Practical 4: Determination of diffusion coefficient
Theory
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
dc
dm = -DA ---- dt ---------------------- (i)
dx
D is the 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
= 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 =1n M0 - x2 /4Dt
or
2.303 x 4D (log10 M0 – 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 M0 – 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 Stokes-Einstein
equation.
D
= kT/6πηa
where k is the Boltzmann
constant 1.38 x 1023 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 4/3 πa3,
thus its weight M is equivalent to 4/3 πa3Nρ (ρ = density).
It
is known that molecular weight M=mN (N is Avogadro’s number 6.023 x 1023 mol-1).
∴ M
= 4/3 πa3Nρ ------------------- (v)
Diffusion
for charged particles, equation (iii) needs 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.
Experiment
Prepare
250m1 agar in Ringer's solution. Divide the agar into six test tubes; allow to cool at room
temperature. Prepare agar in another test tube that has already been added with
1:500,000 crystal violet, this will be used as the standard to measure
the color
distance resulting from the crystal violet diffusion. Prepare solutions of
crystal violet
in distilled water in the concentrations 1:200, 1:400 and 1:600. Place 5ml of
each crystal violet solution
on the gels that was prepared; close to prevent evaporation and store at temperature 28°C and 37°C. Measure accurately the distance between
the interface of this gel
solution with the end of the crystal violet area that has color equivalent to the standard. Obtain the average of
several measurements, this value is x in meter. Record values of x after 2 hours and at suitable time distances up
till 2 weeks (Table 1).
Plot the graph for values of x2 (in M2) against time (in seconds) for each of the concentration used. Calculate the diffusion
coefficient D from the slope of the graph at temperature 28°C and 37°C and calculate also the molecular weight of the crystal violet using
the equation N and V.
Results:
System
|
Time (s)
|
X
(x10-2m)
|
X²
(x10-4m²)
|
Gradient of graph (m²sˉ¹)
|
D (m²hrˉ¹)
|
Temperature
(ºc)
|
Average
of Diffusion Coefficient (m²hrˉ¹)
|
Crystal Violet with dilution 1:200
|
86400
172800
345600
518400
604800
691200
777600
|
1.3
2.0
2.5
3.1
4.0
4.2
4.5
|
1.69
4.00
6.25
9.61
16.00
17.64
20.25
|
(17.64-4.00)x10-4
691200-172800
= 3.055X10-9
|
9.760X10-11
|
28ºc
|
5.76X10-11
|
Crystal Violet with dilution 1:400
|
86400
172800
345600
518400
604800
691200
777600
|
1.3
1.5
2.3
3.0
3.3
3.5
3.7
|
1.69
2.25
5.29
9.00
10.89
12.25
13.69
|
(12.25-2.25) x10-4
691200-172800
= 1.929X10-9
|
6.762X10-11
|
28ºc
|
|
Crystal Violet with dilution 1:600
|
86400
172800
345600
518400
604800
691200
777600
|
0.3
0.4
0.7
0.9
1.1
1.1
1.2
|
0.09
0.16
0.49
0.81
1.21
1.21
1.44
|
(1.21-0.16)x10-4
691200-172800
= 2.025x10-10
|
7.524X10-12
|
28ºc
|
|
Graph x²( m²) in the function of time (s) of different Crystal Violet system at 27˚C
System
|
Time
(s)
|
X
(x10-2m)
|
X²
(x10-4m²)
|
Gradient
of graph (m²sˉ¹)
|
D
(m²sˉ¹)
|
Temperature
(ºc)
|
Average
of
Diffusion Coefficient (m²hrˉ¹)
|
Crystal
Violet with dilution 1:200
|
86400
172800
345600
518400
604800
691200
777600
|
1.7
2.8
3.0
3.3
3.5
4.1
4.5
|
2.89
7.84
9.00
10.89
12.25
