lab 4: Determination of diffusion coefficient

Day of experiment:  Tuesday

Date of experiment: 20/5/2014

Experiment: Determination of diffusion coefficient

Objectives:

To determine the value of diffusion coefficient

Introduction:

Diffusion describes the spread of particles through random motion from regions of higher concentration to regions of lower concentration. The time dependence of the statistical distribution in space is given by the diffusion equation. Fick's laws of diffusion describe diffusion and can be used to solve for the  D. The diffusion coefficient (D) is the amount of a particular substance that diffuses across a unit area in 1 s under the influence of a gradient of one unit. It is usually expressed in the

units . It can be expressed in dm = -DA(dc/dx)dt (i).

Diffusion coefficient depends on size and shape of molecule, interaction with solvent and viscosity of solvent. If a solution which have neutral molecules with concentration, Mo, put in a slim tube next to a water tube, diffusion can be stated as

M = M0 exp(-x²/ 4 Dt)    ……… ()

where M is the concentration at x distance from the level between water and solution that measured at time t.

By changing equation () 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  ……… ()

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 counted.

​If the molecules in the solution are assumed to be a sphere shape, then the size and mass of the molecules can be counted from Stokes-Einstein equation.

D = kT/6пŋa  ……… ()

(D =kT/9  and  9 = 6пŋa )

where 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. The volume for that certain sphere molecule is 4/3пa³, thus the mass of M is equal to 4/3пa³ρ (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 1023mol-1 ).

M = 4/3лa³Nρ ……… ()

Diffusion for molecules with charges, equation () 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 is a semi-solid that can be interfered by water molecules. The water moleculesform a continuous phase in the agar gel. As such, the solute molecules can be diffused freely in the water. Thus, these agar gels provide a supportive system that can be used in the experiment for diffusion of certain molecules in a aqueous medium.

Materials : ​                             ​Apparatus :

Agar powder​                           ​500 mL beaker

Ringer solution                                                     ​5 mL pipette

1 : 500.000 crystal violet solution​       ​Glass rod

1 : 200 crystal violet solution ​   ​ ​14 test tubes with covers

1 : 400 crystal violet solution​​           Hot plate

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

Procedures:

1)

7g of agar powder was weighed and mixed with 420ml of Ringer solution.

2)

The mixture in step 1 was stirred and boiled on a hot plate until transparent yellowish solution.

3)

20ml of the agar solution was pour into each 6 test tubes. The test tubes ware 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 for standardize the color distance that cause by the diffusion of crystal violet.

5)

After the agar solutions in the test tubes were become solidify, 5ml of 1:200, 1:400, 1:600 crystal violet solution were pour into each test tubes and 3 test tubes were to be put in room temperature while another 3 were been put in 37ºc water bath.

6)

The test tubes were been closed immediately to prevent the vaporization of the solutions.

7)

Step 2 to 6 was repeated for Bromothymol Blue solutions.

8)

The data of the distance of solute diffusion was recorded and graph x² versus time was been plotted.

Results:

System (370C)	Time (hr)	X (m)	()	Slope 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	0.032 0.041 0.049 0.058 0.071 0.077 0.082 0.094	0.001024 0.001681 0.002401 0.003364 0.005041 0.005929 0.006724 0.008836	2.1	6.71x10-7	37	9.70x10-7
Crystal Violet with dilution 1:400	0 20 44 68 140 164 188 212	0.042 0.046 0.050 0.055 0.070 0.078 0.085 0.091	0.001764 0.002116 0.002500 0.003025 0.004900 0.006084 0.007225 0.008281	3.0	1.05x10-6	37
Crystal Violet with dilution 1:600	0 20 44 68 140 164 188 212	0.073 0.076 0.078 0.080 0.090 0.096 0.097 0.098	0.005329 0.005776 0.006084 0.006400 0.008100 0.009216 0.009409 0.009604	3.2	1.19x10-6	37

