Fick's Law of diffusion describes diffusion and can be used to solve for the diffusion coefficient,D. Fick's First Law relates the diffusive flux to the concentration under the assumption of steady state. It postulates that the flux goes from regions of high concentration to regions of low concentration, with a magnitude that is proportional to the concentration gradient. Meanwhile Fick's Second Law predicts how diffusion causes the concentration to change with time.
Objective
To determine the diffusion coefficient
Materials
Ringer's solution 250ml
Agar powder
Crystal violet 1:500000, 1:200, 1:400, 1:600
Bromothymol blue 1:500000, 1:200, 1:400, 1:600
Apparatus
Test tubes
Beaker
Glass rod
Heating bath
Test tube caps
Test tube rack
Experiment Procedure
1. 250ml agar is prepared in Ringer's solution.
2. The agar is divided into six test tubes and is allowed to cool at room temperature.
3. The 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 crystal violet diffusion.
4. 5ml of each crystal violet solution is placed on the gels that was prepared and is closed to prevent evaporation and is stored at temperature 28 degree celcius and 37 degree celcius.
5. 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.
6. The average of several measurements are obtained, this value is x in meter.
7. The values of x after 2 hours and at suitable time distances up till 2 weeks are recorded.
8. The graph for values of x2 (in M2) against time (in seconds) for each of the concentration used is plotted.
9. The diffusion coefficient D from the slope of the graph at temperature 28 and 37 degree celcius are calculated.
10. The molecular weight of the crystal violet is calculated using the equation N and V.
11. Steps 3 until step 10 are repeated using bromothymol blue.
Result
Crystal Violet Solution in 28oC
System
|
Day and Time(s)
|
X(m)
|
X2(m2)
|
Temperature(K)
|
Crystal Violet
1:200
|
2(172800)
|
0.019
|
3.610 x 10-4
|
301
|
3(259200)
|
0.022
|
4.840 x 10-4
|
||
6(518400)
|
0.033
|
1.089 x 10-3
|
||
7(604800)
|
0.035
|
1.225 x 10-3
|
||
8(691200)
|
0.038
|
1.444 x 10-3
|
||
9(777600)
|
0.041
|
1.681 x 10-3
|
||
10(864000)
|
0.044
|
1.936 x 10-3
|
||
Crystal Violet
1:400
|
2(172800)
|
0.014
|
1.960 x 10-4
|
301
|
3(259200)
|
0.017
|
2.890 x 10-4
|
||
6(518400)
|
0.029
|
8.410 x 10-4
|
||
7(604800)
|
0.030
|
9.000 x 10-4
|
||
8(691200)
|
0.035
|
1.225 x 10-3
|
||
9(777600)
|
0.037
|
1.369 x 10-3
|
||
10(864000)
|
0.038
|
1.444 x 10-3
|
||
Crystal Violet 1:600
|
2(172800)
|
0.010
|
1.000 x 10-4
|
301
|
3(259200)
|
0.011
|
1.210 x 10-4
|
||
6(518400)
|
0.015
|
2.250 x 10-4
|
||
7(604800)
|
0.017
|
2.890 x 10-4
|
||
8(691200)
|
0.020
|
4.000 x 10-4
|
||
9(777600)
|
0.024
|
5.760 x 10-4
|
||
10(864000)
|
0.027
|
7.290 x 10-4
|
Crystal Violet Solution at 37OC
System
|
Day and Time(s)
|
X(m)
|
X2(m2)
|
Temperature(K)
|
Crystal Violet
1:200
|
2(172800)
|
0.020
|
4.000 x 10-4
|
310
|
3(259200)
|
0.023
|
5.290 x 10-4
|
||
6(518400)
|
0.034
|
1.156 x 10-3
|
||
7(604800)
|
0.038
|
1.444 x 10-3
|
||
8(691200)
|
0.042
|
1.764 x 10-3
|
||
9(777600)
|
0.043
|
1.849 x 10-3
|
||
10(864000)
|
0.045
|
2.025 x 10-3
|
||
Crystal Violet
1:400
|
2(172800)
|
0.017
|
2.890 x 10-4
|
310
|
3(259200)
|
0.021
|
4.410 x 10-4
|
||
