PHT 252 (Practical Notes TA)

PHT 252 Basmah Al-Dossary

SURFACE TENSION ( γ )

Cohesive forces : are the forces that exist between molecules of one phase. Adhesive forces : are the forces that exist between molecules of two different

phases.

Surface tension: is the force per unit length (dyne/cm) that must be applied parallel to the surface so as to counterbalance the net inward pull of molecules of interface together. OR :

Surface tension: is the force per unit length (dyne/cm) on the surface of a liquid which opposes expansion of the surface area.

Methods of Determination of surface tension1. Drop weight method.2. Drop number method.3. Capillary rise method.4. Differential capillary height method.

1. Drop weight method:The γ of a liquid is related to the weight of a drop of that liquid which falls freely from the end of a tube.Method:1. Weigh an empty, clean and dry small beaker.2. Weigh the beaker with 10 drops of water added using the given clean graduated 1

ml pipette. (Pipette should be held vertically while adding the drops of the liquid).3. Repeat steps 1 & 2 three times and calculate the average weight of one drop (m1).

4. repeat steps 1-3 for the provided sample (chloroform , d=1.47 or benzene, d=0.879) using the same pipette and beaker (washed with alcohol and dried) and calculate the surface tension of this sample using the equation:

(Eq. 1)

where: γ1 is the surface tension of water (known reference). γ2 is the surface tension of the sample (unknown). m1 is the mass (weight) of one drop of water. m2 is the mass (weight) of one drop of the sample.

2. Drop number method:The surface tension of unknown solution (γ2) may be obtained by counting the number of drops (n) in a certain volume of liquid (e.g., 0.5 or 1 ml) using a graduated pipette under similar conditions and a liquid of known surface tension (γ1) must be similarly treated using the same pipette under the same conditions.The following equation will then be used:

(Eq. 2)

Method: 1. Count the number of drops (n1) of 0.5 ml of water using a clean 1 ml graduated

pipette. (The pipette should be held vertically).

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2. Clean and dry the same pipette with alcohol, and repeat step # 1 for the provided sample, chloroform (n2).

3. Use the provided data in table 1 and Equation 2 to calculate the surface tension of the sample.

Table 1

Liquid d n γ (dyne/cm)

Water 1 72.8

Chloroform 1.47

Benzene 0.879

3. Capillary rise method:When a capillary tube of a radius (r) is placed in a liquid with density (d) and contained in a beaker, the liquid generally rises up the tube a certain distance (h). By measuring this rise, it is possible to determine the surface tension of the liquid using the following equation:

(Eq. 3)

Where: γ is the surface tension of the liquid (dyne/cm).h is the height of the liquid in the capillary tube (cm).r is the radius of the capillary tube (Cm).d is the density of the liquid (g/Cm3).g is the acceleration due to gravity (980 Cm/sec2).

Method:1. Clean the provided capillary with the solution to be used.2. Attach the capillary to a ruler with a rubber band.3. Place the ruler with the capillary in 50 ml beaker containing water (reference)

perpendicular to its bottom. Be sure that the capillary is just above the bottom of the beaker.

4. Determine the height (h1) after equilibrium. (The height is measured from the surface of the liquid to the mark on the capillary).

5. Calculate the average value of the radius, knowing that the surface tension of

water (γ H2O) =72.8 dyne/Cm at 25 oC, using Equation # 3:

6. Repeat steps 1-4 for the sample of chloroform or benzene using the same capillary tube (washed with alcohol and let dry) and calculate its γ using Equation # 3.

4. Differential capillary height method:Two capillaries with different diameters are used in this experiment. The liquid will rise to height h1 in the narrower tube and to h2 (less than h1) in the wider tube. The difference between h1 and h2 is measured as accurately as possible. This experiment will be performed on solution with known surface tension (γ1) and repeated for solution with unknown surface tension (γ2). Unknown surface tension (γ2) can be calculated from the equation:

(Eq. 4)

Method:

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1. Attach two capillaries of different radius to the ruler.2. Repeat steps 3 & 4 as for the above experiment.3. Measure the difference between h1 and h2 for the two capillaries dipped in water,

then between h1 and h2 when dipped in chloroform or benzene.4. Calculate the surface tension of the sample from Eq. # 4 using the provided data in

table 2.Table 2

Liquid D h1 h2 γ (dyne/cm)

Water 1 72.8

Chloroform 1.47

Benzene 0.879

Requirements of the lab.:

1. Make your calculations.2. Comment on the results including: definitions of surface tension, cohesive and

adhesive forces, and then make a comparison between the γ (H2O) and γ (sample).

N.B., (physical pharmacy, p.446) γ (CCl4) = 26.7 dyne/Cmγ (CHCl3) = 27.1 dyne/Cmγ (C6H6) = 28.9 dyne/Cm

For the comment: If the γ of the sample is less than that of water, this may be due to that the

cohesive forces between the sample molecules are less than the cohesive forces between water molecules. (Increase cohesive forces causes increase in surface tension and decrease in adhesive forces causes an increase in surface tension).

Water has the largest value of γ (72.8 dyne/Cm) because the cohesive forces between water molecules are very strong due to the strong hydrogen bonds existing between water molecules.

Results for the determination of surface tension of chloroform:1. By the drop number method:

n1 (for water) = 12+11+12 / 3 =11.6 =12 dropsn2 (for chloroform) = 46+45+46 /3 = 45 drops

γ2 (CHCl3) = 28.5 dyne/cm.

2. By the differential capillary rise method:

γ2 (CHCl3) = 26.75 dyne/cm.

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INTERFACIAL TENSION ( γA/B)

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Interfacial tension: is the force per unit length existing at the interface between two immiscible liquid phases, and has the unit of dyne/Cm.

Interfacial tensions are less than surface tensions because the adhesive forces between two liquid phases forming an interface are greater than when a liquid and a gas phase exist together (increase adhesive forces cause a decrease in interfacial tension).

If two liquids are completely miscible, no interfacial tension exists between them.

Determination of interfacial tension between two immiscible liquids Determine the interfacial tension between water and chloroform by knowing the

interfacial tension between water and benzene using the drop number method: γ (C6H6 / H2O) = 35 dyne/Cmγ (H2O / CHCl3) = ? dyne/Cmd (C6H6) = 0.879 g/cm 3 , d (CHCl3) = 1.47 g/cm 3

Theory:This method is based on the drop volume (v) of a liquid (density = d1) issuing from a small orifice (pipette) dipped in another liquid (density = d2). The interfacial tension between the two liquids can then be given by:

γ (A/B) = V(d1 –d2) KThe value of K can be determined for the two immiscible liquids 1 and 2 whose interfacial tension is known. This value can then be used to calculate the unknown interfacial tension between a liquid 1 and a third liquid 3, providing the experiment with liquids 1 and 3 are carried out under similar conditions to those for liquids 1 and 2 and using the same pipette (to remove the effect of radius).

Materials and apparatus:Benzene, chloroform, 1 ml graduated pipette and 50 ml beaker.

Procedure:1. Select a clean 1 ml graduated pipette and fill with distilled water (heavier

liquid).2. Dip the pipette in benzene (lighter liquid) and count the number of drops in

0.5 ml (or any fixed volume). The pipette should be held vertically.3. Calculate the volume of each drop. (V = 0.5 / no. of drops)4. knowing that the value of γ (C6H6 / H2O) = 35 dyne/Cm , calculate the K value

from the equation: γ (C6H6 / H2O) = V(d1 –d2) K , where

V is the volume of one dropd1 is the density of heavier liquidd2 is the density of lighter liquidK is a constant

5. Repeat steps 1, 2, and 3 using the same pipette (washed with alcohol) filled with chloroform (heavier liquid) and dipped in a beaker containing water (lighter liquid). Calculate the volume of each drop.

6. Using the value of K obtained from step # 4 and data listed in table 1, calculate the value of γ (H2O / CHCl3).

Table 1

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Liquids d1 –d2 V K γA/B

benzene / H2O 35

H2O / Chloroform

d (C6H6) = 0.879 g/cm 3, d (CHCl3) = 1.47 g/cm 3,d (water) = 1 g/cm3.

