OBJECTIVE: To determine phase diagram for three component system (ethanol/ toluene/ water system theory)
INTRODUCTION:
Ternary or three-component system is a complex type of multi- component system. Ternary systems are more frequently encountered in practice than binary systems. For example, air is often approximated as being composed of nitrogen, oxygen, and argon, while dry natural gas can be rather crudely approximated as being composed of methane, nitrogen and carbon dioxide. Intuitively, having more than two components poses a problem when a pictorial representation is desired. A rectangular coordinate plot, having only two axes, will no longer suffice. Gibbs first proposed the use of a triangular coordinate system.
Ternary phase diagrams are three component systems. To construct a ternary diagram it is necessary to know the three binary systems for the three components. Ternary diagrams have a vertical temperature axis. The actual ternary diagram may be represented as a three dimensional form or more commonly as a two dimensional projection of the liquids surface onto the base of the triangle created when the three binary diagrams are joined together. This irregular triangle is often transformed into an equilateral triangle to facilitate presentation and interpretation. In modern times, we use an equilateral triangle for such a representation. Figure 1 shows an example of a ternary phase diagram. Note that the relationship among the concentrations of the components is more complex than that of binary systems.
A
B
C
C
Figure 1
Features:
- Any point within this triangle represents the overall composition of a ternary system at a fixed temperature and pressure.
- By convention, the lightest component is located at the apex or top of the triangle (A). The heavy and medium components are placed at the left hand corner (B) and right hand corner (C), respectively.
- Every corner represents a pure condition. Hence, at the top we have 100 % B, and at each side, 100 % A and 100 % C, respectively.
- Each side of the triangle represents all possible binary combinations of the three components.
- On any of those sides, the fraction of the third component is zero (0%).
- As you move from one side (0 %) to the 100 % or pure condition, the composition of the given component is increasing gradually and proportionally. At the very center of the triangle, we find 33.33 % of each of the component.
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.
APPARATUS:
Burette, conical flask, retort stand, measuring cylinder, test tubes
 |
| Burette |
 |
| Conical flask and measuring cylinder |
Ethanol, Toluene, Water
 |
| Ethanol |
 |
| Toluene |
 |
| Distilled water |
- 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.
- 20ml of each mixture was prepared by filling a certain volume using a burette accurately.
- Each mixture was titrated until cloudiness is observed due to the existence of a second phase.
- A little water was added and shaken well after each addition.
- The room temperature was measured using the thermometer.
- The percentage based on the volume of each component was calculated when the second phase starts to appear.
- The points were plotted onto a triangular paper to give a triple phase diagram at the recorded temperature. One more measurement was done if necessary.
Ethanol
|
Toluene
|
Water
|
Total volume of Ethanol + Toluene
+ water (mL)
|
|||
Volume (mL)
|
Percentage (%)
|
Volume (mL)
|
Percentage (%)
|
Volume (mL)
|
Percentage (%)
|
|
2
|
9.26
|
18
|
83.33
|
1.6
|
7.41
|
21.6
|
5
|
24.04
|
15
|
72.12
|
0.8
|
3.85
|
20.8
|
7
|
33.82
|
13
|
62.80
|
0.7
|
3.38
|
20.7
|
10
|
45.87
|
10
|
45.87
|
1.8
|
8.26
|
21.8
|
13
|
55.56
|
7
|
29.91
|
3.4
|
14.53
|
23.4
|
15
|
60.00
|
5
|
20.00
|
5.0
|
20.00
|
25.0
|
18
|
56.25
|
2
|
6.25
|
12.0
|
37.50
|
32.0
|
19
|
51.21
|
1
|
2.70
|
17.1
|
46.09
|
37.1
|
CALCULATION:
When the second phase start to appear, the percentage components of:
- 10% ethanol = 2mL ethanol;
% of ethanol = 2mL /(2+18+1.6)mL×100% = 9.26%
% of phenol = 18mL /(2+18+1.6)mL×100% = 83.33%
% of water = 1.6mL /(2+18+1.6)mL×100% = 7.41%
- 25% ethanol = 5mL ethanol;
% of ethanol=5mL/(5+15+0.8)mL×100%=24.04%
% of phenol=15mL/(5+15+0.8)mL×100%=72.12%
% of water=0.8mL/(5+15+0.8)mL×100%=3.85%
- 35% ethanol = 7mL ethano;l
% of ethanol=7mL/(7+13+3.7)mL×100%=33.82%
% of phenol=13mL/(7+13+3.7)mL×100%=62.8%
% of ethanol=0.7mL/(7+13+3.7)mL×100%=3.38%
- 50% ethanol = 10mL ethanol;
% of ethanol=10mL/(10+10+1.8)mL×100%=45.87%
% of phenol=10mL/(10+10+1.8)mL×100%=45.87%
% of water=1.8mL/(10+10+1.8)mL×100%=8.26%
- 65% ethanol = 13mL ethano;l
% of ethanol=13mL/(13+7+3.4)mL×100%=55.56%
% of phenol=7mL/(13+7+3.4)mL×100%=29.91%
% of water=3.4mL/(13+7+3.4)mL×100%=14.53%
- 75% ethanol = 15mL ethanol;
% of ethanol=15mL/(15+5+5)mL×100%=60%
% of phenol=5mL/(15+5+5)mL×100%=20%
% of water=5mL/(15+5+5)mL×100%=20%
- 90% ethanol = 18mL ethanol;
% of ethanol=18mL/(18+2+12)mL×100%=56.25%
% of phenol=2mL/(18+2+12)mL×100%=6.25%
% of water=12mL/(18+2+12)mL×100%=37.5%
- 95% ethanol = 19mL ethanol;
% of ethanol=19mL/(19+1+17.1)mL×100%=51.21%
% of phenol=1mL/(19+1+17.1)mL×100%=2.7%
% of water=17.1mL/(19+1+17.1)mL×100%=46.09%
DISCUSSIONS:
