Phenol-Water System

  1. A.   PURPOSE
  2. Getting a composition curve of phenol – water system agains temperature at constant pressure
  3. Determining the critical temperature of mutual solubility of phenol – water system

 

  1. B.   BASIC THEORY

Phenol- water system is a system that shows the nature of the mutual solubility between phenol and water at a certain temperature and fixed pressure. Called as the number of components of the binary system consisting of a mixture of the two substances are phenol and water. Phenol and water solubility will change when the mixture was added to one of the constituent components of the phenol or water. If the composition of phenol – water mixture is depicted on the temperature will be obtained by a curved :

               
   
 
         
 
   

T0

 

 

 

 

 T

 

T0

 

              t T

                             

                         
   
     
 
     
 
     
       
 
 
 
   

T0

 

 
     

 

 

 

 

 

 

 

 

 

 

 

 

 

L1 is a phenol in water, L2 is the water in phenol, XA and XF are each mole fraction mole of water and phenol, XC is the mole fraction of the components in the critical temperature (TC).

This system has a critical temperature (TC) at a constant pressure is the minimum temperature at the time of the two substances are homogeneously mixed with the composition of the CC. At a temperature of  T1 with a composition between the A1 and B1 or a temperature of T2 with a composition at a temperature between A2 and B2, the system is in two phases (cloudy). Meanwhile, outside the curve (or above the critical temperature, TC), the system is in one phase (clear).

 

  1. C.   EQUIPMENTS AND SUBSTANCE USED
  2. Reaction tube with diametre 4 cm            1 pc
  3. Tube stopper                                             1 pc
  4. Stirrer                                                        1 pc
  5. Beaker glass 400 mL                                1 pc
  6. Tripod and screen                                     1 set
  7. Burner                                                       1 set
  8. Buret 50 mL                                              1 pc

 

  1. Phenol
  2. Aquades

 

  1. D.   PROCEDURE

Weight again until phenol  4 .046 gr

 

Arange the experiment equipment

 

Weight tube

 

                                          fill it                                             fill burete with  

                                      with phenol                                        aquadest

 

               
     
 
 

Repeat the step until 12 data

 

 

Heat this mixture in waterbath, record T1(cloudy to clear), record T2 (cloudy appears again)

 

 

Fill burette with aquades

 

 
 

 

 

 

            If it does not                                 If the solution

                                    turbid                                            is turbid

 

 

 

 

 

 

  1. E.    RESULT AND DISCUSSION

Room temperature: 26oC

Phenol used: 99 % gram/mol

  1. The addition of distilled water until the turbidity occurs first

No.

Aquades (mL)

Observation

T1 (oC)

T2 (oC)

Taverage

1

4

cloudy

60

39

49,5

 

  1. The addition of distilled water after the turbidity

No.

Aquades (mL)

Mass (grams)

Temperature  (oC)

Phenol

Water

T1

T2

Taverage

1

0,1

4,0671

4,1

60

41

50,5

2

0,2

4,0671

4,2

60

45

52,5

3

0,3

4,0671

4,3

61

46,5

53,8

4

0,4

4,0671

4,4

61

47

54

5

0,5

4,0671

4,5

61

48

54,5

6

0,6

4,0671

4,6

61,5

49

55,3

7

0,7

4,0671

4,7

62

51

56,5

8

0,8

4,0671

4,8

61

53

57

9

0,9

4,0671

4,9

60,5

54

57,3

10

1,0

4,0671

5,0

60

52

56

11

1,1

4,0671

5,1

59

48

53,5

12

1,2

4,0671

5,2

59

47

53

 

 

 

 

 

 

 

 

 

 

 

 

This experiment will prove the binary system phenol-water solubility. Phenol and water its dissolve will change when it is added to the mix with one of the constituent components of the phenol and water. Change the color of the solution becomes clear and cloudy from clear to cloudy substance indicates that solubility changes are affected by temperature changes. In this experiment the water component being added and the amount of phenols fixed so that the solution changes from clear to cloudy or otherwise occurred in a changing climate. Changes in temperature depends on the composition or mole fractions of the two substances.
            Of data between the temperature (T) and mole fractions obtained from the experiment can be graphed binary system phenol – water, between the mole fraction vs temperature (T). Graphics should be formed where the parabolic peak is reached at the critical temperature when having a mole fraction of a particular component. In the experiment the critical temperature is 57.3 º C with a composition of the mixture is the mole fraction of phenol 0,136 and 0,864 mole fraction of water. It shows that at a temperature of 57,3 º C, the components inside the system is a two-phase curve and the curve components outside or beyond the critical point of the phase components of a system.

