Sieving

Title: Sieving

Objective:

To determine particle size distribution of powder and the size of solid particle of lactose and microcrystalline cellulose (MCC) by using sieve nest.

Introduction:

Particle size and shape characteristics are critical properties for a large number of pharmaceutical processes. There exist many methods of particle size characterisation of materials. These methods include microscopy, sieve analysis, laser diffraction, backlight imaging and direct imaging.

Sieving is the process of separating particles by size. It is a very commonly used method of particle size analysis is to sieve and sort a large quantity of particles into different size ranges and determine the particle size distribution based on the mass collected in each range. However, it does not provide any particle shape information and therefore can only be truly classed as a particle size analysis process, and not a particle characterisation process. Sieve analysis also cannot be used as an in-line analytical tool, and tends to be less accurate with non-spherical particles.

A sieve test is performed by first assembling a stack of interlocking sieves. In this stack the sieve with the largest openings is at the top each lower sieve will have a smaller opening than the one above it. A pre-weighed sample of the material to be tested is placed in the top sieve. This sieve stack is the shaken until all the material has either been retained on a sieve or passed through. The weight of sediment retained on each sieve is measured and converted into a percentage of the total sediment sample. This method is quick and sufficiently accurate for most purposes. A sieve test analysis or distribution is calculated which shows the proportion of each particle size category in the sample.

Materials:

·         Lactose

·         Microcrystalline cellulose (MCC)

Apparatus:

·         Sieve nest

·         Spatula

·         Weighing boats

·         Electronic balance

Procedure:

1.     

100G lactose was weighed.

2.      The sieve nest is prepared in descending order (largest diameter to the smallest, from top to bottom)

3.      The powder is placed at the uppermost sieve and the sieving process is allowed to proceed for 20 minutes.

4.   Upon completion, the powder collected at every sieve is weighed and the particle size distribution is plotted in the form of a histogram.

5.      The above process is repeated using MCC.

Results and calculation:

Particle size range (µm)	Lactose
	Mass of lactose retained in each sieve nest (g)	% retained = (W sieve/W total )x 100%	Cumulative % retained	% passing = 100% -cumulative % retained
x >425	0.0171	(0.0171/99.5442)x 100% = 0.0172 %	0.0172	100- 0.0172 =99.9828
300<x≤425	0.1443	(0.1443/99.5442)x 100% = 0.145%	0.0172 + 0.145       = 0.1622	100-0.1622 =99.8378
200<x≤300	1.0212	(1.0212/99.5442)x 100% = 1.0259%	0.1622+ 1.0259     = 1.1881	100-1.1881 =98.8119
150<x≤200	32.4183	(32.4183/99.5442)x 100% = 32.5667%	1.1881 + 32.5667  = 33.7548	100- 33.7548 =66.2452
50<x≤150	62.4312	(62.4312/99.5442)x 100% = 62.7171%	33.7548 +62.7171 = 96.4719	100- 96.4719 =3.5281
0<x ≤50	3.5121	(3.5121/99.5442)x 100% = 3.5282%	96.4719 + 3.5282 =100.00	100-100 =0

Initial weight of lactose: 100g

After sieving, weight of lactose:

3.5121+62.4312+32.4183+1.0212+0.1443+0.0171= 99.5442g

Percentage of mass loss during sieving: (100 – 99.5442) / 100 x 100% = 0.4558%

Particle size range (µm)	MCC
	Mass of MCC retained in each sieve nest (g)	% retained = (W sieve/W total )x 100%	Cumulative % retained	% passing = 100% -cumulative % retained
x >425	0.0503	(0.0503/99.559)x 100% = 0.0505%	0.0505	100- 0.0505 =99.9495
300<x≤425	0.1382	(0.1382/99.559)x 100% = 0.1388%	0.0505 + 0.1388 = 0.1893	100-0.1893 =99.8107
200<x≤300	1.7704	(0.17704/99.559)x100% = 1.7782%	0.1893 + 1.7782 = 1.9675	100 – 1.9675 =98.0325
150<x≤200	3.8783	(3.8783/99.559)x 100% = 3.8955%	1.9675 + 3.8955 = 5.8630	100- 5.8630 =94.137
50<x≤150	92.576	(92.576/99.559)x 100% = 92.9816	5.8630 + 92.9816 =98.8446	100- 98.8446 =1.554
0<x ≤50	1.1502	(1.1502/99.559)x 100% = 1.1553%	98.8446 + 1.1553 =99.9993	100-99.9993 =0.0007

Initial weight of MCC: 100g

After sieving, weight of MCC:

0.0503+0.1382+1.7704+3.8783+92.5716+1.1502 = 99.559g

Percentage of mass loss during sieving: (100 – 99.559) / 100 x 100% = 0.441%

Discussion

In this experiment, the particle size distribution of lactose and microcrystalline cellulose (MCC) are determined using sieving method. The sieve nest are arranged in descending order from 425µm, 300µm, 200µm, 150µm to 50µm so that particle with size larger than the diameter of aperture can retained in the sieve nest while the smaller one can pass through it.

