Experiment 2 - Phase Diagram (Part B)

TITLE

Phase Diagram ( Mutual Solubility For Phenol and Water )

DATE

17^th April 2014

OBJECTIVES

1. To determine the solubility of two partially liquid (phenol and water)
2. To construct a mutual solubility for the pair
3. To determine their critical solution temperature

INTRODUCTION

A few liquids are miscible with each other in all proportions.  An example is phenol and water.  Phenol and water can exists in one liquid phase as well as two liquid phase.  Theoretically, the combination of phenol and water above the the temperature 66.8 degree Celsius are completely miscible and yield only one phase.The maximum temperature at which two liquid phase can exist is termed as the critical temperature or known as the upper consolute temperature.  The composition for two layers of liquid in equilibrium state is constant and does not depend on the relative amount of these two phases at any temperature below the critical solution temperature.  The presence of third component will influence the mutual solubility for a pair of partially miscible liquid. 

APPARATUS AND MATERIALS

Phenol

Distilled water

Test tube

Test tube rack

Measuring cylinder

Thermometer

 

Water bath

Dropper

PROCEDURES

1. A tightly sealed tubes containing amounts amounts of phenol and water are given to produce a phenol concentration scale between 8% to 80%.
2. 5 test tubes are labeled as 1, 2, 3, 4 and 5 respectively.
3. Test tube 1 is filled with 8% of phenol, test tube 2 with 11% of phenol, test tube 3 with 37% of phenol, test tube 4 with 63% of phenol and test tube 5 with 82% of phenol.
4. Water is added into all the 5 test tubes until a volume of 10ml is reached.
5. The test tubes are heated. The test tubes are shaken as well.
6. The temperature for each of the test tube at which the turbid liquid becomes clear is observed and recorded.
7. The test tubes are removed from hot water and allowed to cool.
8. The temperature at which the liquid becomes turbid and two layers are separated is recorded.
9. The average temperature for each tube at which two phases are no longer seen or at which two phases exist is determined.

RESULTS

QUESTIONS

1.   Plot the graph of phenol composition (horizontal axis) in the different mixture against temperature at complete miscibility. Determine the critical solution temperatures.

The critical solution temperature is 68.5ºC.

2.      Discuss the diagram with references to phase rule.

The graph above is a two component system containing liquid phases. This system exhibit partial miscibility, it will exist as single or two phases depending on few variables. In order to know and identify the variable affecting the miscibility, phase rule is used. Upon calculating using phase rule (a rule relating the number of variables upon phases that can exist in equilibrium containing a given number of components), it is identified that the degree of freedom are two as F=2+2-2=2. The two degrees of freedom are temperature and percentage of phenol in water. Thus, only these two variables are required to plot the graph of phenol-water system.

From the graph above, increasing composition of phenol in water will decrease the miscibility of phenol    in water as a higher temperature is required to become single phase. However, once the composition of phenol in water exceeds 37%, the miscibility will be increase in increasing composition of phenol in water as lower temperature is needed to make it appears as single phase.

3.      Explain the effects of adding foreign substances and show the importance of this effect in pharmacy.

When foreign material is being added to a binary system, it will results in ternary system. We have to know the solubility of the added material. If the added material is soluble only in one component or the solubility of both liquids is different, the mutual solubility (miscibility) of the two liquids component will decrease. These will results in either lowering of the lower consolute temperature or increasing in the upper consolute temperature.

If the added material is soluble in both liquids component, the mutual solubility of (miscibility) of the two liquids components will increase. This will causes the lowering of the upper consolute temperature or the increasing in lower consolute temperature. This process is called blinding.This effect is important as it will help in extraction of material

DISCUSSIONS

In this experiment, we are going to discuss two-component systems containing liquid phases. A very good example of this two-component system is phenol and water. These liquids are partially miscible in each other. The curve plotted in the graph temperature versus percentage of phenol in water in volume per volume shows the limits of temperature and concentration within which two liquid phases exists in equilibrium. The region outside this curve contains only one liquid phase.

