PRACTICAL 4:
DETERMINATION OF DIFFUSION COEFFICIENT
OBJECTIVE:
To determine the value of diffusion coefficient, D.
INTRODUCTION:
Diffusion is a process of mass transfer
of molecules brought by random molecular motion. It is associated with driving
forces such as concentration gradient which will proceed until equilibrium is
achieved.
Diffusion coefficient can be explained by a
Fick’s Law of diffusion. It also known as diffusivity which measure how easily
a molecule is transversed through a
medium.
|
J= D (dC/dX)
|
J = Flux, D =
diffusion coefficient, dC/dX = concentration gradient ( C is concentration
gradient and X is distance of movement perpendicular to membrane surface )
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
for steady state diffusion in which the concentration within the diffusion point
does not change with respect to time. However, this is not realistic.
|
dC/dt = D (d2C/d2X)
|
Thus, the rate of
change in concentration with time at a point within a diffusional field is
proportional to the rate of change in the concentration gradient at that point.
Fick’s second law is for non-steady state diffusion.
dm = - DA (dc/dx)
dt à
i
D is diffusion coefficient or diffusivity for the solute in unit
m2s-1
If a solution
containing neutral particles with the concentration M0 is placed
within a cylindrical tube next to a water column, diffusion can be stated as
M = M0exp (-x2/4Dt) à ii
Where M is the concentration at distance x from the intersection between
water and solution that is measured at time t.
By changing eq (ii) to its logarithmic form,
we get
ln M = ln M0
–x2/4Dt
or 2.303 x 4D (log M0
– log M)t = x2 à
iii
Thus a plot of x2 against t can produce a straight line that passes
through the origin with the slope 2.303 x 4D (log M0 –log M). From
here D can be calculated.
If the particles in the solution are assumed
to be spherical, their size and molecular weight can be calculated by the
Stokes-Einstein equation.
D = Kt/6(pi)na à
iv
|
where k is the Boltzmann constant 1.38 x 1023 Jk-1 , T is the temperature in Kelvin,
π is the viscosity of the solute, in Nm⁻² and a half diameter of molecule
in M. The volume for that certain sphere molecule is 4/3 πa³, thus the mass
of M is equal to 4/3πa³Nρ (ρ = the density of the molecule).
|
It is known that
molecular weight M =mN (N is Avogadro’s number 6.023 x 1023 ).
M = 4/3 (pi)a3Np à
v
Diffusion for charged
particles, equation (iii) need to be modified to include potential gradient
effect that exists between the solution and solvent. However, this can be
overcome by adding a little sodium chloride into the solvent to prevent the
formation of this potential gradient.
Agar gels contain a
partially strong network of molecules that is penetrated by water. The water
molecules form a continuous phase around the gel. Thus, the molecules of
solutes can diffuse freely in the water if chemical interactions and adsorption
effects do not exist entirely. Therefore, the gel forms an appropriate support
system to be used in diffusion studies for molecules in a medium of water.
MATERIALS
|
APPARATUS
|
Agar powder
Ringer’s solution
1:500,000 crystal
violet solution
1:200 crystal violet
solution
1:400 crystal violet
solution
1:600 crystal violet
solution
1:500,000
bromothymol blue solution
1:200 bromothymol
blue solution
1:400 bromothymol
blue solution
1:600 bromothymol
blue solution
|
500mL beaker
5mL pipette
Glass rod
14 test tubes with
covers
Hot plate
|
EXPERIMENTAL
PROCEDURES:
1.
7g of agar powder was
weighed and mixed with 420ml of Ringer solution in the 500mL beaker.
2.
The mixture in the
beaker was stirred and boiled on a hot plate until a transparent yellowish
solution was obtained.
3.
About 20ml of the agar
solution was poured into each 6 test tubes. The test tubes were then put in the
fridge to let them cool.
4.
An agar test tube
which contained 5ml of 1:500,000 crystal violet was being prepared and it was
used as a standard system to measure the distance of the colour as a result of
the diffusion of crystal violet.
5.
After the agar
solutions in the test tubes solidifying, 5ml of each 1:200, 1:400, 1:600
crystal violet solution were poured into each test tube.
