Isotonicity notes osmosis calculation methods

osmosis
• Osmosis is the diffusion of solvent through a semi-permeable
membrane.
▫ Water always flows from lower solute concentration [dilute solution] to
higher solute concentration until a balance is produced
• Osmotic pressure is the force that cause this diffusion .
• Tonicity is a measure of the osmotic pressure
of two solutions separated by a semi-permeable
membrane.

Types of Tonicity
Hypertonic
isotonic
Hypotonic
NaCl 2%
NaCl 0.9%
NaCl 0.2%
solute ‹ solute
Inside outside
solute =solute
Inside outside
solute › solute
Inside outside
shrinkage
equilibrium
swelling

Isotonicity & route of administration
• Subcutaneous injection:
 not necessarily “small dose” but isotonicity reduce pain.
• Hypodermoclysis
 should be isotonic “Large volume”
• Intramuscular injection
 should be isotonic or slightly hypertonic to increase penetration
• Intravenous injection
 should be isotonic “Large volume ”
 Hypotonic cause haemolysis
 Hypertonic solution may be administered slowly into a vein
 Hypertonic large volume administered through a cannula into large vessels.

Isotonicity & route of administration cont.
• Intrathecal injestion
 Should be isotonic
• Eye drops
 Rapid diluted by tear, but most of it is isotonic to decrease irritation
• Eye lotions
 Preferably isotonic
• Nasal drops
 Isotonic, but not essentially

Classes of adjustment of isotonicity
• Class I
 Adding substace to lower f.p of solution to -0.52º
1. Freezing point depression (FPD) “cryoscopic method”.
2. NaCL equivalent method.
• Class II
 Adding H2O
3. White –Vincent method

Freezing point depression (f.p.d)
• Freezing Pointsolution = Freezing Pointsolvent - ΔTf
• ΔTf =L c L : constant , c : conc.(molarity)
• It is Colligative property
• Depend on concetration
• same f.p.d same conc. same tonicity
• 0.9% NaCl is isotonic i.e. F.p.d = 0.52º

1- Freezing point depression (FPD) “cryoscopic
method”.
• F.P. of blood & tears = - 0.52º
• Any solution have F.P. = - 0.52º is isotonic.
• Any solution have F.P. ›- 0.52º is hypotonic
• - 0.4º hypotonic
• -0.6º hypertonic
• Add solute to hypotonic solution to reach f.p.d of blood (- 0.52º )

How to calculate?
= conc. gm/100 ml of adjusting substance
= f.p.d of 1% of unadjusted substance(table) X percentage strength
= f.p.d of 1% of adjusting substance (table)
b
a
w


52
.
0
%
%
w
a
b

Example I
• How much NaCl is required to render 100 ml of a 1% soln. of
apomorphin HCL isotonic?
• F.p.d of 1%NaCl=0.58º, F.p.d of 1%drug=0.08º
• 1% drug 0.08º (0.52º- 0.08º=0.44º)
• 1% NaCl 0.58º
w% NaCl 0.44º
b
a
w


º
52
.
0
% %
76
.
0
% 
w
º
58
.
0
º
08
.
0
º
52
.
0
%


w
%
76
.
0
% 
w

Example II
• adjust isotonicity of procaine HCl 3% using NaCl ? Fpd of
1%NaCl=0.57º, f.p.d of 1% drug=0.112º
b
a
w


52
.
0
%
576
.
0
)
3
*
112
.
0
(
52
.
0
%


w
NaCl
ml
gm
w 

 100
/
32
.
0
%

2-NaCl equivalent method
• NaCl equivalent “E”
Amount of NaCl that is equivalent to(i.e., has the same osmotic effect (same
f.p.d) as ) 1 gm of drug
• 1st calculate E NaCl
• 2nd add NaCl to reach 0.9%

How to calculate ENaCl ?
drug
drug
drug
drug
V
x
M.wt
.
wt
x
L
Tf iso


NaCl
NaCl
NaCl
(NaCl)
NaCl
V
x
M.wt
.
wt
x
L
Tf iso


aCl
aCl
aCl
aCl
iso
wt
x
L
N
N
N
)
N
(
V
x
M.wt
.
45
.
58
M.wt NaCl
=
drug
(drug)
M.wt
17 iso
NaCl
L
E 
c
L
Tf .


gm
wt 1
.drug  NaCl
E
wt 
NaCl
.
4
.
3
)
( 
NaCL
iso
L
drug
drug
drug
drug
iso
wt
x
L
V
x
M.wt
.
)
(

How to calculate amount of NaCl
NaCl
of
solution
isotonic
9
.
0
drug
of
1gm
weight to
equivalent
NaCl
ml
100
per
gm
in
drug
of
weight
rug%
isotonic)
solution
make
(to
ml
100
per
gm
in
NaCl
of
weight
%




NaCl
E
d
w
)
E
drug%
(
9
.
0
% NaCl



w

Example I
• Calculate ENaCl of drug (M.wt=187, Liso=3.4)?
• How much NaCl needed to make 2% of this drug isotonic?
gm
ENaCl 31
.
0

drug
(drug)
M.wt
17 iso
NaCl
L
E 
)
(
1
)
(
31
.
0 drug
gm
NaCl
gm 
)
(
28
.
0
62
.
0
9
.
0 NaCl
gm


)
(
2
)
(
62
.
0 drug
gm
NaCl
gm 

Example II
• Calculate amount of NaCl needed to adjust 1.5% Atropine SO4 (ENaCl
=0.12gm)
• =0.9 –(W x E)
= 0.9 –(1.5x 0.12)
= 0.72 gm of NaCl should be added

3-White – Vincent method
• Principle:
• 1st Addition of H2O to drug to make it isotonic
• 2nd addition of isotonic vehicle to bring solution to final volume

How to calculate amount of H2O ?
• Suppose preparing 30ml of 1% drug isotonic with body fluid(ENaCl
=0.16gm)
• 1gm 100ml
? 30ml =0.3gm
• Amount of NaCl eq. to 0.3 drug
= 0.3 x 0.16 =0.048gm
• 0.9 gm 100 ml
• 0.048 gm ? ml
=5.3 ml

One step equation
V : volume of H2O
W: weight of drug
111.1= 100/0.9
• Last example
1
.
111


 NaCl
E
w
v
1
.
111
16
.
0
3
.
0 


v
ml
v 3
.
5


II example
Add volume of H2O and then complete with isotonic solution
Phenacaine HCl 0.06 gm (ENaCl=0.16)
Boric acid 0.3 gm (ENaCl=0.5)
sterile distilled H2O up to 100 ml
V = 111.1 x(weight x ENaCl)
V =111.1 x [(0.06x0.16)+(0.3x0.5)] = 17.7 ml H2O

Isotonicity notes osmosis calculation methods

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