            For phenol/water system, it is a two-component system containing liquid phases. Thus, the degrees of freedom are two as F=2-2+2=2. The two degrees of freedom are temperature and percentage of phenol in water in volume. At 5% of phenol in water at 25ºC, single liquid phase is produced. This is due to the less percentage of phenol in water and it is miscible with water completely.From the graph plotted, at 51ºC, a minute amount of a second phase appears. The concentration of phenol and water at which this occurs is 11% by weight of phenol in water. From the graph, we notice that the maximum point is at 68.5ºC. This is known as the upper consolute temperature or critical solution which is the maximum temperature at which the two-phase region exists. All combinations of phenol and water above this temperature are completely miscible. A line drawn across the region containing two phases is known as tie line. The tie line is always parallel to the base line in two-component systems. All systems prepared on the tie line, at equilibrium, will separate into phases of constant temperature. These phases are known as the conjugated phases. Tie line in a phase diagram use to calculate the composition of each phase and the weight of the phases.

            Certain precautions should be taken to acquire an accurate result. The film should be adhered on the mouth of conical flask by placing the thermometer in the middle of the mouth after the addition of phenol. This is because phenol is a volatile chemical. In addition, we have to be careful when handling phenol due to the acidity and carcinogenic properties of phenol. Pipette is used instead of measuring cylinder on order to obtain more accurate volume required.

CONCLUSIONS

The miscibility of phenol and water are depending on the temperature of the mixture and the composition of phenol in water. The critical solution temperature is at 68.5ºC, above it the mixture appears as single phase. The composition of phenol in which high temperature needed to make the mixture miscible is at 37% of phenol in water.

REFERENCES

Patrick J. Sinko, Lippincott Williams and Wilkins. Martin’s Physical Pharmacy and Pharmaceutical Sciences, 5th Edition.

Alexander T Florence and David Attwood. (2006). Physiocochemical Principles of Pharmacy. 4th Ed. Palgrave. USA.

Experiment 4 - Determination of Diffusion Coefficient

TITLE

Determination of Diffusion Coefficient

DATE OF EXPERIMENT

20^th MAY 2014

OBJECTIVE

To determine the Diffusion Coefficient, D

INTRODUCTION

 Diffusion is a process of mass transfer of molecules related to random molecular motion. The driving force for the the diffusion is the existence of a concentration gradient. Diffusion is a result of the kinetics properties of particles of matter. According to Fick’s law, the flowing of material with amount dm in time dt, through a given plane with area A is proportional to the concentration gradient dc/dx.

dm = -DA (dc/dx) dt  ---- (a)

D is the diffusion coefficient or diffusivity for the solute with unit m²s^-1 .

Diffusion coefficient is a useful parameter used to indicate diffusion mobility.

For a solution containing neutral particles with concentration M₀, the diffusion can be stated as below,

2.303 x 4D (log ₁₀ M₀ –log ₁₀ M) t = x  ---- (b)

A graph of X² against t is plotted with the slope 2.303 x 4D (log ₁₀ M₀ –log ₁₀ M). The diffusion coefficient is calculated from the slope.

If the solution contains charged particles instead of neutral particles, we need to modify equation (b) in order to include the potential gradient that exists between the solution and solvent. However, we can add a little sodium chloride into the solvent to prevent the formation of the potential gradient.

In this experiment, agar gels which contain a partially strong network of molecules are used to form a support medium.

MATERIALS

Agar powder

Ringer’s solution 250mL

Crystal violet ( 1:500000, 1:200, 1:400, 1:600 )

Bromothymol blue ( 1:500000, 1:200, 1:400, 1:600 )