Crystal Violet system at 37ºC

System (370C)	Time (hr)	X (m)	()	Slope of Graph (m²hrˉ¹)	D (m²hrˉ¹)	Temperature (ºc)	Average of Diffusion Coefficient (m²hrˉ¹)
Bromothymolblue with dilution 1:200	0 20 44 68 140 164 188 212	0.018 0.021 0.057 0.068 0.081 0.085 0.091 0.098	0.000324 0.000441 0.003249 0.004624 0.006561 0.007225 0.008281 0.009604	4.2	1.34x10-6	37	3.36x10-6
Bromothymolblue with dilution 1:400	0 20 44 68 140 164 188 212	0.040 0.045 0.050 0.056 0.074 0.080 0.087 0.095	0.001600 0.002025 0.002500 0.003136 0.005476 0.006400 0.007569 0.009025	3.3	1.16x10-6	37
Bromothymolblue with dilution 1:600	0 20 44 68 140 164 188 212	0.062 0.063 0.065 0.070 0.080 0.084 0.090 0.093	0.003844 0.003969 0.004225 0.004900 0.006400 0.007056 0.008100 0.008649	2.3	8.55x10-7	37

Bromothymol blue system at 37ºC

System (Room temperature)	Time (hr)	X (m)	()	Slope 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	0.031 0.038 0.045 0.050 0.072 0.076 0.080 0.085	0.000961 0.001444 0.002025 0.002500 0.005184 0.005776 0.006400 0.007225	3.0	9.58x10-7	28	6.87x10-7
Crystal Violet with dilution 1:400	0 20 44 68 140 164 188 212	0.060 0.066 0.070 0.075 0.085 0.087 0.090 0.093	0.003600 0.004356 0.004900 0.005625 0.007225 0.007569 0.008100 0.008649	2.3	8.06x10-7	28
Crystal Violet with dilution 1:600	0 20 44 68 140 164 188 212	0.060 0.062 0.064 0.065 0.069 0.071 0.073 0.074	0.003600 0.003844 0.004096 0.004225 0.004761 0.005041 0.005329 0.005476	8.0	2.97x10-7	28

Crystal Violet system at 28ºC

System (Room temperature)	Time (hr)	X (m)	()	Slope of Graph (m²hrˉ¹)	D (m²hrˉ¹)	Temperature (ºc)	Average of Diffusion Coefficient (m²hrˉ¹)
Bromothymolblue with dilution 1:200	0 20 44 68 140 164 188 212	0.025 0.028 0.032 0.040 0.048 0.060 0.065 0.068	0.000625 0.000784 0.001024 0.001600 0.002304 0.003600 0.004225 0.004624	1.9	6.07x10-7	28	1.824x10-6
Bromothymolblue with dilution 1:400	0 20 44 68 140 164 188 212	0.040 0.048 0.052 0.058 0.070 0.072 0.075 0.082	0.001600 0.002304 0.002704 0.003364 0.004900 0.005184 0.005625 0.006724	2.2	7.71x10-7	28
Bromothymolblue with dilution 1:600	0 20 44 68 140 164 188 212	0.071 0.074 0.077 0.079 0.084 0.086 0.087 0.089	0.005041 0.005476 0.005929 0.006241 0.007056 0.007396 0.007569 0.007921	1.2	4.46x10-7	28

Bromothymol blue at 28ºC

Calculation:

1.

Crystal Violet system with dilution 1:200  at 37ºC :

From graph, gradient

= 2.1x10-5 m²hr-1

         Mo=1:200​​​                          M=1:500,000 (standard)

  =1/200​​​​                 =1/500,000

  =5x10-3​​​​                 =2x10-6

2.303 x 4D (log 10 Mo – log 10 M) = 2.1x10-5 m²hr-1

2.303x4D [log 10 (5x10-3 )-log 10 (2x10-6 )] = 2.1x10-5 m²hr-1

Hence, D = 6.71x10-7 m²hr-1

2.

Crystal Violet system with dilution 1:400  at 37ºC :

From graph, gradient

=3.0x10-5 m²hr-1

         Mo=1:400                                                           M=1:500,000 (standard)​​

  =1/400​​​​                 =1/500,000

  =2.5x10-3​​​​                 =2x10-6

2.303 x 4D (log 10 Mo – log 10 M) = 3.0x10-5 m²hr-1

2.303x4D [log 10 (2.5x10-3 )-log 10 (2x10-6 )] = 3.0 x10-5 m²hr-1

Hence, D = 1.05x10-6 m²hr-1

3.