16.81
20.25
|
(16.81-7.84)
x10-4
691200-172800
= 1.730x10-9
|
5.528X10-11
|
37ºc
|
4.030X10-11
|
Crystal
Violet with dilution 1:400
|
86400
172800
345600
518400
604800
691200
777600
|
1.0
1.3
1.8
2.3
2.8
3.3
3.5
|
1.00
1.69
3.24
5.29
7.84
10.89
12.25
|
(10.89-1.69)
x10-4
691200-172800
= 1.775X10-9
|
6.204X10-11
|
37ºc
|
|
Crystal
Violet with dilution 1:600
|
86400
172800
345600
518400
604800
691200
777600
|
0.1
0.5
0.6
0.7
0.9
1.0
1.1
|
0.01
0.25
0.39
0.49
0.81
1.00
1.21
|
(1.00-0.25)
x10-4
691200-172800
= 1.447X10-10
|
5.376X10-12
|
37ºc
|
|
Graph x²(m²) in the function
of time (hour) of different Crystal Violet system at
37˚C
|
|||||||
System
|
Time
(s)
|
X (x10-2m)
|
X²
(x10-4m²)
|
Gradient
of graph (m²sˉ¹)
|
D
(m²sˉ¹)
|
Temperature
(ºc)
|
Average
of
Diffusion Coefficient (m²sˉ¹)
|
Bromothymol
blue with
dilution
1:200
|
86400
172800
345600
518400
604800
691200
777600
|
1.3
1.6
2.1
2.7
3.0
3.3
3.5
|
1.69
2.56
4.41
7.29
9.0
10.89
12.25
|
(10.89-2.56)
x10-4
691200-172800
= 1.607X10-9
|
5.134X10-11
|
27ºc
|
4.845X10-11
|
Bromothymol
blue with
dilution
1:400
|
86400
172800
345600
518400
604800
691200
777600
|
1.1
1.5
2.1
2.7
2.9
3.0
3.1
|
1.21
2.25
4.41
7.29
8.41
9.00
9.61
|
(9.00-2.25)
x10-4
691200-172800
= 1.302X10-9
|
4.564X10-11
|
27ºc
|
|
Bromothymol
blue with
dilution
1:600
|
86400
172800
345600
518400
604800
691200
777600
|
0.5
0.7
1.1
1.4
1.6
1.7
1.8
|
0.25
0.49
1.21
1.96
2.56
2.89
3.24
|
(2.89-0.49)
x10-4
691200-172800
= 4.630X10-10
|
4.838X10-11
|
27ºc
|
|
Graph x²( m²) in the function
of time (hour) of different Bromothymol Blue system at 27˚C
|
|||||||
System
|
Time
(hr)
|
X
(x10-2m)
|
X²
(x10-4m²)
|
Gradient
of graph (m²sˉ¹)
|
D
(m²sˉ¹)
|
Temperature
(ºc)
|
Average
of
Diffusion Coefficient (m²sˉ¹)
|
Bromothymol
blue with
dilution
1:200
|
86400
172800
345600
518400
604800
691200
777600
|
1.4
1.7
2.3
2.7
3.1
3.2
3.3
|
1.96
2.89
5.29
7.29
9.61
10.24
10.89
|
(10.24-2.89)
x10-4
691200-172800
= 1.418X10-9
|
4.530X10-11
|
37ºc
|
2.409X10-11
|
Bromothymol
blue with
dilution
1:400
|
86400
172800
345600
518400
604800
691200
777600
|
0.9
1.4
1.7
2.0
2.1
2.2
2.4
|
0.81
1.96
2.89
4.0
4.41
4.84
5.76
|
(4.84-1.96)
x10-4
691200-172800
= 5.550X10-10
|
1.945X10-11
|
37ºc
|
|
Bromothymol
blue with
dilution
1:600
|
86400
172800
345600
518400
604800
691200
777600
|
0.3
0.4
0.7
1.0
1.1
1.1
1.2
|
0.09
0.16
0.49
1.0
1.21
1.21
1.44
|
(1.21-0.25)
x10-4
691200-172800
= 2.020X10-10
|
2.409X10-11
|
37ºc
|
Graph x²( m²)
in the function of time (hour) of different Bromothymol
Blue system at 37˚C
Calculations :
1(1) Crystal Violet system with dilution 1:200 (28ºC)
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 =
3.055X10-9 m²/s
M=1:500,000
(standard) Mo=1:200
=2x10-6 =5x10-3
2.303 x 4D [log 10 (5x10-3
) - log (2x10-6 )] = 3.055X10-9 m²/hour
D = 9.760X10-11 m²/s
2 (2) Crystal Violet system with dilution 1:400 (27ºC)
Gradient of the graph =
1.929X10-9 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 )] = 1.929X10-9 m²/hour
D = 6.762X10-11 m²/hour
3(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 )] = 2.025X10-10 m²/hour
D =7.524X10-12 m²/hour
4(4) Average of Diffusion Coefficient, m²/hour for
Crystal Violet system at 27ºC
= [(9.760X10-11 m²/s)
+ (6.762X10-11 m²/s) + (7.524X10-12 m²/s]/3
=5.76X10-11 m²/s
5(5) Crystal Violet system with dilution 1:200 (37ºC)
