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ISOTONICITY
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ISOTONICITY
Body fluids, including blood and tears, have the same osmoticpressure as that of a 0.9% w/v sodium chloride solution.
Solutions having the same osmotic pressure as that of 0.9%w/v NaCl solution are said to be isotonic with blood.
Solutions with a higher osmotic pressure than body fluids arecalled hypertonic and those with a lower osmotic pressureare called hypotonic.
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Isotonic solutions calculations :
CALCULATION OF DISSOCIATION (i) FACTOR
Since osmotic pressure depends upon the number of particlesof solute(s) in solution, the osmotic pressure of an electrolyteis directly proportional to the degree (or extent) ofdissociation.
The dissociation factor, symbolized by the letter i, can becalculated by dividing the total number of particles (whichinclude undissociated molecules and ions) in a solution by thenumber of particles before dissociation, i.e.,
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CALCULATION OF DISSOCIATION ) FACTOR
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CALCULATION OF DISSOCIATION ) FACTOR
What is the dissociation factor of NaCl, having 80%dissociation in water?
Assume that we have 100 particles of NaCl prior todissociation. Upon
80% dissociation, 100 molecules of sodium chloride yield:
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CALCULATION OF DISSOCIATION ) FACTOR
What is the dissociation factor of zinc chloride, having 80%dissociation in water?
Assume that we have 100 particles of zinc chloride prior todissociation.
Upon 80% dissociation, 100 molecules of zinc chloride yield:
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SODIUM CHLORIDE EQUIVALENTS OF DRUG SUBSTANCES
The sodium chloride equivalent of a chemical is defined as theamount of sodium chloride (in grams or grains) that has thesame osmotic pressure as that of 1 g of the chemical.
The sodium chloride equivalents are symbolized by the letter E.
The quantities of two substances that are isotonic equivalentsare proportional to the molecular weight of each multiplied bythe i value of the other.
Thus, if the molecular weight and i value of a given chemicalare known, one can calculate the sodium chloride equivalent, E,of that chemical as follows:
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SODIUM CHLORIDE EQUIVALENTS OF DRUG SUBSTANCES
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SODIUM CHLORIDE EQUIVALENTS OF DRUG
SUBSTANCES
Calculate the sodium chloride equivalent of a 1% solution ofpilocarpine nitrate.
Pilocarpine nitrate has a molecular weight of 271 and i of 1.8.
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SODIUM CHLORIDE EQUIVALENTS OF DRUG
SUBSTANCES
Calculate the sodium chloride equivalent of a 1% boric acid.
Boric acid has a molecular weight of 62 and i of 1.
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ISOTONICITY ADJUSTMENTS BY SODIUM
CHLORIDE EQUIVALENT METHOD
The sodium chloride equivalent method is the mostfrequently used method in the calculation of the amount ofsodium chloride needed to prepare isotonic drug solutions.
The sodium chloride equivalent of any drug substance, as
discussed earlier, is the amount (in grams) of sodium chloridethat is osmotically equivalent to 1 g of the drug.
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In the prescription above, 1% atropine sulfate is ordered. The sodium chloride equivalent of atropine sulfate is 0.13 .
This means that 1% solution of atropine sulfate has sameosmotic pressure as that of 0.13% solution of sodium
chloride. This solution is hypotonic. Addition of 0.77 g (i.e., 0.9 0.13 = 0.77) of sodium
chloride per 100 mL of the 1% solution of atropine sulfateresults in an isotonic solution.
To determine them amount of sodium chloride required torender a given solution isotonic, the following steps maybe used:
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Step 1: Find how much sodium chloride is needed to render theformulation isotonic with body fluids. (Remember isotonicity
refers to 0.9% or 0.9 g/100 mL).
Step 2: Find the amount of sodium chloride represented by theingredients in the prescription by multiplying the quantity of
each ingredient by itsE
value. Add up all the values obtained.This is the total amount of sodium chloride represented by allthe ingredients in the prescription.
