To determine the value for diffusion coefficient, D.
Introduction
Diffusion, which is the spontaneous
movement of solutes from an area of high concentration to an area of low
concentration can be explained by Fick’s law which states that the flux of
material (amount dm in time dt) across a given plane (area A) is proportional
to the concentration gradient dc/dx.
D is the diffusion coeeficient 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 satted as
M=M0 exp(-x2/4Dt)--------------------(ii)
Where M is the concentration at distance x
from the intersection between water and solution that is measured at time t.
By changing equation (ii) to its
logarithmic form, we get
ln M= ln M0-x2/4Dt
Or 2.303 x 4D (log10-log10
M) t=x2-------------------(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 (log10-log10 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 Stoke-Einstein equation.
D=Kt/6πƞa
Where k is the Boltzman constant 1.38x1023
Jk-1, T temperature in Kelvin, π the viscosity of the solvent in Nm-2s
and a the radius of particle in M. The volume of a spherical particle is 4/3πa3
, thus its weight M is equivalent to 4/3πa3Nρ(ρ= density).
It is known that molecular weight M=mN (N
is Avogadro’s number 6.023x1023mol-1).
... M=4/3πa3Nρ-------------------(v)
Diffusion for charged particles, equation
(iii) needs to be modifie 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 gels forms an appropriate support system to be used in
diffusion studies for molecules in a medium of water.
Procedures
- 250ml agar was prepared in Ringer’s solution. The agar was divided into six test tubes and allowed to cool at room temperature.
- Agar was prepared in another test tube that has already been added with 1:500000 crystals violet, this will be used as the standard to measure the colour distance resulting from the crystal violet diffusion.
- Solutions of crystal violet were prepared in distilled water in the concentration 1:200, 1:400, and 1:600.
- 5ml of each crystal violet solution on the gels that was prepared; close to prevent evaporation and was stored at temperature 28⁰c and 37⁰c.
- The distance between the interfaces if this gel solution with the end of the crystal violet area that has colour equivalent to the standard was accurately measured.
- The average of several measurement was obtained, this value is x in meter.
- The value of x after 2 hours was recorded and the suitable time distance up till 2 weeks.
- Diffusion graph for value of x² (in M²) against time (in second) was plotted for each of the concentration used. The diffusion coefficient D was calculated from the slope of the graph at temperature 28⁰cand 37⁰c.
- The molecular weight of the crystal violet was calculated using the equation N and V.
- This experiment was repeated using bromothymol blue. Sufficient alkali has to be added into this solution to obtain the colour completely from this dye.
Results
Calculations
|
System
|
Time (hr)
|
X (m)
|
X² (m²)
|
Gradient of
graph (m²hrˉ¹)
|
D (m²hrˉ¹)
|
Temperature
(ºc)
|
Average
of Diffusion
Coefficient (m²hrˉ¹)
|
|