Requirements of the lab.:1. Make your calculations.2. Comment on the results including: definitions of interfacial tension, cohesive

and adhesive forces, and then make a comparison between the γ (H2O / CHCl3) and γ (C6H6 / H2O) .

N.B., (from physical pharmacy) γ (H2O / CHCl3) = 32.8 dyne/Cm

For the comment: If the γ (H2O / CHCl3) is less than γ (C6H6 / H2O), this means that the adhesive

forces between water and chloroform molecules are greater than those between benzene and water molecules.

Results:No. of drops of water (0.5 ml) in benzene = 5 dropsV of one drop of water = 0.5 / 5 = 0.1ml

γ (C6H6 / H2O)= V(d1 –d2) K35 = 0.1 (1-0.87)K

K= 2692.3

No. of drops of chloroform (0.5 ml) in water = 19 dropsV of one drop of chloroform = 0.5 / 19 = 0.026 ml

γ (H2O / CHCl3) = V(d1 –d2) K = 0.026 (1.47-1) x2692.3 = 32.8 dyne/Cm

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DETERMINATION OF THE CRITICAL MICELLE CONCENTRATION (CMC) OF A SURFACE ACTIVE AGENT BY SURFACE TENSION

METHOD

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A surface active agent: is a molecule or ion that has a certain affinity for both polar and non polar solvents; therefore, they arrange themselves at the interface and lower the surface tension. An alternative expression is surfactant or amphiphile.

Surfactants when present in a liquid medium at low concentrations exist separately and are of such a size to be subcolloidal. As the concentration is increased, aggregation occurs over a narrow range of concentration. These aggregates are called micelles.

The critical micelle concentration (CMC): is the concentration at which micelles are formed. Or (it is the concentration of the surfactant above which it will migrate to the bulk and start to form micelles and show the minimum value of surface tension).

At low surfactant concentration, the surface tension of the solution decreases with the increase in concentration up to the formation of micelles. At CMC and higher concentrations, the surface tension shows no significant changes.

Procedure:1. Select 7 clean test tubes. (Wash with distilled water).2. Place in each tube different concentration of sodium lauryl sulfate solution

ranging from 0.005% to 0.5%, (using the provided 1% stock solution of the surfactant. The total volume in each tube is to be 10 ml. Consider the density

of the surfactant solution and water is equal (so the equation

becomes ).

N.B., For calculation of the volumes of the SAA and water, use the relation:C1V1 = C2V2

e.g., 0.005% x 10 = 1 x V2

V2 = 0.05 ml of surfactant (taken by a graduated pipette) then add 9.95 ml of dist water to complete to 10 ml total in the test tube.And so on for the rest of concentrations.

3. After preparation of the tubes, mix and count the number of drops in 0.5 ml of the mixture (taken by a clean pipette washed with the solution) for each concentration. Start from the lower conc to the higher.

4. Record the results in a table form (see below) and calculate the value of surface tension obtained for each conc of SAA.

5. Plot the relation between the surface tension and conc of SAA, and comment on the results.

No.Conc of SAA

mls of SAA

mls of water

No. of drops in 0.5 ml solution

1 0 0 10 11 72.8 dyne /cm

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2 0.005% 0.05 9.95 12 66.733 0.01% 0.1 9.9 14 57.2

4 0.05% 0.5 9.5 17 47.1

5 0.2% 2 8 20 406 0.3% 3 7 20 407 0.5% 5 5 20 40

For the comment: Definition of SAA: as mentioned above. Definition of CMC: as mentioned above. Explanation of the curve: the surface tension value decreases with increasing the

conc of SAA (due to increase in adhesion forces) until a certain point (CMC) at which no further change in surface tension value is observed with increasing the concentration of SAA (this effect is due to the formation of micelles).

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SOLUBILIZATION OF SALICYLIC ACID IN NON-IONIC SURFACTANT (TWEEN 80)

Non-ionic surfactants may act as wetting, emulsifying , and solubilizing agents.

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One of the properties of surfactants is their ability to increase the solubility of a sparingly soluble compound through micelle formation.

Below the CMC, as the concentration of surfactant increases, the solubility of the compound shows slower increase. At the CMC range, a sharp increase in the solubility can be observed due to micelle formation (which entraps the undissolved molecules and get them dissolved) and then level off. i.e., Micellar solubilization.

Among the pharmaceutical applications of solubilization are the steroid hormones, oil-soluble vitamins and antimicrobial agents.

Materials and apparatus:Salicylic acid (solubility 1 g in 550 ml water, i.e., slightly soluble (1:100 to 1: 1000)), tween 80, N/10 NaOH solution, phenolphthalein indicator, 100 ml beaker, 250 ml conical flasks, 10 ml pipettes, 50 ml burettes, filter paper and funnels.

Procedure:1. Select 8 clean conical flasks, give each flask a number, and place exactly 1 gm of

salicylic acid in each flask.2. As shown in the table, add different amounts of tween 80 and distilled water to

each of the 8 flasks so that: The total volume in each flask is 50 ml. The conc of SAA ranges from 1-8% v/v.

3. Shake all of the flasks for 15 minutes (shaking should be efficient).4. Filter the mixture of each flask.5. Titrate 10 ml of each filtrate with N/10 NaOH solution using phenolphthalein as

indicator. (1 ml 0.1 N NaOH ≡ 13.8 mg salicylic acid). End point: colourless to faint pink.

6. Record the results in the table and calculate the amount of salicylic acid dissolved in 50 ml.

7. Plot the amount of salicylic acid dissolved vs. tween 80 concentrations, and comment on the results.

No.Conc of

SAA(% v/v)

mls of SAA

mls of water

End point (mls of 0.1 N

NaOH)

Amount of salicylic acid dissolved

= mg/50 ml

1 0 --- 50 1.7 110.5

2 1 0.5 49.5 2.2 143

3 2 1 49 2.5 162.5

4 3 1.5 48.5 3 195

5 4 2 48 3.4 221

6 5 2.5 47.5 3.8 247

7 6 3 47 5 325

8 8 4 46 5.2 338

For the comment: Definition of SAA: as mentioned before. Definition of CMC: as mentioned before.

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Definition of micellar solubilization: which is the solubility due to the formation of micelles.

Explanation of the curve: at low conc of SAA, there is a slow increase in the solubility of salicylic acid, until a certain point (CMC) at which there is a sharp increase in the solubility of salicylic acid due to the formation of micelles.

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PREPARATION AND STABILITY OF COLLOIDS

Dispersed systems consist of particulate matter known as the dispersed phase, distributed throughout a continuous phase, or dispersion medium.

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Dispersed systems are classified on the basis of mean particle diameter of the dispersed material:

1. Molecular dispersions (less than 1 nm (mμ)). E.g., O2 molecules, ions.2. Colloidal dispersions (0.5 μm to 1 nm (mμ)). E.g., colloidal silver sol,

natural and synthetic polymers.3. Coarse dispersions (> 0.5 μm). E.g., emulsions and suspensions.

Types of colloidal systems:1. Lyophilic colloids: (solvent-loving) Colloidal particles interact to an appreciable extent with the dispersion medium. Attraction between the dispersed phase and the dispersion medium leads to

salvation (formation of a solvent sheath around the dispersed phase). It consists of organic molecules e.g., gelatin, acacia, albumin. Their stability depends on the presence of solvent sheath . if removed, the sol

will precipitate (unstable).2. Lyophobic colloids: (solvent-hating)

There is little attraction between the dispersed phase and the dispersion medium.

There is no solvent sheath around the particles. They composed of inorganic particles e.g., silver, gold, sulfur, and silver

iodide. Their stability depends on the presence of the surface charges (the like charges

produce repulsion which prevent coagulation of particles). The neutralization of these charges (by an electrolyte) will cause instability of the sol (ppt or discoloration).

