Lecture 3 Buffer and Isotonic solutions Lecture 4

Lecture 3

Buffer and Isotonic solutions

Lecture 4

Isotonicity PHT 434

osmosis • Osmosis is the diffusion of solvent through a semipermeable 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 Hypotonic Na. Cl 0. 2% solute Inside solute ›outside swelling isotonic Na. Cl 0. 9% solute Inside solute =outside equilibrium Hypertonic Na. Cl 2% solute Inside solute ‹outside shrinkage

Why using isotonic solutions?

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 Ø Adding substace to lower f. p of solution to -0. 52º 1. Freezing point depression (FPD) “cryoscopic method”. 2. Na. CL equivalent method. • Class II Ø Adding H 2 O 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% Na. Cl 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)

Example I • How much Na. Cl is required to render 100 ml of a 1% soln. of apomorphin HCL isotonic? • F. p. d of 1%Na. Cl=0. 58º, F. p. d of 1%drug=0. 08º • 1% drug • 1% Na. Cl w% Na. Cl 0. 08º (0. 52º- 0. 08º=0. 44º) 0. 58º 0. 44º

Example II • adjust isotonicity of procaine HCl 3% using Na. Cl ? Fpd of 1%Na. Cl=0. 57º, f. p. d of 1% drug=0. 112º

2 -Na. Cl equivalent method • Na. Cl equivalent “E” Amount of Na. Cl that is equivalent to(i. e. , has the same osmotic effect (same f. p. d) as ) 1 gm of drug • 1 st calculate E Na. Cl • 2 nd add Na. Cl to reach 0. 9%

How to calculate ENa. Cl ? =

How to calculate amount of Na. Cl

Example I • Calculate ENa. Cl of drug (M. wt=187, Liso=3. 4)? • How much Na. Cl needed to make 2% of this drug isotonic?

Example II • Calculate amount of Na. Cl needed to adjust 1. 5% Atropine SO 4 (ENa. Cl =0. 12 gm) • =0. 9 –(W x E) = 0. 9 –(1. 5 x 0. 12) = 0. 72 gm of Na. Cl should be added

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

How to calculate amount of H 2 O ? • Suppose preparing 30 ml of 1% drug isotonic with body fluid(ENa. Cl =0. 16 gm) • 1 gm 100 ml ? 30 ml =0. 3 gm • Amount of Na. Cl eq. to 0. 3 drug = 0. 3 x 0. 16 =0. 048 gm • 0. 9 gm 100 ml • 0. 048 gm ? ml =5. 3 ml

One step equation V : volume of H 2 O W: weight of drug 111. 1= 100/0. 9 • Last example

example II Add volume of H 2 O and then complete with isotonic solution Phenacaine HCl 0. 06 gm (ENa. Cl=0. 16) Boric acid 0. 3 gm (ENa. Cl=0. 5) sterile distilled H 2 O up to 100 ml V = 111. 1 x(weight x ENa. Cl) V =111. 1 x [(0. 06 x 0. 16)+(0. 3 x 0. 5)] = 17. 7 ml H 2 O