6(518400)
|
0.030
|
9.000 x 10-4
|
||
7(604800)
|
0.035
|
1.225 x 10-3
|
||
8(691200)
|
0.037
|
1.369 x 10-3
|
||
9(777600)
|
0.039
|
1.521 x 10-3
|
||
10(864000)
|
0.040
|
1.600 x 10-3
|
||
Crystal Violet 1:600
|
2(172800)
|
0.013
|
1.690 x 10-4
|
310
|
3(259200)
|
0.014
|
1.960 x 10-4
|
||
6(518400)
|
0.017
|
2.890 x 10-4
|
||
7(604800)
|
0.020
|
4.000 x 10-4
|
||
8(691200)
|
0.023
|
5.290 x 10-4
|
||
9(777600)
|
0.026
|
6.760 x 10-4
|
||
10(864000)
|
0.029
|
8.410 x 10-4
|
Bromotymol Blue Solution at 28OC
System
|
Day and Time(s)
|
X(m)
|
X2(m2)
|
Temperature(K)
|
Bromotymol Blue
1:200
|
2(172800)
|
0.016
|
2.560 x 10-4
|
301
|
3(259200)
|
0.019
|
3.610 x 10-4
|
||
6(518400)
|
0.029
|
8.410 x 10-4
|
||
7(604800)
|
0.033
|
1.089 x 10-3
|
||
8(691200)
|
0.034
|
1.156 x 10-3
|
||
9(777600)
|
0.039
|
1.521 x 10-3
|
||
10(864000)
|
0.042
|
1.764 x 10-3
|
||
Bromotymol Blue
1:400
|
2(172800)
|
0.012
|
1.440 x 10-4
|
301
|
3(259200)
|
0.015
|
2.250 x 10-4
|
||
6(518400)
|
0.025
|
6.250 x 10-4
|
||
7(604800)
|
0.029
|
8.410 x 10-4
|
||
8(691200)
|
0.032
|
1.024 x 10-3
|
||
9(777600)
|
0.035
|
1.225 x 10-3
|
||
10(864000)
|
0.037
|
1.369 x 10-3
|
||
Bromotymol Blue
1:600
|
2(172800)
|
0.010
|
1.000 x 10-4
|
301
|
3(259200)
|
0.011
|
1.210 x 10-4
|
||
6(518400)
|
0.014
|
1.960 x 10-4
|
||
7(604800)
|
0.016
|
2.560 x 10-4
|
||
8(691200)
|
0.018
|
3.240 x 10-4
|
||
9(777600)
|
0.020
|
4.000 x 10-4
|
||
10(864000)
|
0.024
|
5.760 x 10-4
|
Bromotymol Blue Solution at 37oC
System
|
Day and Time(s)
|
X(m)
|
X2(m2)
|
Temperature(K)
|
Bromotymol Blue
1:200
|
2(172800)
|
0.018
|
3.240 x 10-4
|
310
|
3(259200)
|
0.020
|
4.000 x 10-4
|
||
6(518400)
|
0.030
|
9.000 x 10-4
|
||
7(604800)
|
0.034
|
1.156 x 10-3
|
||
8(691200)
|
0.039
|
1.521 x 10-3
|
||
9(777600)
|
0.040
|
1.600 x 10-3
|
||
10(864000)
|
0.044
|
1.936 x 10-3
|
||
Bromotymol Blue
1:400
|
2(172800)
|
0.015
|
2.250 x 10-4
|
310
|
3(259200)
|
0.018
|
3.240 x 10-4
|
||
6(518400)
|
0.028
|
7.840 x 10-4
|
||
7(604800)
|
0.033
|
1.089 x 10-3
|
||
8(691200)
|
0.035
|
1.225 x 10-3
|
||
9(777600)
|
0.038
|
1.444 x 10-3
|
||
10(864000)
|
0.039
|
1.521 x 10-3
|
||
Bromotymol Blue
1:600
|
2(172800)
|
0.011
|
1.210 x 10-4
|
310
|
3(259200)
|
0.013
|
1.690 x 10-4
|
||
6(518400)
|
0.015
|
2.250 x 10-4
|
||
7(604800)
|
0.017
|
2.890 x 10-4
|
||
8(691200)
|
0.021
|
4.410 x 10-4
|
||
9(777600)
|
0.024
|
5.760 x 10-4
|
||
10(864000)
|
0.027
|
7.290 x 10-4
|
At
concentration 1:200
Gradient = 2.303x4D(log10Ma-log10 M)
(14.44-5.20)x10-4 =1.514x10-9= 2.303x4D(log10Ma-log10 M)
(8-3)(86400)
2.14x10-9 = 2.303x4D[log10(1/200)-log10(1/500
000)]
D =
6.84x10-11 m2s-1
At
concentration 1:400
Gradient = 2.303x4D(log10Ma-log10 M)
(11.60-1.60)x10-4 =1.157x10-9= 2.303x4D(log10Ma-log10 M)
(8-2)(86400)
1.93x10-9 = 2.303x4D[log10(1/400)-log10(1/500
000)]
D =
6.77x10-11 m2s-1
At
concentration 1:600
Gradient = 2.303x4D(log10Ma-log10 M)
(3.60-1.00)x10-4 =7.234x10-10= 2.303x4D(log10Ma-log10
M)
(8-2)(86400)
5.02x10-10 = 2.303x4D[log10(1/600)-log10(1/500 000)]
D =
1.86x10-11 m2s-1
Average
diffusion coefficient: 5.16 x 10-11 m2s-1
At
concentration 1:200
Gradient = 2.303x4D(log10Ma-log10 M)
(16.80-5.60)x10-4 =1.514x10-9= 2.303x4D(log10Ma-log10 M)
(8-3)(86400)
2.60x10-9 = 2.303x4D[log10(1/200)-log10(1/500