3. Association colloids: These colloids consist of amphiphiles or surface active agents.

Preparation of lyophobic colloids:

1. Ferric hydroxide Fe(OH)3 sol ( + ve charge): Heat 150 ml of distilled water on a hot plate for boiling. Add 50 ml of 2% FeCl3 dropwise (to avoid formation of coarse aggregates)

until a clear dark red sol is obtained. 2. silver tannate sol (-ve charge): To 500 ml of distilled water, add 20 ml 0.1 N AgNO3 (pipette) and then add

10 ml of 1% tannic acid solution. Heat this solution to 80 oC . Add 10 ml of 1% Na2CO3 solution (pipette) dropwise with continuous stirring

until a tea-colored sol is obtained.

Preparation of lyophilic colloids: 1. gelatin sol: Soak 5 gm of gelatin powder in 50 ml water for 5 minutes. Add 175 ml of water and heat the mixture on a hot plate with stirring until

complete dissolving of gelatin. Cool, and adjust the volume to 250 ml with water.

Stability of colloids

1. Fe(OH)3 sol (using 4 M NaCl):

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No. 1 2 3 4 5 6

Fe(OH)3 sol 5 ml 5 ml 5 ml 5 ml 5 ml 5 ml

Distilled water 5 ml 4 ml 3 ml 2 ml 1 ml --

4 M NaCl -- 1 ml 2 ml 3 ml 4 ml 5 ml

Notice the test tube in which precipitation or discoloration may occur, why?Precipitation may occur due to the neutralization of charge on Fe(OH)3 sol.N.B., if an electrolyte with a higher valency than NaCl is used (e.g., Na2SO4), less volume will be required to induce the discoloration.

2. Silver tannate sol (using 0.1 M NaCl):

No. 1 2 3 4 5 6

Silver tannate sol 5 ml 5 ml 5 ml 5 ml 5 ml 5 ml

Distilled water 5 ml 4 ml 3 ml 2 ml 1 ml --

0.1 M NaCl -- 1 ml 2 ml 3 ml 4 ml 5 ml

Notice the test tube in which precipitation or discoloration may occur, why?Precipitation may occur due to the neutralization of charge on silver sol.N.B., if an electrolyte with a higher valency than NaCl is used (e.g., CaCl2 or AlCl3), less volume will be required to induce the discoloration.

3. Gelatin sol (using ethanol):

No. 1 2 3 4

Gelatin sol 5 ml 5 ml 5 ml 5 ml

Ethanol 5 ml 10 ml 15 ml 20 ml

Notice the test tube in which precipitation may occur, why?Precipitation may occur due to the removal of the solvent (water) sheath around gelatin sol. This sol can be redispersed again by the addition of water.

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PHARMACEUTICAL SUSPENSIONS

A pharmaceutical suspension may be defined as a coarse dispersion of finely divided insoluble material randomly distributed in a liquid medium or available in dry form to be distributed in the liquid when desired.

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An ideal suspension should be:1. Easily redispersed by shaking,2. Should remain suspended long enough to withdraw an accurate dosage, and3. Should have the desired flow properties (i.e., viscosity), so it is pourable,4. The particles in suspension should be small and uniform in size so that the

product is free from a gritty texture.

Flocculated suspension system:Is the one in which the repulsive surface charges of the suspended particles have been chemically neutralized. Once these repulsive surface charges have been neutralized, the attractive "Van der Waals" becomes dominant. Under these conditions, the particles may approach each other more closely and form loose aggregates, termed ' flocs'.

Deflocculated suspension system:Are characterized as dispersions in which the particles exist as single entities with high repulsive surface charges. In contrast to flocculated systems, a deflocculated system exhibits well dispersed particles which settle singly but more slowly. The particles have a tendency to form a sediment or cake that is difficult to redisperse.

Characteristics of a "Flocculated system":1. Particles form loose aggregates.2. Rate of sedimentation is rapid. Effect of gravity is dominant.3. Suspension is somewhat unsightly due to the presence of an obvious

supernatant layer above a coarse sediment layer. This unsightly appearance can be minimized by manipulating the formulation to achieve the largest possible sediment volume. Ideally, the sediment volume should encompass the entire volume of the suspension.

Characteristics of a "Deflocculated system":1. Particles settle slowly.2. Particles form dense, cake-like sediment which may be difficult to redisperse.3. Suspension has pleasing appearance.

Compounding:The first steps in preparation are critical to making a good (no lumps) suspension. Weigh and place insoluble powders in mortar. Triturate to break up particles, include suspending agent if a powder at this step. Now with trituration, add the wetting agent, i.e., glycerin or 10-20 ml of the vehicle if the wetting agent and/or suspending agent are already prepared, continue until smooth mucilage is formed. Add any flocculating or deflocculating agents next. Triturate well. Gradually, add about 3/4 of the vehicle, pour suspension into graduate and wash mortar with portions of the vehicle to QS to final volume.

Evaluation of suspending agentsA demonstration will be set up to measure the effectiveness of several suspending agents against a control in distilled water. The agent with the target sediment height is a superior agent.

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Control: Calamine 1.25gDistilled water q.s. to 25 ml

Suspending agent%

Sediment height (cm)at each time (min)

10 15 20 30 45 60 9024 h.

1 w.

Water (control)---

F

Acacia

2%

F

4%

F

Methyl cellulose 5%2%

F

Tragacanth2%

F

Bentonite3%

F

N.B., Sediment volume ratio (F) =

Questions: 1. What is the major disadvantage of higher concentrations of suspending

agents?

2. Does addition of suspending agents ensure redispersibility after settling? Explain.

Redispersibility of suspensions

In groups, prepare the four formulations as shown below. Place in 4 clean bottles or cylinders and observe.

Ingredient A B C D

Sulfadiazine 1.2 1.2 1.2 1.2

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AlCl3 solution -- 1 -- 1KH2PO4 solution -- -- 1 --Na CMC 7 MP (2%) -- -- -- 0.5%0.2% DSS solution q.s. to 60 60 60 60

F (0.5 h)

F (1h)

F (1 w)

Flocculation

Caking

Redispersion

Appearance

Explanation

N.B., Sulfadiazine is a – ve charged particles (negative zeta potential). AlCl3 solution is a (cationic) + ve charged flocculating agent.KH2PO4 solution is a (anionic) – ve charged flocculating agent.DSS is a wetting agent (dioctyl sodium sulfosuccinate; aerosol O.T.).Na CMC is a suspending agent.

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DISPENSING OF PHARMACEUTICAL SUSPENSIONS

A. Magnesium Trisilicate Mixture (BP, 1980)Rx Magnesium trisilicate 50 g 2.5 g

Light magnesium carbonate 50 g 2.5 g

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Sodium bicarbonate 50 g 2.5 gConcentrated peppermint emulsion 25 ml 1.25 mlDouble strength chloroform water 500 ml 25 mlWater to produce 1000 ml 50 ml

Send: 50 ml.Sig. 15 ml to be taken when necessary.

Calculations and procedure:1. Calculate the amounts required for 50 ml of the mixture. The insoluble

materials of magnesium trisilicate, light magnesium carbonate and sodium bicarbonate are diffusible powders (easily wetted with water), so no need to add a thickening (suspending) agent to this formula.

F = 50/1000 = 0.05 Multiply the Rx by Factor. Vehicle (water) =50 – (25+1.25) = 23.75 ml.

2. In a mortar, mix thoroughly the fine powders of magnesium trisilicate, light magnesium carbonate and sodium bicarbonate.

3. Pour gradually with trituration about half of the amount of vehicle to the powder mixture in step # 2 in order to prepare at first a smooth paste and then a fine suspension devoid of any lumps.

4. dilute the formed paste with the amount of conc. Peppermint emulsion and chloroform water until it becomes pourable.

5. Transfer to a cylinder. Wash the mortar with successive small amounts of water left, and mix the washings with the mixture prepared. Adjust to volume.

6. Transfer the mixture to a suitable bottle and write the label.

Role of each ingredient:1. Magnesium trisilicate, Light magnesium carbonate and Sodium bicarbonate

are diffusible materials that act as antacids.2. Concentrated peppermint emulsion is a flavoring agent.3. Double strength chloroform water is a flavoring agent, sweetening agent and a

preservative.

General use: as antacid preparation.