Crystal Violet
with dilution 1:200
|
0
20
44
68
140
164
188
212
|
0.020
0.025
0.029
0.033
0.055
0.060
0.067
0.074
|
0.000400
0.000625
0.000841
0.001089
0.003025
0.003600
0.004489
0.005476
|
0.0045-0.0012
190-50
= 2.357X10-5
|
4.083X10-7
|
27ºc
|
3.154X10-7
|
|
Crystal Violet
with dilution 1:400
|
0
20
44
68
140
164
188
212
|
0.015
0.018
0.021
0.023
0.033
0.038
0.042
0.048
|
0.000225
0.000324
0.000441
0.000529
0.001089
0.001444
0.001764
0.002304
|
0.00176-0.0004
190-45
= 9.379X10-6
|
3.213X10-7
|
27ºc
|
|
|
Crystal Violet
with dilution 1:600
|
0
20
44
68
140
164
188
212
|
0.011
0.012
0.014
0.016
0.020
0.021
0.023
0.025
|
0.000121
0.000144
0.000196
0.000256
0.000400
0.000441
0.000529
0.000625
|
0.0006-0.0002
210-45
= 2.424X10-6
|
2.167X10-7
|
27ºc
|
Graph x²( m²) in the function of time (hour) of different Crystal Violet
system at 27˚C
|
System
|
Time (hr)
|
X (m)
|
X² (m²)
|
Gradient of graph (m²hrˉ¹)
|
D (m²hrˉ¹)
|
Temperature
(ºc)
|
Average
of Diffusion Coefficient (m²hrˉ¹)
|
|
Crystal Violet with dilution 1:200
|
0
20
44
68
140
164
188
212
|
0.016
0.028
0.033
0.039
0.052
0.061
0.069
0.078
|
0.000256
0.000784
0.001089
0.001521
0.002704
0.003721
0.004761
0.006084
|
0.0048-0.0012
190-50
= 2.571X10-5
|
5.057X10-7
|
37ºc
|
4.072X10-7
|
|
Crystal Violet with dilution 1:400
|
0
20
44
68
140
164
188
212
|
0.014
0.020
0.027
0.035
0.050
0.055
0.059
0.063
|
0.000196
0.000400
0.000729
0.001225
0.002500
0.003025
0.003481
0.003969
|
0.0030-0.0010
160-50
= 1.818X10-5
|
3.754X10-7
|
37ºc
|
|
|
Crystal Violet with dilution 1:600
|
0
20
44
68
140
164
188
212
|
0.012
0.013
0.014
0.020
0.030
0.032
0.034
0.037
|
0.000144
0.000169
0.000196
0.000400
0.000900
0.001024
0.001156
0.001369
|
0.0012-0.0004
190-60
= 6.154X10-6
|
3.406X10-7
|
37ºc
|
|
|
| |||||||
Graph x²(m²) in the function of time (hour) of different Crystal Violet system at 37˚C
|
System
|
Time (hr)
|
X (m)
|
X² (m²)
|
Gradient of graph (m²hrˉ¹)
|
D (m²hrˉ¹)
|
Temperature
(ºc)
|
Average
of Diffusion Coefficient
(m²hrˉ¹)
|
|
Bromothymol blue with
dilution
1:200
|
0
20
44
68
140
164
188
212
|
0.020
0.023
0.028
0.035
0.048
0.056
0.062
0.067
|
0.000400
0.000529
0.000784
0.001225
0.002304
0.003136
0.003844
0.004489
|
0.0039-0.0020
200-100
= 1.90X10-5
|
1.418X10-7
|
27ºc
|
1.01X10-7
|
|
Bromothymol blue with
dilution
1:400
|
0
20
44
68
140
164
188
212
|
0.013
0.015
0.018
0.022
0.031
0.035
0.040
0.046
|
0.000169
0.000225
0.000325
0.000484
0.000961
0.001225
0.001600
0.002116
|
0.0016-0.0005
190-60
= 8.46X10-6
|
1.052X10-7
|
27ºc
|
|
|
Bromothymol blue with
dilution
1:600
|
0
20
44
68
140
164
188
212
|
0.011
0.012
0.014
0.015
0.020
0.021
0.023
0.024
|
0.000121
0.000144
0.000196
0.000225
0.000400
0.000441
0.000529
0.000576
|
0.0005-0.0002
190-50
= 2.143X10-6
|
5.573X10-8
|
27ºc
|
|
|
|
|||||||
Graph x²( m²) in the function of time (hour) of different Bromothymol Blue system at 27˚C
|
System
|
Time (hr)
|
X (m)
|
X² (m²)
|
Gradient of graph (m²hrˉ¹)
|
D (m²hrˉ¹)
|
Temperature
(ºc)
|
Average
of Diffusion Coefficient
(m²hrˉ¹)
|
|
Bromothymol blue with
dilution
1:200
|
0
20
44
68
140
164
188
212
|
0.020
0.026