000)]
D = 8.31 x10-11 m2s-1
At
concentration 1:400
Gradient = 2.303x4D(log10Ma-log10 M)
(13.69-2.89)x10-4 =1.157 x 10-9= 2.303x4D(log10Ma-log10
M)
(8-2)(86400)
2.08
x 10-9 =
2.303x4D[log10(1/400)-log10(1/500 000)]
D = 7.30 x10-11 m2s-1
At
concentration 1:600
Gradient = 2.303x4D(log10Ma-log10 M)
(3.60-1.69)x10-4 =7.234 x 10-10= 2.303x4D(log10Ma-log10
M)
(6-2)(86400)
5.53x10-10 = 2.303x4D[log10(1/600)-log10(1/500 000)]
D = 2.06x10-11 m2s-1
Average
diffusion coefficient: 5.89 x 10-11 m2s-1
At
concentration 1:200
Gradient = 2.303x4D(log10Ma-log10 M)
(11-4)x10-4 =1.514x10-9= 2.303x4D(log10Ma-log10 M)
(8-3)(86400)
1.62x10-9 = 2.303x4D[log10(1/200)-log10(1/500
000)]
D =
5.18 x10-11 m2s-1
At
concentration 1:400
Gradient = 2.303x4D(log10Ma-log10 M)
(9.80-2.25)x10-4 =1.157 x 10-9= 2.303x4D(log10Ma-log10
M)
(8-3)(86400)
1.75
x 10-9 = 2.303x4D[log10(1/400)-log10(1/500
000)]
D =
6.13 x10-11 m2s-1
At
concentration 1:600
Gradient = 2.303x4D(log10Ma-log10 M)
(3.24-1.00)x10-4 =7.234 x 10-10= 2.303x4D(log10Ma-log10
M)
(8-2)(86400)
7.234
x 10-10 =
2.303x4D[log10(1/600)-log10(1/500 000)]
D =
1.61x10-11 m2s-1
Average
diffusion coefficient: 4.31 x 10-11 m2s-1
At
concentration 1:200
Gradient = 2.303x4D(log10Ma-log10 M)
(14.00-2.25)x10-4 =1.514x10-9= 2.303x4D(log10Ma-log10 M)
(8-2)(86400)
2.27x10-9 = 2.303x4D[log10(1/200)-log10(1/500
000)]
D = 7.25
x10-11 m2s-1
At
concentration 1:400
Gradient = 2.303x4D(log10Ma-log10 M)
(12.25-3.24)x10-4 =1.157 x 10-9= 2.303x4D(log10Ma-log10
M)
(8-3)(86400)
2.09
x 10-9 =
2.303x4D[log10(1/400)-log10(1/500 000)]
D =
7.31 x10-11 m2s-1
At
concentration 1:600
Gradient = 2.303x4D(log10Ma-log10 M)
(3.2-1.21)x10-4 =7.234 x 10-10= 2.303x4D(log10Ma-log10
M)
(7-2)(86400)
4.61
x 10-10 =
2.303x4D[log10(1/600)-log10(1/500 000)]
D =
1.71x10-11 m2s-1
Average
diffusion coefficient: 5.42 x 10-11 m2s-1
From
the experiment value for D28, estimate the value of D37 using the following equation
D28° C T28°
C
--------- =
---------
D37°C T37°C
where η1 and η2
is the viscosity of water at
temperature 28°C and 37°C.
5.16 x 10-11 m2s-1 28 + 273
--------- =
---------
D37°C 37 + 273
D37°C = 5.31 x 10-11 m2s-1
1. 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 difference between the calculated molecular weight with the real molecular weight?
Yes,because during the experiment, error might occur,that
is why the weight reading can be different.Maybe it involve parallax error that
occur when student measure the length of solubility in the agar.Not only
that,error also occur when transferring the agar or solution in another
container that may cause change in the actual measurement. Besides that the scientific
error occur cause by the molecular weight itself.
The molecular weight of a molecule is
the weight of such molecules. The
molecular weight of a molecule is equal to the sum of the atomic weights of its
constituting atoms.The weight in grams of a single atom is very small. It is
only a matter of convenience to give the weight of atoms, instead of a single atom, in order to
have a convenient measurable value. There is nothing more than that. The
weight of a single atom can always be calculated back from the atomic weight by
dividing by Avogadro's number. It is similarly true for molecules.