Step 3: Subtract the total value obtained in Step 2 from theamount of sodium chloride required to render the formulationisotonic (i.e., the value obtained in Step 1). The value obtained inthis step represents the amount of sodium chloride required torender the solution isotonic.
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TONICITY AGENTS OTHER THAN SODIUM CHLORIDE
If one desires to use a chemical other than sodium chloride,such as dextrose or boric acid, the quantity of that chemicalcan be calculated by dividing the value obtained in Step 3(i.e., the amount of sodium chloride needed to render the
solution isotonic with body fluids) with the E value of thatchemical.
A proportion can be set up which can be treated as Step 4 inaddition to the three steps described earlier.
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Find the quantity of boric acid (in grams) to be used incompounding the following prescription.
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ISOTONICITY ADJUSTMENTS BY CRYOSCOPIC METHOD
Since osmotic pressure of a solution is not a readilymeasurable quantity, other easily measurable colligativeproperties such as thefreezing point depression is used in theisotonicity calculations.
The normal freezing (or melting) point of a pure compoundis the temperature at which the solid and the liquid phasesare in equilibrium at a pressure of 1 atm.
Pure water has a freezing point of 0C. When solutes areadded to water, its freezing point is lowered.
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ISOTONICITY ADJUSTMENTS BY CRYOSCOPIC METHOD
The freezing point depression (or lowering) of a solvent isdependent only on the number of particles in the solution.Blood plasma has a freezing point of 0.52 {or freezingpoint depression of 0.52, i.e., ( [0.52])}.
If freezing point depression value of a chemical in certainconcentration is known, one can calculate the concentrationof that chemical required for isotonicity by setting aproportion as follows:
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Thus, 0.9% sodium chloride has the same osmotic pressure andthe same
freezing point depression of 0.52 as that of blood plasma, redblood cells, and tears.
Drug solutions which have a freezing point depression of 0.52
are, therefore, isotonic with blood.
A list of freezing point depression values of selected compounds at1% concentration .
These _Tfvalues may be used to calculate the concentration oftonicity agents, such as sodium chloride or boric acid, needed torender a hypotonic drug solution isotonic with blood plasma. Thefollowing steps may be used to find the percentage concentrationof NaCl required to render hypotonic drug solutions isotonic with
blood plasma:
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Step 1: Find the value of freezing point depression of the drug at 1%
concentration, _T1%f from Table 8.2.
Step 2: Subtract _T1%f of the drug from the value of freezingpoint. depression of 0.9% sodium chloride solution, i.e., 0.52.This difference may be symbolized as _Tf, which is the freezingpoint lowering needed for isotonicity.
Step 3: Since 0.9% sodium chloride has a freezing point depressionof 0.52, one can calculate the percentage concentration of sodiumchloride required to lower the difference in freezing points, i.e.,the value obtained.
in Step 2, _Tf, by the method of proportion.The calculations involved in this method are explained best by
following examples.
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Compound the following prescription.
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Step 1: Freezing point depression (_Tf) of 1% atropine solution is0.07.
Step 2: Find _Tfby subtracting the _Tfvalue of 1% atropine sulfate
from the _Tfof blood plasma, i.e., 0.52 0.07 = 0.45. This means,
sufficient sodium chloride must be added to lower the freezing pointby an additional 0.45.
Step 3: Find the percentage concentration of sodium chloriderequired by setting up the proportion as follows:
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It is observed that 1% solution of sodium chloride has afreezing point lowering of 0.58. Therefore, one can alsoexpress the proportion as:
answer: solving forX, we get: (0.45/0.58) 1 = 0.78%
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VOLUMES OF ISO-OSMOTIC SOLUTIONS
White-Vincent Method White and Vincent2 provided a method for readily finding
the correct volume of water in which to dissolve a drug toproduce a solution iso-osmotic with tears, followed by the
addition of an isotonic vehicle to bring the solution to thefinal volume.
The volume (V) of isotonic solution that can be preparedfrom any given drug is obtained by the equation
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