Saturday 13 December 2014

Practical 3: Phase Diagram Part A

Date

3rd November 2014

Objective

To determine the phase diagram for Ethanol/ Toluene/ Water System

Introduction

In this system, there are 3 components but only 1 phase exists. Thus, by using the  phase rule, we can find out that the degree of freedom is 4 which are temperature, pressure, and concentrations of the two of the three components.

At constant temperature and pressure, the composition of a ternary system can be presented in the form of triangular coordinates.

Each peak of the triangle will represent 100% of different component respectively. Each side of the triangle represents a binary mixture where two components mixed together and any line drawn parallel to one of the sides shows the percentage of a particular component. For example, line FG means there are 50% of B. the intercept point K, showing that there are combination of 20% of A, 50% of B and 30% of C.

 Addition of the third component into one pair of miscible liquids can change their solubility. If this third component is more soluble in either one from the two components, the solubility of both components will reduce. But if the third component is soluble in both components at the same time, the solubility increases. Thus, when ethanol is added into a mixture of benzene and water, the solubility of  these two components will increase until a point is reached, where the mixture become homogeneous. This application can be used in formulations of solutions. Examples of three-component liquid system that have been tested, are castor oil or alcohol or water; peppermint oil or polypropylene, glycol or water; peppermint oil or polyethylene glycol or water.

Apparatus

Conical flask

 retort stand and clamp

 burette

 pipette

 measuring cylinder

 test tube with rack

 dropper


Chemicals

Ethanol

 toluene

distilled water


Procedure

1. The mixtures of ethanol and toluene were prepared in a sealed containers measuring 100cm3 containing the following percentages of ethanol: 10, 25, 35,50, 65, 75, 90 and 95 (percent,%)


2. 20ml of each mixture was prepared by filling a certain volume accurately by using a burette.
3. A burette was filled with distilled water
4. Each mixture was titrated with water until cloudiness was observed as there was an existence of a second phase.

5. A little of water was added and shaken well after each addition.
6. Room temperature was measured.
7. The volume of the water used was recorded.
8. Steps 1-6 were repeated for a second titration.

9. The volume of water required for complete titration of each mixture was recorded.
10. Average volume of water used was calculated.

11. The percentage based on the volume of each component when a second phase starts to appear or separate was calculated.
12. The points are plotted onto a triangular paper to give a triple phase diagram at the recorded temperature.
13. A few more measurements are done if necessary.


Results


      Test     tube
%Ethanol

     (v/v)
Total volume
(average v + 20 ml)
           Ethanol
          Toluene
            Water
Volume
    (ml)
Percentage volume (%)
Volume
   (ml)
Percentage
volume(%)
Volume
   (ml)
Percentage volume (%)
A
      10
   22.9
  18.0
     78.6
     2.0
     8.7
    2.9
   12.7
   B
      25
   21.0
  15.0
     71.4
     5.0
     23.8
    1.0
    4.8
C
      35
   21.2
  13.0
     61.3
     7.0
    33.0
    1.2
    5.7
   D
      50
   22.5
  10.0
     44.4
   10.0
    44.4
    2.5
   11.1
   E
      65
   23.0
   7.0
     30.4
   13.0
    56.5
    3.0
   13.0
   F
      75
   24.3
   5.0
     20.6
   15.0
     61.7
    4.3
   17.7
  G
      90
   28.6
   2.0
      7.0
   18.0
     62.9
    8.6
   30.1
  H
      95
   34.8
   1.0
      2.9
   19.0
     54.6
  14.8
   42.5



Discussion

The ethanol / toluene / water system in this experiment involves adding water (as a third component) into a miscible mixture of ethanol and toluene. Toluene is soluble in ethanol while it is not soluble with 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. This is known as solvent effect. However, when the ethanol, toluene and water are mixed together, only a partial miscibility can be reached.

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).

For ethanol/toluene/water system, we have 3 components and 1 liquid phase. F= 3-1+2= 4. Hence, 4 degrees of freedom are required. They are temperature, pressure, and the concentrations of two of the three components. Only concentrations of two components are required because concentration of the third component can be obtained by further calculation. The experiment was carried out at a constant temperature and pressure. The temperature when this experiment is carried out is 27°C.

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 appearance of cloudiness indicates that a two-phase system is created. This proved that the ethanol had increased the miscibility of the other two components and more water is needed to break the homogeneity.

A second phase will only separate out when the proportion of water added exceed the theoretical percentage where three components are partially miscible. As for the lower percentage of ethanol compared to toluene, the volume of water needed to titrate the mixture until cloudiness is observed is lower. Having higher percentage of toluene in the mixture than ethanol, little water is needed to disrupt the homogeneity.

On the other hand, the triple phase diagram showed the concentrations of the three components in the system. The curve drawn in the triangular phase diagram is normally known as binomial curve or binomial. Normally, the area bounded by the binomial curve indicates a two-phase region while the area unbound by the curve is a single-phase region. However, from the triple phase diagram above, the binomial curve is incomplete and a tie line is not obtained.

In our experiment, one of the errors is the random error due to different person doing the observation. Different degree of cloudiness was achieved. Hence, the volume of water added to the solution was not accurate enough. The values obtained were varied. The percentage by volume and the curve was affected. Besides, the conical flask must be shaked well after each addition of water. The purity of the liquids may affect the result obtained. Some water might be added throughout the experiment into the apparatus when the apparatus used is not completely dry after washing. This may result in slight dilution of the mixture.

Practice

1. Does the mixture containing the 70% ethanol, 20% water, and 10% toluene (volume) appear clear or does it form two layers.

=The mixture appears clear and 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?

     Ethanol: 1 part mixture x 70% ethanol = 1 x 70/100 = 0.7 part of ethanol
Water:    1 part mixture x 20% water = 1 x 20/100 = 0.2 part of water
Toluene: 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 under the area of the binodal curve. Hence, a 2 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. Hence, 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. Hence, a clear single liquid phase of solution is formed.

Conclusion

Phase diagram for ternary systems-ethanol?toluene?water system is represented using a triangle. As having higher percentage of toluene in the mixture than ethanol, little water is needed to disrupt the homogeneity, water and toluene will form two-phase system because they are only slightly miscible. As 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, ethanol and water will miscible with both toluene and water.

References

1. http://www.facebook.com/l.php?u=http%3A%2F%2Fchemed.chem.purdue.edu%2
Fgenchem%2Ftopicreview%2Fbp%2Fch14%2Fphase.php&h=1AQFdqXmK

2. http://www.chemguide.co.uk/physical/phaseeqia/phasediags.html

3. http://www.csun.edu/~jeloranta/ CHEM355L/experiment5 .pdf

4. Martin's Physical Pharmacy and Pharmacautical Science, Sixth Edition, Patrick J. Sinko,       Wolters Kluwer, Lippincott Williams & Wilkins.

5. m.youtube.com/watch?v=gGYHXhcKM5s

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