Transcript PHT-224 Lectures 4

Measurements of dissolution profiles (solubility):
•
Finely divided powder of known , uniform particle size in
excess is suspended in water and stirred continuously.
•
Samples are taken at time intervals and filtered to remove
any undissolved drug, then analysed to determine
concentration of dissolved drug.
•
Plot Ct against time of sampling
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When all of the drug is dissolved i.e., solution is saturated
with drug the drug concentration will not change as a
function of time any further and this is Cs.
Measurements of dissolution profiles (solubility):
Cs
Concentration
mg/ml
Time (hr)
Sink and non-sink conditions:
dm/dt =KS (Cs-Ct)
If Ct is very small (Ct >0.1 Cs), then dissolution rate is directly
proportional to the saturation solubility (Cs). This assumes
that S is constant which is only true at the very beginning
of the experiment when only a small amount (>0.5%) of
the solid drug has dissolved.
When these two requirements are obeyed, then the system is
under sink condition. If either requirements is violated,
then non-sink condition operate and rate of dissolution is
not directly proportional to saturation solubility.
Ct
Sink
Time (hr)
Non-sink
Solubility Product:
For slightly soluble electrolytes (eg AgCl & Al(OH)3) dissolved
in water to form saturated solutions, their solubility is
referred to as solubilityKproduct (Ksp).
Formation of saturated aqueous solution is an equilibrium
process:
K
AgCl (solid)
Ag+ + Cl-
[Ag+] [Cl-]
K=
[AgCl] solid
[AgCl] solid is considered constant because there is always
some solid present.
K = [Ag+] [Cl-]
Ksp = [Ag+] [Cl-]
Ksp is solubility product
Example:
A saturated solution of AgCl in water at 20oC has a
concentration of 1.12X10-5 M of AgCl ie., Cs of AgCl =
1.12X10-5 M.
Calculate the solubility product Ksp.
Ksp = [Ag+] [Cl-]
Ksp = (1.12X10-5 ) (1.12X10-5 ) = 1.25 X 10-10
For salts that the cations and anions have different number of
charges e.g., Al(OH)3 you have to raise each
concentration to a power corresponding to the
stoichiometric number of ion.
K
Al(OH)3 (solid)
Ksp = [AL+3] [OH-]3
AL+3 + 3OH-
II. Effect of Salts on solubility of solid in liquid:
Common ion effect:
a) Salting out:
K
AgCl (solid)
Ag+ + Cl-
If an ion common with Ag+ or Cl- is added to the solution of
AgCl (e.g. NaCl), the equilibrium will be shifted towards
the left and more AgCl will precipitate leading to reduced
solubility.
b) Salting in:
When the added common ion makes a complex with the
salt whereby the net solubility is increased because
solubility of new salt is more than solubility of AgCl.
Energetics of Solubility
Solubility and heat of solution
Solubility as a function of temperature for non electrolytes or
weak electrolytes or strong electrolytes in non ideal solution can
be calculated according to the equation:
∆H solution
ln (C= /C- ) =
(T= - T-)
X
R
(T= T-)
Where C= is concentration at temperature two
C- is concentration at temperature one
∆H solution is heat of solution (Cal/ mol) and defined as
heat
gained when one mole of solute is dissolved in
solvent
R is universal gas constant (1.9872 Cal/ mol.deg)
T= is temperature two in Kelvin = Co +273
T- is temperature one in Kelvin = Co +273
Non electrolytes:
Substances that do not yield ions when dissolved in water and
therefore do not conduct an electric current through the
solution e.g., sucrose or glycerin in water
Strong electrolytes:
Substances that yield ions by completely ionizing in water
when dissolved in water leading to strong conduction of
electric current. e.g., hydrochloric acid and sodium
sulfate.
Weak electrolytes:
Substances that yield ions by partially ionizing in water when
dissolved in water leading to weak conduction of electric
current. e.g., ephedrine and phenobarbital.
Ideal solution:
A solution where contribution of different components in a liquid
mixture to vapour pressure is according to the molar
fraction of the different components.
Example:
The solubility of urea (Mwt. = 60.6 gm/ mol) in water
at 25oC is 1.20 gm/ gm water. The ∆H solution
for urea in water at 25oC is 2820 Cal/ mol. What
is solubility in gm/ gm at 5oC.
Solubility in gm/ gm at 5oC is unknown
T- = 5 + 273 = 278 oK
T= = 25 + 273 = 298 oK
∆H solution = 2820 Cal/ mol
C= = 1.20 gm/ gm
R = 1.9872 Cal/ deg. Mol
C- = 
ln (C= /C- ) = (∆H / R) X (T= - T- ) / (T= T-)
lnC=- ln C- = (∆H / R) X (T= - T- ) / (T= T-)
lnC=- [(∆H / R) X (T= - T- ) / (T= T-)] = ln C-
Example:
The solubility of urea (Mwt. = 60.6 gm/ mol) in water
at 25oC is 1.20 gm/ gm water. The ∆H solution
for urea in water at 25oC is 2820 Cal/ mol. What
is solubility in gm/ gm and at 5oC.
lnC=- [(∆H / R) X (T= - T- ) / (T= T-)] = ln CLn1.2-[(2820/1.9872)X(298-278 )/(298x278))] = ln Cln C- = -0.161
C- = 0.851 gm/gm
 Solubility at 5oC = 0.851 gm/ gm
 Solubility at 25oC = 1.2 gm/ gm
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