From the experiment above, most particle size of lactose are in the range of 50-150 µm which is 62.7171% followed by 150-200µm which is 32.5667%, while for MCC, most particles size are in the range of 50 -150 µm which is 92.9816%. This is because the particle size of MCC is finer than lactose as more MCC powder can pass through the sieve nest with 150 µm aperture. The physical vibration that exerted on the particles having greater effect on MCC compared to lactose. Besides, different materials have different physical properties which include the surface hardness of solids that will influence the reduction of size particles.

Initially, the mass of both lactose and MCC is 100g while after sieving, the mass collected are 99.5442g and 99.559g respectively. The weight after sieving is slightly less than the before which is 0.4558% for lactose and 0.441% for MCC.  This may because some particles left in the sieve nest and some spilled out from the container during preparation and sieving. Air movement in the laboratory may also be one of the reasons causing the error of experiment as the particles is light and fluffy. Hence, the particle size distribution obtained for both lactose and MCC may not be accurate. To improve the accuracy of the result, various precautions should be carried out such as clean the sieve nest with brush to remove any residue in it before using and set up the machine correctly and carefully to reduce the spilling out of particles.

Questions

1.      Explain in brief the various statistical methods that you can use to measure the diameter of a particle.

·         Laser Diffraction

A laser beam is passed through a sample of particle. Particles will scatter light at an angle that is directly related to their size. Large particles scatter light at narrow angles with high intensity while small particles scatter at wider angles with low intensity.

·         Dynamic Light Scattering

This method is a non-invasive and sensitive technique used for measuring the size of molecules and particles in the submicron region. It can measure particles, emulsions or molecules, which have been dispersed or dissolved in a liquid. The Brownian motion of particles or molecules in suspension causes laser light to be scattered at different intensities. Analysis of these intensity fluctuations yields the velocity of the Brownian motion and hence the particle size using the Stokes-Einstein relationship.

·         Sedimentation

Andreason pipette or centrifuges and X-rays can be used in this method. The main advantage of this technique is that it determines particle size as a function of settling viscosity. However, as the density of the material is needed, this method is not suitable for emulsions where the material does not settle or for dense material that settles too quickly. It is also based on spherical particles, so can give large errors for particles large aspect ratio.

·         Image Analysis

This technology generates data by capturing direct images of each particle, providing users with the ultimate sensitivity and resolution. Image analysis systems are capable of high-resolution sizing ranging from 0.5µm – 1000µm. Subtle differences in particle size and shape can be accurately characterised using this method.

·         Coulter counter

This instrument measures particle volume. The number and size of particles suspended in an electrolyte is determined by causing them to pass through an orifice an either side of which is immersed an electrode. The changes in electric impedance (resistance) as particles pass through the orifice generate voltage pulses whose amplitude are proportional to the volumes of the particles.

2.      State the best statistical method for each of the samples that you have analysed.

Conclusion

Size of particles can be determined using sieving process. After sieving, 62.7171% of lactose are in the range of 50 - 150µm while as for MCC, 92.9816% are in the range of 50 - 150µm. This shows that most particles in MCC is smaller than those of lactose.

Reference

1.      Particle size: Sieves. http://www.cscscientific.com/particle-size/sieves

2.      Particle characterisation. http://www.innopharmalabs.com/tech/particle-characterisation

3.      An Overview of the Different Particle Size Measurement Techniques. October 8th, 2010. http://www.atascientific.com.au/blog/2010/10/08/overview-particle-size-measurement-techniques/

4.      Dynamic Light Scattering (DLS) http://www.malvern.com/en/products/technology /dynamic-light-scattering/

Mutual Solubility Curve for Phenol and Water

OBJECTIVES:

1. to determine the solubility of two partially liquids (phenol – water solution)
2. to construct a mutual solubility for the pair
3. to determine their critical solution temperature.

INTRODUCTION:

A few liquids are miscible with each other in all proportion, for example: ethanol and water. Others have miscibility in limited proportions in other liquids, for example: etherwater, phenol-water. (Here phenol is not really liquid but is considered to be so since the addition of the first part of water reduces the solid's melting point under room temperature to produce a liquid-liquid system).

Generally both liquids become more soluble with rising temperatures until the critical solution temperature or consolute point is attained, and above this point the liquids completely miscible. There is a big possibility that any pair of liquids can form a closed system where both upper and lower critical solution temperatures exist, however it is not easy to determine both the temperatures (before the substance freezes or evaporates) except for nicotine and water.

