Practical
4: Determination of diffusion coefficient
INTRODUCTION:
Diffusion is the movement of molecules from region with higher concentration gradient to lower concentration gradient, where it will continuosly happen until it reaches equilibrium.
Fick’s first law of diffusion states that the
rate of transfer of diffusing substance through unit area of a section is
proportional to the concentration gradient. This law is fot steady state
diffusion.
Where,
J=
flux
D=
diffusion coefficient
dC = concentration gradient
dx
C=
concentration
x=
distance of movement perpendicular to the membrane surface
When
concentration at a point is changing with time, use Fick’s Second Law of
diffusion:
Diffusion
coefficient or also known as diffusion constant (D) is expressed in units of
area per time. The larger the value of D, the faster the molecule diffuse.
APPARATUS:
500 mL beaker, 5 mL pipette, glass rod, 14 test tubes with
cover, hot plate
MATERIALS:
Agar powder, Ringer’s solution, 1:500000 crystal violet
solution, 1:200 crystal violet solution, 1:400 crystal violet solution, 1:600 crystal
violet solution, 1:500000 bromothymol blue solution, 1:200 bromothymol blue
solution, 1:400 bromothymol blue solution, 1:600 bromothymol blue solution
EXPERIMENTAL
PROCEDURES:
1) 250 mL agar in Ringer’s solution is prepared.
1) 250 mL agar in Ringer’s solution is prepared.
2) The agar solution is heated.
3) The
agar is divided into six test tubes and allowed to cool at room temperature.
4) Agar is prepared in another test tube that has
already been added with 1:500000
crystal violet, this will be used as the
standard to measure the color distance resulting
from the cystal violet diffusion.
5) Solutions of crystal violet in distilled water
in the concentrations 1:200, 1:400 and
1:600 are prepared.
6) 5mL
of each crystal violet solution is placed on the gel that was prepared and it
is
closed to prevent evaporation and stored at
temperature 28 ̊C and
37 ̊C.
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) The
average of several measurements is obtained and this value is x in meter.
9) The
value of x after 2 hours and at suitable time distances up till 2 weeks are
recorded
(table 1).
10) The graph for values of x² (in M²) against
time (in hours) for each of the concentration
used is plotted.
11) The
diffusion coefficient D from the slope of the graph at temperature 28 ̊C and
37
̊C is calculated.
The molecular weight of the crystal violet is also calculated using
the equation N and V.
12) This experiment is repeated using
bromothymol blue. However, sufficient alkali has
to be added into this solution to obtain
the color completely from the dye.
RESULTS:
a) Bromothymol Blue at 28⁰C
b)Bromothymol Blue in 37⁰C
c)Crystal Violet in 28⁰C
d)Crystal Violet in 37⁰C
Calculations:
1) Bromothymol Blue
system with dilution 1:200 (28ºC)
Gradient
of the graph = ((8 - 6) / (90 - 70)) x 10-4 m²/hour
= 1 x10-5
m²/hour
M=1:500,000
(standard) Mo=1:200
=2x10-6 =5x10-3
2.303
x 4D [log 10 (5x10-3
) - log (2x10-6 )] = 1
x10-5 m²/hour
D =
3.19 x 10-7 m²/hour
Gradient
of the graph = ((9.2 - 2) / (140 - 40)) x 10-4 m²/hour
= 7.2 x
10-6 m²/hour
M=1:500,000
(standard) Mo=1:400
=2x10-6 =2.5x10-3
2.303
x 4D [log 10 (2.5x10-3) - log 10 (2x10-6)]
= 7.2 x 10-6 m²/hour
D =
2.52 x 10-7 m²/hour
Gradient
of the graph = ((4 - 2) / (100 - 55)) x 10-4 m²/hour
= 4.44 x
10-6 m²/hour
M=1:500,000
(standard) Mo=1:600