Label: white

Shake Before UseMagnesium Trisilicate Mixture

One tablespoonful to be taken when necessary.Name: Date:

B. Phenacetin suspensionRx Phenacetin iss 6 g 3 g

Glycerin i 4 g 2 g /1.25 mlMethyl cellulose fl VII 28 ml 14 mlAromatic elixir to fl z III 90 ml 45 ml

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.M.ft. Mist. Mette. Dim.Sig. fl to be taken every four hours for pain.

Calculations and procedure:1. Calculate the metric equivalents and prepare half the amount (F= 1/2).2. Phenacetin is a sparingly soluble powder so it is prepared as suspension using

a dispersion stabilizer.3. Methyl cellulose is used as dispersion stabilizer and is very often used in the

form of mucilage in prescription compounding.So the preparation can be handled by:a) Mixing methyl cellulose thoroughly with 1/3 of the required amount of

water as hot water (80 to 90 oC).b) After complete dispersion cool and immerse in a beaker that contains

crushed ice.c) Add the remainder of water as cold water drop by drop with continuous

stirring till a clear gel is obtained.4. Phenacetin powder is weighed and triturated thoroughly in a mortar with

glycerin into a fine paste.5. Add methyl cellulose mucilage gradually with trituration. Add aromatic elixir

and complete the volume.6. Transfer to a clean bottle and write the label.

Label : white

Shake Before UsePhenacetin Suspension

One teaspoonful to be taken every four hours for pain.Name: Date:

General use: analgesic and antipyretic preparation.

Role of each ingredient: Phenacetin: analgesic and antipyretic, active ingredient. Glycerin: sweetening agent, increase viscosity,and levigating (wetting) agent. Methyl cellulose: suspending agent. Aromatic elixir: flavoring and sweetening agent.

C. Calamine lotion BP 1988: Rx Calamine 150 g 7.5 g

Zinc oxide 50 g 2.5 gBentonite 30 g 1.5 gSodium citrate 5 g 0.25 g

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Liquefied phenol 5 ml 0.25 mlGlycerol 50 ml 2.5 mlPurified water to 1000 ml 50 ml

Send: 50 ml.Sig. MDUCalculations and procedure:

1. Calculate the amounts required for 50 ml of the lotion. F = 50/1000 = 0.05 Multiply the Rx by F. Water = 50 – (2.5+0.25) = 47.25 ml

2. Triturate the calamine, zinc oxide and bentonite with a solution of sodium citrate in about 35 ml of purified water.

3. Add the liquefied phenol, and glycerol.4. Add sufficient water to complete to the final volume.

Label : red

Shake Before UseCalamine Lotion

To be used as directed.Name: Date:

General use: antipruritic , soothing and emollient preparation.

Role of each ingredient: Calamine: soothing, and antipruritic agent. Zinc oxide: astringent. Bentonite: suspending agent. Liquefied phenol: preservative. Glycerol: emollient.

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OINTMENTS

Ointments: are semisolid preparations for external application to the body.

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Therapeutically, ointments function as protectives and emmolient for the skin, but are used primarily as vehicle or bases for the topical application of medicinal substances.

Creams: are semi-solid emulsions containing suspension or solution of medicinal agents for external application. Creams (O/W type and W/O type) are employed as emollients.

Pastes: are ointment-like preparations with thick, viscous and less greasy and more absorpitive than ointments, due to higher proportions of powdered ingredients such as starch, zinc oxide calcium carbonate, and talc in the base.

OINTMENT BASES

Ointment bases are classified into four groups according to their physical properties:1. Oleaginous ointment bases:

These bases are fats, fixed oils, hydrocarbons or silicones. They are anhydrous, greasy, non-washable, does not absorb water, and

occlusive (form a film on the skin so it increases skin hydrationby reducing the rate of loss of surface water).

They should not be applied to infected skin. They are water free, and aqueous preparations are incorporated into them

in small amounts and with great difficulty. They are useful primarily for their emollient and occlusive effects. Examples: Petrolatum NF and White ointment USP.

2. Absorption ointment bases: They are anhydrous but hydrophilic ointment bases, they can absorb

several times their weight of water to form water-in-oil emulsion. They are non washable, not water soluble, compatible with most

medications and heat stable. Their composition variea, but usually contain a lanolin fraction and

petrolatum. The addition of fatty alcohols to these bases, particularly stearyl and cetyl alcohols increases the water number of almost any absorption base.

These bases are useful as emollients, and as such they can help incorporation into oleaginous bases. The aqueous solution is first incorporated into the absorption base to form W/O emulsion and the resulting semi-solid can then be incorporated into oleaginous base.

Examples: Anhydrous lanolin, Hydrophilic Petrolatum USP, Aquaphor .3. Water-soluble bases:

These bases are anhydrous, water soluble, absorb water and water washable.

They are either carbowaxes (PEG) or hydrated gums (Bentonite, gelatin, cellulose derivatives).

They are frequently used in the incorporation of non aqueous or solid substances, as they soften too easily for the incorporation of aqueous liquids.

Examples: Polyethylene glycol ointment USP.4. Emulsion ointment bases:

a. Water-in-oil emulsion bases: These are hydrous, hydrophilic, absorbs water, non water removable, with

low thermal conductivity and occlusive.

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They have the same properties as the absorption bases. They are used primarily as emollients. Examples: Cold Cream.

b. Oil-in-Water emulsion bases: These are hydrous, hydrophilic, absorbs water and water removable. They are more acceptable by patients because they are greasless, easily

washable and with good texture. These bases may be used solely for their ability to absorb discharges from

dermatological lesions. Non ionic surfactants are often used with these emulsions because they are

less irritated, less toxic and produce more stable emulsions than the ionic surfactants.

Glycerin, propylene glycol or PEG must be added to slow the evaporation of water from aqueous phase.

Also, fatty alcohols are usually added to improve the texture and stabilize the emulsion.

Examples: Hydrophilic ointment USP.

Preparation of ointmentsOintments are prepared by two general methods:

1. Mechanical incorporation and2. Fusion.

The choice of the method depends upon the medication and the physical properties of the constituents of the base.

1. Preparation by Mechanical incorporation: Mechanical incorporation is performed by trituration in a mortar, or on a glass

slab with a spatula. Insoluble powder is first triturated with small amount of levigating agent. The levigating agents could be an oil (mineral oil) when preparing oleaginous ointment or glycerin and propylene glycol when preparing hydrophilic ointments. Water soluble salts can be dissolved in water and incorporated with anhydrous lanolin before being incorporated with oleaginous base.

2. Preparation by Fusion: When wax, spermaceti or other hard, fusible bodies are to be incorporated with

soft, oleaginous materials, they are melted on water bath to avoid excessive temperatures, beginning with the material possessing the highest fusion point and adding the other ingredients in order of decreasing values, until the softer oleaginous and perhaps liquid ingredients have all been thoroughly incorporated by stirring. The ointment should be stirred until it congeals to insure a homogenous preparation. Powdered medicaments should be incorporated after the base congeals using a small portion of the base as levigating agent.

AN EXAMPLE OF OLEAGINOUS OINTMENT BASES

(White Ointment USP)

Rx White bees wax 50 gWhite petrolatum 950 g

19


Send 25 g. (F = 25/1000 = 0.025, multiply by F)

Preparation:1. Melt white bees wax (higher melting point) in a suitable dish on water bath. 2. Add white petrolatum (lower melting point), warm until liquefied, then

discontinue the heating.3. Remove from water bath, stir the mixture until it congeals.4. Transfer to a clean container (cup) and fix the label.

Label: RedWhite Ointment USP


General use:Emollient and a vehicle for medicinal substances.

AN EXAMPLE OF ABSORPTION OINTMENT BASES(Hydrophilic Petrolatum USP)

Rx Cholesterol 30 gStearyl alcohol 30 gWhite bees wax 80 gWhite petrolatum 860 g


Preparation:1. Melt white bees wax (higher melting point), stearyl alcohol and petrolatum in a

suitable dish on steam bath. 2. Add cholesterol and stir until it completely dissolves.3. Remove from water bath, stir the mixture until it congeals.4. Transfer to a clean container (cup) and fix the label.

Label: RedHydrophilic Petrolatum USP


General use:Emollient and protective.

