Donor Receiver

CdC1

C2

Cr

h

D

C0 Donorsolution P

erm

ea

ble

me

mb

ran

e

Blo

od

K

J

dM

dt S

D C C

h

1 2Where, dM = change in mass transferred dt in change of time t

D = diffusion constantC1 = concentration in donor compartmentC2 = concentration in receptor compartmentS = surface area of membrane

Since K = C1 = C2

Cd Cr

dM

dt

SDK C C

hd r

Under sink conditions and rearranging all constants,

M PSC tr d Where, P = permeability constant

6. 1 Fick’s second law of diffusion6. 1 Fick’s second law of diffusion

Factors effecting diffusion

• Donor Concentration

• Surface area

• Partition coefficient

Donor Concentration

How is the concentration expressed?MolarityNormalityMolalityMole Fraction

Molarity

solution

Normality

Normality (Contd.)

Molality

Mole Fraction

Sample problems1. Molarity is defined as

A. gram molecular weight of solute in 1L of solution

B. gram equivalent weight of solute in 1L of solution

C. saturated solution of solute in 1L of solution

D. ml of solvent required to dissolve 100g of solute

E. gm molecular weight of solute in 1000g of solution.

• An aqueous solution of exsiccated ferrous sulfate was prepared by adding 41.50g of FeSO4 to enough water to make 1000ml of solution at 18oC. The density of the solution is 1.0375 and the molecular weight of FeSo4 is 151.90.

2. Calculate the molarity of FeSO4.

3. The Normality of the above solution is:

• An aqueous solution of glycerin , 7.00% by weight is prepared. The solution is found to have a density of 1.0149gm/cm3 at 20 oC. the molecular weight of glycerin is 92.0473 and its density is 1.2609 g/cm2 at 20 oC.

4. The molality of the solution is

5. A non-electrolyte is a substance whichA. Forms at least two ions when dissolved in water

B. Forms more than two ions when dissolved in water

C. Has the same molarity as water

D. Does not form ions when dissolved

• Dissolved 10.0g of KNO3 (MW 101) in 100 ml of water

6. Determine the mole fraction of KNO3

7. The molarity of the solution is:

Factors effecting diffusion

• Donor Concentration

• Surface area

• Partition coefficient

Partition coefficient

• Most of the drugs are either weak acids or weak bases

• Drug molecules are absorbed at the biological barriers such as intestinal mucosa (oral drug administration) and lung epithelium (inhalation drug deliver) mostly in their unionized form.

• The partition coefficient of unionized drug molecules is higher than the ionized species

Sample questions1) What are the important conditions (assumptions) of Fick’s first lawNote: assumptions are the conditions that don’t appear in the equationa) Steady state and the concentration in the receiver compartment is zerob) Steady state and the concentration in the donor compartment is zeroc) Surface area is 1 cm2 and thickness of the skin is zerod) Diffusion coefficient is 10-5 and log partition coefficient is 1e) None of the above2) Increase in one of the following factors will decrease the flux of a drug across the skina) Surface areab) Diffusion coefficientc) Thicknessd) Partition coefficiente) Donor concentration3) Compound X is a weak base (pKa = 7.8). If applied as a cream, which of the following conditions will best

enhance its permeability across the skin:a) Solubility = 10 mg/ml, pH = 7.8 b) Solubility = 20 mg/ml, pH = 5.8c) Solubility = 10 mg/ml, pH = 9.8d) Solubility = 20 mg/ml, pH = 9.8e) Solubility = 5 mg/ml, pH 7.44) It is clear from the course material that Flux (J) = (DKC)/h, where:D = Diffusion coefficient; K = Partition coefficient; C = Donor concentration; h = Skin thickness. Based on your understanding, DK/h isa) Activity coefficientb) Solubility parameterc) Permeability coefficientd) Diffusion constant e) Fick’s first law

Factor other than the solution pH that influences ionization

Strong electrolytes

Solubility of drugs

Polymorphism determines the solubility of solids in liquids

Crystalline solids• Solids are arranged in fixed geometric patterns or lattices

and have an orderly shape and arrangement. Solids are mostly incompressible. Crystalline solids show very sharp melting points.

• Six crystal forms are possible, cubic (sodium chloride), tetragonal (urea), hexagonal (iodoform), rhombic (iodine), monoclinic (sucrose) and triclinic (boric acid).

• In the crystal lattice, ions or atoms can be held together by Vanderwaals, hydrogen bonding or by covalent linkage

Amorphous solids• Amorphous solids may be also termed as super cooled liquids in

which molecules are arranged in a random manner as in liquid state.

• Glass, Synthetic plastics are good examples

• Amorphous substances have a higher aqueous solubility than the crystalline solids. As a consequence their absorption across cellular barriers my be higher

• Crystalline Novobiocin acid is poorly absorbed while the amorphous form is readily absorbed

• It is difficult to distinguish between crystalline and amorphous substances based on the physical appearance. Beexwax and paraffin might look amorphous, but they are crystalline.

• Differential scanning calorimetry can distinguish them based on the melting points. Crystalline solids have definite melting points

Polymorphism

• Substances such as sulfur may exist in more than one crystalline form

• Polymorphs have generally have different solubilities and melting points

• Incase of slightly soluble drugs, low solubility of some polymorphs might effect their therapeutic activity eg. Chloramphenicol palmitate

• Polymorphism is involved in suspension stability eg. Cortisone acetate exists in 5 polymorphic forms. However only 1 of those is stable in water. All the other unstable forms will be converted the stable form upon storage

Example: Cocoa butter (theobroma oil)

• Cocoa butter exists in 4 polymorphic forms• The unstable gamma forms melts at 18 oC• Alpha form melts at 22 oC • Beta prime form melts at 28 oC • The most stable beta form melts at 34.5 oC• If cocoa butter is heated beyond 35 oC, the beta

form nuclei are completely destroyed and the resultant melt will not cool above 15 oC