Lab 5 if an excess liquid or solid is added to a mixture of two immiscible or partially miscible liquids it will distribute itself bet The two phases so that each become saturated If the sub Is added in amount insufficient to saturate the ID: 914048
Download Presentation The PPT/PDF document "Partition coefficient p.c." is the property of its rightful owner. Permission is granted to download and print the materials on this web site for personal, non-commercial use only, and to display it on your personal computer provided you do not modify the materials and that you retain all copyright notices contained in the materials. By downloading content from our website, you accept the terms of this agreement.
Slide1
Partition coefficientp.c.
Lab 5
Slide2if an excess liquid or solid is added to a mixture of two immiscible, or partially miscible liquids it will distribute itself bet. The two phases so that each become saturated.
If the sub. Is added in amount insufficient to saturate the
solu
. It still become distributed bet. The two layers in a definite conc. ratio .
So equilibrium expression
K =C₁ / C₂
Where
C₁ & C₂
is
equilibrium conc.
Of the sub. In solvent 1& solvent2, respectively.
While K
is equilibrium constant also known
as
distribution ratio, distribution coefficient or partition coefficient
at constant temp.
Slide3example/
when
boric acid distributed between water & amyl
alch
. At 25 c⁰, the conc. In water is (0.0510 mole /L) & in amyl
alch
. Is (0.0155 mole/L) what is the distribution coefficient?
K = C H₂O /c
alch
. = 0.0510/0.0155 =3.29
Or
K = c
alch
. / C H₂O = 0.0155/ 0.0510 =0.304
Slide4Importance of P.C. to the pharmacist:-
1)the
preservation of oil / water system.
2)absorption& distribution of drug through the body.
3)In the extraction process to remove a particular solute From
homogeneous
system.
4)In dosage form formulation.
preservation
-
it will
be
explained by
knowing the effect of ionic dissociation & molecular association on P.C
.
as the solute can exist partly or wholly as
associated molecule
in one phase or it may
dissociate into ions
in either of the liquid phases.
the distribution low applies only to the conc. Of the species common to both phases , namely, monomers or simple molecule of the solute
.
e.g. when benzoic acid used as preservative it is distributed between oil phase & water phase & the distribution law applies only when it is neither
associated in oil phase
nor
dissociated in aqueous phase
.
Slide6extraction:-
To
remove a particular solute from homogeneous solution by adding another solvent that is immiscible with the first one .
3) Absorption & distribution of drug
through the body :-
The passage of d. through the lipid membrane & interaction with the receptor site sometimes correlates with the P.C.
4)Dosage form design :-
In oily type supp. Base which melt at body temp. , the release of the drug depends on P.C.
Experimental work :-
Determination
the partition coefficient of iodine between water and chloroform.
Procedure:-in dry Stoppered conical flask (iodine flask) put 20 ml of 1% iodine in chloroform (use burette).
Add 50 ml D.W. to it
.
The flask is the thoroughly shaken from time to time half hour after equilibrium is established ,allow to stand for complete phase separation, this need another half an hour
.
10 ml of the sample are taken from the upper aqueous layer, care is taken to avoid touching the chloroforming layer. Then titrate against 0.02 N sodium
thiosulphate
.the end point is the disappearance of light brownish color.
5 ml are taken from the organic layer (lower layer) .the inside wall of the pipette must be kept dry as it passes through aqueous phase by placing the finger tightly over the upper end of the pipette. Then titrate against 0.1 N sodium
thiosulphate
.
Before titration, add 5 ml of 10% pot. Iodide to
facillate
extraction of I2 from the organic layer and it's titration with aqueous sodium
thiosulphate
. The end point is the disappearance of the brownish color
.
Slide10Calculation:
Iodine
distributed between the aqueous phase and
chloroformic
phase.
Aqueous phase:-
the no. of ml of sodium
thiosulphate
(0.02N) consumed in the titration is equivalent to the amount of Iodine present.
(Na
2
S
2
O
3
) V1 X C1 = V2 X C2 (iodine)
E.P X 0.02 N =10 X N2
N2 =conc. Of iodine in water
Chloroformic
phase:-
the no. of ml of sodium
thiosulphate
(0. 1N) consumed in the titration is equivalent to the amount of Iodine present.
(Na
2
S
2
O
3
) V1 X C1 = V2 X C2 (iodine)
E.P
2
X 0.1N = 5 X N
2
N2 =conc. Of iodine in chloroform
Conc. Of iodine in CHCl
3
Partition coefficient = ---------------------------------------
Conc. Of iodine in water