APPARATUS

Beaker

Test tubes

Glass rod

Test tube rack

Measuring cylinder

Hot plate

PROCEDURES

1. 7g of agar powder is mixed with 420 ml of Ringer solution. The solution is stirred and heated on a hot plate.
2. Then, the agar solution is poured into six test tubes with 15 ml of agar solution in each test tube.
3. Then, the test tubes is left in the fridge to be cool.
4. Another test tube that has already been added with 1:500000 crystal violet is filled with agar. This test tube will be used as the standard to measure the color distance resulting from the crystal violet diffusion.
5. After the agar solution solidify, 5 ml of 1:200, 1:400, 1:600 of crystal violet are poured into the test tubes respectively.
6. Three test tubes are stored at temperature 28 ºC and another three test tubes are stored at 37 ºC water bath.
7. The distance between the interface of this gel solution with the end of the crystal violet area that has color equivalent to the standard is measured accurately.
8. This value is assigned as x in meter and the average is obtained from several measurements.
9. The values of x are recorded after 2 hours and at suitable time distances up till one week.
10. The graph for values of x² (in M²) against time ( in seconds) for each of the concentration used is plotted.
11. The diffusion coefficient, D from the gradient of the graph at temperature 28 ºC and 37 ºC are calculated.
12. The molecular weight of the crystal violet is calculated using the equation N and V.
13. Step 4 until step 12 is repeated by using bromothymol blue.

RESULTS

1    Crystal violet at 28^oC

System	Day/ time(s)	X(m)	X2(m2)
1:200	2 (172800)	0.017	2.89×10-4
	3 (259200)	0.022	4.84×10-4
	4 (345600)	0.035	1.23×10-3
	6 (518400)	0.048	2.30×10-3
	7 (604800)	0.055	3.03×10-3
	8 (691200)	0.062	3.84×10-3
	9 (777600)	0.075	5.63×10-3
1:400	2 (172800)	0.015	2.25×10-4
	3 (259200)	0.016	2.56×10-4
	4 (345600)	0.033	1.09×10-3
	6 (518400)	0.040	1.60×10-3
	7 (604800)	0.054	2.92×10-3
	8 (691200)	0.065	4.23×10-3
	9 (777600)	0.070	4.90×10-3
1:600	2 (172800)	0.012	1.44×10-4
	3 (259200)	0.015	2.25×10-4
	4 (345600)	0.025	6.25×10-4
	6 (518400)	0.030	9.00×10-4
	7 (604800)	0.035	1.25×10-3
	8 (691200)	0.042	1.76×10-3
	9 (777600)	0.050	2.50×10-3

2     Crystal violet at 37^oC

System	Day/ time(s)	X(m)	X2(m2)
1:200	2 (172800)	0.019	3.61×10-4
	3 (259200)	0.023	5.29×10-4
	4 (345600)	0.030	9.00×10-4
	6 (518400)	0.044	1.94×10-3
	7 (604800)	0.046	2.12×10-3
	8 (691200)	0.050	2.50×10-3
	9 (777600)	0.060	3.60×10-3
1:400	2 (172800)	0.015	2.25×10-4
	3 (259200)	0.019	3.61×10-4
	4 (345600)	0.024	5.76×10-4
	6 (518400)	0.037	1.37×10-3
	7 (604800)	0.038	1.44×10-3
	8 (691200)	0.040	1.60×10-3
	9 (777600)	0.045	2.03×10-3
1:600	2 (172800)	0.007	0.49×10-4
	3 (259200)	0.010	1.00×10-4
	4 (345600)	0.012	1.44×10-4
	6 (518400)	0.020	4.00×10-4
	7 (604800)	0.024	5.76×10-4
	8 (691200)	0.028	7.84×10-4
	9 (777600)	0.035	1.23×10-3

3     Bromothymol Blue at 28^oC

System	Day/ time(s)	X(m)	X2(m2)
1:200	2 (172800)	0.017	2.89×10-4
	3 (259200)	0.023	5.29×10-4
	4 (345600)	0.025	6.25×10-4
	6 (518400)	0.030	9.00×10-4
	7 (604800)	0.033	1.09×10-3
	8 (691200)	0.037	1.37×10-3
	9 (777600)	0.045	2.03×10-3
1:400	2 (172800)	0.016	2.56×10-4
	3 (259200)	0.019	3.16×10-4
	4 (345600)	0.022	4.84×10-4
	6 (518400)	0.025	6.25×10-4
	7 (604800)	0.029	8.41×10-4
	8 (691200)	0.030	9.00×10-4
	9 (777600)	0.032	1.02×10-3
1:600	2 (172800)	0.013	1.69×10-4
	3 (259200)	0.015	2.25×10-4
	4 (345600)	0.019	3.61×10-4
	6 (518400)	0.020	4.00×10-4
	7 (604800)	0.021	4.41×10-4
	8 (691200)	0.023	5.29×10-4
	9 (777600)	0.025	6.25×10-4