6.
The test tubes were
closed immediately to prevent the vaporization of the solutions.
7.
Three test tubes were
put in room temperature, 28 ºC while another three were put in 37ºC water bath.
8.
The distance between
the agar surface and the end of crystal violet where that area has the same
color as in the indicator was measured accurately.
9.
Average of the
readings were obtained, this value is x in meter.
10.
The x values were
recorded after 2 hours and at appropriate intervals for 1 week.
11.
Procedures 3 to 10
were repeated for bromothymol blue solutions.
12.
Graph of x² values (in
m²) versus time (in hours) was potted.
The diffusion coefficient, D was determined
from the graph gradient for both 28 ºC and 37 ºC, the molecular
mass of crystal violet and bromothymol blue were also determined by using
N and V equation.
RESULTS:
Crystal violet at room temperature (280C)
System
|
Day
|
x
(m)
|
x2(x
10-4 m2)
|
Temperature
(K)
|
Crystal violet 1:200
|
1
|
0.015
|
2.25
|
301
|
2
|
0.018
|
3.24
|
||
3
|
0.020
|
4.00
|
||
4
|
0.032
|
10.24
|
||
5
|
0.034
|
11.56
|
||
6
|
0.036
|
12.96
|
||
7
|
0.038
|
14.44
|
||
Crystal violet 1:400
|
1
|
0.011
|
1.21
|
301
|
2
|
0.015
|
2.25
|
||
3
|
0.017
|
2.89
|
||
4
|
0.022
|
4.84
|
||
5
|
0.025
|
6.25
|
||
6
|
0.027
|
7.29
|
||
7
|
0.029
|
8.41
|
||
Crystal violet 1:600
|
1
|
0.003
|
0.09
|
301
|
2
|
0.005
|
0.25
|
||
3
|
0.007
|
0.49
|
||
4
|
0.009
|
0.81
|
||
5
|
0.011
|
1.21
|
||
6
|
0.013
|
1.69
|
||
7
|
0.014
|
1.96
|
Crystal
violet in water bath (370C)
System
|
Day
|
x
(m)
|
x2(
x10-4 m2)
|
Temperature
(K)
|
Crystal violet 1:200
|
1
|
0.017
|
2.89
|
310
|
2
|
0.020
|
4.00
|
||
3
|
0.026
|
6.76
|
||
4
|
0.035
|
12.25
|
||
5
|
0.035
|
12.25
|
||
6
|
0.036
|
12.96
|
||
7
|
0.037
|
13.69
|
||
Crystal violet 1:400
|
1
|
0.012
|
1.44
|
310
|
2
|
0.015
|
2.25
|
||
3
|
0.018
|
3.24
|
||
4
|
0.027
|
7.29
|
||
5
|
0.029
|
8.41
|
||
6
|
0.031
|
9.61
|
||
7
|
0.034
|
11.56
|
||
Crystal violet 1:600
|
1
|
0.006
|
0.36
|
310
|
2
|
0.008
|
0.64
|
||
3
|
0.013
|
1.69
|
||
4
|
0.018
|
3.24
|
||
5
|
0.020
|
4.00
|
||
6
|
0.023
|
5.29
|
||
7
|
0.025
|
6.25
|
Bromothymol blue at room temperature (280C)
System
|
Day
|
x
(m)
|
x2
(x 10-4 m2)
|
Temperature
(K)
|
Bromothymol blue 1:200
|
1
|
0.010
|
1.00
|
301
|
2
|
0.015
|
2.25
|
||
3
|
0.018
|
3.24
|
||
4
|
0.030
|
9.00
|
||
5
|
0.032
|
10.24
|
||
6
|
0.035
|
12.25
|
||
7
|
0.038
|
14.44
|
||
Bromothymol blue 1:400
|
1
|
0.008
|
0.64
|
301
|
2
|
0.010
|
1.00
|
||
3
|
0.014
|
1.96
|
||
4
|
0.028
|
7.84
|
||
5
|
0.030
|
9.00
|
||
6
|
0.033
|
10.89
|
||
7
|
0.035
|
12.25
|
||
Bromothymol blue 1:600
|
1
|
0.005
|
0.25
|
301
|
2
|
0.015
|
2.25
|
||
3
|
0.017
|
2.89
|
||
4
|
0.020
|
4.00
|
||
5
|
0.023
|
5.29
|
||
6
|
0.026
|
6.76
|
||
7
|
0.028
|
7.84
|
Bromothymol blue in water bath (370C)
System
|
Day
|
x
(m)
|
X2
(x 10-4 m2)
|
Temperature
(K)
|
Bromothymol blue 1:200