AN EXAMPLE OF WATER SOLUBLE BASES(Polyethylene glycol Ointment)

Rx Polyethylene glycol 4000 400 gPolyethylene glycol 400 600 g

20



Preparation:1. Melt PEG 4000 (higher melting point) in a suitable dish on water bath. 2. Add PEG 400.3. Remove from water bath, stir the mixture until it congeals.4. Transfer to a clean container (cup) and fix the label.

Label: RedPolyethylene glycol Ointment USP


General use:Emollient and a vehicle for other medicinal substances.

AN EXAMPLE OF W/O EMULSION OINTMENT BASES(Cold Cream USP)

Rx Spermaceti 125 gWhite bees wax 120 gMineral oil 560 gSodium borate 5 gPurified water 190 g


Preparation:1. Reduce the spermaceti and the wax into small pieces.2. Melt them with the mineral oil in a suitable dish on steam bath. 3. Dissolve sodium borate in the purified water, warm to 70 oC, (covered).4. Gradually add the warm solution to the melted mixture, stirring rapidly and

continuously. 5. Remove from water bath; continue stirring until the mixture congeals.6. Transfer to a clean container (cup) and fix the label.

Label: RedCold Cream


General use:Emollient and cleansing cream.

Notes: The emulsifying agent in this cream is the sodium soap formed by the reaction of

sodium borate with the free fatty acids present in bees wax. Spermaceti acts as a stiffening (solidifying) agent.

21


Cold cream is called so due to the cooling sensation provided by the slow evaporation of water contained in the emulsion when the cream is rubbed on the skin.

Cold cream is usually applied at night.

AN EXAMPLE OF O/W EMULSION OINTMENT BASES(Vanishing Cream, Hydrophilic Ointment USP)

Rx Methyl paraben 0.25 gPropyl paraben 0.15 gSodium lauryl sulfate 10 gPropylene glycol 120 gStearyl alcohol 250 gWhite petrolatum 250 gPurified water 370 g


Preparation:1. Melt stearyl alcohol and white petrolatum on steam bath. 2. Gradually, add the other ingredients, previously dissolved in water and warmed

to 75 oC, (covered).3. Remove from water bath; continue stirring until the mixture congeals.4. Transfer to a clean container (cup) and fix the label.

Label: RedVanishing Cream


General use:Emollient cream.

Notes: The emulsifying agent in this cream is sodium lauryl sulfate (non ionic

surfactant). Propylene glycol is hygroscopic material that slows the evaporation of water from

the base and prevent rapid dryness of the cream. Stearyl alcohol improves the texture of the cream and act as a stabilizer for the

emulsion. Methyl and propyl barabens are preservatives. Vanishing cream is called so due to the disappearance of the cream after rubbing

into the skin (no shine effect). Vanishing cream is usually applied during the day time.

Another Vanishing cream formulaRx

Stearic acid 200 g

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Pot.hydroxide 14 gGlycerin 40 gWater 746 gPerfume q.s.Preservative q.s.


Preparation:1. Melt stearic acid on water bath at 75-80 oC.2. Mix glycerin with water and then dissolve KOH in the mixture, and heat to the

same temperature.3. Add #2 to #1 gradually with stirring on water bath for 5 min for complete

emulsification.4. Remove from water bath and continue stirring until cooling.5. Transfer to a clean container (cup) and fix the label.

Comparison between cold and vanishing creams

properties Cold cream Vanishing cream

Emulsion Type w/o o/w

Oil content 45-80 % 15-35 %

Water content 25 % 80 %

Emulgent Bees wax, glyceryl monostearate Alkali soap

Time of use Night Day time

Appearance Shiny (not easily disappear) Matty (disappear rapidly)

Humectants Not incorporated Incorporated

Cooling effect Provides cooling sensation No cooling effect

***********************************************

METHODS FOR PREPARING EMULSIONS

1. Theoretical considerations:

23


An emulsion is a dispersed system containing at least two immiscible liquid phases in which one liquid phase is dispersed in form of small globules throughout the other phase. The dispersed liquid is known as the internal phase, whereas, the dispersion medium is known as the external or continuous phase.

Types of emulsions:When oil is the dispersed phase and an aqueous solution is the external phase, the system is designed as an oil-in-water (O/W) emulsion. Conversely, when water or an aqueous solution is the dispersed phase and oil is the continuous phase, the system is designed as water-in-oil (W/O) emulsion.

Due to the increase in the total surface area of the internal phase, the surface free energy will be increases. Therefore, emulsions are thermodynamically unstable and the dispersed droplets strive to coalesce (to come together) to reduce the surface free energy. In order to minimize this effect, a third component, the emulsifying agent is added to the system to improve its stability.

There are three classes of emulsifying agents:1) Natural emulsifying agents: e.g., acacia, tragacanth, gelatin, eg yolk,….2) Synthetic emulsifying agents: these include the anionic, cationic and nonionic

surfactants, e.g., Tweens, Spans, sodium lauryl sulfate.3) Finely divided solids: e.g., Bentonite, magnesium hydroxide, aluminum

hydroxide and magnesium trisilicate.The proper choice of emulsifying agent is frequently critical in developing a successful emulsion. The choice of the emulsifying agent is critical to the preparation of an emulsion possessing optimum stability.

Methods of emulsion preparation:a. Dry gum method or Continental method:

The emulsifier for the continental method is most often powder acacia, although other natural gums have been used. The success of this method lies in the speed with which an emulsion may usually be formed. The skill required with this method is generally less than that required with the wet gum method, and making the primary or mother emulsion is the only crucial step in the technique. For both the wet and dry gum methods the ratio of oil/water/gum is 4/2/1 for fixed oils and 2/2/1 for volatile oil. In case of fixed oil, the powdered acacia (1 part) is lightly triturated with four parts of fixed oil (or two parts of volatile oil) in a dry mortar. When acacia is evenly dispersed, two parts of water are added all at once with rapid trituration. The trituration is continued at high speed, using a spiral motion of the pestle, until a thick primary emulsion is formed. A snapping sound is heard when a good stable primary emulsion has been prepared. It is during this process that the droplet size is reduced to its minimum dimensions and therefore, the time spent in this operation yields dividends in stability of the final product. Finally, the remainder of the aqueous phase is added slowly with trituration.

b. Wet gum method or English method: The technique of emulsion formation is suitable for preparing emulsions with mucilages or dissolved gums as the emulsifying agents and uses the 4:2:1 ratio as in the dry gum method. It is necessary to use this method, although it is slower than and not as reliable as the dry method, if the emulsifying agent is available only in solution or if it must be dissolved before being used, as with perhaps methylcellulose. In this method, a viscous mucilage of the one part of gum is made with two parts of water

24


and the oil is added in small amounts, with thorough, rapid trituration. When all of the oil has been added, the mixture is brought to volume with water.

c. Forbesor bottle method of emulsification: Emulsions which are to be made form volatile oils (turpentine oil) or other volatile components (chloroform) are most often made in a bottle. The main advantage is that the emulsion remains relatively fluid, and the vigorous shaking of a closed bottle impartssufficient shear-force to break down the oil droplet size.

d. HLB system (Griffin's approach): In the HLB system, surface active agents and oils are assigned a number (HLB

number). This number reflects the balance between the hydrophilic and lipophilic portions of the compound's molecule. The higher the number, the more hydrophilic the compound. The actual number has no significance, except as a means of ranking the surfactants. This ranking was based on extensive experimentations, as a result of which Griffin was able to assign an application to the preparation of stable emulsion.