=2x10-6 = 1.67 x 10-3
2.303
x 4D [log 10 (1.67 x 10-3 ) - log 10 (2x10-6
)] = 4.44 x 10-6 m²/hour
D =
1.65 x 10-7 m²/hour
Average of Diffusion
Coefficient, m²/hour for Bromothymol Blue system at 28ºC
= [(3.19 x 10-7 m²/hour)
+ (2.52 x 10-7 m²/hour) + (1.65 x 10-7 m²/hour)]/3
= 2.45 x 10-7 m²/hour
Gradient
of the graph = ((22 - 17) / (150 - 120)) x 10-4 m²/hour
= 1.67 x
10-5 m²/hour
M=1:500,000
(standard) Mo=1:200
=2x10-6 =5x10-3
2.303
x 4D [log 10 (5x10-3 )- log 10 (2x10-6 )] = 1.67 x 10-5
m²/hour
D =
5.34 x 10-7 m²/hour
5) Bromothymol Blue
system with dilution 1:400 (37ºC)
Gradient
of the graph = ((15 - 10) / (150 - 110)) x 10-4 m²/hour
= 1.25 x
10-5 m²/hour
M=1:500,000
(standard) Mo=1:400
=2x10-6 =2.5x10-3
2.303
x 4D [log 10 (2.5x10-3)-log 10 (2x10-6)]
= 1.25 x 10-5 m²/hour
D =
4.38 x 10-7 m²/hour
6) Bromothymol Blue
system with dilution 1:600 (37ºC)
Gradient of the graph = ((6 - 3) / (120 - 70)) x 10-4 m²/hour
= 6 x10-6
m²/hour
M=1:500,000
(standard) Mo=1:600
=2x10-6 = 1.67 x 10-3
2.303
x 4D [log 10 (1.67 x 10-3) - log 10(2x10-6)]
= 6 x 10-6 m²/hour
D =
2.23 x 10-7 m²/hour
= [(5.34 x 10-7 m²/hour)
+ (4.38 x 10-7 m²/hour) + (2.23 x 10-7 m²/hour)]/3
=3.98 x 10-7 m²/hour
7) Crystal
Violet system with dilution 1:200 (28ºC)
From
equation: 2.303 x 4D (log 10 Mo – log 10 M) t = X²
Hence
the gradient of the graph = 2.303 x 4D (log 10 Mo - log 10
M)
Gradient of the graph
= ((26 - 14) / (140 - 90)) x 10-4 m²/hour
= 2.4 x
10⁻⁵m²/hour
M=1:500,000
(standard) Mo=1:200
=1/5000,000 =1/200
=2x10-6 =5x10-3
2.303
x 4D (log 10 Mo – log 10 M) = 2.4 x 10⁻⁵
m²/hour
2.303x4D
[log 10 (5x10-3) - log 10 (2x10-6)]
= 2.4 x 10⁻⁵ m²/hour
D =
7.67 x 10-7 m²/hour
8) Crystal
Violet system with dilution 1:400 (28ºC)
Gradient
of the graph = ((14 - 4) / (150 - 60)) x 10-4 m²/hour
=1.11 x10-5
m²/hour
M=1:500,000
(standard) Mo=1:400
=2x10-6 =2.5x10-3
2.303 x 4D [log 10
(2.5x10-3 ) - log 10
(2x10-6 )] = 1.11 x10-5 m²/hour
D =
3.89 x10-7 m²/hour
Gradient
of the graph = ((3 - 2) / (130 - 80)) x 10-4 m²/hour
= 2 x 10-6 m²/hour
M=1:500,000
(standard) Mo=1:600
=2x10-6 =1.67x10-3
2.303
x 4D [log 10 (1.67x10-3 ) -log 10 (2x10-6
)] = 2 x 10-6 m²/hour
D =
7.43 x 10 -8 m²/hour
=
[(7.67 x 10-7 m²/hour) +
(3.89 x10-7 m²/hour) + (7.43 x 10-8 m²/hour)]/3
= 4.10 x 10-7 m²/hour
10) Crystal Violet system with dilution
1:200 (37ºC)
Gradient of the graph = ((22 - 7) / (110 - 50)) x 10-4 m²/hour
= 2.5 x10-5
m²/hour
M=1:500,000
(standard) Mo=1:200
=2x10-6 =5x10-3
2.303
x 4D [log 10 (5x10-3) - log 10 (2x10-6)]
= 2.5 x 10-5 m²/hour
D =
7.99 x 10-7 m²/hour
Gradient
of the graph = ((17 - 11) / (150 - 100)) x 10-4 m²/hour
=1.2 x 10-5
m²/hour
M=1:500,000
(standard) Mo=1:400
=2x10-6 =2.5x10-3
2.303
x 4D [log 10 (2.5x10-3 )-log 10 (2x10-6 )] = 1.2 x 10-5 m²/hour
D =
4.21 x 10-7 m²/hour
Gradient
of the graph = ((4 - 3) / (130 - 90)) x 10-4 m²/hour
= 2.5 x
10-6 m²/hour
M=1:500,000
(standard) Mo=1:600
=2x10-6 =1.67x10-3
2.303
x 4D [log 10 (1.67x10-3 )- log 10 (2x10-6 )] = 2.5 x 10-6 m²/hour
D =
9.29 x 10-8 m²/hour
= [(7.99 x 10-7 m²/hour)
+ (4.21 x 10-7 m²/hour) + (9.29 x 10-8 m²/hour)]/3
= 4.38 x 10-7 m²/hour
PRACTICE
1) From the experiment value for D28, estimate the value of D37 by using the following equation:
Is the calculated value of D37˚C the same as the value from the experiment? Give some explanation if it is different. Is there any different between the calculated molecular weight with the real molecular weight?