|
1
|
0.015
|
2.25
|
310
|
2
|
0.017
|
2.89
|
||
3
|
0.020
|
4.00
|
||
4
|
0.030
|
9.00
|
||
5
|
0.032
|
10.24
|
||
6
|
0.035
|
12.25
|
||
7
|
0.037
|
13.69
|
||
Bromothymol blue 1:400
|
1
|
0.010
|
1.00
|
310
|
2
|
0.014
|
1.96
|
||
3
|
0.015
|
2.25
|
||
4
|
0.027
|
7.29
|
||
5
|
0.030
|
9.00
|
||
6
|
0.033
|
10.89
|
||
7
|
0.035
|
12.25
|
||
Bromothymol blue 1:600
|
1
|
0.008
|
0.64
|
310
|
2
|
0.010
|
1.00
|
||
3
|
0.012
|
1.44
|
||
4
|
0.020
|
4.00
|
||
5
|
0.022
|
4.84
|
||
6
|
0.025
|
6.25
|
||
7
|
0.028
|
7.84
|
CALCULATION
:
Graph of x2
against time for Crystal Violet at 280C
At
concentration 1:200
Gradient = 2.303 x 4D (Iog M0 – log M)
(12-6)x 10-4 / (5.6-3.0)86400 = 2.671 x 10-9
= 2.303(4D)(log 1/200 – log 1/500000)
D = 8.533 x 10-11 m2 s-1
At
concentration 1:400
Gradient = (7.2-4.0) x 10-4 / (6.0-3.4)86400 =
1.425 x 10-9
1.425 x 10-9 = 2.303(4D) (log 1/400 – log
1/500000)
D = 4.995 x 10-11 m2s-1
At
concentration 1:600
Gradient = (1.6-0.4) x 10-4 / (6.0-2.6)86400 = 4.085
x 10-10
4.085 x 10-10 = 2.303(4D) (log 1/600 –log
1/500000)
D = 1.518 x 10-11 m2s-1
Average
diffusion coefficient = 5.015 x 10-11 m2s-1
Graph of x2
against time for Crystal Violet at 370C
At concentration
1:200
Gradient = (14.0-8.0) x 10-4 / (6.4-3.4)86400 =
2.315 x 10-9
2.315 x 10-9 = 2.303(4D) (log 1/200 – log
1/500000)
D = 7.396 x 10-11 m2 s-1
At
concentration 1:400
Gradient = (11.6-4.4) x 10-4 / (7.0-3.0)86400 =
2.083 x 10-9
2.083 x 10-9 = 2.303(4D) (log 1/400 – log
1/500000)
D = 7.301 x 10-11 m2s-1
At
concentration 1:600
Gradient = (6.0-2.0) x 10-4 / (6.8-3.0)86400 =
1.218 x 10-9
1.218 x 10-9 = 2.303(4D) (log 1/600 – log
1/500000)
D = 4.527 x 10-11 m2 s-1
Average
diffusion coefficient = 6.408 x 10-11 m2s-1
Graph of x2
against time for Bromothymol Blue at 280C
At
concentration 1:200
Gradient = (12.8 – 6.0) x 10-4 / (6.2-3.4)86400 =
2.811 x 10-9
2.811 x 10-9 = 2.303(4D) (log 1/200 – log
1/500000)
D = 8.980 x 10-11 m2s-1
At
concentration 1:400
Gradient = (12.8 – 4.8) X10-4 / (7.0-3.4)86400 =
2.572 X 10-9
2.572 X 10-9 = 2.303(4D) (log 1/400 – log
1/500000)
D = 9.015 x 10-11 m2s-1
At
concentration 1:600
Gradient = (7.2-4.0) x 10-4 / (6.4-3.8)86400 =
1.425 x 10-9
1.425 x 10-9 = 2.303(4D) (log 1/600 – log
1/500000)
D = 5.296 x 10-11 m2s-1
Average
diffusion coefficient =7.764 x 10-11 m2s-1
Graph of x2
against time for Bromothymol Blue at 370C
At concentration 1:200
Gradient = (12.0 – 6.0) x 10-4
/ (6.0-3.2)86400 = 2.480 x 10-9
2.480 x 10-9 = 2.303(4D)
(log 1/200 – log 1/500000)
D = 7.923 x 10-11 m2s-1
At concentration 1:400
Gradient = (10.0 – 6.0) x 10-4
/ (5.8-3.8)86400 = 2.315 x 10-9
2.315 x 10-9 = 2.303(4D)
(log 1/400 – log 1/500000)
D = 8.115 x 10-11 m2s-1
At concentration 1:600
Gradient = (6.0-2.0) x 10-4
/ (5.8-2.6)86400 = 1.447 x 10-9
1.447 x 10-9 = 2.303(4D)
(log 1/600 – log 1/500000)
D = 5.378 x 10-11 m2s-1
Average diffusion coefficient = 7.139 x 10-11 m2s-1