4      Bromothymol Blue at 37^oC

System	Day/ time(s)	X(m)	X2(m2)
1:200	2 (172800)	0.017	2.89×10-4
	3 (259200)	0.026	6.76×10-4
	4 (345600)	0.030	9.00×10-4
	6 (518400)	0.033	1.09×10-3
	7 (604800)	0.040	1.60×10-3
	8 (691200)	0.041	1.68×10-3
	9 (777600)	0.045	2.03×10-3
1:400	2 (172800)	0.013	1.69×10-4
	3 (259200)	0.017	2.89×10-4
	4 (345600)	0.022	4.84×10-4
	6 (518400)	0.024	5.76×10-4
	7 (604800)	0.030	9.00×10-4
	8 (691200)	0.035	1.23×10-3
	9 (777600)	0.040	1.60×10-3
1:600	2 (172800)	0.011	1.21×10-4
	3 (259200)	0.015	2.25×10-4
	4 (345600)	0.018	3.24×10-4
	6 (518400)	0.020	4.00×10-4
	7 (604800)	0.025	6.25×10-4
	8 (691200)	0.031	9.61×10-4
	9 (777600)	0.035	1.23×10-3

 Graphs and calculations :

Calculation for the gradients:

Remarks :  The value for time will not be days but in seconds

At concentration of 1:200,

Gradient = diffusion coefficient, D

 (38.40-23.00)×10^-4 m²   = 2.302× 4D(log₁₀ 1/200 – log₁₀ 1/500000)

  (691200-518400) s

15.40×10^-4 m² = 2.302× 4D(log₁₀ 1/200 – log₁₀ 1/500000)

 172800 s

 2.85×10^-10m²s^-1 = D

 At concentration of 1:400,

Gradient = diffusion coefficient ,D

  (16.00-10.89)×10^-4 m²  = 2.302× 4D(log₁₀ 1/400 – log₁₀ 1/500000)

   (518400-345600) s

   5.11×10^-4 m²    = 2.302× 4D(log₁₀ 1/400 – log₁₀ 1/500000)

   172800 s

 1.04×10^-10 m²s^-1 = D

At concentration of 1:600,

Gradient =  diffusion coefficient ,D

  (25.00-12.25)×10^-4 m²  = 2.302× 4D(log₁₀ 1/600 – log₁₀ 1/500000)

   (777600-604800) s

    12.75×10^-4 m²   = 2.302× 4D(log₁₀ 1/600 – log₁₀ 1/500000)

   172800 s

  2.74×10^-10 m²s^-1 =D

Therefore, average diffusion coefficient

= (2.85×10^-10 + 1.04×10^-10 + 2.74×10^-10) m²s^-1

                             3

= 2.21×10^-10 m²s^-1

Calculation for the gradients:

Remarks :  The value for time will not be days but in seconds

At concentration of 1:200,

Gradient = diffusion coefficient, D

   (19.36-5.29)×10^-4 m²     = 2.302× 4D(log₁₀ 1/200 – log₁₀ 1/500000)

   (518400-259200) s

    14.07×10^-4 m²   = 2.302× 4D(log₁₀ 1/200 – log₁₀ 1/500000)

    259200 s

   1.74×10^-10 m²s^-1 = D

 At concentration of 1:400,

Gradient = diffusion coefficient, D

  (13.69-3.61)×10^-4 m²    = 2.302× 4D(log₁₀ 1/400 – log₁₀ 1/500000)

