Practical 4:
Determination of Diffusion Coefficient
Objective:
To determine the value of diffusion
coefficient, D
Introduction:
Diffusion coefficient is proportionality constant between the molar flux due to molecular diffusion and the gradient in the concentration of the species. Diffusivity is encountered in Fick's law which 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 across a given plane(area A) is proportional to the concentration gradient dc/dx.
Diffusion coefficient is proportionality constant between the molar flux due to molecular diffusion and the gradient in the concentration of the species. Diffusivity is encountered in Fick's law which 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 across a given plane(area A) is proportional to the concentration gradient dc/dx.
dm
= -DA(dc/dx)dt ……… (i)
where
D is the diffusion coefficient or diffusivity for the solute that has unit as
m²s⁻¹.
If a solution which have neutral particles
with concentration, MO, is placed within a cylindrical tube next to a water
column, diffusion can be stated as
M = M0 exp (-x2/4Dt) ……… (ii)
where
M is the concentration at distance x from the intersection between water and
solution that measured at time t.
By changing equation (ii) to logarithmic form, we can
obtain
ln M
= (ln M0 )(-x²/4Dt)
or 2.303 x 4D (log 10 M0 –log
10 M) t = x² ……… (iii)
Thus, one x² versus t graph can produce a
straight line which cross the origin with its gradient 2.303 x 4D (log 10 M0–log
10 M). From here, D can be calculated.
If the particles in the solution are
assumed to be a spherical, then the size and weight of the molecules can be
counted from Stokes-Einstein equation.
D =
kT/6 πŋa ……… (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).
As we know that the molecular mass of M = mN
(N is the Avogadro’s number 6.02 x 1023 mol-1 ).
M = 4/3πa³Nρ
……… (v)
Diffusion for molecules with charges,
equation (iii) has to be
changed to insert the gradient force effect that exists between the solution
and the solvent. However, this can be overcome by adding a little of sodium
chloride into the solvent to prevent the forming of this gradient force.
Agar gel contains a semi-solid molecular
net that can be interfering by water molecules. The water molecules will form a
continuous phase in the agar gel. By this, the solute molecules can be diffused
freely in the water, if not there will be no chemical interaction and diffusion
occur. Thus, these agar gels provide a suitable supportive system that can be
used in the experiment for diffusion of certain molecules in a aqueous medium.
Material:
Agar powder, Ringer 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
Apparatus:
500
mL beaker, measuring cylinder, glass rod, 16 test tubes with covers, hot plate.
Procedure:
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 pour into each 8 test tubes. The test tubes were
then 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 pour into each test tubes.
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 weeks.
11. Procedures
3 to 10 were repeated for Bromothymol Blue solutions.
12. Graph
of x² values ( in M²) versus time ( in hours) was potted.13. The
diffusion coefficient , D was determined from the graph gradient for both 28ºC and 37ºC ; the molecular mass of crystal violet and
bromothymol blue were also determined by
using N and V equation.
Results:
Calculation:
1. Crystal Violet system with dilution 1:200 (28ºC)
1. Crystal Violet system with dilution 1:200 (28ºC)
From equation 2.303 x 4D (Log 10 MO – log 10 M) t = X²
Therefore the gradient of the graph = 2.303
x 4D (log 10 Mo - log 10 M)
Gradient of the graph= (18.49-0.81)/(330-18) x 10^-1m2/hour
= 5.67X10⁻⁶ m2/hour
M = 1:500000
Ma = 1:200
= 1 / 500000
= 1 / 200
= 2 x 10-6
=
5 x 10-3
2.303 x 4D (log 10 Mo – log 10 M) =5.67X10⁻⁶ m2/hour
2.303x4D [log 10 (5x10-3 )-log 10 (2x10-6 )]
= 5.67X10⁻⁶ m2/hour
D = 1.81x10⁻⁷m2/hour
2. Crystal Violet system with
dilution 1:400 (28ºC)
Gradient = (12.25-0.16)/(258-18)x 10^-1m2/hour
= 5.03
x 10⁻⁶ m²hour⁻¹
M =
1:500000
Ma = 1:400
= 1 /
500000 = 1 / 400
= 2 x 10-6 =
2.5 x 10-3
2.303 x 4D (log 10
Mo – log 10 M) = 5.03X10⁻⁶ m2/hour
2.303x4D [log 10 (2.5x10-3 )-log 10 (2x10-6 )]
= 5.03X10⁻⁶ m2/hour
D= 1.76 x10⁻⁷m2/hour
3.