Question
(D28ºC)/(D37ºC)=
(T28ºC )/(T37ºC)
where ŋ1and ŋ2 are viscosity of water at 28ºC and 37ºC.
For crystal violet,
D28 ºC = 5.015 x 10-11 m2s-1
D28ºC / D37ºC = T28ºC / T37ºC
5.015 x 10-11 m2s-1 / D37ºC = (28+273) / (37+273)
D37ºC = 5.16 x 10^(-11) m^2 s^(-1)
For Bromothymol Blue,
D28 ºC = 7.764 x 10-11 m2s-1
D28ºC / D37ºC =
T28ºC / T37ºC
7.764 x 10-11 m2s-1 / D37ºC = (28+273) / (37+273)
D37ºC
= 7.996 x 10^(-11) m^2 s^(-1)
(2) Is the calculated value of D37ºC 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?
No, from the experiment , D37ºC for crystal violet is 6.408 x 10-11 m2s-1 while D37ºC for bromothymol blue is 7.139 x
10-11 m2s-1. The difference of value obtained from
experiment and estimated value may due to certain errors occur throughout the experiment.
Inconsistency and inaccuracy in measuring the length of dye by different person
as the colour is shown clear obviously. The
viscosity of the agar produced along the test tube which is not uniform may cause
the difference in the rate of diffusion of the dye. Temperature which is not
constant all the time (370c) during the experiment may give influence to the accuracy
of result obtained.
The shape of the particle is assumed to be spherical
Volume, V=
Weight of a particle = volume × density
= 4/3 πa^3×p
Weight of 1 molecule, M = weight of a particle × Avogrado’s
constant
= 4/3 πa^3×p ×n
(3) Between Crystal Violet and Bromothymol Blue, which one
diffuse quicker? Explain if there are any differences in the diffusion
coefficient values?
The crystal violet will diffuse faster than bromothymol
blue. This is because crystal violet have smaller molecular size (408 g/mol)compared
to bromothymol blue (624g/mol) . From the experiment, the diffusion coefficient
calculated for D of bromothymol blue is greater than D of crystal violet . Some mistakes have been done will cause the
inaccuracy of result as crystal violet is supposed to have diffusion
coefficient greater than bromothymol blue. The higher the diffusion coefficient, the faster the diffusion rate will be. Crystal diffuse
faster than bromothymol blue.
Discussion
Diffusion refers to the process by which molecules
intermingle as a result of their kinetic energy of random motion. In this process,
the molecules from a highly concentrated area move to a place where there are
fewer molecules due to concentration difference which is reduced by a
spontaneous flow of matter. Agar is the
inert medium used to investigate diffusion through in this experiment. It is
made to be by adding small amounts of sodium hydroxide. This experiment is
carried out to determine the diffusion coefficient of the crystal violet and
bromothymol blue. The controlled variables in this experiment are the size of
the particles and also the temperature. Factors such as viscosity and
concentration of agar gel may affect the rate of diffusion.