   (518400-259200) s

   10.08×10^-4 m²  = 2.302× 4D(log₁₀ 1/400 – log₁₀ 1/500000)

   259200 s

  1.36×10^-10 m²s^-1 = D

At concentration of 1:600,

Gradient = diffusion coefficient, D

   (7.84-1.44)×10^-4 m²           = 2.302× 4D(log₁₀ 1/600 – log₁₀ 1/500000)

   (691200-345600) s

    6.40×10^-4 m²   = 2.302× 4D(log₁₀ 1/600 – log₁₀ 1/500000)

   345600 s

   6.88×10^-11 m²s^-1  = D

Therefore, average diffusion coefficient

= (1.74×10^-10 + 1.36×10^-10 + 6.88×10^-11) m²s^-1

                             3

= 1.26×10^-10 m²s^-1

Calculation for the gradients:

Remarks :  The value for time will not be days but in seconds

At concentration of 1:200,

Gradient = diffDiffusion coefficient, D

   (13.69-9.00)×10^-4 m^{2       =} 2.302× 4D(log₁₀ 1/200 – log₁₀ 1/500000)

   (691200-518400) s

   4.69×10^-4 m²       = 2.302× 4D(log₁₀ 1/200 – log₁₀ 1/500000)

   172800 s

   8.68×10^-11 m²s^-1  = D

 At concentration of 1:400,

Gradient = diffusion coefficient, D

   (4.84-2.56)×10^-4 m²           = 2.302× 4D(log₁₀ 1/400 – log₁₀ 1/500000)

   (345600-172800) s

   2.28×10^-4 m²       = 2.302× 4D(log₁₀ 1/400 – log₁₀ 1/500000)

   172800 s

   4.62×10^-11 m²s^-1 = D

At concentration of 1:600,

Gradient = diffusion coefficient, D

   (6.25-4.41)×10^-4 m²         = 2.302× 4D(log₁₀ 1/600 – log₁₀ 1/500000)

   (777600-604800) s

    1.84×10^-4 m²         = 2.302× 4D(log₁₀ 1/600 – log₁₀ 1/500000)

   172800 s

   3.94×10^-11 m²s^-1  = D

Therefore, average diffusion coefficient

= (8.68×10^-11 + 4.62×10^-11 + 3.94×10^-11) m²s^-1

                             3

= 5.75×10^-11 m²s^-1

Calculation for the gradients:

Remarks :  The value for time will not be days but in seconds

At concentration of 1:200,

Gradient = diffusion coefficient, D

  (16.00-10.89)×10^-4 m²  = 2.302× 4D(log₁₀ 1/200 – log₁₀ 1/500000)

   (604800-518400) s

   5.11×10^-4 m²   = 2.302× 4D(log₁₀ 1/200 – log₁₀ 1/500000)

   86400 s

  1.89×10^-10 m²s^-1 = D

 At concentration of 1:400,

Gradient = diffusion coefficient, D

   (12.25-9.00)×10^-4 m²      = 2.302× 4D(log₁₀ 1/400 – log₁₀ 1/500000)

   (691200-604800) s

    3.25×10^-4 m²   = 2.302× 4D(log₁₀ 1/400 – log₁₀ 1/500000)

    86400 s

   1.32×10^-10 m²s^-1  = D

At concentration of 1:600,

Gradient = diffusion coefficient, D

   (9.61-6.25)×10^-4 m²          = 2.302× 4D(log₁₀ 1/600 – log₁₀ 1/500000)

   (691200-604800) s

    3.36×10^-4 m²  = 2.302× 4D(log₁₀ 1/600 – log₁₀ 1/500000)

   86400 s

= 1.45×10^-10 m²s^-1

Therefore, average diffusion coefficient

= (1.89×10^-10 + 1.32×10^-10 + 1.45×10^-10) m²s^-1

                             3

= 1.55×10^-10 m²s^-1

Molecular weight of Crystal Viol

=4/3πa²Nr

From the experiment value for D₂₈, estimate the value of D₃₇ using the following equation

