Thursday, December 1, 2016

Practical 3: Phase Diagram (Part A)

TITLE :

Determination of Phase Diagram for Ethanol/Toluene/Water System Theory (Three-Component Systems)

OBJECTIVE :

To identify the miscibility of the mixtures of three components which are Ethanol, Toluene and Water.

DATE OF EXPERIMENT :

7 November 2016


INTRODUCTION :


       Figure above is known as the triangular or ternary phase diagram at which three components are involved in a system. The phase diagram states the coordinate of compositions of the three components with temperature and pressure as the fixed variables. Each apex of the triangle: A, B and C represents 100% pure for a component and the same apex will represent 0% for other two components. Meanwhile, each side of the triangle represents two-component mixtures such that side AB contains only mixture of A and B and so on while the area within the triangle itself consists of the three-component system. Any line parallel to a side of the triangular phase diagram shows constant percentage of a component. For example, any point lies on the line CD in the figure above shows constant percentage of 30% B with varying composition of A and C.

       The addition of third component to a pair of miscbile liquids can change their mutual solubility such that the third component may increase or decrease the mutual solubility. Further increasing the mutual solubility of the liquid pair may result in what we called homogenous mixture, a single-phase system where three liquids are miscible together. Thus, the determination of compositions for each component by using triangular phase diagram is important.

APPARATUS :

Measuring cylinder, beaker, burette, conical flask, retort stand and clamp.

MATERIALS :

Ethanol, toluene and distilled water.

PROCEDURE :

1. 20 mL mixtures of ethanol and toluene containing the following percentages of ethanol are prepared. (10%, 25%, 35%, 50%, 65%, 75%, 90% and 95% of ethanol)
2. Each mixture is then titrated with water until cloudiness is observed due to the existence of second phase. Little water is added and the mixture is shaken well after each addition.
3. The room temperature is recorded.
4. The percentage based on the volume of each component when the second phase starts to appear is calculated.
5. Points of percentage are plotted onto a triangular paper to give a triple phase diagram at the recorded temperature.
6. The experiment is repeated and the average volume is calculated.



RESULTS :

% of ethanol
(v/v)
Volume of ethanol (mL)
Volume of toluene (mL)
Titration of water I (mL)
Titration of water II (mL)
Average titration (mL)
10
2
18
1.7
1.0
1.35
25
5
15
0.7
1.0
0.85
35
7
13
0.7
0.7
0.70
50
10
10
1.4
1.3
1.35
65
13
7
2.5
2.3
2.40
75
15
5
4.5
4.2
4.35
90
18
2
10.4
10.9
10.65
95
19
1
16.1
22.9
19.50

Total volume of mixture (mL)
Composition of Ethanol
Composition of Toluene
Composition of Water
Volume (mL)
%
Volume (mL)
%
Volume (mL)
%
21.35
2
9.37
18
84.31
1.35
6.32
20.85
5
23.98
15
71.94
0.85
4.08
20.7
7
33.82
13
62.80
0.70
3.38
21.35
10
46.84
10
46.84
1.35
6.32
22.4
13
58.04
7
31.25
2.40
10.71
24.35
15
61.61
5
20.53
4.35
17.86
30.65
18
58.72
2
6.53
10.65
34.75
39.5
19
48.10
1
2.53
19.50
49.37




DISCUSSION :

      Based on the experiment, ternary phase diagram contains three component which are water, ethanol and toluence. A phase diagram shows the phase existing in equilibrium at any given condition. According to the phase rule, a maximum of four intensive variables such as pressure, temperature and the concentration of two of the three component must be specified to completely defined the state of a three component. In systems containing three component may possess four degrees of freedom.


F=C-P+2
=3-1+2
=4
F= degree of freedom, C= component, P= phase

       A simplification is achieved by choosing one of the intensive variables as a parameter in this experiment. If concentration is chosen as a parameter the condition of equilibrium can be represented by using a triangular prisms each corresponding to a different concentration that shown in the diagram above.This graph accounts for the fact that only two variables are required.Along the phase boundary only one variable is required.


