL)
|
A
|
10
|
2
|
18
|
B
|
25
|
5
|
15
|
C
|
35
|
7
|
13
|
D
|
50
|
10
|
10
|
E
|
65
|
13
|
7
|
F
|
75
|
15
|
5
|
G
|
90
|
18
|
2
|
H
|
95
|
19
|
1
|
RESULTS:
VOLUME OF DISTILLED WATER USED
VOLUMETRIC
FLASK
|
VOLUME OF
DISTILLED WATER USED (mL)
|
||
TITRATION I
|
TITRATION II
|
AVERAGE
|
|
A
|
1.4
|
1.2
|
1.3
|
B
|
0.9
|
0.5
|
0.7
|
C
|
1.2
|
0.5
|
0.9
|
D
|
2.8
|
1.4
|
2.1
|
E
|
2.9
|
2.5
|
2.7
|
F
|
3.9
|
3.9
|
3.9
|
G
|
9.8
|
11.0
|
10.4
|
H
|
17.3
|
16.5
|
16.9
|
FINAL VOLUME AND PERCENTAGE OF EACH COMPONENT IN THE
MIXTURE
VOLUMETRIC
FLASK
|
ETHANOL
|
TOLUENE
|
DISTILLED
WATER
|
TOTAL
|
|||
mL
|
%
|
mL
|
%
|
mL
|
%
|
mL
|
|
A
|
2.0
|
9.39
|
18.0
|
84.51
|
1.3
|
6.10
|
21.3
|
B
|
5.0
|
24.15
|
15.0
|
72.46
|
0.7
|
3.39
|
20.7
|
C
|
7.0
|
33.49
|
13.0
|
62.20
|
0.9
|
4.31
|
20.9
|
D
|
10.0
|
45.25
|
10.0
|
45.25
|
2.1
|
9.50
|
22.1
|
E
|
13.0
|
57.27
|
7.0
|
30.84
|
2.7
|
11.89
|
22.7
|
F
|
15.0
|
62.76
|
5.0
|
20.92
|
3.9
|
16.32
|
23.9
|
G
|
18.0
|
59.21
|
2.0
|
6.58
|
10.4
|
34.21
|
30.4
|
H
|
19.0
|
51.49
|
1.0
|
2.71
|
16.9
|
45.80
|
36.9
|
DATA
PRESENTATION
DISCUSSION
Ethanol is completely
miscible with Toluene and thus the mixtures of the two solutions prepared in
the conical flask earlier were in one phase. However, toluene is not miscible
with water and the result from the titration of the content in the conical
flask with distilled water, at certain point will becomes completely
immiscible. Solubility of a component in a system can differs or changes when
another foreign component is added into the system since it can affect the
mutual solubility of the system. In the case of our system, since different
percentage composition of Toluene and Ethanol is prepared, we will discuss more
on the effect of the percentage composition of toluene with the volume of water
needed to form a two-phase system. Generally, we can say that as the percentage
composition of Ethanol in the conical flask is higher, the volume of water
needed to turn the system into a two-phase system is also higher.
Based on the basis of
the understanding on the solubility of a component can change when different
component is added, adding water into the system when it is immiscible with
toluene will cause the mutual solubility of the system to decrease. Since
ethanol is miscible with the distilled water added, this explains the reason of
the conical flask with higher ethanol composition needed a higher amount of
distilled water for a two-phase system to be formed. As a reversal, if the
third component added into the system is miscible with both toluene and ethanol
then the mutual solubility of the system will increase and thus the system will
remains as a one-phase system. The end point of the titration of the mixture of
ethanol and toluene with water is recorded when the solution was turned into
the two-phase system and physically put as the solution turned cloudy.
To prove the
explanation above, we can compare the system in conical flask A and H. In
conical flask A where the percentage composition of Ethanol is the lowest;
which is at 10%, the volume of water needed is also the lowest of all which is
only 1.3mL of distilled water is needed to turn the system into a two-phase
system. In conical flask H where the percentage composition of ethanol is the
highest; which is at 95%, the volume of water needed is also the highest of all
which is 16.9mL is needed in order to form a two-phase system. On the ternary
triangular diagram, each vertices of the triangle represents the pure percentage
composition of each component involved in the three system mixture (after
titration with distilled water).