D₂₈^oC     =       T₂₈^oC    

D₃₇^oC              T₃₇^oC 

Where n1 and n2 is the viscosity of water at temperature 28^oC and 37^oC

2.21×10^-10 m²s^-1   =       28+273  

D₃₇^oC                             37+273

6.42×10^-9 m²s^-1   =       301  

D₃₇^oC                             310

D₃₇^oC  = 2.28×10^-10 m²s^-1  

DISCUSSIONS

Diffusion is a process of mass transfer of molecules brought by random molecular motion. It is associated with driving forces such as concentration gradient. It will continue to proceed until equilibrium is achieved This experiment has been carried out to determine the diffusion coefficient of the crystal violet and bromothymol blue. The manipulated variables in this experiment are the size of the particles, the concentration and the temperature. Factors such as viscosity and concentration of agar gel may affect the rate of diffusion too. When discussing about what factors are going to influence the diffusion rate, Strokes Einstein Law can be used to explain it.

D=kT/6 ทa,
where D = diffusion coefficient , T = temperature , ท = viscosity and a = particle size

From the equation above, we can see that D is directly proportional to the temperature, inversely proportional to the viscosity and particle size. However, in this experiment, we are eager to identify how temperature affecting diffusion coefficient. Theoretically, as the temperature increases, the diffusion rate will also increase as high energy will be provided to the molecules making it vibrate more vigorously and diffuse relatively faster than at low temperature. As we can see from this experiment, diffusion coefficient will be decreased as the particle size increase. blue. To get all this theory right, it must be ensure that no adsorption or chemical reaction occur. Other than that, it is important to ensure that its diffusion rate through the agar medium is almost equivalent to its diffusion rate in its own solution.

This experiment has proven that different particle size will have different effect towards diffusion coefficient. This is because the diffusion coefficient of bromothymol blue is lower than crystal violet when at 28oC. When comparing the particle size of bromothymol blue and crystal violet, it is found out that crystal violet is smaller than bromothymol blue. However at 37oC, bromothymol blue has slightly higher diffusion coefficient which shows that some errors might have occur when conducting this experiment.

As for the temperature the diffusion coefficient for bromothymol blue increases as the temperature increases. At 37oc it has a D value of 1.45X10^-10 while at 28oc it was only 5.75x10^-11. However again in crystal violet at 37oc it possess a slightly higher D value compared to in 28oC which is 1.26x10^-10 and 2.21x10^-10. This proves that an error might have occurred in crystal violet experiment.

Moving to the concentration, according to the Fick’s first law diffusion will occur from a region of high concentration to lower concentration. Thus, as the concentration is increase, the diffusion coefficient will also increase. Again we have problem with the crystal violet where it do not obey this law. The errors that have occur might be due to the different people who have been taking measurements. The measurement might be read differently and this leads in inconsistency in the reading. The colour of the dyes which are not that obvious also causes the measurement to be taken wrongly and make the process of taking measurement inaccurate and difficult. Some precautions need to be taken in which the level of the eye of the observer must be parallel in order to avoid parallax error. Lastly, the test tube which contain agar solution need to be immediately closed once the crystal violet solution has been added to avoid vaporization.

QUESTIONS

CONCLUSION

The diffusion coefficient of bromothymol blue higher than crystal violet at 28oC is 2.21x10^-10 m2s-1 meanwhile at 37oC is 1.26x10^-10 m2s-1.  As for bromothymol blue it is lower than crystal violet due to its bigger particle size. At 28oC, its diffusion coefficient is 5.75x10^-11 m2s-1 and at 37oC1.55×10-10 m2s-At higher temperature the diffusion coefficient of the particles is increase and so do those which in high concentration.

REFERENCES

Patrick J. Sinko, Lippincott Williams and Wilkins. Martin’s Physical Pharmacy and Pharmaceutical Sciences, 5th Edition.

Alexander T Florence and David Attwood. (2006). Physiocochemical Principles of Pharmacy. 4th Ed. Palgrave. USA.