The fundamental to the utility of the HLB concept for preparing emulsion is the fact that the HLB values are algebraically additive. Thus, by using surfactant with a low HLB with one having a high HLB, it is possible to prepare blends having HLB values intermediate between those of the two individual emulsifiers that match the required HLB value for the oil phase of a given emulsion. Naturally, the required HLB value differs depending on wether the final emulsion is o/w or w/o. If an o/w emulsion is required, the formulator should use emulsifiers with an HLB in the range of 8-18. On the other hand, if an w/o emulsion is desired, the formulator should use emulsifiers with an HLB in the range of 3-6.Calculation of the amount of the emulsifying agents using HLB system:

The preparation of an emulsion using the HLB system involves the determination of the required HLB of the oil phase, selection of twe emulsifiers (one of which has HLB value lower than the desired HLB while other has HLB value higher than the desired HLB ) and algebraically calculate a blend of the two emulsifiers that have any HLB value equal to that of the oil phase "the required HLB value".To illustrate this concept, the following example is given:

Rx Mineral oil 25%Olive oil 25%Emulsifier 2%Water q.s. 60 ml

Prepare 60 mls of o/w emulsion.

i. Knowing that the HLB values of mineral oil and olive oil are 11 and 8 respectively, then the required HLB value should be :

= 11 x 0.5 + 8 x 0.5 = 9.5 ii. Since there is no commercially available emulsifier with HLB value of 9.5, one

can use a blend of Tween 80 (HLB =15) and Span 80 (HLB =4.3) to give the desired HLB.

iii. Using the process of alternate allegation one can blend these 2 emulsifiers in a mathematical proportion in such a way as to obtain a product whose HLB is 9.5.

Tween 80 (HLB) = 15 5.2

25


9.5

Span 80 (HLB) = 4.3 5.5-----10.7

% of Tween 80 to be used = (5.2/10.7) x 100 =48.6%% of Span 80 to be used = (5.5/10.7) x 100 =51.4%Therefore, the amount of Tween 80 = (60x2/100) x 0.486 = 0.583 g

the amount of Span 80 = (60x2/100) x 0.514 = 0.617 g

Preparation of emulsion using the HLB system:Griffin suggested the following experimental procedure for the preparation of stable emulsions:a. Group the ingredients on the basis of their solubilities in the aqueous and non

aqueous phases.b. Determine the type of emulsion required and calculate an approximate HLB value.c. Blend a low HLB emulsifier and a high HLB emulsifier to the calculated value.

For experimental formulations, use a higher concentration of emulsifier (e.g., 10-30% of the oil phase) than that required to produce a satisfactory product. Emulsifiers should, in general, be chemically stable, non toxic, and suitably low in color, odor, and taste. The emulsifier is selected on the basis of these characteristics, on the type of equipment being used to blend the ingredients, and on the stability characteristics of the final product. Emulsions should not coalesce at room temperature, when frozen and thawed repeatedly, and at elevated temperatures of up to 50 oC. Mechanical energy input varies with the type of equipment used to prepare the emulsion. The more the energy input, the less the demand on the emulsifier. Both process and formulation variables can affect the stability of an emulsion.

d. Dissolve the oil-soluble ingredients and the emulsifiers in the oil. Heat, if necessary, to an approximately 5-10 oC over the melting point of the highest melting ingredient or to a maximum temperature of 70 to 80 oC.

e. Dissolve the water-soluble ingredients (except acids and salts) in a sufficient quantity of water.

f. Heat the aqueous phase to a temperature which is 3 to 5 oC higher than that of the oil phase.

g. Add the aqueous phase to the oily phase with a suitable agitation.h. If acids or salts are employed, dissolve them in water and add the solution to the

cold emulsion.i. Examine the emulsion and make adjustments in the formulation if the product is

unstable. It may be necessary to add more emulsifier, to change to an emulsifier with a slightly higher or lower HLB value, or to use an emulsifier with different chemical characteristics.

PREPARATION OF EMULSIONS CONTAINING FIXED OILS AND VOLATILE OIL USING WET AND DRY GUM METHODS

1. a. Emulsion containing fixed oil (Liquid paraffin Emulsion):

26


Rx Liquid paraffin 20 mlSaccharine sodium (1% solution) 0.018 gAmaranth solution 0.1 mlSodium benzoate 0.6 gAcacia powder q.s.Water to 60 ml

i. Prepare the emulsion using dry gum (continental method):ii. Prepare the emulsion using wet gum method (English):Comment on the ease of preparation, viscosity, and color of each sample. Hand in the samples. Label them as: dry or wet gum method.

Calculations:1. To prepare the primary emulsion (mother emulsion), use the ratio:

oil : water : gum4 : 2 : 1 (for fixed oils)2 : 2 : 1 (for volatile oils)

Since the preparation contains a fixed oil (mineral oil or liquid paraffin), then the ratio for the primary emulsion will be:

oil : water : gum4 : 2 : 1

Or 20 : 10 : 5 2. Amount of saccharine sodium solution to be used = (1/100) x 0.018 = 1.8 ml.3. Amount of vehicle (water) to be used = 60 – (20 + 1.8 + 0.1) = 38.1 ml.

The amount of water is to be divided into two portions: 10 ml for the primary emulsion, and 28.1 ml for the dilution of the formed primary emulsion.

Procedure:1. preparation of the primary emulsion:

a. By dry gum method: In a dry mortar, triturate the powder of acacia (5 g) with the 20 ml of liquid

paraffin until a smooth mixture is obtained. Add the 10 ml of water all at once with rapid and strong trituration in one

direction until a thick, creamy mass is produced and a snapping (clicking) sound can be heard.

b. By wet gum method: In a clean mortar, triturate the amount of acacia (5 g) with the 10 ml of water

until smooth fine mucilage is formed. Measure the amount of oil (20 ml) in a dry cylinder and add it gradually (drop

wise) to the formed mucilage of acacia with trituration after each addition in one direction , until a thick, creamy mass is produced and a snapping sound can be heard.

2. dilution of the formed primary emulsion:

In a small beaker, dissolve sodium benzoate in the rest amount of water (≈ 28 ml), then add saccharine solution and amaranth solution.

27


Dilute the formed primary emulsion , gradually , by the above solution with trituration after each addition.

Transfer the final emulsion to a clean, dry bottle and fix the label.

Labeling: White Shake Before Use

Liquid Parrafin EmulsionTo be taken as directed.

Name: Date:

General use: liquid paraffin emulsion is to be used internally as a laxative preparation in the treatment of constipation. It is to be taken all at once. (Liquid paraffin acts as a lubricant).

Role of each ingredient: Liquid paraffin: laxative, active ingredient. Saccharine sodium: sweetening agent. Amaranth solution: coloring agent. Sodium benzoate: preservative. Acacia: a natural emulsifying agent.

1. b. Emulsion containing fixed oil (Castor Oil Emulsion):Rx Castor oil 16 ml

Gum acacia Q.S.Water to 80 ml

Fiat: Emulsion. Sig. MDSPrepare the emulsion using the dry gum (continental) method.

Calculations:1. castor oil is a fixed oil , therefore, the primary emulsion formula to be used is:

oil : water : gum4 : 2 : 1 (for fixed oils)16 : 8 : 4

2. Amount of vehicle (water) to be used = 80 – (8) = 72 ml.The amount of water is to be divided into two portions: 8 ml for the primary emulsion, and 72 ml for the dilution of the formed primary emulsion.

The dry gum method or wet gum method can either be used but the former is more suitable.

Procedure:1. preparation of the primary emulsion by dry gum method:

In a dry mortar, place 4 g of very finely pulverized acacia. Measure 16 ml of castor oil in a dry measure, and pour it on the gum in the

mortar allowing time for the measure to drain. Triturate the oil and the gum together for few moments only (prolonged

trituration at this stage is undesirable, since excessive of the oil and gum coats the gum particles so thoroughly with oil that the water cannot penetrate to them.

28


For the same reason, failure may follow a long delay between mixing the oil and gum and adding water).

Measure exactly 8 ml of water in another dry measure. Begin to triturate the oil and the gum again using a whipping motion, not

grinding action, and while triturating, add the water all at once. As water reaches the oil-gum mixture increase the rate of trituration , taking care

to work in one direction only and to maintain a whipping motion. Continue trituration until the primary emulsion is formed, which is indicated when the mixture becomes white in colour, and a clicking or cracking sound is heared.

Continue the trituration for a little longer time before attempting to dilute.

2. Dilution of the formed primary emulsion: Measure 10 ml of water and add it dropwise, with continuous trituration, to the

primary emulsion and then dilute with about 10 ml of water and transfer to a measure.

Rinse the mortar with more water and adjust to volume. Transfer to a clean dry bottle and fix the label.


Castor Oil EmulsionTo be taken as directed.