 The components are in a single phase when mixed with soluble homogeneous (clear), while the two components are in phase when the addition of water to produce two layers (cloudy). Graphics are formed in this experiment is less than perfect because of its curves are not symmetrical and more dominant on the left. At least it tends to form a parabolic curve. This curve is the curve of solubility of phenol in water and showed no mutual solubility of phenol to water.

Shape of the curve obtained less according to the theory, this might be due to the following :

  1. less conscientious practitioner at trial, for example, when reading a thermometer or when adding water from the burette scale reading quite right
  2. the addition of excess water that is too little
  3. the validity of the tools used
  4. error analysis of the data
  5. Insufficiently accurate determine temperature upon changed solution from cloudy to clear and from clear to cloudy back because change of from cloudy to clear is short in time
  6. Fault when phenol weighing, because phenol   oxidize so easy volatile

 

  1. F.    TASK
  2. Write the chemical formula of phenol and the molecular mass (Mr)!

Answer :

Phenol has the chemical formula C6H5OH,

The molecular mass is :

12(6) + 6(1) + 1(16) = 72 + 6 + 16 = 94

 

The formula is as follows :

           OH   

 
   

 

 

 

 

 

  1. If phenol content used levels of 95% (w/w) and a weighed mass of phenol is 5,140 grams, calculate the number of moles of phenol!

Answer :

mass of phenol :  x 5,140 = 4,883 grams

moles of phenol :  = =  0,052 moles

  1. Briefly, what is meant by phase ? Is there any difference with the form ?

Answer :

Phase is homogeneous part of a substance that can be separated by mechanical and uniform in physical and chemical properties, while a form is a form of substance at a particular temperature. Substances at different temperatures may have different manifestations. Suppose the water at a temperature of -10 º C its form solid, while at a temperature of 10 º C molten his form.

 

  1. G.   ANSWER QUESTIONS
  2. What is the composition of a mixture of phenol and water in % (w/w) at the critical temperature of the solution ?

Answer :

Mass of phenol = 4,026 g ; mole fraction of phenol = 0,136

Mass of water = 4,9 g    ; mole fraction of water = 0,864

Composition of a mixture %

Phenol =  x 100 % = 45,104%

 

Water =  x 100% = 54, 896%

 

  1. What is the composition of a mixture of phenol and water in units of mole fraction at 50oC in which the system is in single phase and two phase ?

Answer :

The composition of a mixture at temperature 50oC (at average tmperature is 50,5oC)

Xf   = 0,841

X= 0,159

Two phase at temperature : 57,3oC

Single phase at temperature : 57,3oC

 

  1. H.   CONCLUSSION AND SUGGESTION
  2. Conclussion
    1. Binary system composition curve phenol-water can be made by measures phenol mixture temperature water upon formed two phase (solution becomes cloudy while T1) and while changed as single phase (solution becomes unclouded while T2) then accounted by its average temperature.
    2. It critical temperature gets to be seen by curve which is while reach its top point.
    3. Binaries system  solubility phenol – water have critical temperature 57,3 ºC.
    4. On critical temperature mole fraction of phenol 0,136 and mole fractions of water 0,864.
    5. Phenol binary system of phenol – water shows properties of solubility  among phenol and water on given temperature and constant pressure.
    6. a mixture of phenol and water gets mutually dissolve, one that its amount a lot of as dissolving as conversely.
    7. Suggestion
      1. Praktikan should be accurate in observation.
      2. Praktikan must be accurate in especially while determine early its appearance  cloudy and  change temperature turbid and clear.

 

 

 

 

 

 

 

 

 

 

 

 

 

 

  1. I.      REFERENCE

Wahyuni, Sri.2003.Buku Ajar Kimia Fisika 2.Semarang:UNNES.

Wahyuni, Sri, and Team Lecturer of Physical Chemistry.2011.Practicum Guide of Physical Chemistry.Semarang:UNNES.

