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Isotonicity PHT 434. osmosis Osmosis is the diffusion of solvent through a semi- permeable membrane. ▫Water always flows from lower solute concentration.

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Presentation on theme: "Isotonicity PHT 434. osmosis Osmosis is the diffusion of solvent through a semi- permeable membrane. ▫Water always flows from lower solute concentration."— Presentation transcript:

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 HypertonicisotonicHypotonic NaCl 2%NaCl 0.9%NaCl 0.2% solute ‹ solute Inside outside solute = solute Inside outside solute › solute Inside outside shrinkageequilibriumswelling

4 Why using isotonic solutions?

5 Isotonicity & route of administration Subcu taneous 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.

6 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

7 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

8 Freezing point depression (f.p.d) Freezing Point solution = Freezing Point solvent - ΔT f ΔT f =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

9 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 )

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 0.08º (0.52º- 0.08º=0.44º) 1% NaCl 0.58º w% NaCl 0.44º

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 1 st calculate E NaCl 2 nd add NaCl to reach 0.9%

14 How to calculate E NaCl ? =

15 How to calculate amount of NaCl

16 Example I Calculate E NaCl of drug (M.wt=187, L iso =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 SO 4 (E NaCl =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: ▫1 st Addition of H2O to drug to make it isotonic ▫2 nd addition of isotonic vehicle to bring solution to final volume

19 How to calculate amount of H 2 O ? Suppose preparing 30ml of 1% drug isotonic with body fluid(E NaCl =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

20 One step equation V : volume of H 2 O W: weight of drug 111.1= 100/0.9 Last example

21 II example Add volume of H 2 O and then complete with isotonic solution Phenacaine HCl 0.06 gm (E NaCl =0.16) Boric acid 0.3 gm (E NaCl =0.5) sterile distilled H 2 O up to 100 ml V = 111.1 x(weight x E NaCl ) V =111.1 x [(0.06x0.16)+(0.3x0.5)] = 17.7 ml H 2 O


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