D28oC = 4.10 x 10-7
D28oC = T28oC
D37oC = T37oC
4.10 x 10-7
= 28 + 273.15
D37oC = 37 + 273.15
D37oC =
4.10 x 10-7 x 310.15
301.15
= 4.22 x 10-7 m2
hr -1
No, this may due to some errors that occur during the experiment. There might be inaccuracy when we measure the length of dye diffused because the colour is hard to be seen. Besides, the temperature around the test tubes may not be constant. Other than that the viscosity of gel formed maybe not uniform which may affect the length of dye diffused.
2) Between crystal violet and bromothymol
blue, which will diffuse quiker? Explain if there are any differences in the
diffusion coefficient values.
The molecular weight of crystal violet is
408 g mol-1 whereas the molecular weight for bromotyhmol blue is 624 g mol-1.
The crystal violet will diffuse faster than the bromothymol blue because
crystal violet has a smaller molecular mass compared with bromothymol blue.
DISCUSSION:
Diffusion involves the movement of molecules from a region of high concentration to a region of low concentration until an equilibrium is reached. The diffusion process can be affected by the differences in temperature, concentration gradient and pressure.
In this experiment, the diffusion process involves the addition of crystal violet and bromothymol blue on the gel solutions respectively. The factors that affect the rate of diffusion are the concentration gradient and the difference in temperature. Based on the graph plotted in this experiment, when the temperature and the pressure are fixed, the increase in the concentration gradient for both crystal violet and bromothymol blue will definitely increase the rate of diffusion.
In this experiment, the diffusion process involves the addition of crystal violet and bromothymol blue on the gel solutions respectively. The factors that affect the rate of diffusion are the concentration gradient and the difference in temperature. Based on the graph plotted in this experiment, when the temperature and the pressure are fixed, the increase in the concentration gradient for both crystal violet and bromothymol blue will definitely increase the rate of diffusion.
This obeys the Fick’s first law of diffusion. The Fick’s first law of diffusion states that the rate of transfer of diffusing substance through unit area of a section is proportional to the concentration gradient. When the concentration gradient of crystal violet and bromothymol blue is constant, the increase in the temperature also increase the rate of diffusion. This happens because the kinetic energy of the molecules increases when the temperature increases. The molecules will be able to overcome the intermolecular force and the Van der Waals forces in a shorter time and able to diffuse from the region of higher concentration to a region of lower concentration in a shorter time.
There is some errors may arise and causing inaccuracy to the final result. During the process of making agar, the agar powder is not completely dissolved in the Ringer solution. If the agar not complete solidify, this indicators will diffuse faster and the real diffusion coefficient of each indicators is not accurate anymoreVolume of used agar in each test tubes may have some differences.Besides, the temperature, T around the test tube is always not constant, such as the temperature 270 oC, room temperate is not kept constant for the day and night. Indirectly this will influence the diffusion coefficient and so diffusion rate.
CONCLUSION:
The diffusion coefficient of crystal violet is generally higher than bromothymol blue because of the difference inmolecular weight. In term of temperature, diffusivity at 37°C is much higher than diffusivity at 28°C. This is due to the rapid movement of the solute particles as the temperature increases.
REFERENCES:
1. Martin’s Physical Pharmacy and Pharmaceutical Sciences, 5th Edition, Patrick J. Sinko, Lippincott Williams and Wilkins
2. http://highered.mcgrawhill.com/sites/0072495855/student_view0/chapter2/animation__how_diffusion_works.html
3. http://en.wikipedia.org/wiki/Fick%27s_laws_of_diffusion
4. http://www.pojman.com/mg_materials/Diffusion/Diffusion.html
1. Martin’s Physical Pharmacy and Pharmaceutical Sciences, 5th Edition, Patrick J. Sinko, Lippincott Williams and Wilkins
2. http://highered.mcgrawhill.com/sites/0072495855/student_view0/chapter2/animation__how_diffusion_works.html
3. http://en.wikipedia.org/wiki/Fick%27s_laws_of_diffusion
4. http://www.pojman.com/mg_materials/Diffusion/Diffusion.html
