UKM PHYSICAL PHARMACY LAB REPORT: Practical 4: Determination of diffusion coefficient

Title:

Determination of Diffusion Coefficient

Objective:

To determine the value for diffusion coefficient, D.

Theory:
Diffusion is a process where the spontaneous movement of certain molecules from a high concentration gradient to a low concentration gradient. This phenomenon can be explained by Fick’s law of diffusion. Fick’s law state that the flowing of a substance (amount, dm in time, dt) through certain dimensions (area A) is directly proportional to concentration gradient dc/dx.

dm = -DA(dc/dx)dt

where D is the diffusion coefficient or diffusion force for solute that has unit as m²sֿ¹.

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 eksp ^{(-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 (II) to logarithm 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 ₁₀M₀–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 4/3пa³р where p is the density of the molecule.

As we know that the molecular mass of M = mN, where N is the Avogadro’s number 6.02 x 10²³ mol^-1.

M = 4/3пa³Nр

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.

Procedures:

1. 7g of agar powder was weighed and mixed with 420 ml of Ringer solution.

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

3. 14ml 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 colour 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:

28⁰C (room temperature)

System	Time (s)	X (m)	X² (m²)	Gradient From graph (m²s^-1)	D (m²s^-1)	Temp (ºC)	Average diffusion Coefficient (m²s^-1)
Crystal violet with dilution 1:200	059400 080100 325800 423000 509400 576600 667800	0.01 0.012 0.019 0.022 0.023 0.026 0.028	0.000100 0.000144 0.000316 0.000484 0.000529 0.000676 0.000784	1.125x10^-9	3.59x10^-11	28 28 28 28 28 28 28	3.149x10^-11
Crystal violet with dilution 1:400	059400 080100 325800 423000 509400 576600 667800	0.009 0.010 0.016 0.020 0.022 0.024 0.026	0.000081 0.000100 0.000256 0.000400 0.000484 0.000576 0.000676	9.644x10^-10	3.380x10^-11	28 28 28 28 28 28 28
Crystal violet with dilution 1:600	059400 080100 325800 423000 509400 576600 667800	0.005 0.006 0.010 0.014 0.020 0.022 0.023	0.000025 0.000036 0.000100 0.000196 0.000400 0.000484 0.000529	6.667x10^-10	2.477x10^-11	28 28 28 28 28 28 28

37⁰C (water bath)

System	Time (s)	X (m)	X² (m²)	Gradient From graph (m?s^-1)	D (m²s^-1)	Temp (ºC)	Average diffusion Coefficient (m?s^-1)
Crystal violet with dilution 1:200	059400 080100 325800 423000 509400 576600 667800	0.01 0.011 0.022 0.026 0.027 0.028 0.029	0.000100 0.000121 0.000484 0.000676 0.000729 0.000784 0.000841	1.385x10^-9	4.425x10^-11	37 37 37 37 37 37 37	3.954x10^-11
Crystal violet with dilution 1:400	059400 080100 325800 423000 509400 576600 667800	0.007 0.008 0.011 0.024 0.025 0.026 0.027	0.000049 0.000064 0.000121 0.000576 0.000625 0.000676 0.000729	9.644x10^-10	3.380x 10^-11	37 37 37 37 37 37 37
Crystal violet with dilution 1:600	059400 080100 325800 423000 509400 576600 667800	0.005 0.006 0.014 0.022 0.023 0.024 0.025	0.000025 0.000036 0.000196 0.000484 0.000529 0.000576 0.000625	6.667x10^-10	2.477x 10^-11	37 37 37 37 37 37 37

28⁰C (room temperature)

System	Time (s)	X (m)	X² (m²)	Gradient From graph (m?s^-1)	D (m²s^-1)	Temp (ºC)	Average diffusion Coefficient (m?s^-1)
Bromothymol blue with dilution 1:200	059400 080100 325800 423000 509400 576600 667800	0.015 0.016 0.020 0.022 0.025 0.027 0.030	0.000225 0.000256 0.000400 0.000484 0.000625 0.000729 0.000900	1.286x10^-9	4.081x10^-11	28 28 28 28 28 28 28	3.192x10^-11
Bromothymol blue with dilution 1:400	059400 080100 325800 423000 509400 576600 667800	0.013 0.014 0.016 0.018 0.020 0.022 0.025	0.000169 0.000196 0.000256 0.000324 0.000400 0.000484 0.000625	8.75x10^-10	3.067x10^-11	28 28 28 28 28 28 28
Bromothymol blue with dilution 1:600	059400 080100 325800 423000 509400 576600 667800	0.011 0.012 0.014 0.015 0.017 0.020 0.022	0.000121 0.000144 0.000196 0.000225 0.000289 0.000400 0.000484	6.538x10^-10	2.429x10^-11	28 28 28 28 28 28 28