        From the diagram, each corner represent 100% of ethanol, 100% of toluence and 100% of water while other two component are 0%. YZ line represent the water-toluence system, YX line represent water-ethanol system and XZ line represent ethanol-toluence system.In going along a line bounding the triangle so as to represent the concentration in two-component system, it does not matter whether we proceed in clockwise or counterclockwise direction around the triangle. The more usual convention is clockwise and has been adopted here.Hence, as we move along YX in the direction of X we are signifying system of water and ethanol containing increasing of ethanol and correspondingly the smaller amount of water.Moving along the XZ toward Z will represent system of X and Z. At the same time, the concentration of ethanol are decreasing toward the Z system while concentration of toluence is higher. The closer we approach toluence on the line of YZ, the greater will be the concentration of water in system of Y and Z.The area within the triangle represent all the possible combination of water, ethanol and toluence to give three component system.For example, the conical flask A, the system consist of 6.32% of water, 84.31% of toluence and 9.37% of ethanol.If a line is drawn through any apex to a point on the opposite side, then all the systems represented by points on such a line have a constant ratio of two components. Any line drawn parallel to one side of the triangle represent ternary system in which the proportion (or precent by weight)of one component is constant . The curve of the plotted graph is termed as binomial curve. The region bounded by the curve shows the presence of two phases. The mixture within this region is cloudy in appearance due to phase separation as amount of ethanol are not sufficient for a homogenous solution.

        Based on the result obtained, when there is a higher percentage of ethanol compared to percentage of toluence in the mixture, the volume of water needed to titrate the mixture until cloudiness is observed is higher. This proves that the ethanol has increased the miscibility of the other two component and more water is needed to break homogeneity.

        There are some errors made during the experiment. One of the errors is due to the degree of cloudiness. Different person did the observation each time will cause the different degree of cloudiness of mixture achieved. Another error is parallax error. Our eyes must be perpendicular to the scale of measurement. This cause inaccurate measurement of liquid. Ethanol and toluence are very volatile liquids. Their volume may be less than actual needed value as some of them have been evaporated. Avoid expose ethanol and toluence with air for too long. This is because to prevent it vaporize to the surrounding. The contaminated apparatus may affect the result as well. The glass wares were not completely dried after washing will cause slight dilution of the mixture. rinse the the glassware before use to remove impurities as the impurities may affect the result.

QUESTIONS & ANSWERS :

Questions & Answers
1.       Does the mixture containing 70% ethanol, 20% water and 10% toluene (volume) appear clear or does it form two layers?
-          The mixture appears clear which is 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?

-          Ethanol   :     70/100 x 1 = 0.7 

-          Water      :     20/100 x 1 = 0.2

-          Toluene   :    10/100 x 1 = 0.1

 (a)    Add 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%

-          Two liquid phase will form and mixture become cloudy.

(b)     Add 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%

-          A single liquid phase is formed.

(c)    Add 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%


-          A clear single liquid phase is formed.

CONCLUSION :

       In conclusion, ethanol,toluence and water system is a ternary system with one pair of partially miscible( toluence and water). The addition of sufficient amount of ethanol to the toluence water-system would produce a single liquid phase in which all the three component are miscible and the mixture is homogenous. As the number of volume of ethanol by percentage increase and number of volume of toluence by percentage decrease, the volume of water will increase. The two phase system only shown if the water added exceed the theoritical percentage where the three component are partially miscible. Data also show that when percentage of toluence is higher compared to ethanol, only small amount of water needed to break the homogenity.

REFERENCES :

1. Alexander T.& David A., 2006, Physicochemical Principles of Pharmacy Fourth Edition [pg. 221], Pharmaceutical Press, London, USA.

2. Ternary Phase Diagrams, retrieved from https://brocku.ca/earthsciences/people/gfinn/petrology/ternary2.htm

3. 2012, Ternary Phase Diagrams, ASM International, retrieved from http://www.asminternational.org/documents/10192/1849770/05342G_Sample_BuyNow.pdf/230f1568-32a2-4669-a03d-b33278ca8825

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