  Rohman, Ijang.2004.Kimia Fisika I. Bandung:UPI.

 

 

                                                                       

                                                                                                                 Semarang, 18  September 2012

            Praktikan,                                                                                Dosen pengampu,

 

 

           Falasifah Aulia                                                                         Sri wahyuni

 

APPENDIX

 

 

  1. Mole of phenol

Mass of phenol = 4,0671 grams

% phenol = 99%

Mr phenol = 94

Mr water = 18

Density water = 1 g/mL

Mass of phenol = 99% x 4,0671 = 4,026 grams

Mole of phenol =  = 0,043 mole

 

  1. Mole of water

ᵨ = 1g/mL

mwater = 4,1

nw = 0,228

mwater = 4,2

nw = 0,233

mwater = 4,3

nw = 0,239

mwater = 4,4

      nw = 0,244

mwater = 4,5

      nw = 0,25

mwater = 4,6

      nw = 0,256

mwater  = 4,7

      nw = 0,261

mwater = 4,8

      nw = 0,267

mwater = 4,9

      nw = 0,272

mwater = 5,0

      nw = 0,278

mwater = 5,1

      nw = 0,283

mwater = 5,2

      nw = 0,289

  1. Mole fraction of phenol
    1. Xf1   = 0,159
    2. Xf2   = 0,155
    3. Xf3   = 0,153
    4. Xf4   = 0,149
    5. Xf5   = 0,147
    6. Xf6   = 0,144
    7. Xf7   = 0,141
    8. Xf8   = 0,139
    9. Xf9   = 0,136
    10. Xf10 = 0,134
    11. Xf11 = 0,133
    12. Xf12 = 0,129
    13. Mole fraction of water
      1. Xa1   = 1 – 0,159  = 0,841
      2. Xa2   = 1 – 0,155 = 0,845
      3. Xa3   = 1 – 0,153  = 0,847
      4. Xa4   = 1 – 0,149  = 0,851
      5. Xa5   = 1 – 0,147  = 0,853
      6. Xa6   = 1 – 0,144  = 0,856
      7. Xa7   = 1 – 0,141  = 0,859
      8. Xa8   = 1 – 0,139  = 0,861
      9. Xa9   = 1 – 0,136  = 0,864
      10. Xa10 = 1 – 0,134  = 0,866
      11. Xa11 = 1 – 0,133  = 0,868
      12. Xa12 = 1 – 0,129  = 0,871
      13. Mass percent of phenol

% W=  

ρ water = 1 g/mL

a.    % WP1 = 49.79 %

b.    % WP2 = 49.19 %

c.    % WP3 = 48.61 %

d.    % WP4 = 48.03 %

e.    % WP5 = 47.47 %

f.     % WP6 = 46.92 %

g.    % WP7 = 46.39 %

h.    % WP8 = 45.86%

i.      % WP9 = 45.35 %

j.      % WP10 = 44.85 %

k.    % WP11 = 44.36 %

l.      % WP12 = 43.88 %

 

6.    Mass percent of water

a.    % WW1 = 100 % – % WP1 = 100 % – 49.79 % = 50.21 %

b.    % WW2 = 100 % – % WP2 = 100 % – 49.19 % = 50.81 %

c.    % WW3 = 100 % – % WP3 = 100 % – 48.61 % = 51.39 %

d.    % WW4 = 100 % – % WP4 = 100 % – 48.03 % = 51.97 %

e.    % WW5 = 100 % – % WP5 = 100 % – 47.47 % = 52.53 %

f.     % WW6 = 100 % – % WP6 = 100 % – 46.92 % = 53.08 %

g.    % WW7 = 100 % – % WP7 = 100 % – 46.39 % = 53.61 %

h.    % WW8 = 100 % – % WP8 = 100 % – 45.86 % = 54.14 %

i.      % WW9 = 100 % – % WP9 = 100 % – 45.35 % = 54.65 %

j.      % WW10 = 100 % – % WP10 = 100 % – 44.85 % = 55.15 %

k.    % WW11 = 100 % – % WP11 = 100 % – 44.36 % = 55.64 %

l.      % WW12 = 100 % – % WP12 = 100 % – 43.88 % = 56.12 %

 

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