37⁰C (water bath)

System	Time (s)	X (m)	X² (m²)	Gradient From graph (m²s^-1)	D (m²s^-1)	Temp (ºC)	Average diffusion Coefficient (m²s^-1)
Bromothymol blue with dilution 1:200	059400 080100 325800 423000 509400 576600 667800	0.018 0.020 0.035 0.045 0.046 0.047 0.049	0.000324 0.000400 0.001225 0.002025 0.002116 0.002209 0.002401	3.846x10^-9	1.229x10^-10	37 37 37 37 37 37 37	9.728x10^-11
Bromothymol blue with dilution 1:400	059400 080100 325800 423000 509400 576600 667800	0.015 0.016 0.026 0.035 0.036 0.037 0.038	0.000225 0.000256 0.000676 0.001225 0.001296 0.001369 0.001444	2.411x10^-9	8.451x10^-11	37 37 37 37 37 37 37
Bromothymol blue with dilution 1:600	059400 080100 325800 423000 509400 576600 667800	0.014 0.015 0.028 0.034 0.035 0.036 0.037	0.000196 0.000225 0.000784 0.001156 0.001225 0.001296 0.001369	2.273x10^-9	8.444x10^-11	37 37 37 37 37 37 37

Graph:
i. Graph of x²(x10^-4 m²) in the function of time (hour) of different crystal violet system at 28⁰c.

ii. Graph of x²(x10^-4 m²) in the function of time (hour) of different crystal violet system at 37⁰C.

iii. Graph of x²(x10^-4 m²) in the function of time(s) of different Bromotymol blue system

at 28⁰C.

iv. Graph of x²(x10^-4 m²) in the function of time(s) of different Bromotymol blue system

at 37⁰C.

Calculation:

(a) Crystal violet system with dilution 1:200 at 28⁰c

Gradient from graph = (6.8-1.4)x10^-4 m² / (600000-120000)s

= 5.4x10^-4 m² / 480000 s

= 1.125x10^-9 m² s^-1

By using formula,
2.303 x 4 D ( log₁₀M₀ – log₁₀ M) t = x²

M= 1:500000 = 2x10^-6
M₀= 1:200 = 5x10^-3

2.303 x 4 D ( log₁₀M₀ – log₁₀ M) = 1.125x10^-9

D = (1.125x10^-9 ) / [ 2.303 x 4 ( log₁₀(5x10^-3) – log₁₀ (2x10^-6)) ]

= 3.59x 10^-11 m²s^-1

(b) Crystal violet system with dilution 1:400 at 28⁰c

Gradient from graph = (6.2-0.8)x10^-4 m² / (640000-80000)s

= 5.4x10^-4 m²/ 560000 s

= 9.644x10^-10 m² s^-1

By using formula,
2.303 x 4 D ( log₁₀M₀ – log₁₀ M) t = x²

M= 1:500000 = 2x10^-6
M₀= 1:400 = 2.5x10^-3

2.303 x 4 D ( log₁₀M₀ – log₁₀ M) = 9.644x10^-10

D = (9.644 x 10^-10) / [ 2.303 x 4 ( log₁₀(2.5x10^-3) – log₁₀ (2x10^-6)) ]

= 3.380 x 10^-11 m²s^-1

Gradient from graph = (4.0-0.8)x10^-4 m² / (560000-80000)s

= ( 3.2x10^-4 m²) / 480000 s

= 6.667x10^-10 m² s^-1

By using formula,
2.303 x 4 D ( log₁₀M₀ – log₁₀ M) t = x²

M= 1:500000 = 2x10^-6
M₀= 1:600 = 1.67x10^-3

2.303 x 4 D ( log₁₀M₀ – log₁₀ M) = 6.667x10^-10

D = (6.667x10^-10 ) / [2.303 x 4 ( log₁₀(1.67x10^-3) – log₁₀ (2x10^-6))]

