1. PHYSICAL pharmacy
By
Dr. Ahmed R. Gardouh
Lecturer of Pharmaceutics
And Pharmaceutical Technology
Suez Canal University
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3. IV- Spreading Coefficient
When a substance is placed on the surface of water, it will show one of two
behaviors:
1. Spread as a film if the force of adhesion between the substance molecules and the
water molecules is greater than the cohesive force between the substance molecules
themselves such as oleic acid on water.
2. Do not spread and form separate spots if the cohesive force between the substance
molecules is greater than the adhesion force between the substance molecules and the
water molecules such as mineral oil on water.
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4. The work of adhesion, which is the energy required to break the attraction between
the unlike molecule (e.g. oleic acid and water) can be given by the equation:
Work = Surface tension x Unit area change
If we consider a hypothetical cylinder of cross–sectional area of 1cm2 of a sub-layer
liquid “S” (water), Overlaid with a similar section of the spreading liquid “L” (Oleic
acid).
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5. The work required to separate the two sections of liquid each with a cross–sectional area of 1 cm2, is
therefore numerically related to the surface or interfacial tension involved, because the area increment is
unity. Accordingly, the work done equal to the newly created surface tensions, (L) and ( S) minus the
interfacial tension ( LS) that has been destroyed in the process. The work of adhesion is:
The work of cohesion, required to separate the molecules of the spreading liquid so that it can flow
over the sub-layer, is obtained by the same way: No interfacial tension exists between the like molecules
of the liquid, and when the hypothetical 1 cm2 cylinder is divided, two new surfaces are created, each
with a surface tension of L. The work of cohesion is
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6. With reference to the spreading on water surface, spreading occurs if: the work of
adhesion (attraction force between substance and water) is greater than the work of
cohesion (attraction force between the substance molecules).
The term (Wa – Wc) is known as the spreading coefficient “S”.
Where,
S: is the surface tension of the sub-layer liquid,
L: is the surface tension of the spreading liquid,
LS: is the interfacial tension between them.
N.B. If “S” is positive, spreading will occur. If negative no spreading occurs.
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7. If the surface tension of the sub-layer is greater than the sum of the surface tension of
the spreading layer and the interfacial tension between the two layers, S will be positive
and spreading will occur.
If (L + LS) is greater than S the substance forms globules or floating lens and fail to
spread over the surface, an example is mineral oil on water.
The previous discussion was involving the initial spreading. When equilibrium is
reached and the water surface become saturated with the spreading material, the
Spreading Coefficient (S) may be reduced or may even become negative. This means that
although initial spreading may occur on the liquid substrate, it can be followed by
coalescence of the excess material into a lens if the spreading coefficient becomes
negative.
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8. N.B. In this circumstances, s become less than (l + ls)in which ls does not changed since the
interfacial tension is determined under conditions of mutual saturation
It is important to consider the type of molecular structures that lead to high spreading
coefficient, oil may spreads over water if it contains polar group such as COOH or
OH, But when the carbon chain of an acid increases, the ratio of polar/non polar
decreases and the spreading coefficient on water decreases.
Example:
If the surface tension of water “γS” is 72.8 dynes/cm at 20°C, the surface
tension of benzene “γL” is 28.9 dynes/cm, and the interfacial tension between
benzene and water “γLS“ is 35.0 dynes/cm, what is the initial spreading coefficient
“S”? Following equilibration, S is 62.2 dynes/cm and L is 28.8. What is the final
spreading coefficient?
Solution:
S = 72.8 – (28.9 + 35.0) = 8.9 dynes/cm
S`= 62.2 – (28.8 + 35.0) = - 1.6 dynes/cm
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9. V- Adsorption at Liquid Interfaces:
Surface–active agents (SAA):
Certain molecules and ions, when dispersed in the liquid, move to the interface. First, They
concentrated at the interface then exceeds their concentration in the bulk of the liquid. The
adsorption of the molecules at the interface reduces the surface free energy and the surface tension
of the system. Such a phenomenon, where the added molecules are partitioned in favor of the
interface, is termed positive adsorption. Other materials such as inorganic electrolytes are
partitioned in favor to the bulk, leading to negative adsorption.
Adsorption differ from absorption in that , the former is solely surface effect while the second is
penetration of the absorbent into the capillary space of the absorbing medium.
Molecules and ions that are adsorbed at interfaces are termed surface–active agents (SAA) or
surfactants. They are also known as amphiphiles or amphipathic, which suggests that the
molecules or ions have certain affinity for both polar and nonpolar solvents. Depending on the
number and nature of the polar and nonpolar groups present in their molecules, the amphiphile
may be:
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10. a. Predominantly hydrophilic (water–loving)
b. Lipophilic (oil–loving)
c. Well balanced between these two extremes
For example the straight–chain alcohols, amines and acids are amphiphiles which are:
Example:
*Ethyl alcohol: is miscible with water in all proportions.
*Amyl alcohol: The aqueous solubility is much reduced
*Cetyl alcohol may be said to be strongly lipophilic and insoluble in water.
The Amphiphile Nature:
The amphiphile nature of the surface–active agents is the character that cause them to be adsorbed at
the interfaces (liquid/gas or liquid/liquid). Their should be a suitable balance between the hydrophilic
and lipophilic groups to ensure surface activity.