One mole is a quantity of atoms or
molecules. Therefore, the atomic weight is the weight of one mole of atoms, and
the molecular weight is the weight of one mole of molecules.The molecular
weight Mw of a molecule, multiplied by the number of moles n, is equal to the
total weight W of the molecules:
W = n∙Mw
When a quantity of atoms or molecules is
given in moles, then their absolute overall number can be calculated by
multiplying by Avogadro's number .Since the numerical value of the atomic
weight of a chemical isotope is (nearly) equal to its number of nucleons
(protons and neutrons). Thus, the atomic weight of hydrogen with one proton is
(nearly) one gram, and of carbon, with six protons and six neutrons, is twelve
grams. The atomic weight of a chemical element is usually a non-integral number
because it relates to its natural abundance of isotopes.The real value can be
difference as the average mass is usually not identical to the mass of any
single molecule. The actual numerical difference can be very small when
considering small molecules and the molecular mass of the most common
isotopomer in which case the error only matters to physicists and a small
subset of highly specialized chemists; however it is always more correct,
accurate and consistent to use molar mass in any bulk stoichiometric
calculations. The size of this error becomes much larger when considering
larger molecules or less abundant isotopomers. The molecular mass of a molecule
which happens to contain heavier isotopes than the average molecule in the
sample can differ from the molar mass by several mass units.
2. Between the crystal violet and
bromothymol blue, which diffuse quicker?Explain if there are any differences in
the diffusion coefficient values?
The molecular
weight of crystal violet is 408 g mol-1 and average diffusion
coefficient: 5.16 x 10-11 m2s-1 whereas the molecular weight bromotyhmol
blue is 624 g mol-1. The crystal violet diffuses faster than bromothymol blue
as crystal violet has a smaller molecular mass as compared with bromothymol
blue and average diffusion coefficient: 4.31 x 10-11 m2s-1.Although
there is difference value of diffusion coefficient and bromothymol average
diffusion coefficient is lower than crystal violet ,we should kmow that these
two dyes diffuse in the agar according to the pH.
Bromothymol blue acts as a weak
acid in solution. It can thus be in protonated or deprotonated form, appearing
yellow or blue respectively. It is bluish green in neutral solution. The
deprotonation of the neutral form results in a highly conjugated structure,
accounting for the difference in color.
Crystal violet and its the colour of the dye depends
on the acidity of the solution. At a pH of 1.0 the dye is green with absorption
maxima at 420 nm and 620 nm while in a strongly acidic solution (pH of -1), the
dye is yellow with an absorption maximum at 420 nm.
In the yellow form all three nitrogen atoms carry a
positive charge, of which two are protonated, while the green colour
corresponds to a form of the dye with two of the nitrogen atoms positively
charged. At neutral pH both extra protons are lost to the solution leaving only
one of the nitrogen atoms positive charged. The pKa’s for the loss of the two
protons are approximately 1.15 and 1.8.
In alkaline solutions, nucleophilic hydroxyl ions
attack the electrophilic central carbon to produce the colourless
triphenylmethanol or carbinol form of the dye. Some triphenylmethanol is also
formed under very acid condition when the positive charges on the nitrogen
atoms lead to an enhancement of the electrophilic character of the central
carbon which allows the nucleophilic attack by water molecules. This effect
produces a slight fading of the yellow colour. Thus,crystal violet diffuse
quicker compare to the bromothymol.
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. 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.
Errors :
Errors :
- Volume of used agar in each test tubes may have some differences.
- Agar prepared may not good enough as we over heated it.
- Measurement taken is not much accurate as some of us who were assigned to take the measurement wrongly assumed the end point.
- We sometimes inverted the test tubes which may affect the actual reading.
Precaution :
- Students must wear safety goggles while working in the lab. These safety glasses protect anything from entering their eye.
- Students should make sure read the bottle labels carefully and only use what are supposed to use during an experiment. Students should not try out anything on their own unless they are very sure of what reagents or chemicals need to use and their results.
- Students should rinse their apparatus with distilled water to avoid any contamination present.
- Student should make sure their eye must be perpendicular to the reading scale of the apparatus to avoid parallax error while obtaining their readings.
- Students should let the agar soldify first before proceeding with the experiment to obtain accurate result as the dyes agent used need to diffuse inside the medium.
Conclusion.
As obtained from this experiment data it is calculated that crystal violet diffusion coefficient is 5.16 x 10-11 m2s-1 while as for bromothymol blue, it is 4.31 x 10-11 m2s-1. Some factor do affect this value. Firstly, crystal violet diffuse quicker compare to the bromothymol as it have smaller molecular mass. Temparature also play a role in diffusion as it will increase the rate when the temperature was to be increased too. Besides, other factor that affect rate of diffusion is the concentration of the crsytal violet and bromothymol used where as the concentration increasese as more diluted ones' were used.
References :
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