Crystal Violet system with dilution 1:600  at 37ºC :

From graph, gradient

= 3.2x10-5 m²hr-1

          Mo=1:600​​​                          M=1:500,000 (standard)

   =1/600​​​​                 =1/500,000

  =1.67x10-3​​​​                 =2x10-6

2.303 x 4D (log 10 Mo – log 10 M) = 3.2x10-5 m²hr-1

2.303x4D [log 10 (1.67x10-3 )-log 10 (2x10-6 )] = 3.2x10-5 m²hr-1

Hence, D = 1.19x10-6 m²hr-1

4.

Average diffusion coefficient, m²hr-1  for Crystal Violet system at 37ºC

​=[(6.71x10-7 m²hr-1 ) + (1.05x10-6 m²hr-1) + (1.19x10-6 m²hr-1)]/3

​=9.70x10-7 m²hr-1

5.

Bromothymol Blue system with dilution 1:200  at 37ºC :

From graph, gradient

=4.2x10-5 m²hr-1

         Mo=1:200​​​                          M=1:500,000 (standard)

  =1/200​​​​                 =1/500,000

  =5x10-3​​​​                 =2x10-6

2.303 x 4D (log 10 Mo – log 10 M) = 4.2x10-5 m²hr-1

2.303x4D [log 10 (5x10-3 )-log 10 (2x10-6 )] = 4.2x10-5 m²hr-1

Hence, D = 1.34x10-6 m²hr-1

6.

Bromothymol Blue system with dilution 1:400  at 37ºC :

From graph, gradient

= 3.3x10-5 m²hr-1

         Mo=1:400                                                           M=1:500,000 (standard)​​

  =1/400​​​​                 =1/500,000

  =2.5x10-3​​​​                 =2x10-6

2.303 x 4D (log 10 Mo – log 10 M) = 3.3x10-5 m²hr-1

2.303x4D [log 10 (2.5x10-3 )-log 10 (2x10-6 )] = 3.3x10-5 m²hr-1

Hence, D = 1.16x10-6 m²hr-1

7.

Bromothymol Blue system with dilution 1:600  at 37ºC :

From graph, gradient

= (8.40-1.50) x 10-4 / (150-27)m²hr-1

= 2.3x10-5 m²hr-1

         Mo=1:600​​​                          M=1:500,000 (standard)

  =1/600​​​​                 =1/500,000

  =1.67x10-3​​​​                 =2x10-6

2.303 x 4D (log 10 Mo – log 10 M) = 2.3x10-5 m²hr-1

2.303x4D [log 10 (1.67x10-3 )-log 10 (2x10-6 )] = 2.3x10-5 m²hr-1

Hence, D = 8.55x10-7 m²hr-1

8.

Average diffusion coefficient, m²hr-1 for Bromothymol Blue system at 37ºC

​=[(1.34x10-6 m²hr-1 ) + (1.16x10-6 m²hr-1) + (8.55x10-7 m²hr-1)]/3

​=3.36x10-6 m²hr-1

9.

Crystal Violet system with dilution 1:200  at 28ºC :

From graph, gradient

=3.00x10-5 m²hr-1

         Mo=1:200                                                           M=1:500,000 (standard)

  =1/200​​​​                 =1/500,000

  =5x10-3​​​​                 =2x10-6

2.303 x 4D (log 10 Mo – log 10 M) = 3.00x10-5 m²hr-1

2.303x4D [log 10 (5x10-3 )-log 10 (2x10-6 )] = 3.00x10-5 m²hr-1

Hence, D = 9.58x10-7 m²hr-1  

10.

Crystal Violet system with dilution 1:400  at 28ºC :

From graph, gradient

=2.30x10-5 m²hr-1

         Mo=1:400                                                           M=1:500,000 (standard)​​

  =1/400​​​​                 =1/500,000

  =2.5x10-3​​​​                 =2x10-6

2.303 x 4D (log 10 Mo – log 10 M) = 2.30x10-5 m²hr-1

2.303x4D [log 10 (2.5x10-3 )-log 10 (2x10-6 )] = 2.30x10-5 m²hr-1

Hence, D = 8.06x10-7 m²hr-1

11.