= 2.477x 10^-11 m²s^-1

(d) Crystal violet system with dilution 1:200 at 37⁰c

Gradient from graph = (8.8-1.6)x10^-4 m² / (640000-120000)s

= ( 7.2x10^-4 m²) / 520000 s

= 1.385x10^-9 m²s^-1

By using formula,
2.303 x 4 D ( log₁₀M₀ – log₁₀ M) t = x²

M= 1:500000 = 2x10^-6
M₀= 1:200 = 5x10^-3

2.303 x 4 D ( log₁₀M₀ – log₁₀ M) = 1.385x10^-9

D = ( 1.385x10^-9) / [ 2.303 x 4 ( log₁₀(5x10^-3) – log₁₀ (2x10^-6))]

= 4.425x 10^-11 m²s^-1

(e) Crystal violet system with dilution 1:400 at 37⁰c

Gradient from graph = (6.2-1.8)x10^-4 m²/ (560000-160000)s

= ( 4.4x10^-4 m²) / 400000 s

= 1.1x10^-9 m²s^-1

By using formula,
2.303 x 4 D ( log₁₀M₀ – log₁₀ M) t = x²

M= 1:500000 = 2x10^-6
M₀= 1:400 = 2.5x10^-3

2.303 x 4 D ( log₁₀M₀ – log₁₀ M) = 1.1x10^-9
D = (1.1x10^-9) / [2.303 x 4 ( log₁₀(2.5x10^-3) – log₁₀ (2x10^-6))]

= 3.855x 10^-11 m²s^-1

(f) Crystal violet system with dilution 1:600 at 37⁰c

Gradient from graph = (6.2-0.8)x10^-4 m² ) / (640000-80000)s

= ( 5.4x10^-4 m² ) / 560000 s

= 9.643x10^-10 m²s^-1

By using formula,
2.303 x 4 D ( log₁₀M₀ – log₁₀ M) t = x²

M= 1:500000 = 2x10^-6
M₀= 1:600 = 1.67x10^-3

2.303 x 4 D ( log₁₀M₀ – log₁₀ M) = 9.643x10^-10

D = ( 9.643x10^-10) / [2.303 x 4 ( log₁₀(1.67x10^-3) – log₁₀ (2x10^-6))]

= 3.582x 10^-11 m²s^-1

∴ Crystal violet system at 28⁰c

Average value of diffusion coefficient

= (3.59x10^-11) + (3.380x10^-11) + (2.477x10^-11) m² / 3
= 3.149x10^-11m²s^-1

∴ Crystal violet system at 37⁰c

Average value of diffusion coefficient

= (4.425x10^-11) + (3.855x10^-11) + (3.582x10^-11) m² / 3

= 3.954x10^-11m²s^-1

(a) Bromothymol blue system with dilution 1:200 at 28⁰c

Gradient from graph = (8.2-1.0)x10^-4 m² / (640000-80000)s

= ( 7.2x10^-4 m²) / 560000 s

= 1.286x10^-9 m²s^-1

By using formula, 2.303 x 4 D ( log₁₀M₀ – log₁₀ M) t = x²

M= 1:500000 = 2x10^-6
M₀= 1:200 = 5x10^-3

2.303 x 4 D ( log₁₀M₀ – log₁₀ M) = 1.286x10^-9

D = (1.286x10^-9) / [ 2.303 x 4 ( log₁₀(5x10^-3) – log₁₀ (2x10^-6))]

= 4.108x 10^-11 m²s^-1

(b) Bromothymol blue system with dilution 1:400 at 28⁰c

Gradient from graph = (5.2-1.0)x10^-4 m²/ (600000-120000)s

= ( 4.2x10^-4 m²) / 480000 s
= 8.75x10^-10 m²s^-1

By using formula,
2.303 x 4 D ( log₁₀M₀ – log₁₀ M) t = x²

M= 1:500000 = 2x10^-6
M₀= 1:400 =2.5x10^-3

2.303 x 4 D ( log₁₀M₀ – log₁₀ M) = 8.75x10^-10

D =( 8.75x10^-10) / [2.303 x 4 ( log₁₀(2.5x10^-3) – log₁₀ (2x10^-6))]

= 3.067x 10^-11 m²s^-1

Gradient from graph = (4.2-0.8)x10^-4 m² ) / (640000-120000)s

= 3.4x10^-4 m²/ 520000 s
= 6.538x10^-10 m²s^-1

By using formula, 2.303 x 4 D ( log₁₀M₀ – log₁₀ M) t = x²

M= 1:500000 = 2x10^-6
M₀= 1:600 = 1.67x10^-3

2.303 x 4 D ( log₁₀M₀ – log₁₀ M) = 6.538x10^-10

D =( 6.538x10^-10) / [ 2.303 x 4 ( log₁₀(1.67x10^-3) – log₁₀ (2x10^-6))]