With reference to the
triangle in the data presentation part, the top vertices represent the
percentage composition of the ethanol, the right vertices on the bottom part
represent the percentage composition of distilled water and the left vertices
on the bottom represents the percentage composition of Toluene. To plot the
graph, the value of each point is taken from the percentage composition of each
component (toluene, distilled water and ethanol) after the addition of water (the
data can be referred in the result section). The area bounded below the graph
is when the system turned into a three-phase system after the addition of
foreign substance which changes the mutual solubility of the system.
A phase rule is a useful device for relating
the effect of the least number of independent variables (temperature,
concentration, density) that might affect the result of the experiment to
improve the accuracy of the data obtained. Knowing the number of the degrees of
freedom for the variables mentioned above can help to minimize possible errors
that might arise in this experiment since we can fix the involving variables. For
a system at equilibrium based on the phase rule:
F = C – P + 2
F = number of
phases that can coexist
C = number of
components making up the phases
F = degree of
freedom
In this
experiment, the number of components that that was involved in the system is 3
and the number of phases that it owned was 2. Thus, the value of the degree of
freedom is:
F = 3 – 2 + 2
= 3
From the degree of freedom obtained, we can say that
there are three factors that can affect the experimental data.
From the graph plotted,
noticed that there was a single point left unconnected since the point lies a
bit deviated from the rest of the points and we conclude that some errors might
arise while experimenting. The main
error that can affect the overall result is due to the contaminated or unclean
apparatus. Some apparatus used in the lab were practically used by another
student before, and there was a high possibility of the apparatus for not being
clean thoroughly and some residue from the previous experiment might be left in
the apparatus. Thus, the precaution step that must be taken in overcoming this
source of error is to clean and rinse the apparatus beforehand. However, second
error might arise due to the precaution step by cleaning the apparatus. Since
our experiment involve distilled water as the manipulated variable, presence of
water from the cleaning process can greatly influenced the result obtained
since the mixture in the conical flask might get cloudy even before the
titration begin.
To overcome this, the
apparatus used must be wiped well to remove excess water from the cleaning
process. Apart from that, error might also occur from the degree of the
cloudiness observed. There are variety degrees of cloudiness that can be
observed and thus, the same person must observe the phase change of the system
to avoid error in determining the end point of the titration. Besides, the
volatile properties of ethanol and toluene were also contributing to the error
in this experiment. The mixture prepared must be titrated quickly to minimize
the amount of vaporization that can cause the volume of the mixture to be lower
than it supposedly to be. Parallax error was also included in response to the
inaccurate data obtained. This error can occur when there is some distance
between the measuring scale and the indicator used to obtain a measurement and
the observer eyes was not squarely aligned with the calibration, causing the
reading may be too high or low. To overcome this, the observer eyes must be
directly perpendicular with the calibration reading.
CONCLUSION
In a nutshell, as the
percentage composition of ethanol increase, the volume of distilled water
needed to titrate the mixture is also increasing. Understanding the principle
of mutual solubility when a foreign substance is added into the system is the
basis of doing this experiment. Before water was added into the system, both
toluene and ethanol is miscible with each other and the mixture exist in
one-phase. When distilled water is added, it was immiscible towards one of the
mixture component which is the toluene and this causes the mutual solubility of
the mixture to decrease and addition of enough water can cause the formation of
a two-phase system. Since ethanol is miscible with water, conical flasks with
higher percentage composition of ethanol needed more distilled water to form a
two-phase system. Some errors have affected the overall result obtained and to
overcome this, several precautions step have been done to minimize the
inaccurate reading.
QUESTIONS
1.
Does the mixture containing 70% ethanol, 20% water
and 10% toluene (volume) appear clear or does it form 2 layers?
From the graph plotted,
at these concentrations, the solution will appear clear (one phase) as the plot
is outside the two-phase boundary.
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
Thus, 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 curve. Therefore, a
clear single liquid phase of solution will be form.
(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 in the area of the curve. Therefore, a two
liquid phase will form and the mixture is cloudy.
(c)
1 part of mixture + 4 parts of ethanol
Ethanol
= (0.7 + 4 / 1 + 4) x 100% = 94%
Water
= (0.2 / 1 + 4) x 100% = 4%
Toluene
= (0.1 / 1 + 4) x 100% = 2%
From
the phase diagram, this mixture is outside the area of the curve. Therefore, a
single liquid phase of solution will be form and appear clear.
REFERENCE
1.
Chang/Goldsby.2016.Chemistry Twelfth Edition.McGraw-Hill
Education, USA:2 Penn Plaza, New York
2.
Wikipedia, Ternary Plot, updated on 15 November 2016 at 18:28
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