Crystal Violet system with dilution 1:600  at 28ºC :

From graph, gradient

=8.00x10-6 m²hr-1

         Mo=1:600​​​                          M=1:500,000 (standard)

  =1/600​​​​                 =1/500,000

  =1.67x10-3​​​​                 =2x10-6

2.303 x 4D (log 10 Mo – log 10 M) = 8.00x10-6 m²hr-1

2.303x4D [log 10 (1.67x10-3 )-log 10 (2x10-6 )] = 8.00x10-6 m²hr-1

Hence, D = 2.97x10-7 m²hr-1

12.

Average diffusion coefficient, m²hr-1 for Crystal Violet system at 28ºC

​= [(9.58x10-7 m²hr-1) + (8.06x10-7 m²hr-1) + (2.97x10-7 m²hr-1)]/3

​= 6.87x10-7 m²hr-1

13.

Bromothymol Blue system with dilution 1:200  at 28ºC :

From graph, gradient

=1.90x10-5 m²hr-1

         Mo=1:200​​​                          M=1:500,000 (standard)

  =1/200​​​​                 =1/500,000

  =5x10-3​​​​                 =2x10-6

2.303 x 4D (log 10 Mo – log 10 M) = 1.90x10-5 m²hr-1

2.303x4D [log 10 (5x10-3 )-log 10 (2x10-6 )] = 1.90x10-5 m²hr-1

Hence, D = 6.07x10-7 m²hr-1

14.

Bromothymol Blue system with dilution 1:400  at 28ºC :

From graph, gradient

= 2.20x10-5 m²hr-1

         Mo=1:400                                                           M=1:500,000 (standard)​​

  =1/400​​​​                 =1/500,000

  =2.5x10-3​​​​                 =2x10-6

2.303 x 4D (log 10 Mo – log 10 M) = 2.20x10-5 m²hr-1

2.303x4D [log 10 (2.5x10-3 )-log 10 (2x10-6 )] = 2.20x10-5 m²hr-1

Hence, D = 7.71x10-7 m²hr-1

15.

Bromothymol Blue system with dilution 1:600  at 28ºC :

From graph, gradient

=1.2x10-5 m²hr-1

         Mo=1:600​​​                          M=1:500,000 (standard)

  =1/600​​​​                 =1/500,000

  =1.67x10-3​​​​                 =2x10-6

2.303 x 4D (log 10 Mo – log 10 M) = 1.2x10-5 m²hr-1

2.303x4D [log 10 (1.67x10-3 )-log 10 (2x10-6 )] = 1.2x10-5 m²hr-1

Hence, D = 4.46x10-7 m²hr-1

16.

Average diffusion coefficient, m²hr-1 for Bromothymol Blue system at 28ºC

​=[(6.07x10-7 m²hr-1 ) + (7.71x10-7 m²hr-1) + (4.46x10-7 m²hr-1)]/3

​=1.824x10-6 m²hr-1

Questions:

1)      From the experiment value for D_28, the value of D₃₇ using the following equation are:

D₂₈^oC/ D₃₇^oC = T₂₈^oC/ T_37oC

A)    Crystal violet

i.                    1:200

(9.58 x 10^-5m²/hour)/ D₃₇^oC = 28/37

   D₃₇^oC                     =   1.266x 10^-4 m²/hour

Experimental value      =  6.71 x 10^-7 m²/hour

of    D₃₇^oC              

ii.                  1:400

(2.3 x 10^-5m²/hour)/ D₃₇^oC = 28/37

 D₃₇^oC                     =   3.039x 10^-5 m²/hour

Experimental value      =  1.05 x 10^-6 m²/hour

of    D₃₇^oC              

                               

