Practical 4: Determination of Diffusion Coefficient

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.

                                            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)

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                      M_a = 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                      M_a = 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                      M_a = 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

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                       M_a = 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

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

M = 1:500000                      M_a = 1:400

     = 1 / 500000                        = 1 / 400

     = 2 x 10^-6                            = 2.5 x 10^-3

2.303 X4D [log₁₀M₀ -log₁₀M) = 7.27 X10⁻⁶m²/hour
2.303 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                      M_a = 1:600

     = 1 / 500000                         = 1 / 600

     = 2 x 10^-6                                   = 1.67 x 10^-3

2.303 X4D [log₁₀M₀ -log₁₀M) = 1.42 X10⁻⁶m²/hour
2.303 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                       M_a = 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/hour

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                      M_a = 1:400

     = 1 / 500000                        = 1 / 400

     = 2 x 10^-6                            = 2.5 x 10^-3

2.303 X4D [log₁₀M₀ -log₁₀M) = 5.56x10⁻⁶m²/hour
2.303 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

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]= 4.16x10-6 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 D_28ºC, estimate the value of   D_37ºC by using the following equation:

_{D28ºC  =   T28ºC}
D37_ºC    =  T37_ºC

where ŋ₁and ŋ₂ are viscosity of water at 28ºC and 37ºC.

      

Is the calculated value of D_37ºC  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?

D28_ºC=1.285x10⁻⁷ m²hour⁻¹                        

                         1.285x10⁻⁷    =    28+273.15

                             D37_ºC              37+273.15  

                             D37_ºC     =1.32x10⁻⁷m²hour⁻¹

The value of D37_ºC 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 D37_ºC 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 M_o concentration, the agar medium is considered homogeneous and with constant concentration, hence, the diffusion can be express as,

   M = M_o eksp ^{(-x2/ 4Dt)}  ---------(2)

Change the equation (2) to logarithm form we have

ln M  =  ln M_o –x²/4Dt

            2.303 x 4 D ( log₁₀ M_o – 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 M_o – 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, 3^rd  Edition. Macmillan Press Ltd.

2.Physical Pharmacy, by Alfred Martin, 4^th Edition.