Isotonicity PHT 434

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% solute ‹ solute Inside outside solute =solute solute › solute shrinkage equilibrium swelling

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

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

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

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.

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 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

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

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 ? =

How to calculate amount of NaCl

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

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 Last example V : volume of H2O W: weight of drug 111.1= 100/0.9 Last example

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