iii.                1:600

(8.00 x 10^-6m²/hour)/ D₃₇^oC = 28/37

       D₃₇^oC                     =    1.057 x 10^-5 m²/hour

Experimental value      =  1.19 x 10^-6 m²/hour

of    D₃₇^oC  

Average D_28ºC = 6.87x10^-7 m²hr^-1

                                                     D_{28ºC /} D_{37ºC   =}   T_28ºC/T_37ºC

6.87x10^-7 m²hr^{-1 /} D_37ºC    ⁼           28 / 37                                                                                                                     

  D_37ºC  = 9.078x10^-7 m²hr^-1

Experimental value of D_37ºC  =  9.7 x 10^-7 m²hr^-1

            The D_37ºC value is 9.078x10^-7 m²hr^-1 while the experimental value is 9.7x10^-7 m²hr^-1. There is a slightly difference between these two values, where it is less of 6.22x10^-8   m²hr^-1of D_37ºC value from the experimental value.

B)    Bromothymol blue

i.                    1:200

(6.07 x 10^-7m²/hour)/ D₃₇^oC = 28/37

               D₃₇^oC=     8.021 x 10^-7 m²/hour

Experimental value of  D₃₇^oC =   1.34 x 10^-6 m²/hour

ii.                  1:400

(7.71 x 10^-7m²/hour)/ D₃₇^oC = 28/37

D₃₇^oC        1.019 x 10^-6 m²/hour

Experimental value of D₃₇^oC   = 1.16 x 10^-6 m²/hour

iii.                1:600

(4.46 x 10^-7m²/hour)/ D₃₇^oC = 28/37

D₃₇^oC        5.894 x 10^-7  m²/hour

Experimental value of D₃₇^oC   = 8.55 x 10^-7 m²/hour

Average D_28ºC = 2.824x10^-6 m²hr^-1

                                                     D_{28ºC /} D_{37ºC   =}   T_28ºC/T_37ºC

1.824x10^{-6 /} D_37ºC    ⁼           28 / 37                                                                                                                    

    D_37ºC  = 2.41x10^-6 m²hr^-1

Experimental value of D_37ºC  =  3.36 x 10^-6 m²hr^-1

            The D_37ºC value is 2.41x10^-6 m²hr^-1 while the experimental value is 3.36x10^-6 m²hr^-1. There is a slightly difference between these two values, where it is less of 9.5x10^-7   m²hr^-1of D_37ºC value from the experimental value.

2.Between Crystal Violet and Bromothymol Blue, which one diffuse much faster? Explain if there are any differences in these diffusion coefficients.

M = 4/3 πa²Nρ

a³ = 3M/4πNρ

a  = cube root M/4πNρ

where:

M = molecular weight,

a = size of particle,

ρ = density,

N = Avogadro’s number 6.023 x 10²³

From the formula above, we can say that the size of particle is proportional to the molecular weight. So, the smaller the size of particle, the easier the particle to diffuse into the medium.

Based on the formula, the diffusion coefficient, D is inversely proportional to the molecular weight. Since crystal violet has a higher diffusion coefficient than that in bromothymol blue, it diffuse quickly in the agar medium than bromothymol blue. Besides, the molecular weight of crystal violet is smaller, i.e 407.99 gmol^-1 when compared to that in bromothymol blue that is 624.38 gmol^-1. So, it facilitates the diffusion of the particles into the solid agar medium.

Discussion:

In this experiment, some errors had been done and this caused inaccuracy to the final result. The first error is the agar that we form are not solid enough. This may be due to some of the agar powder that was not dissolved completely in the Ringer’s solution while heating. When the agar powder do not dissolve completely, the agar is difficult to be formed as it cannot solidify. The result of our experiment is therefore affected as the bromothymol blue and crystal violet solution can diffuse easily through the agar. Besides, parallax error may occur due to the difficulty in determining the colour of 1:600 bromothymol blue and crystal violet solution. These two dilute solution are very light in colour and when they diffuse through the agar, it is difficult to determine the actual distance the chemical solution had diffused. Furthermore, as different students are taking the readings in turns within the two weeks, the inconsistency of result may arise. For example the method and standard in measuring the diffusion distance is different between each students. Hence, there is a high possibility that error may occur and resulting in inaccurate data. The errors formed can be proved while plotting the graphs and also while calculating the values using D₂₈^oC/ D₃₇^oC = T₂₈^oC/ T_37oC. From the calculations, it is shown that the values of the calculated D₃₇^oC is different from the experimental data. From the graph, there are many points that the straight line does not pass through. Since the Crystal Violet particles diffuse spontaneously, so the x value taken is from personal judgement and estimation. It varies in different people. Moreover, the room temperate is not kept constant for the day and night. This will indirectly affect the diffusion coefficient and also the result of experiment.

