Transcript PHT-224 Lectures 3

The Solution Process Model
• Whether or not a substance will dissolve in another substance
depends on the magnitude of the forces between molecules.
 Cohesive forces: Forces between similar molecules
 Adhesive forces: Forces between dissimilar molecules
Solute molecules
cohesive
Solvent molecules
cohesive
Solute + solvent
adhesive
The Solution Process Model
• Step # 1:
• Removal of solute molecule from the pure solute structure, this
results in loss of energy due to breakage of cohesive forces.
The Solution Process Model
•
•
Step # 2:
Creation of a hole in solvent molecules by disrupting intermolecular
forces between solvent molecules. This results in loss of energy due to
breakage of cohesive forces:
The Solution Process Model
•
•
Step # 3:
Introduction of the solute molecule into the hole in the solvent structure
and subsequent solvation of the solute molecule by the solvent. This
results in gain of energy from formation of new adhesive forces :
In general:
Overall gain of energy < loss of energy, solubility will be very high.
Overall gain of energy > loss of energy, solubility will be very low.
I. Effect of particle size on solubility of solid in liquid
• Dissolution rate is a measure of how fast a solid drug
dissolves in a liquid solvent.
• Dissolution is defined as the process of dissolving a
solute to form a homogeneous solution as described
by Noyes Whitney equation.
• When a particle of a drug is dissolved in water, the
molecules at the very surface of the particles dissolve
and saturate the diffusion layer.
• The dissolved drug particles then diffuse out of the
diffusion layer through the less tightly bound water to
the bulk (stirred) solution.
• The rate of dissolution is given by Noyes-Whitney equation
which takes into account the two processes necessary for
dissolution to occur which are: Saturation of the diffusion layer
and diffusion of drug molecules into solvent.
dm/dt =KS (Cs-Ct)
K = D/h
dm/dt =(D/h)S (Cs-Ct)
Where:
dm/dt is dissolution rate which is the change in mass of solute per
time, unit is mg/second.
K = Dissolution rate constant (cm/sec).
S = Surface area of exposed solid in square cm.
D = Diffusion coefficient (square cm/ sec) of the drug in solvent ( a
measure of how fast the drug molecules move or diffuse
through the solvent.
h = Thickness of the diffusion layer ( >0.05 mm thick).
Cs = Saturation solubility (Molar or mg/ml).
Ct = Solubility at any time t (Molar or mg/ml).
•
•
•
Example:
A preparation of drug granules weighing 5.5 gm and having a total surface area
of 2800 cm2 is allowed to dissolve in a 500 ml of water at 25oC. After the first
minute, 0.76 gm have dissolved. The saturation solubility (Cs) of the drug is 15
mg/ml.
a) Calculate the dissolution rate constant (K)
dm/dt =KS (Cs-Ct)
dm/dt =0.76gm / 60 seconds = 0.01267 gm/sec X 1000= 12.67 mg/sec
S = 2800 cm2
Cs = 15 mg/ml
Ct = 0.76 gm / 500 ml = 0.00152 gm/ ml X 1000 = 1.52 mg/ml
12.67 mg/sec = K (2800 cm2) (15 mg/ml - 1.52 mg/ml)
K = 0.000336 cm /sec
b) If the diffusion layer thickness (h) is 0.005 cm, calculate the diffusion
coefficient (D).
K = D/h
D=KXh
D = 0.000336 cm /sec X 0.005 cm = 1.6 X10-6 cm2/ sec
c) Suppose that surface area was increased to 5000 cm2, what would be the
dissolution rate.
dm/dt =KS (Cs-Ct) = (0.000336 cm /sec)(5000 cm2)(15- 1.52 mg/ml)
dm/dt = 22.65 mg/sec
•
•
When surface area is 2800 cm2:
When surface area is 5000 cm2:
i.e.,
dm/dt = 12. 67 mg/sec
dm/dt = 22.65 mg/sec
surface area leads to increasing dissolution rate.
How to utilize Noyes Whitney equation to enhance solubility:
dm/dt =(D/h)S (Cs-Ct)
1)
Increase surface area by decreasing particle size.
Effective surface area is area in direct contact with water.
Reduced particle size leads to increased surface area leading to
increased effective surface area and increased solubility.
2)
Mechanical stirring leads to reduced diffusion layer thickness
( reduced drug concentration in diffusion layer) leading to increased
solubility.