0.031
0.037
0.050
0.056
0.063
0.068
|
0.000400
0.000676
0.000961
0.001369
0.002500
0.003136
0.003969
0.004624
|
0.0035-0.001
170-40
= 1.923X10-5
|
3.195X10-7
|
37ºc
|
2.055X10-7
|
|
Bromothymol blue with
dilution
1:400
|
0
20
44
68
140
164
188
212
|
0.015
0.019
0.024
0.029
0.039
0.043
0.047
0.050
|
0.000225
0.000361
0.000576
0.000841
0.001521
0.001849
0.002209
0.002500
|
0.0023-0.0007
200-50
= 1.07X10-5
|
1.753X10-7
|
37ºc
|
|
|
Bromothymol blue with
dilution
1:600
|
0
20
44
68
140
164
188
212
|
0.012
0.014
0.017
0.019
0.027
0.028
0.030
0.032
|
0.000144
0.000196
0.000289
0.000361
0.000729
0.000784
0.001024
|
0.0009-0.0003
190-50
= 4.286X10-6
|
1.237X10-7
|
37ºc
|
Graph x²( m²) in the function of time (hour) of different Bromothymol Blue system at 37˚C
1)
Crystal Violet system with
dilution 1:200 (27ºC)
From equation: 2.303 x 4D (log 10 Mo – log 10 M ) t = X²
Gradient of the graph = 2.303 x 4D (log 10 Mo - log 10 M )
Gradient of the
graph = 2.357X10-5 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 )] = 2.357X10-5
m²/hour
D = 1.418X10-7
m²/hour
2)
Crystal Violet system with
dilution 1:400 (27ºC)
Gradient of the
graph = 9.379X10-6 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 )] = 9.379X10-6 m²/hour
D = 1.052X10-7 m²/hour
3)
Crystal Violet system with
dilution 1:600 (27ºC)
Gradient of the
graph = 2.424X10-6 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 )] = 2.424X10-6 m²/hour
D =5.573X10-8 m²/hour
4)
Average of Diffusion
Coefficient, m²/hour for Crystal Violet system at 27ºC
= [(1.418X10-7
m²/hour)+ (1.052X10-7 m²/hour)+( 5.573X10-8 m²/hour]/3
=1.01X10-7
m²/hour
5)
Crystal Violet system with
dilution 1:200 (37ºC)
Gradient of the graph = 2.571X10-5
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
)] = 2.571X10-5m²/hour
D = 3.195X10-7
m²/hour
6)
Crystal Violet system with
dilution 1:400 (37ºC)
Gradient of the graph = 1.818X10-5 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 )] = 1.818X10-5 m²/hour
D = 1.753X10-7
m²/hour
7)
Crystal Violet system with
dilution 1:600 (37ºC)
Gradient of the graph = 6.154X10-6 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 )] = 6.154X10-6
m²/hour
D = 1.237X10-7 m²/hour
8)
Average of Diffusion
Coefficient, m²/hour for Crystal Violet system at 37ºC
= [(3.195X10-7
m²/hour)+( 1.753X10-7m²/hour)+( 1.237X10-7 m²/hour)]/3
= 2.055X10-7
m²/hour
9)
Bromothymol Blue system with
dilution 1:200 (27ºC)
From equation: 2.303 x 4D (log 10 Mo – log 10 M ) t = X²
Gradient of the graph = 2.303 x 4D (log 10 Mo - log 10 M )
1.90X10-5 m²/hour
= 2.303 x 4D (log 10 Mo - log 10 M )
M=1:500,000 (standard) Mo=1:200
=1/500,000 =1/200
=2x10-6 =5x10-3
2.303 x 4D (log 10 Mo – log 10 M ) = 1.90X10-5
m²/hour
2.303x4D [log 10 (5x10-3 )-log 10
(2x10-6 )] = 1.90X10-5 m²/hour
D = 4.083X10-7 m²/hour
10) Bromothymol Blue system with dilution 1:400 (27ºC)
Gradient of the graph = 8.46X10-6 m²/hour
M=1:500,000 (standard) Mo=1:400
=2x10-6 =2.5x10-3
2.303 x 4D (log 10 Mo – log 10 M ) = 8.46X10-6
m²/hour
2.303 x 4D [log 10 (2.5x10-3 ) - log 10 (2x10-6 )] = 8.46X10-6m²/hour
D = 3.213X10-7 m²/hour
11)