Fick’s Law states that the diffusion coefficient is affected the temperature. For the diffusion coefficient of the liquid, we can use the Stokes-Einstein equation:

D_T1/ D_T2 = T₁V_T2 / T₂V_T1

where:

T₁ and T₂ denote temperatures 1 and 2, respectively

D is the diffusion coefficient (cm²/s)

T is the absolute temperature (K),

V is the dynamic viscosity of the solvent (Pa·s)

From the above equation we can conclude that the diffusion coefficient is directly proportional to the temperature. The higher the temperature, the higher the tendency the particles gain energy and collide with each other  more vigorously. Therefore, it results in higher diffusion rate through the agar medium as compared to the one with  lower temperature(28 ^oC ) This can also be shown from the experiment’s results. In the crystal violet experiment, when we compare the average diffusion coefficient for 28°C and 37°C, we can see that the diffusion coefficient for 37°C (9.70x10^-7) is higher than 28°C (6.87 x10^-7). For the bromothymol blue, the diffusion coefficient for 37°C (3.36x10^-6) is also higher than 28°C (1.824 x10^-6).

The rate of absorption is directly proportional with diffusion coefficient. The smaller the particle the faster it dissolves. Crystal violet have a smaller molecular weight compared to bromothymol blue thus it’s supposed to have higher diffusion coefficient. But based from the result shows that bromothymol blue have a higher diffusion coefficient and this is due to the error that occur in this experiment.

The rate of diffusion obeys Fick’s Law of Diffusion too. It states that the rate of change in concentration of solute with the time is directly proportional to the concentration difference between the two sides of diffusion layer. Assuming the solute is moving in x direction, then the solvent molecules will moving in opposite direction, which is –x direction. Hence, if concentration of solute is high, this create a net flow of solute in the x direction.  In this practical, the equation given is:

ln M = ln M_o –x²/4Dt

or         2.303 x 4D (log₁₀ M_o – log₁₀ M) t = x²

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 leads to the diffusion would be larger and the rate of diffusion process will increase. Therefore, 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.

For the crystal violet experiment at 37°C, our data is inaccurate as the diffusivity increases ascendingly form 1:200 to 1:600. However, for the crystal violet experiment at 28°C,we can see that as the concentration M_o decrease from 1:200, 1:400 and 1:600, the diffusivity of crystal violet decrease. For the bromothymol blue experiment at 37°C, the data is considered accurate because as the concentration M_o decrease from 1:200, 1:400 and 1:600, the diffusivity of bromothymol blue also decreases.

But for the result at 37°C we have found that the diffusion coefficient for concentration 1:400 is higher than 1:200 when it supposed to be lower. This may be due to the error during the experiment where the agar is still not hard enough when we pour the bromothymol blue in test tube 1:400. This causes the solute to easily diffuse through the agar and produce an incorrect reading.

Conclusion:

Diffusion coefficient for Crystal Violet system at 28ºC , D28ºC  is 6.87x10-7 m²hr-1 while at 37ºC, D37ºC is 9.7x10-7 m²hr-1.The diffusion coefficient for Bromothymol Blue system at 28ºC,D28ºC is 1.824x10-6 m²hr-1while at 37ºC , D37ºC is 3.36x10-6 m²hr-1. The temperature, concentration and molecular mass of diffusing molecules are the main factors affecting the diffusion coefficient.

References:

Patrick J. Sinko, 2006, Martin’s Physical Pharmacy and Pharmaceutical Science 5th  Ed. Lippincott Williams and Wilkin.

A.T.Florence and D.Attwood. (1998). Physicochemical Principals of Pharmacy, 3rd  Edition. Macmillan Press Ltd.

Michael E. Aulton, 2007, Aulton’s Pharmaceutics’s: The Design and Manufacture of Madicines,Churchill Livingstone Elsevier.