Name: Date:

General use: castor oil administered orally as a laxative.

2. Emulsion containing volatile oil (Oil of Turpentine Emulsion):

Rx Oil of turpentine 8 mlAcacia powder q.s.Water ad 60 ml

Fiat: Emulsion. Sig. MDUPrepare the emulsion using wet gum method (English).

Calculations:1. Turpentine oil is a volatile oil, therefore, the primary emulsion (mother

emulsion) formula to be used is, oil : water : gum2 : 2 : 1 (for volatile oils)8 : 8 : 4

2. Amount of vehicle (water) to be used = 60 – (8) = 52 ml.The amount of water is to be divided into two portions: 8 ml for the primary emulsion, and 44 ml for the dilution of the formed primary emulsion.

Procedure:1. preparation of the primary emulsion by wet gum method:

29


In a clean mortar, triturate the amount of acacia (4 g) with the 8 ml of water until smooth fine mucilage is formed.

Measure the amount of oil (8 ml) in a dry cylinder and add it gradually (drop wise) to the formed mucilage of acacia with trituration after each addition in one direction , until a thick, creamy mass is produced and a snapping sound can be heard.

2. dilution of the formed primary emulsion: In a cylinder, put the rest amount of water (≈ 44 ml). Dilute the formed primary emulsion, gradually, by the rest amount of water with

trituration after each addition. Adjust to volume. Transfer the final emulsion to a clean, dry bottle and fix the label.

Labeling: Red Shake Before Use

Oil of Turpentine EmulsionTo be used as directed.

Name: Date:

General use: oil of turpentine emulsion is to be used topically as a counterirritant to increase the blood supply.

PREPARATION OF TURPENTINE LINIMENT

Liniments are alcoholic or oleaginous solutions or emulsions of various medicinal substances intended for external application to the skin, generally by rubbing.

Liniments may contain insoluble materials or they may consist of mixtures of immiscible liquids. When such conditions exist, it is best that an emulsion to be formed, if possible, to prevent rapid separation of the ingredients and to improve the appearance of the product.

Emulsification of liniments is usually accomplished by the formation of a soap which acts as the emulsifying agent.

Turpentine Liniment BP 1988:

Rx Soft soap 75 g 3.75 gCamphor 50 g 2.5 gTurpentine oil 650 ml 32.5 mlPurified water 225 ml 11.25 ml

Fiat: liniment Mitte: 50 mlSig. MDU

Calculations:F = 50 / 1000 = 0.05

Procedure: This is an alkali soap type of emulsion.1. Finely powder the camphor in a fairly large mortar, add the soap and mix well.

30


2. Add the oil in small amounts, mixing well after each addition. When lumps have been dispersed and a smooth suspension produced, the rest of the oil may be added quickly.

3. Transfer the suspension to a beaker for ease of pouring. Place all of the water in the formula into a bottle at least 50% larger than the final volume of the product.

4. Add the oily suspension in small quantities, shaking vigorously after each addition.

5. Note that if the oily suspension is placed in the bottle and the water is added, the emulsion will invert.

6. It is necessary to wait for the air generated by shaking to disperse before transferring the appropriate quantity to a calibrated bottle.

Labeling: Red Shake Before Use

Turpentine Liniment To be used as directed.

Name: Date:

General use: turpentine oil and camphor act as rubefacient and counterirritants.

31


PREPARATION OF EMULSIONS USING HLB SYSTEM(BOTTLE METHOD)

Mineral oil Emulsion:

Rx Mineral oil 50%Emulsifier 2%Water q.s. ad 60 ml

Using a blend of Tween 80 (HLB = 15) and Span 20 (HLB = 8.6):i. Prepare this emulsion assuming the required HLB = 9, or

ii. Prepare this emulsion assuming the required HLB = 11, or iii. Prepare this emulsion assuming the required HLB = 13.

Calculations:1. Amount of mineral oil to be used = (50/100) x 60 = 30 ml.2. Amount of emulsifier to be used = (2/100) x 60 = 1.2 g.

Tween 80 (HLB) = 15 0.4

9

Span 20 (HLB) = 8.6 6-----6.4

Therefore, the amount of Tween 80 = (1.2 x 0.4) / 6.4 = 0.075 gthe amount of Span 20 = (1.2 x 6) / 6.4 = 1.125 g

And so on for the rest of the required HLB values.3. The amount of water to be used = 60 –(30 +1.2) = 28.8 ml.

Procedure:1. Mix the amount of oil with the calculated amount of span 20 in a small beaker,

then transfer the mixture to a clean, dry bottle.2. Mix the calculated amount of Tween 80 with that of water in a small beaker.3. Add #2 to #1 gradually (drop wise) with vigorous shaking after each addition.4. Fix the label.


Mineral Oil EmulsionTo be taken as directed.

Name: Date:

General use: this emulsion is to be used internally as a laxative preparation in the treatment of constipation.

**************************************************

32


DETERMINATION OF THE REACTION RATE CONSTANT OF HYDROLYSIS OF ETHYL ACETATE IN THE PRESENCE OF AN EQUAL

QUANTITY OF SODIUM HYDROXIDE

CH3COOC2H5 + NaOH -------------- CH3COONa + C2H5OH

The reaction depends on the concentration of both ethyl acetate and sodium hydroxide.

In this experiment, the concentration of both reactants are equal (a = b). The decomposition of ethyl acetate follows 2nd order reaction kinetics:

, where

a is the initial concentration of either reactant,x is the concentration of the decomposed reactant,a-x is the concentration of the remaining reactant,K is the second order reaction rate constant, and t is the time.

Procedure:1. In two separate conical flasks, place 50 ml of N/10 ethyl acetate and 50 ml of

N/10 sodium hydroxide.2. Put 10 ml of N/10 HCl and few drops of phenolphthalein in a titration flask.3. Fill a burette with N/20 NaOH.4. Add the solutions of # 1 to one another, shake and quickly withdraw 10 ml to the

titration flask and titrate against N/20 NaOH.5. Cover the reaction flask and repeat the withdrawal and titration at the specified

time intervals taking 10 ml at each time.

Calculations: HCl is used to prevent further hydrolysis is N/10, wheras NaOH titrant is N/20.

Thus, the volume of NaOH will be double that of HCl. Calculate the remaining sodium hydroxide after each specified time interval.

a = moles / liter.

x = moles / liter.

Plot a curve for against time and calculate the K value. The K value

equals to the slope of the obtained line. The unit of K value is (mole. L -1.min-1).

Table of results:

33


Time (min)

EP (ml)

EP x f a(moles/L)

x (moles/L)

a(a-x)

0 13.4 12.73 0.036 0 1.29 x 10-3 0

5 17.8 16.91 0.036 0.0209 5.43 x 10-4 38.44

10 18.2 17.29 0.036 0.0238 4.75 x 10-4 47.97

15 18.8 17.86 0.036 0.0256 3.72 x 10-4 68.70

20 19 18.05 0.036 0.0266 3.38 x 10-4 78.6

25 19.3 18.34 0.036 0.0280 2.88 x 10-4 97.22

30 19.3 18.34 0.036 0.0280 2.88 x 10-4 97.22

Slope =

Slope = K, then K = 4.06 mole. L-1.min-1

***************************************************

EFFECT OF TEMPERATURE ON THE RATE OF HYDROLYSIS OF ASPIRIN

34


Aspirin is stable in acidic medium and unstable in alkaline or aqueous medium.Since aspirin is insoluble in water, sodium citrate can be used to dissolve aspirin. The hydrolysis of 3 molecules of aspirin in sodium citrate solution yield 1 molecule of citric acid and 3 molecules of acetic acid.

Procedure:1. In a clean, dry stoppered measuring (volumetric) flask, place 50 ml of 5% sodium

citrate solution taken by a burette.2. Put the flask in a constant temperature water bath adjusted at 50 oC.3. Weigh accurately 1 g of aspirin.4. When sodium citrate solution reaches the assigned temperature (check by

thermometer), dissolve the amount of aspirin weighed. Shake vigorously to dissolve aspirin.