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11. In an aqueous dispersion of SAA:
a. The polar groups are able to associate with water molecules.
b. The non-polar portion is rejected because the adhesive force it can develop with water molecules is
smaller than the cohesive force between the adjacent water molecules.
c. As a result the amphiphile molecule is adsorbed at the interface.
*In order for the amphiphile to be concentrated at the interface, it must be balanced with the proper
amount of water–soluble and oil–soluble groups (e.g. Glyceryl monostearate).
*If the molecule is too hydrophilic, it will remain within the body of the aqueous phase and exerts no
effect at the interface (e.g. Glycerin).
*Likewise, if it is too lipophilic, it will dissolve completely in the oil phase and little appears at the
interface (e.g. Glyceryl tristearate).
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12. A- Systems of Hydrophile – Lipophile Classification:
Griffin devised an arbitrary scale of values to serve as a measure of the hydrophilic –
Lipophilic balance (HLB) of surface-active agents.
*This system establishes an HLB
range of optimum efficiency for
each class of surfactant.
*The higher the HLB of an Hydrophilic
agent, the more hydrophilic it is.
*The spans, sorbitan esters,
are Lipophilic and have low
HLB values of 1.8 to 8.6.
*The tweens, polyoxyethylene Lipophilic
derivatives of the spans are
hydrophilic and have HLB
values of 9.6 to 16.7.
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Classification of surface active agent
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13. There are several formulae for calculating HLB values of non-ionic surfactants:
* The HLB for polysorbates and sorbitan esters (spans) from the following
formula:
HLB = (E+P) / 5
Where: E, is the percentage by weight of oxyethylene chain and P , , is the percentage by weight of
polyhydric alcohol groups (glycerol or sorbitol)
*If the surfactants contain only polyoxyethylene as the hydrophilic group then the
formula become
HLB = E / 5
*Alternatively, we can calculate the HLB values from the chemical formula as the
following:
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14. HLB = 7 + Ʃ (hydrophilic group numbers) - Ʃ (lipophilic group numbers)
*The HLB of a number of polyhydric alcohol fatty acid esters, such as glyceryl
monostearate, may be estimated by using the formula:
In which S is the saponification number of the ester and A is the acid number of the
fatty acid.
Example: Calculate the HLB of polyoxyethylene sorbitan monolurate (tween 20),
where, S = 45.5 and A = 276, is:
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15. *In addition, it has been suggested that certain emulsifying agents of a given HLB
value appear to work best with a particular oil phase, and this has given rise to the
concept of a required HLB value for any oils or combination of oils. However, this does
not necessarily mean that every surfactant having the required HLB value will produce a
good emulsion. For this reason, mixture of surfactants give more stable emulsions than
when used singly. The HLB of a mixture of surfactants consisting of fraction x of A and
(1-x) of B, is assumed to be an algebraic mean of the two HLB numbers:
HLB mixt. = x HLB A + (1-x) HLBB
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16. Formulation by the HLB method
It has already been shown that physically stable emulsions are best achieved by the presence of a
condensed layer of emulsifying agent at the oil / water interface, and that the complex interfacial films
formed by a blend of an oil-soluble emulsifying agent with a water-soluble one produces the most
satisfactory emulsions. A useful method has been devised for calculating the relative quantities of these
agents necessary to produce the most physically stable emulsion for a particular oil /water type. This is
called the hydrophile-lipophile balance (HLB) method.
Calculate the total required HLB for the following emulsion:
Liquid paraffin 35%
Wool fat 1%
Cetyl alcohol 1%
Emulsifier system 5%
Water to 100%
Solution:
The total percentage of oil phase is 37 and the proportion of each is:
Liquid paraffin 35/37 x 100 = 94.6%
Wool fat 1/37 x 100 = 2.7%
Ceytl alcohol 1/37 x 100 = 2.7%
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17. The total required HLB number is obtained as follows:
Liquid paraffin (HLB 12) 94.6/100 x 12 = 11.4
Wool fat (HLB 10) 2.7/100 x 10 = 0.3
Cetyl alcohol (HLB 15) 2.7/100 x 15 = 0.4
Total required HLB = 12.1
Assuming that a blend of sarbitan mono-oleate (HLB 4.3) and polyoxyrthylene sorbitan
mono-oleate (HLB 15) is to be used as the emulsifying system, the proportions of each to be
added to the emulsion to provide an HLB of I2.1 are calculated as follows. Let “A” be the
percentage concentration of the hydrophilic and “B” the percentage of the hydrophobic
surfactants required to give a blend having an HLB value of “X”. Then:
100(x - HLB of B)
A=
(HLB of A - HLB of B)
and B = 100 – A
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18. In our example, therefore:
100(12.1 – 4.3)
A= = 72.9
(15 – 4.3)
B = 100 – 72.9 = 27.1
Because the total percentage of emulgent blend in the formulation is 5, the percentage of
each emulsifier will be:
Sorbitan mono-oleate 5 x 27.1/100 = 1.36
Polyoxyethylene sorbitan mono-oleate 5 - 1.36 = 3.64
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