Bromothymol Blue system with
dilution 1:600 (27ºC)
Gradient of the
graph = 2.143X10-6 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 )] = 2.143X10-6
m²/hour
D = 2.167X10-7
m²/hour
12)
Average of Diffusion
Coefficient, m²/hour for Bromothymol Blue system at 27ºC
= [(4.083X10-7 m²/hour)+ (3.213X10-7 m²/hour)+ (2.167X10-7
m²/hour)]/3
= 3.154X10-7 m²/hour
13) Bromothymol Blue system with dilution 1:200 (37ºC)
Gradient of the
graph = 1.923X10-5 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 )] = 1.923X10-5 m²/hour
D = 5.057X10-7 m²/hour
14) Bromothymol Blue system with dilution 1:400 (37ºC)
Gradient of the
graph = 1.071X10-5m²/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 )] = 1.071X10-5 m²/hour
D = 3.754X10-7
m²/hour
15) Bromothymol Blue system with dilution 1:600 (37ºC)
Gradient of the graph =4.286X10-6 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 )] = 4.286X10-6m²/hour
D = 3.406X10-7 m²/hour
16)
Average of Diffusion Coefficient, m²/hour for Bromothymol
Blue system at 37ºC
= [(5.057X10-7
m²/hour)+( 3.754X10-7 m²/hour)+( 3.406X10-7 m²/hour)]/3
= 4.072X10-7
m²/hour
Questions
- From the final value for D27ºc , estimate the value
of D37ºc by using following equation .
D27ºC
= T27ºC
D37ºC T37ºC
Where
ŋ1 and ŋ2 are viscosity of water at 27ºC and 37ºC .
Will
the value of D37ºc is the same as final value? Give some explanation if they
are different.
D27ºC
= 3.154X10-7 m²/hour
D27ºC = T27ºC
D37ºC T37ºC
3.154X10-7
= 27+273.15
D37ºc 37+273.15
D37ºc = 3.259X10-7 m²/hour
The D37ºC value is 3.259X10-7 m²/hour while the trial value is 4.389X10-7
m²/hour . There is slightly differences between these two values . This is
because there is error occurred during the trial . For example parallax error
when reading the measure , temperature at the room are not constant at 27ºC ,
the viscosity of agar in the test tube are not uniform , the “broken“ agar
medium formed and others .
- Between Crystal Violet and Bromothymol Blue ,
which will diffuse much faster . Explain it .
Crystal Violet will diffuse much faster than Bromothymol
Blue because the diffusion coefficient of Crystal Violet is larger than
diffusion coefficient of Bromothymol Blue.
From
the formula , M = 4/3Πa³NP
a³ = 3M / 4ΠNP
a
= ³√ 3M / 4ΠNP ----------------
equation (1)
where
M = molecular weight
N = Avogadro’s number
(6.02X10²³molˉ¹)
P
= density
a
= ratio of particles
The size of particles , a is proportional to
the molecular weight ( M ) . The smaller of size of the particle , the easier
for the particle to diffuse through the agar medium . It is proved through the
following equation :
D
= KT
6Πŋa
------------------------- equation (2)
= KT_________
6Πŋ³√ 3M / 4ΠNP
= KT³√4 ΠNP ___
6Πŋ³√ 3M
where K = Boltzmamn
constant ( 1.38X10²³JKˉ¹ )
T = Temperature in Kelvin
Ŋ = viscosity of solvent ( Nmˉ²s )
a
= radius of a particle
D = diffusion coefficient
Through equation shown above , it shows that
D increases when M decreases . This means that D is inversely proportional to M
. Therefore , a smaller molecular weight of Crystal Violet diffuse faster than Bromothymol
Blue .