= 2.429x 10^-11 m²s^-1

(d) Bromothymol blue system with dilution 1:200 at 37⁰c

Gradient from graph = [ (25-5)x10^-4 m²] / (640000-120000)s

= 20x10^-4 m² / 520000 s
= 3.846x10^-9 m²s^-1

By using formula,
2.303 x 4 D ( log₁₀M₀ – log₁₀ M) t = x²

M= 1:500000 = 2x10^-6
M₀= 1:200 = 5x10^-3

2.303 x 4 D ( log₁₀M₀ – log₁₀ M) = 3.846x10^-9

D = (3.846x10^-9) / [2.303 x 4 ( log₁₀(5x10^-3) – log₁₀ (2x10^-6))]

= 1.229x 10^-10 m²s^-1

(e) Bromothymol blue system with dilution 1:400 at 37⁰c

Gradient from graph = (15,5-2)x10^-4 m² / (640000-80000)s

= 13.5x10^-4 m²/ 560000 s
= 2.411x10^-9 m²s^-1

By using formula,
2.303 x 4 D (log10M₀ – log₁₀ M) t = x²

M= 1:500000 = 2x10^-6
M₀= 1:400 = 2.5x10^-3

2.303 x 4 D (log10M₀ – log₁₀ M) = 2.411x10^-9

D = ( 2.411x10^-9) / [ 2.303 x 4 (log102.5x10^-3) – log₁₀ (2x10^-6))]

= 8.451x 10^-11 m²s^-1

(f) Bromothymol blue system with dilution 1:600 at 37⁰c

Gradient from graph = (14-4)x10^-4 m²/ (640000-160000)s

= 10x10^-4 m²/ 480000 s
= 2.273x10^-9 m²s^-1

By using formula,
2.303 x 4 D (log10M₀ – log₁₀ M) t = x²

M= 1:500000 = 2x10^-6
M₀= 1:600 = 1.67x10^-3

2.303 x 4 D (log10M₀ – log₁₀ M) = 2.273x10^-9

D = ( 2.273x10^-9) / [2.303 x 4 ( log₁₀(1.67x10^-3) – log₁₀ (2x10^-6))]

= 8.444x 10^-11 m²s^-1

∴ Bromothymol blue system at 28⁰c

Average value of diffusion coefficient

= (4.081x10^-11) + (3.067x10^-11) + (2.429x10^-11) m²/ 3

= 3.192x10^-11m²s^-1

∴ Bromothymol blue system at 37⁰c

Average value of diffusion coefficient

= (1.229x10^-10) + (8.451x10^-11) + (8.444x10^-11) m²/ 3

= 9.728x10^-11m²s^-1

Question:

1. From the experiment value for D₂₈, estimate the value of D₃₇ using the following equation

D₂₈^o_c / D₃₇^o_c = T₂₈^o_c / T₃₇^o_c

where n1 and n2 are the viscosity of water at temperature 28⁰c and 37⁰c.

Is the calculated value of D₃₇^o_c 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^o_c = 3.149x10^-11 m²s^-1

D₂₈^o_c / D₃₇^o_c = T₂₈^o_c / T₃₇^o_c

(3.149x10^-11 ) / D₃₇^o_c = 28 + 273.15 / (37 + 273.15)

D₃₇^o_c= ( 310.15 x 3.149x10^-11) / ( 301.15 ) = 3.243x10^-11 m²hr^-1

The D₃₇^o_c value is 3.243x10^-11 m²s^-1while the trial value is 3.954 x10^-11 m²s^-1.There is a slightly difference between these two values. The final value of D₃₇^o_cis slightly higher than the calculated D₃₇^o_cvalue. The difference between these two values is due to the errors which occurred during conducting the experiment. First of all, the temperature is not kept constant at all the time (37⁰c).There might be a fluctuation in temperature while the experiment is carried out. Furthermore, the measurement the “x” value is just an approximation due to the different observations made by student. Misreading might occur. Moreover, the viscosity of agar along the test tube is not uniform do affect the rate of diffusion of the dye.

2. Between the crystal violet and bromothymol blue, which diffuse quicker? Explain if there are any differences in the diffusion coefficient values.