The concentration varies for each crystal violet and
bromothymol blue solution. Rate of diffusion is higher for higher concentration
of solution. The substances diffuse faster between two compartments of greater
concentration difference. This could be explained by Fick’s first law. Fick’s
first law equation describes the diffusion process under conditions of steady
state; that is the concentration gradient does not change with time.
In this experiment, the diffusion particles are neutral with
Mo concentration, the agar medium is considered homogenous and with constant
concentration, hence the diffusion can be expressed as
M = Mo eksp (x²/4Dt) ………….(II)
Change the equation II to logarithm form, we have
ln M = ln Mo – (x²/4Dt)
2.303 x 4D (log 10 Mo- log 10 M) t = x² ………….(III)
Thus, when a graph x² against time t is plotted, a straight
line is obtained with the gradient of 2.303 x 4D (log 10 Mo- log 10 M) . From
here D can be calculated. We can know the both 28ºC and 37 ºC system, the rate
of diffusion from the result that is 1:200 > 1:400 > 1:600.
M is the system with the dilution 1:500,000. It is used to
be a standard system during the experiment. When Mo is increased, (log 10 Mo-
log 10 M) will increased. This causes
the concentration gradient become larger, therefore the driving force for the
occurrence of diffusion would be larger and the diffusion process will become
faster.
This experiment is done in two different temperature of 28oC
water bath and 37oC room temperature. Diffusion rate is higher t 37oC than that
of 28oC. This is because the amount of energy available for diffusion is
increased with the increase of temperature. The increase of kinetic energy
causing molecules to move faster and there will be more spontaneous spreading
of the material which means that diffusion occurs quicker. This can be
explained by the Stokes-Einstein equation—the relationship between the radius
of the drug molecules and its diffusion coefficient in which the particles are spherical.
D = kT/6пŋa
This equation shows that the diffusion coefficient is
directly proportional to the T, temperature. Thus, rate of diffusion is higher
at higher temperature. For bromothymol blue,
the diffusion coefficient at 28oC is higher than that of 37oC. Some mistakes
might occur which cause the inaccuracy of this result. This might due to the
inconsistency of readings as different person are in charge of taking the
measurement for different days. It might be different for the technique of
different person observing the change of the dyes which the colour shown on it
is not very obvious.
Molecular size of particles is factor affecting the rate of
diffusion. It is found that crystal violet diffuse faster than bromothymol blue
solution in this experiment. Crystal
violet with molecular formula C25N3H30Cl has molecular weight of 407.979 g
mol-1 while bromothymol blue solution with molecular formula C27H28Br2O5S has
molecular weight of 624.38 g mol−1. It would be easier for the molecules of
smaller size to move into the restricted space between agar molecules. The
bromothymol blue molecules of bigger size might trap at the small space and
fail to move forward. The molecular weight of crystal violet is smaller than
that of bromothymol blue. Thus, the diffusion rate of the crystal violet is
faster than the bromothymol blue.
Conclusion
The diffusion coefficient of crystal violet at 28oC is =
5.015 x 10-11 m2s-1 while at 37oC is 6.408 x 10-11 m2s-1. Diffusion coefficient
of bromothymol blue at 28oC is 7.764 x 10-11 m2s-1while at 37oC is 7.139 x
10-11 m2s-1the factor that influence the rate of diffusion is concentration,
temperature and molecular weight since the surface area, Permeability is kept
constant. diffusion rate is faster in the concentration of diffusing molecules
1:200> 1:400> 1:600.
Refence
1. A.T.Florence and D.Attwood. (1998). Physicochemical
Principals of Pharmacy, 3rd Edition. Macmillan Press Ltd.
2. Physical Pharmacy and Pharmaceutical Sciences, by Patrick
J. Sinko, 5th Edition
3.http://www.coursework.info/GCSE/Chemistry/Aqueous_Chemistry/Factors_affecting_the_rate_of_Diffusion_L38515.html
