1. Isotonicity
PHT 434

2. 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.

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

4. Why using isotonic solutions?

5. 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.
subcutaneous infusion, is the subcutaneous administration of fluids to the body. This would often be in the form of a saline or glucose solution

6. Isotonicity & route of administration cont.
Intrathecal injection
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

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

8. 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
Colligative properties are properties of solutions that depend on the number of molecules in a given volume of solvent and not on the properties (e.g. size or mass) of the molecules.

9. 1- Freezing point depression (FPD) “cryoscopic method”.
F.P. of blood & tears = º
Any solution have F.P. = º is isotonic.
Any solution have F.P. › º is hypotonic
- 0.4º hypotonic
-0.6º hypertonic
Add solute to hypotonic solution to reach f.p.d of blood (- 0.52º )

10. 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)

11. 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 º (0.52º- 0.08º=0.44º)
1% NaCl º
w% NaCl º

12. Example II
adjust isotonicity of procaine HCl 3% using NaCl ? Fpd of 1%NaCl=0.57º, f.p.d of 1% drug=0.112º

13. 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%

14. How to calculate ENaCl ? =

15. How to calculate amount of NaCl

16. Example I Calculate ENaCl of drug (M.wt=187, Liso=3.4)?
How much NaCl needed to make 2% of this drug isotonic?

17. 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

18. 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

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

20. One step equation Last example V : volume of H2O W: weight of drug
111.1= 100/0.9
Last example

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