TITLE
Determination of Diffusion Coefficient
DATE
OF EXPERIMENT
20th 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
- 7g of agar powder is mixed with 420 ml of Ringer solution. The solution is stirred and heated on a hot plate.
- Then, the agar solution is poured into six test tubes with 15 ml of agar solution in each test tube.
- Then, the test tubes is left in the fridge to be cool.
- 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.
- After the agar solution solidify, 5 ml of 1:200, 1:400, 1:600 of crystal violet are poured into the test tubes respectively.
- Three test tubes are stored at temperature 28 ºC and another three test tubes are stored at 37 ºC water bath.
- 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.
- This value is assigned as x in meter and the average is obtained from several measurements.
- The values of x are recorded after 2 hours and at suitable time distances up till one week.
- The graph for values of x2 (in M2) against time ( in seconds) for each of the concentration used is plotted.
- The diffusion coefficient, D from the gradient of the graph at temperature 28 ºC and 37 ºC are calculated.
- The molecular weight of the crystal violet is calculated using the equation N and V.
- Step 4 until step 12 is repeated by using bromothymol blue.
RESULTS
1 Crystal violet
at 28oC
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 37oC
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 28oC
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 37oC
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 m2 = 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
D28oC =
T28oC
D37oC T37oC
Where
n1 and n2 is the viscosity of water at temperature 28oC and 37oC
2.21×10-10
m2s-1 = 28+273
D37oC 37+273
6.42×10-9
m2s-1 = 301
D37oC 310
D37oC = 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
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.
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