5. After complete dissolution of aspirin, immediately, remove 5 ml sample (by pipette). Return the flask to the water bath to keep the temperature of the solution constant throughout the analysis time.

6. Add the withdrawn sample to a conical flask containing 30 ml of distilled water and few drops of phenolphthalein indicator. Mix and titrate against N/10 NaOH. The end point shows a change in the color of the solution from colourless to faint pink.

7. Repeat withdrawal of samples at the specified time intervals and treat each sample as for the first sample.

8. Carry out the above experiment at different temperatures, e.g., 60 and 70 oC. Thus three experiments are run together each one at either of the three temperatures.

9. Record the results of each experiment in a table form. Calculate the % remaining aspirin, plot the possible graph and estimate the reaction rate constant at each temperature (i.e., K 50 oC , K 60 oC, and K 70 oC ).

10. In order to make the Arrhenius plot for the reaction, calculate the corresponding log K at each temperature and the absolute temperature for the considered 3 values then 1/T.

11. Plot 1/T against log K. extrapolate the line obtained to room temperature. Estimate the shelf life of aspirin solution at 25 oC.

Calculations: Aspirin hydrolysis follows 1st order reaction rate kinetics:

Log Ct = log Co – Kt / 2.303Then slope = -K / 2.303 And K = slope x (-2.303)

Aspirin remaining at any time t (Ct) = 2X-Y

% Aspirin remaining at any time t (Ct%) = , where

X is the titer at time = 0Y is the titer at time = t.

Tabulate the results, then plot log Ct% against time and calculate the K value at the assigned temperature.

35


Shelf life (t90): is the time for the drug to reach 10% decomposition or 90% remaining.Half life (t1/2): is the time for the drug to reach 50% decomposition.For 1 st order reactions: t90 = 0.105 / K

t 1/2 = 0.693 / K

Arrhenius equation:

Arrhenius plot involves plotting log K against 1/T. The estimation of the shelf life of aspirin at 25 oC require the extrapolation of the

line in the arrhenius plot to the absolute temperature corresponding to 25 oC. the value of log K at 25 oC is read and K value is estimated. Then the t90 is calculated from the equation: t90 = 0.105 / K .

Results

1. The rate of hydrolysis of aspirin at 50 oC:

Slope =

K50 =


Time (min)

EP (ml) Ct Ct% Log Ct%

0 4.8 4.8 100 2

10 4.9 4.7 97.9 1.99

20 5 4.6 95.8 1.98

30 5.3 4.3 89.58 1.95

45 5.5 4.1 85.42 1.93

60 5.7 3.9 81.25 1.91

Time (min)


0 5 5 100 2

10 5.3 4.7 94 1.97

20 5.5 4.5 90 1.95

30 5.7 4.3 86 1.93

45 6 4 80 1.90

60 6.5 3.5 70 1.84

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Slope=

K60 =


Slope=

K70 =

Arrhenius plot:

Temp. (oC)

T (oK) 1/T K (min-1) Log K

50 323 3.09 x 10-3 3.42 x 10-3 -2.47

60 333 3.00 x 10-3 5.63 x 10-3 -2.25

70 343 2.9 x 10-3 0.011 -1.96

Calculation of t90 of aspirin solution at 25 o C: 25 oC = 298 oK. Then 1/T = 3.35 x 10-3 which is to be extrapolated to

the line:Then log K25 = -3.2

K25 =6.31 x 10-4 min-1.

Therefore, t90 = 0.105 / K = 0.105 / 6.31 x 10-4 = 166.4 min = 2.77 h.

**************************************************

Time (min)


0 5.1 5.1 100 2

10 5.5 4.7 92.15 1.96

20 6.1 4.1 80.39 1.90

30 6.5 3.7 72.54 1.86

45 7 3.2 62.74 1.79

60 7.6 2.6 50.98 1.70

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EFFECT OF pH ON THE RATE OF HYDROLYSIS OF ASPIRIN AT 50oC

Method:1. Accurately weigh 100 mg (0.1 g) of aspirin.2. Place 100 ml the buffer solution (pH 3, 6, 8, or 10) in a constant temperature

water bath adjusted at 50 oC. 3. When the buffer reaches the assigned temperature add to it the weighed aspirin.

Shake to dissolve. 4. When complete dissolution is achieved, immediately remove 1 ml sample (with a

pipette), transfer it to a dry cuvette, add 5 ml colouring reagent and read the absorbance at 540 nm (in a colorimeter) against a blank consisting of the buffer solution (1 ml) and 5 ml of the colouring reagent.

5. Samples are removed at the following time intervals 0, 5, 10, 15, 30, 45, 60, and 90 minutes from the beginning of the experiment. Each sample is to be analyzed as for the first sample.

6. Tabulate the results then plot log Ct% versus time and calculate the K value at the tested pH.

7. Repeat the above experiment using buffer solution with different pHs. Calculate the K value at each pH, and then plot the pH – rate profile for aspirin. Estimate the pH for maximum stability of aspirin.

Notes: Aspirin is more stable in acidic pH than in alkaline pH, since it is an acidic drug.

(it is most stable at pH 2.5 as shown in the pH – rate profile, Conner's). Hydrolysis of one molecule of aspirin (acetyl salicylic acid) yields one molecule

of salicylic acid and one molecule of acetic acid. i.e.,

C9H8O4 C7H6O3

m.wt aspirin m.wt salicylic acid 180 mg aspirin 138 mg salicylic acid or 1.304 mg aspirin 1 mg salicylic acid

The colouring agent (FeNO3) react with decomposition product (salicylic acid) to

give a color (violet) that can be measured colorimetrically. The intensity of the color increases with increasing the amount of salicylic acid in the sample.

A universal buffer (No. 13) can be used for this experiment. The pH range of this buffer form 2-12.

Calculatios: Each absorbance is converted to concentration of salicylic acid by means of the

slope of the standard calibration curve of salicylic acid. This slope equals to 0.019. Upon hydrolysis each molecule of aspirin yields one molecule of salicylic acid.

Therefore, each one mg salicylic acid represents the hydrolysis of 1.304 mg of aspirin.

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Results:1. Hydrolysis rate in buffer solution pH 4 at 50 oC:

Tim

e (m

in)

Abs

. at 5

40 n

m

Abs

t-Abs

0 Conc of salicylic acid =

Asp

. Hyd

roly

zed=

con

c of

sal

icyl

ic x

1.3

04

Asp

. Rem

aine

d (C

t) =

10

0-hy

drol

yzed

Ct%

= (

Ct/1

00)

x 10

0

log

Ct%

0 0.005 0 0 0 100 100 2

10 0.01 0.005 0.263 0.343 99.65 99.65 1.998

15 0.015 0.01 0.526 0.686 99.31 99.31 1.997

30 0.02 0.015 0.789 1,029 98.97 98.97 1.995

45 0.03 0.025 1.315 1.715 98.28 98.28 1.992

60 0.035 0.03 1.578 2.058 97.94 97.94 1.990

Slope = -1.73 x 10-4

K pH 4 = 2.303 x slope = 3.99 x 10-4 min-1.

2. Hydrolysis rate in buffer solution pH 8 at 50 oC:

Tim

e (m

in)

Abs

. at 5

40 n

m

Abs

t-Abs

0 Conc of salicylic acid =

Asp

. Hyd

roly

zed=

con

c of

sal

icyl

ic x

1.3

04

Asp

. Rem

aine

d (C

t) =

10

0-hy

drol

yzed

Ct%

= (

Ct/1

00)

x 10

0

log

Ct%

0 0.015 0 0 0 100 100 2

10 0.02 0.005 0.263 0.343 99.65 99.65 1.998

15 0.03 0.015 0.789 1.029 98.97 98.97 1.995

30 0.055 0.04 2.105 2.745 97.25 97.25 1.987

45 0.08 0.065 3.421 4.461 95.53 95.53 1.980

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60 0.11 0.095 5 6.52 93.48 93.48 1.970

Slope = -4.375 x 10-4

K pH 8 = 2.303 x slope = 1.007 x 10-3 min-1.

Therefore, it can be noticed that aspirin is more stable in acidic pH (low K value) than in alkaline pH.

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PHT 252 (Practical Notes TA)

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