At any temperature below critical solution temperature, the composition for two layers of liquids in equilibrium state is constant. The mutual solubility for a pair of partially miscible liquids in general is extremely influenced by the presence of a third component.

APPARATUS:
Boiling tubes, test tube rack, measuring cylinder, 5 mL pipette, 1 mL pipette, water bath, thermometer.

MATERIALS:
Distilled water, phenol.

PROCEDURE:

1. Eight boiling tubes were prepared and labeled as A,B,C,D,E,F, G and H respectively. The boiling tubes was filled with different amount of phenol and water. Boiling tube A filled with 8%, B 11%, C 25%, D 35%, E 50%, F 63% G 72% and H 80% of phenol respectively and distilled water was added until every boiling tube contains 20 mL of the mixtures. Volume of distilled water and phenol are as follows:

Water (mL)		18.4	17.8	15.0	13.0	10.0	7.4	5.6	4.0
Phenol (mL)		1.6	2.2	5.0	7.0	10.0	12.6	14.4	16.0

2. Heated the boiling tubes with the water bath. The boiling tubes were swirled and shaken well. 
3. Once the mixtures becomes a single phase (totally clear), the temperatures are recorded. 
4. The tubes were removed from the hot water and the temperature was allowed to decrease. Then the temperature were recorded when the liquid become turbid and two separate layers were formed again. 
5. Steps 1 to 4 are repeated.
6. By using the temperatures obtained, the average temperature for each tubes at which the two phases are seen as a single phase or at which two phases are separate were determined. 

RESULTS:

Test tube	Single-phase (°C)	Two-phase (°C)
A	55	X
B	59	46
C	74	64
D	77	72
E	79	70
F	70	60
G	65	X
H	55	X

DISCUSSION

The graph shows the temperature against the percentage by weight of phenol in water. Phase diagram is divided into two regions by the transition curve.In the region inside the curve two phases are separate and outside it only single phase exists.

In phenol-water system,the maximum temperature (79 ˚C  at which the two phase region exists Is called the critical solution temperature or upper consolute temperature.

The phase rule is P + F = C + 2, where

• P = number of phases that can coexist, to
• C = number of components making up the phases, and
• F = degrees of freedom.

Phase rule can also be expressed as F=C-P+2. In this experiment, we have two components which is the phenol and water and phases depend on the conditions in which the experiment was conducted. When phenol and water are miscible with each other at a particular condition, the degree of freedom, F = 2 − 1 + 2 = 3. Since the pressure is fixed for this system, F is reduced to 2. This means that we need to fix both temperature and concentration to define this system. When phenol and water are immiscible with each other, the degree of freedom is calculated by F = 2 − 2 + 2 = 2. Since this is a condensed system which means the pressure of the system is fixed, thus the F is reduced to 1. This means that we need to fix the temperature only to completely define the system.

Based on the graph shows the temperature at complete miscibility against the percentage by weight of phenol in water. The temperature when the two liquid phases become one liquid phase (complete miscibility) increase as the composition of phenol in water increase until the maximum temperature; also known as the critical temperature. Phenol can be soluble in water at certain temperature or when the composition of the phenol is much higher than the water.

At room temperature, a tube contains two liquid phases, one more dense than the other. The tube is heated in a water bath until the two phases merge. The temperature at which they merge is the "clearing temperature," also known as the "cloud" temperature, and lies on the liquid-liquid coexistence line. By using several sample tubes, one obtains several points on the coexistence line. 

Critical temperature in this experiment can be define as the maximum temperature at which the liquid phase can be converted into a liquid by an increase in temperature. From the results obtained, the critical temperature is 79 ˚C. The result shows a little bit higher than the theoretically value which is 68 ˚C. This happened due to some errors that we might get into during conducting this experiment. One of those errors is when warming the test tubes in the water bath, the exact temperature at which the two phases become one could not be seen immediately due to sealing film blocking the thermometer. Also the sealing film is not applied properly and leaking may have occurred.

i.   The glass wares must be rinsed before used.

ii. The temperature must be taken immediately to get accurate results.

CONCLUSION:
The critical solution temperature peaks at 79°C. A single phase is observed at that temperature. the solubility of phenol in water is depending on the temperature and the percentage of phenol in water by volume.

REFERENCES:
1. Sinko, Patrick J., Martin's Physical Pharmacy and Pharmaceutical Sciences,  5th ed.,  Lippincott Williams & Wilkins, 2005
2. http://jeplerts.wordpress.com/2008/12/21/partially-miscible-liquids-determination-of-mutual-solubility-of-phenol-water/