Crystal violet system with dilution 1:600 (28˚C)
Gradient of the graph= (2.25-0.04)/(306-18)x 10^-1m2/hour
= 7.67 x 10 ⁻⁷ m²/hour
M =
1:500000
Ma = 1:600
= 1 /
500000
= 1 / 600
= 2 x 10-6 =
1.67 x 10-3
2.303 X4D [log₁₀M₀ -log₁₀M) = 7.67 X10⁻⁷m²hour⁻¹
2.303 X 4D [log₁₀(1.67X10⁻³)-log₁₀(2X10⁻⁶)]=7.67X10⁻⁷ m²hour⁻¹
D =2.85 x 10⁻⁸ m²hour⁻¹
Average of Diffusion Coefficient, m²hour⁻¹ for Crystal Violet
system at 28ºC
=(1.81 x10⁻⁷+1.76 x10⁻⁷+2.95 x10⁻8) m²hour⁻¹
=1.285 x10⁻⁷
m²hour⁻¹
5. Crystal violet system with dilution 1:400 (37˚C)
Gradient of the graph= (18-2)/ (300-80)
=7.27 x 10⁻⁶ m²/hour
4. Crystal violet system with dilution 1:200 (37˚C)
Gradient of the graph= (16-7) / (220-120)
= 9.0 x10⁻⁶ m²/hour
=5×10-3 m²/hour
M =
1:500000
Ma = 1:200
= 1 /
500000
= 1 / 200
= 2 x 10-6 =
5 x 10-3
2.303 X4D [log₁₀M₀ -log₁₀M) = 9.0 X10⁻⁶m²/hour
2.303 X 4D [log₁₀(5X10⁻³)-log₁₀(2X10⁻⁶)]=9.0X10⁻⁶ m²/hour
D = (9.0×10-6) / 2.303 x 4 (log10 (5x10-3) – log10 (2x10-6))
=2.87 x 10⁻⁷ m²/hour
Gradient of the graph= (18-2)/ (300-80)
=7.27 x 10⁻⁶ m²/hour
M =
1:500000
Ma = 1:400
= 1 /
500000 = 1 / 400
= 2 x 10-6 =
2.5 x 10-3
2.303 X4D [log₁₀M₀ -log₁₀M) = 7.27 X10⁻⁶m²/hour2.303 X 4D [log₁₀(2.5X10⁻³)-log₁₀(2X10⁻⁶)]=7.27X10⁻⁶ m²/hour
D = (7.27×10-6)/ 2.303 x 4 ( log10 (2.5x10-3) – log10 (2x10-6))
= 2.55 x10⁻⁷ m²/hour
6. Crystal violet system with dilution 1:600 (37˚C)
Gradient of the graph= (2-0)/(160-20)m²/hour
= 1.42 x 10⁻⁶ m²/hour
=1.67 x 10⁻³m²/hour
M =
1:500000
Ma = 1:600
= 1 /
500000
= 1 / 600
= 2 x 10-6 =
1.67 x 10-3
2.303 X4D [log₁₀M₀ -log₁₀M) = 1.42 X10⁻⁶m²/hour2.303 X 4D [log₁₀(1.67X10⁻³)-log₁₀(2X10⁻⁶)]=1.42 X10⁻⁶m²/hour
D = (1.42×10-6)/ 2.303 x 4 (log10 (1.67x10-3) – log10 (2x10-6))
=5.28 x 10 ⁻⁸ m²/hour
Average of Diffusion Coefficient, m²/hour for Crystal Violet system at 37ºC
= ( 2.87x 10⁻⁷ m²/hour +2.55x 10⁻⁷ m²/hour +5.28x 10⁻⁸ m²/hour)/3
= 1.98x 10⁻⁷ m²/hour
7. Bromothymol Blue system with dilution 1:200 (28˚C)
Gradient of the graph= (8-1)/(140-20)
= 5.83 x 10⁻⁶ m²/hour
M =
1:500000
Ma = 1:200
= 1 /
500000
= 1 / 200
= 2 x 10-6 =
5 x 10-3
2.303 X4D [log₁₀M₀ -log₁₀M) = 5.83 X10⁻⁶m²/hour
2.303 X 4D [log₁₀(5X10⁻³)-log₁₀(2X10⁻⁶)]=5.83X10⁻⁶ m²/hour
D = 5.83x10-⁶/ 2.303 x 4 ( log10 (5x10-3) – log10 (2x10-6))
=1.86x10⁻⁷m²/hour2.303 X 4D [log₁₀(5X10⁻³)-log₁₀(2X10⁻⁶)]=5.83X10⁻⁶ m²/hour
D = 5.83x10-⁶/ 2.303 x 4 ( log10 (5x10-3) – log10 (2x10-6))
8. Bromothymol blue system with dilution 1:400 (28˚C)