The system contains 3 components (ethanol, toluene, and water) but only one phase, F = 3-1+2 =4 for a non-condensed system. The four degrees of freedom are temperature, pressure and the concentrations of two of the three components. Only concentration of two components are required because the sum of these subtracted from the total will give the concentration of the third component. In this experiment, we regard the system as condensed and hold the temperature constant, then F=2. Each of three corners or apexes of the triangle represent 100% by weight of one component A represent for ethanol, B represents water while C represents toluene. As a result, that same apex will represent 0% of the other two components. The three lines joining the corner points represent two-component mixtures of the three possible combinations of A, B and C. Thus the lines AB, BC and CA are used for two-component mixtures .By dividing each line into 100 equal units, the location of a point along the line can be directly related to the per cent concentration of one component in a two-component system. In going along a line bounding the triangle so as to represent the concentration in a two-component system, it does not matter whether we proceed in a clockwise or counter clockwise direction around the triangle, provided we are consistent. The more usual convention is clockwise and has been applied in this experiment. Hence, as we move along A to B, we are signifying systems of A (ethanol) and B (water) containing increasing concentrations of B, and correspondingly smaller amounts of A.
In this experiment, water and toluene form a two-phase system because they are only slightly miscible while ethanol is completely miscible with both toluene and water. But, as these three components were mixed until certain proportion, all three components would be completely miscible. This experiment is carried out by first making a solution of ethanol and toluene which will be completely miscible and addition of water where at first it will make up two phase. As we continue to add more water until appropriate amount, it will result to one phase system. However when conducting this experiment, we must not clean the apparatus with distilled water so that it will not effect the mixture become cloudy when ethanol and toluene is pour into the apparatus. This is shown in the triple phase diagram that has been plotted on the triangular diagram.
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. The region under the graph shows that there are two phase system form which consist of water and toluene since the solubility of water with toluene is weaker than that with ethanol. Meanwhile the region above the graph shows homogenous mixture. All the experimental mixtures should all plotted within the triangle theoretically. From the triangle above, the points are deviated a bit from theoretical points which are aligned in parallel line. So, there are might be some errors were occurs when conducting the experiment.
One of the causes of error is because ethanol and toluene are volatile liquids which can evaporate easily and so they will vaporize as it is left longer and exposed to the air. This caused the measured volume is less than the actual one as some of them already evaporated and thus affected the volume of water needed for titration. Secondly, the cloudiness was hard to be judged because there was no specific range of degree of cloudiness in each of the experiment .This might affect the volume of water added to the system and hence causes deviation of the final result. Next, the eyes of the observer is not perpendicular to the meniscus level of the apparatus so parallax error happened. This caused inaccurate reading or measurement of liquids and thus affecting the curve.
Thus to overcome this problem, a few precaution must be taken when doing this experiment to obtain the good result .First of all, the mixtures of ethanol and toluene must be closed immediately with rubber stopper when poured to the conical flask to avoid the solution to evaporated. The eye of the observer must be perpendicular to the meniscus of the liquids to avoid parallax error to obtain accurate volume of liquids. Besides that, we have to the same student to observe the cloudiness throughout the experiment so that the results will be more accurate.
QUESTIONS:
1. Does the mixture containing 70% ethanol,
20% water and 10% toluene (volume) appear clear or does it form 2 layers?
The mixture
appear as clear solution
2.
What will happen if you dilute 1 part of
the mixture with 4 parts of (a) water (b) toluene (c) ethanol?
(a) water: two
phases are formed
(b) toluene: two
phases are formed
(c) ethanol: the
mixture remains clear
CONCLUSION:
Phase diagram for ternary systems which contain ethanol/toluene/water is represented using a triangle diagram. The real curve was determined in this experiment. Water and toluene form a two-phase system because they are only slightly miscible. Ethanol is completely miscible with both toluene and water. The rule of the triangle fully explained the three-component system. However, we obtained
CONCLUSION:
Phase diagram for ternary systems which contain ethanol/toluene/water is represented using a triangle diagram. The real curve was determined in this experiment. Water and toluene form a two-phase system because they are only slightly miscible. Ethanol is completely miscible with both toluene and water. The rule of the triangle fully explained the three-component system. However, we obtained
REFERENCE:
1) http://pubs.acs.org/doi/abs/10.1021/j150093a005
2) http://chemwiki.ucdavis.edu/Physical_Chemistry/Physical_Properties_of_Matter/Phases_of_Matter/Phase_Transitions/Phase_Diagrams
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