1. h C1 Cd Cr C2 Donor Receiver K Permeable membrane D C0 Blood Donor
solution
Permeable membrane
Blood K
Donor
Receiver Cd C1 C2 Cr

2. 6. 1 Fick’s second law of diffusion
Where, dM = change in mass transferred
dt in change of time t
D = diffusion constant
C1 = concentration in donor
compartment
C2 = concentration in receptor
S = surface area of membrane
Since K = C1 = C2
Cd Cr
Under sink conditions and rearranging all constants,
Where, P = permeability constant

3. Factors effecting diffusion
Donor Concentration
Surface area
Partition coefficient

4. Donor Concentration How is the concentration expressed? Molarity
Normality
Molality
Mole Fraction

5. Molarity
solution

6. Normality

7. Normality (Contd.)

8. Molality

9. Mole Fraction

10. Sample problems 1. 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 and the molecular weight of FeSo4 is
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 gm/cm3 at 20 oC. the molecular weight of glycerin is and its density is g/cm2 at 20 oC.
4. The molality of the solution is
5. A non-electrolyte is a substance which
A. 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
Determine the mole fraction of KNO3
The molarity of the solution is:

11. Factors effecting diffusion
Donor Concentration
Surface area
Partition coefficient

12. 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

13. Sample questions
1) What are the important conditions (assumptions) of Fick’s first law
Note: assumptions are the conditions that don’t appear in the equation
a) Steady state and the concentration in the receiver compartment is zero
b) Steady state and the concentration in the donor compartment is zero
c) Surface area is 1 cm2 and thickness of the skin is zero
d) Diffusion coefficient is 10-5 and log partition coefficient is 1
e) None of the above
2) Increase in one of the following factors will decrease the flux of a drug across the skin
a) Surface area
b) Diffusion coefficient
c) Thickness
d) Partition coefficient
e) Donor concentration
3) 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 = 5.8
c) Solubility = 10 mg/ml, pH = 9.8
d) Solubility = 20 mg/ml, pH = 9.8
e) Solubility = 5 mg/ml, pH 7.4
4) 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 is
a) Activity coefficient
b) Solubility parameter
c) Permeability coefficient
d) Diffusion constant
e) Fick’s first law

15. Factor other than the solution pH that influences ionization

16. Strong electrolytes

17. Solubility of drugs Crystalline solids
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

18. 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

19. 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

20. 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