Gradient of the graph= (13-3)/(260-80)
=5.56x10⁻⁶ m²/hour
M =
1:500000
Ma = 1:400
= 1 /
500000 = 1 / 400
= 2 x 10-6 =
2.5 x 10-3
2.303 X4D [log₁₀M₀ -log₁₀M) = 5.56x10⁻⁶m²/hour2.303 X 4D [log₁₀(2.5X10⁻³)-log₁₀(2X10⁻⁶)]=5.56X10⁻⁶ m²/hour
D = (5.56x10⁻⁶ ) / 2.303 x 4 ( log10 (2.5x10-3) – log10 (2x10-6))
=1.95x10⁻⁷m²/hour
9. Bromothymol blue system with dilution 1:600 (28˚C)
Gradient of the graph= (6-1)/ (220-100)x 10-4 m²/hour
= 4.16x10-6 m²/hour
= 1/ (500,000) = 1 /600
= 2 x 10 -6 = 1.67 x 10 -3
2.303 x4D [log10Mo - log10M]= 4.16x10-6 m²/hour
2.303 x 4D [log10(1.67x 10-3 )-log10(2x 10-6)]=4.16x 10-6 m²/hour
D = 4.16x10-6 / (2. 03 x 4 (log10 (1.67x10-3) – log10 (2x10-6))
=1.55 x10-7 m²/hour
Average of Diffusion Coefficient,m2hour-1, for Bromothymol Blue system at 28ºC
=(1.86x 10-7+ 1.95x 10-7+1.55x 10-7 )m²/hour / 3
=5.36x 10-7m²/hour
10. Bromothymol Blue system with dilution 1:200 (37˚C)
Gradient of the graph= (13-2)/(220-4) x 10-4 m²/hour
= 6.11x10-6 m²/hour
M = 1: 500,000 Mo = 1: 200
= 1/ (500,000) = 1 / 200
= 2 x 10 -6 = 5x 10 -3
2.303 x4D[log10Mo - log10M] = 6.11 x 10-6 m²/hour
2.303 x 4D [log10(5x10-3)-log10(2x10-6)]=6.11x 10-6 m²/hour
D = 6.11x10-6 / (2.303 x 4 (log10 (5x10-3) – log10 (2x10-6))
=1.95x10-7 m²/hour
11. Bromothymol Blue system with dilution 1:400 (37˚C)
Gradient of the graph= (13-4)/ (260-100) x 10-4m²/hour
= 5.63x10-6 m²/hour
M = 1: 500,000 Mo = 1: 400
= 1/ (500,000) = 1 /400
= 2 x 10 -6 = 2.5x 10 -3
2.303 x4D[log10Mo - log10M] = 5.63 x 10-6 m²/hour
2.303 x 4D [log10 (2.5x10⁻³)-log₁₀(2x10⁻⁶)]=5.63x10-6
D = 5.63x10 -6 / (2.303 x 4 (log10 (2.5x10-3) – log10 (2x10-6))
=1.97x10-7 m²/hour
12. Bromothymol Blue system with dilution 1:600 (37˚C)
Gradient of the graph= (6-1)/ (220-100) x10-4 m²/hour
= 4.17x10-6 m²/hour
M = 1: 500,000 Mo = 1: 600
= 1/ (500,000) = 1 /600
= 2 x 10 -6 = 1.67 x 10 -3
2.303 x4D [log10Mo - log10M] = 4.17 x10-6 m²/hour
2.303 x 4D [log10 (1.67x 10-3 )-log10(2x 10-6)]=4.17x10-6 m²/hour
D = 4.17 x10 -6 / (2.303 x 4 (log10 (1.67x10-3) – log10 (2x10-6))
=1.55x10-7 m²/hour
Average of Diffusion Coefficient, m2hour-1, for Crystal Violet system at 37ºC
=(1.95x10-7+1.97x10-7+1.55x10-7)m²/hour/3
=5.47x10-7m²/hour
Questions:
(1)
From the experiment value for D28ºC, estimate the value of D37ºC by using the
following equation:
D28ºC = T28ºC
D37ºC = T37ºC
D37ºC = T37ºC
where ŋ1and ŋ2 are
viscosity of water at 28ºC and 37ºC.
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?