Bromothymol Blue will diffuse faster than Crystal Violet as the diffusion coefficient of Bromothymol Blue is larger than diffusion coefficient of Crystal Violet. The smaller size of molecule, the easier it penetrates through agar.

Discussion:

Diffusion is the spontaneous net movement of particles from an area of high concentration to an area of low concentration in a given volume of fluid (either liquid or gas) down the concentration gradient. For example, diffusing molecules will move randomly between areas of high and low concentration but because there are more molecules in the high concentration region, more molecules will leave the high concentration region than the low concentration one. Therefore, there will be a net movement of molecules from the high to low concentration. Initially, a concentration gradient leaves a smooth decrease in concentration from high to low which will form between the two regions. As time progresses, the gradient will grow increasingly shallow until the concentration are equalized. The solvent molecules on the other hand, move in a reverse direction (that is of lower chemical potential, to higher chemical potential). Considering the solute move to x direction, then the solvent molecules will move to a reverse direction, that is –x. If the solute have a higher concentration , there is a net flow of solute in a direction of x.

Diffusion process can be expressed by Fick's First Law of Diffusion. This law relates the flow of material to the concentration gradient, dc/dx. According to this law, the concentration gradient does not change with time provided that diffusion process occurs in steady condition:

dm = -DA dc/dx . dt

where D is the diffusion coefficient of a solute in m²s^-1. The negative sign show that diffusion occurs in the opposite direction which in the low concentration direction.

In this practical, diffusion particles involved are considered as neutral with concentration Mo, and the agar medium is assumed to be homogenous and with constant concentration. Hence, the relationship between the diffusion coefficient and concentration of solute at a distance from the agar boundary can be expressed as:

M = Mo e ^(-x2/4Dt)

or 2.303 x 4D ( log₁₀M₀ – log₁₀ M) t = x²

Graph of x² in the function of t can be plotted and use to calculate the value of D because the gradient of graph is 2.303 x 4 D ( log₁₀M₀ – log₁₀ M).From the practical, we found out that in both of the 28⁰c and 37⁰c system, diffusion is faster in 1:200>1:400>1:600 since the value of D decrease in this sequence is the system with dilution 1:5000000 and it function as a standard in the experiment. When Mo increase, (log10M₀ – log₁₀ M)also increase. Its increase resulting in the increases of concentration gradient and thus the force for diffusion becomes bigger and consequently the diffusion process becomes faster and more favorable.

The value of diffusion coefficient is affected by temperature. The systems conducted at 37⁰c has a higher D value compare to 28⁰c system. When temperature increases, diffusion coefficient will also increase because more energy is supplied to the diffuse into the agar medium. Other than that, the size of the solute particles also influences the diffusion rate. In this experiment, the diffusion rate in bromothymol blue system is faster than crystal violet system. Thus we can conclude that the size particle of bromothymol blue system is smaller than the size particle of crystal violet system.

Agar gel is a network of molecules which has the ability to trap water. These water molecules will form a continuous phase which enables the solute to diffuse freely. When the concentration of agar increases, the rate or diffusion also increases because the solute particles are less freely to pass through the agar.

Furthermore, when the crystallinity of the gel medium increases, the diffusion will also decrease because the crystalline region of the gel medium present an impenetrable barrier to the solute particles’ movement. Hence, the movement of diffusion molecules becomes slower and lead to the decrease in rate of diffusion.

In conclusion, diffusion is important in the diffusion of drug particles from dosage form, through the biological membrane. This application is essential for the absorption of drug particles effectively in patients.

Conclusion:

The value of D for crystal violet at 28⁰c and 37⁰c are 3.194x10^-11m²s^-1 and3.954 x10^-11 m²s^-1 respectively. The value of D for bromothymol blue at 28⁰c and 37⁰c are 3.192x10^-11 m²s^-1 and 9.728x10^-11 m²s^-1 respectively. The value of diffusion coefficient is affected by temperature and concentration of diffusing molecules. Diffusion system using bromothymol blue will have higher diffusion coefficient than crystal violet due small particles size.

References:

A.T. Florence and D. Attwood. 1998. Physicochemical Principles of Pharmacy. 2nd Edition. Maemillan Press LTD.

Mostinsky, I.L. 2011. Thermopedia. Diffusion Coefficient. http://www.thermopedia.com/content/696/ [2 February 2011].

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Tuesday, 6 May 2014

Practical 4: Determination of diffusion coefficient

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