Discussion
Diffusion is a
passive process which is driven by the difference of the gradient of the
chemical potential, where the solute will spontaneously diffuse from a region
of higher chemical potential (concentration gradient) to a region of lower
chemical potential concentration gradient whilst the solvent molecules will
move in reverse direction. If we assume the solute is moving in x direction,
then the solvent molecules will moving in opposite direction, which is –x
direction. Hence, this creates a net flow of solute in the x direction, if
concentration of solute is high.
In this practical, the equation given is:
ln M = ln Mo –x2/4Dt
or 2.303 x 4D (log10 Mo –
log10 M) t = x2
The graph of x2 versus t
is plotting and a straight line should be obtained which is the gradient of the
graph, showing 2.303 x 4D (log10 Mo – log10 M).
Hence, we can know that in both system of 28oC and 37oC,
diffusion is faster in 1: 200 than 1: 400, followed by 1:600 through the
decreasing value of D in this sequence. M is the system with dilution 1:
500,000 which acts as standard during the experiment. When Mo
increases, (log10 Mo – log10 M) also increase,
causing the concentration gradient to be bigger, then the driving force for the
occurrence of diffusion would be bigger, and diffusion becomes faster and
favorable.
Temperature is an important factor influencing the diffusion rate.
The diffusion rate at 37oC is faster than 28oC. We can
use Strokes-Enstein Law to explain this, that is
D = kT/6 ทa
since the D is
directly proportional to T, temperature, this leads to higher diffusion
coefficient by increasing temperature. It is due to the energy provided from
temperature causing the molecules vibrate vigorously and diffuse faster
relatively through the agar medium.
From the equation, the diffusion coefficient, D is also influenced
by particle size, a(inversely proportional), and viscosity, ท(inversely proportional – less space can pass through by increasing
viscosity of medium).
However,
to get all these theories right, some conditions are considered. For example,
the solutes should have low molecular weight, its diffusion rate through the
agar medium is almost equivalent to its diffusion rate in its own solution, and
there is no chemical reaction or adsorption to occur. Besides, to reduce
ionization of charged ionic groups of solute(not in this case of agar),
suitable electrolyte like NaCl is added into the gel medium to avoid adsorption
and exchange of ions, which will slow down the diffusion.
In conclusion, if we want to increase the diffusion rate, we need to
increase the diffusion coefficient, D that is lowering the viscosity of agar
medium, decreasing the size of particle that need to pass through the agar and
increasing the temperature surrounding the test tube.
In this experiment, 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. This lead to the agar cannot
completely crystallize and some suspension of agar powder will occur leads to
the concentration of agar solid is not equal. If the agar not complete
solidify, this indicators will diffuse faster and the real diffusion
coefficient of each indicators is not accurate anymore. As refer to the graph
we plotted, there are many points that the straight line does not pass through.
This can be explained by parallax errors when taking the x, diffusion distance.
Since the Crystal Violet particles diffuse spontaneously, so the x value taken
is from personal judgement and estimation. Besides, the temperature, T around
the test tube is always not constant, such as the temperature 270C,
room temperate is not kept constant for the day and night. Indirectly this will
influent the diffusion coefficient and so diffusion rate.
Conclusion
The value of D for Crystal Violet at 27ºc is 3.154X10-7 m²/hour.
The value of D for Crystal Violet 37ºc is 4.072X10-7 m²/hour.
The value of D for Bromothymol Blue at 27ºc is 1.01X10-7 m²/hour.
The value of D for Bromothymol Blue at 37ºc is 2.055X10-7 m²/hour.
The temperature and concentration of diffusing molecules will affect the value of D ( diffusion coefficient ) .
References
- Physicochemical Principles of Pharmacy , 2nd Edition ( 1988 ) A.T Florence and D.ettwood
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