Give some explanation if it is different. Is there any difference between the calculated molecular weight with the real molecular weight?
D28ºC=1.285x10⁻⁷ m²hour⁻¹
1.285x10⁻⁷ = 28+273.15
D37ºC 37+273.15
D37ºC =1.32x10⁻⁷m²hour⁻¹
The value of D37ºC is 1.32x10⁻⁷m²hour⁻¹ while the trial value is 1.98x10⁻⁷m²hour⁻¹.There is a slightly difference between these two values, where it is less of 6.6x10-⁸ of D37ºC value from the trial value. The differences may due to the some mistakes during the practical. For example:
· the room temperature ie. the laboratory
temperature is not constant at 28 ºC
· parallax error might occur when taking the readings
· the viscosity of agar in the test tube is not
uniform, causing the diffrerence of the rate of diffusion of the dye.
(2) Between Crystal Violet and Bromothymol
Blue, which one diffuse quicker? Explain if there are any differences in the
diffusion coefficient values?
Crystal
Violet diffuses faster than Bromothymol Blue as the diffusion coefficient
of Crystal Violet is larger that of
Bromothymol Blue. This is because crystal violet has a smaller molecular weight
(408 g/mol) as compared to the molecular weight of Bromothymol Blue (624g/mol).
According to the equation,
M = 4/3лa³Nρ ……(v)
a³ = 3 M/ 4лNρ
a = ( 3 M/ 4лNρ )
⅓
where
M = molecular weight, a = radius of particles, N = Avogadro’s number (6.02x10²³
mol-1 ), ρ= density, we can deduce that the size of molecule, a is
directly proportional to the molecular weight,M. The smaller the size
of.molecules, the faster they diffuse through agar , which can be proved by
following equation:
D = kT/6лŋa
…… (iv)
where D = diffusion coefficient, k is the
Boltzmann constant 1.38 x 1023 Jk-1 , T is the
temperature in Kelvin, ŋ is the viscosity of the solute, in Nm-2s and a half
diameter of molecule in M. It is shown that the diffusion coefficient is
inversely proportional to the size of molecules. As a result, the smaller the
molecular weight, the larger the diffusion coefficient,ie. the faster the
molecules diffuse through agar.
Discussion:
Diffusion is one of the fundamental processes by which
material moves. Diffusion is a consequence of the constant thermal motion of
atoms, molecules, and particles, and results in material moving from areas of
high to low concentration. . The speed of mixing by diffusion depends on three
main parameters: temperature, size (mass) of the diffusing particles, and
viscosity of the environment.