Gradient of the graph = 1.730X10-9 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 )] = 1.730X10-9m²/s
D = 5.528X10-11 m²/s
6(6) Crystal Violet system with dilution 1:400 (37ºC)
Gradient of the graph =
1.775X10-9 m²/s
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 )] = 1.775X10-9 m²/s
D = 6.204X10-11 m²/s
7(7) Crystal Violet system with dilution 1:600 (37ºC)
Gradient of the graph =
1.447X10-10 m²/s
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 )] = 1.447X10-10 m²/s
D = 5.376X10-12 m²/s
8(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
= 4.030X10-11 m²/s
9(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) = 1.607X10-9 m²/s
2.303x4D [log 10 (5x10-3 )-log 10 (2x10-6 )]
= 1.607X10-9 m²/s
D = 5.134X10-11 m²/s
1(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 )] = 1.302X10-9m²/s
D = 4.564X10-11 m²/s
1(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 )] = 1.302X10-9 m²/s
D = 4.838X10-11 m²/s
1(12) Average of Diffusion Coefficient, m²/hour for
Bromothymol Blue system at 27ºC
= [(4.838X10-11 m²/s) + (4.564X10-11 m²/s) +
(5.134X10-11 m²/s)]/3
= 4.845X10-11 m²/s
1(13) Bromothymol Blue system with dilution 1:200
(37ºC)
Gradient of the graph =
1.418X10-9 m²/s
M=1:500,000 (standard) Mo=1:200
=2x10-6 =5x10-3
2.303 x 4D [log 10 (5x10-3)
- log 10 (2x10-6 )] = 1.418X10-9 m²/s
D= 4.530X10-11 m²/s
1(14) Bromothymol Blue system with dilution 1:400
(37ºC)
Gradient of the graph =
5.550X10-10m²/s
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 )] = 5.550X10-10 m²/s
D = 1.945X10-11 m²/s
1(15) Bromothymol Blue system with dilution 1:600
(37ºC)
Gradient of the graph
=2.02X10-10 m²/s
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 )] = 2.02X10-10m²/s
D = 7.505X10-12 m²/s
1(16) Average of Diffusion Coefficient, m²/hour
for Bromothymol Blue system at 37ºC
= [(7.505X10-12 m²/s)
+ (1.945X10-11 m²/s) + (4.530X10-11 m²/s)]/3
= 2.409X10-11 m²/s
Questions :
1. From the final value for D28ºc , estimate the
value of D38ºc by using following equation .
D28ºC
= T28ºC
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.
For Crystal Violet:
D28ºC = 5.760X10-11 m²/s
D28ºC
= T27ºC
D37ºC
T37ºC
5.76X10-11 = 28+273.15
D37ºc
37+273.15
D37ºc = 5.93X10-11 m²/s
For Bromothymol Blue;
D28ºC = 4.845X10-11 m²/s
D28ºC = T28ºC
D37ºC
T37ºC
4.845X10-11 = 28+273.15
D37ºc
37+273.15
D37ºc =
4.99X10-11 m²/s
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 .
2. 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 280C, room temperate is not kept
constant for the day and night. Indirectly this will influent the diffusion
coefficient and so diffusion rate.
put the agar into the fridge
the condition of agar before become liquid form
Conclusion
The value of D for
Crystal Violet at 27ºc is 5.76X10-11 m²/s
The value of D for
Crystal Violet 37ºc is 4.03X10-11 m²/s.
The value of D for
Bromothymol Blue at 27ºc is 4.845X10-11 m²/s
The value of D for
Bromothymol Blue at 37ºc is 2.409X10-11 m²/s
The temperature and
concentration of diffusing molecules will affect the value of D ( diffusion
coefficient ) .
References :
1. Physicochemical Principles of Pharmacy , 2nd Edition
( 1988 ) A.T Florence and D.ettwood
2. Connors, Kenneth Antonio. 1932. A Textbook of
Pharmaceutical Analysis. Canada: John Wiley & Sons, Inc.