Fick's
first law is used in steady-state diffusion, i.e., when the concentration
within the diffusion volume does not change with respect to time. The
expression which relates the flow of material to the concentration of gradient
is referred to as Fick’s first law equation. Fick's laws of diffusion describe diffusion and can be used to solve for the diffusion coefficient D.This
equation states that the amount of the dm substance diffuse in the direction x at the time dt go through an area, A
is proportional with the concentration gradient dc/dx in the area .While D is the diffusion coefficient.
dm= -DA (dc/dx) dt ---------(1)
In this experiment, the diffusion
particles are neutral with Mo
concentration, the agar medium is considered homogeneous and with constant
concentration, hence, the diffusion can be express as,
M
= Mo eksp (-x2/ 4Dt) ---------(2)
Change the equation (2) to
logarithm form we have
ln
M =
ln Mo –x²/4Dt
2.303 x
4 D ( log10 Mo – log10 M) t = x2
---------(3)
Therefore when a graph x² against time, t is
plotted, a straight line is obtained with the gradient of 2.303 x 4D ( log10 Mo – log10 M).
From here, D can be calculated. We
can know that both of the 28˚C and 37˚C system, the rate of diffusion from the
result that is 1:200 > 1:400 > 1:600
For this practical, there are three
main factors affecting the rate of diffusion or diffusion coefficient.
First of all, they are concentration
of the solutes. M is the system with the dilution 1:500,000. which function as
a standard system during this practical. When Mo is increased, the value of(log
10 Mo- log 10 M)
will increase. This causes the concentration gradient become larger and thus
the driving force that cause the diffusion to occur would be larger and the
rate of diffusion process will increase. Hence, when concentration of solutes
increases, the diffusion coefficient increases. Therefore, for both Crystal
Violet and Bromothymol Blue system at 28 ºC and 37ºC, the diffusion is the
fastest in 1:200 system followed by 1:400 and the slowest is 1:600.
Besides that, size or molecular
weight of the solute particles. In this practical, Crystal Violet diffuses
faster than Bromothymol Blue due to smaller size or molecular weight of Crystal
Violet particles. Another things is temperature of the system. The diffusion
rate at 37ºC is higher than at 28ºC. This can be explained by the Stokes-Einstein
equation in which the particles are spherical.
D = kT/6пŋa
According
to this equation, we can deduce that the diffusion coefficient,D is directly
proportional to the temperature, T. This is because when the temperature is
increased, the particles gain more energy, hence they can diffuse more rapidly
through the agar medium.
Besides that, the agar gel also can
influence the rate of the diffusion. When the concentration of the gel
substances increase, the size of the hole will decrease and the diffusion rate
will decrease too as the size of hole same with the size of the diffuse
molecule. Moreover, the viscosity of the solution in the hole is also can
influence the diffusion rate. When the crystallinity of the gel medium
increased, the diffusion rate will decrease. The larger the volume fraction of
crystalline material, the slower the movement of diffusion molecule. This can
be happened because crystalline regions of the gel medium represent an
impenetrable barrier to the movement of the solute particles where it have to
circumnavigate through it.
Conclusion:
The diffusion coefficient, D value for crystal violet system at 28˚C is
1.285x10⁻⁷m²hour⁻¹ whereas 1.98x10⁻⁷m²hour⁻¹at 37˚C. The
diffusion coefficient, D value for bromothymol blue system at 28˚C is 5.36x10⁻⁷m²hour⁻¹ whereas 5.47x10⁻⁷m²hour⁻¹ at 37˚C. The diffusion system by using crystal violet will have higher
diffusion coefficient than Bromothymol Blue due to lower molecular weight. As
temperature increase, D will also increase. As concentrarion of solute
increase, the value of D will decrease. The diffusion rate is decreasing in the
concentration of diffusion molecule 1:200 >1:400> 1:600.
Reference:
1.A.T.Florence and D.Attwood. (1998).
Physicochemical Principals of Pharmacy, 3rd Edition. Macmillan Press Ltd.
2.Physical Pharmacy, by Alfred Martin, 4th
Edition.