Parenteral Solution part 1: Isotonic Solutions

University of the Philippines- ManilaIndustrial Pharmacy 121

Parenteral Products• sterile preparations

that are intended for use by injection under or through one or more layers of the skin or mucous membrane

Requirements of Parenterals Products

• Sterility• Absence of Pyrogens (fever causing

agents)• Absence of Particulate Matter• Isotonicity• pH

Osmosis

2 solutions of different concentrations are separated by a semi-permeable membrane (only permeable to the solvent) the solvent will move from the solution of lower conc. to that of higher conc.

Osmotic pressure

is the pressure that must be applied to the solution to prevent the passage of the solvent through a perfect semipermeable membrane.

Osmotic Pressure

• pressure responsible for the movement of the solvent through a semipermeable membrane

Osmotic pressure

• Colligative property

• Depends on the number of particles in the solution

• Substance that dossociate has greater number of particles and therefore greater osmotic pressure

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Nonelectrolytes• dissolve as

molecules in water• do not produce ions

in water• Osmotic pressure will

vary only due to concentration of solute

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A weak electrolyte• dissociates only slightly in water• in water forms a solution of a few ions

and mostly undissociated molecules

Weak Electrolytes

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Strong electrolytes • dissociate in water, producing positive

and negative ions• OP will vary not only in concentration

but also in degree of dissociation

Strong Electrolytes

Classification of Solutes in Aqueous Solutions

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Isosmotic

• solutions having the same osmotic pressure

• required of solutions intended to be mixed with body fluids– for greater comfort , efficacy, and safety

Isotonic

• Isotonic – same osmotic pressure as a specific body fluid (equal

tone with that body fluid)

– Blood and the body fluids of the eye, nose and bowel

– Products which require the same tonicity biologic fluids: ophthalmic, nasal, parenteral and some rectal preparations

Hypotonic and Hypertonic• Hypotonic

– lower osmotic pressure than that of the body fluid

• Hypertonic– higher osmotic pressure than that of the body fluid

Fate of cells

Clinical Consideration of Tonicity

Ophthalmic and Parenteral preparation-Isotonic

Edematous tissues- Hypertonic

Dehydration- Hypotonic

METHODS TO MAKE SOLUTIONS ISOTONIC

Method I: NaCl Equivalent

Step 1: CALCULATION OF DISSOCIATION (i) FACTOR

• Since osmotic pressure depends upon the number of particles of solute(s) in solution, the osmotic pressure of an electrolyte is directly proportional to the degree (or extent) of dissociation.

Calculation of dissociation factor (i)

Example

1. What is the dissociation factor of NaCl, having 80% dissociation in water? Assume that we have 100 particles of NaCl prior to dissociation.

Since strong electrolyte

Example

• What is the dissociation factor of zinc chloride, having 80% dissociation in water? Assume that we have 100 particles of zinc chloride prior to dissociation.

Answer

Dissociation Factors

• Nonelectrolytes and substances of slight dissociation ; 1.0

• Substances that dissociate into 2 ions; 1.8• Substances that dissociate into 3 ions; 2.6• Substances that dissociate into 4 ions; 3.4• Substances that dissociate into 5 ions; 4.2

Step 2: Know the Sodium Chloride Equivalent

• The sodium chloride equivalent of a chemical is defined as the amount of sodium chloride (in grams or grains) that has the same osmotic pressure as that of 1 g of the chemical.

Sodium Chloride Equivalent of Substance

Example

• Calculate the sodium chloride equivalent of a 1% solution of pilocarpinenitrate. Pilocarpinenitrate has a molecular weight of 271 and i of 1.8

Example 2

• Calculate the sodium chloride equivalent of a 1% boric acid. Boric acid has a molecular weight of 62 and i of 1

ISOTONICITY ADJUSTMENTS BY NaCl EQUIVALENT

• most frequently used method in the calculation of the amount of sodium chloride needed to prepare isotonic drug solutions.

Fill the Prescription

Step 1

• how much sodium chloride is needed to render the formulation 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 the ingredients in the prescription by multiplying the quantity of each ingredient by its E value.

• Add up all the values obtained. This is the total amount of sodium chloride represented by all the ingredients in the prescription

Step 2

• The sodium chloride equivalent of atropine sulfate is 0.13 .

• This means that 1% solution of atropine sulfate has same osmotic pressure as that of 0.13% solution of sodium chloride.

• This solution is hypotonic.

Step 3

• Subtract the total value obtained in Step 2 from the amount of sodium chloride required to render the formulation isotonic

• The value obtained in this step represents the amount of sodium chloride required to render the solution isotonic.

Step 3

0.9 − 0.13 = 0.77 of sodium chloride per 100 mL of the 1% solution of atropine sulfate results in an isotonic solution.

Final answer: Addition of 0.77 g

TONICITY AGENTS OTHER THAN SODIUM CHLORIDE

• such as dextrose or boric acid• A proportion can be set up which can be

treated as Step 4 in addition to the three steps described earlier

• the quantity of that chemical can be calculated by dividing the amount of sodium chloride needed (step 3) with the E value of that chemical.

• Find the quantity of boric acid (in grams) to be used in compounding the following prescription. Boric acid E value= 0.52

• Find the quantity of boric acid (in grams) to be used in compounding the following prescription. But instead of Boric acid use dextrose with E= 0.16

ISOTONICITY ADJUSTMENTS BY CRYOSCOPIC METHOD

• The normal freezing (or melting)point of a pure compound is the temperature at which the solid and the liquidphases are in equilibrium at a pressure of 1 atm.

• Pure water has a freezing point of 0°C. When solutes are added to water, its freezing point is lowered.

The freezing point depression

• dependent only on the number of particles in the solution.

• Blood plasma has a freezing point of −0.52 {or freezing point depression of 0.52, i.e., (− [−0.52])}.

Cryoscopic Method

• 1% NaCl has a freezing point of 0.576 c. What is the percentage concentration of NaCl required to make isotonic saline solution?

• 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 at 1% concentration .

• Tf values may be used to calculate the concentration of tonicity agents, such as sodium chloride or boric acid, needed to render a hypotonic drug solution isotonic with blood plasma.

Example

• Compound the prescription

Steps

• Step 1:Find the value of freezing point depression of the drug at 1%concentration

Based on the example

• Freezing point depression (Tf) of 1% atropine solution is 0.07.

Step 2:Subtract T1% f of the drug from the value of

freezing point. depression of 0.9% sodium chloride solution, i.e., 0.52.

This difference may be symbolized as T′f, which is the freezing point lowering needed for isotonicity.

Step 2 of example

• 0.52 − 0.07 = 0.45.

• This means, sufficient sodium chloride must be added to lower the freezing point by an additional 0.45°

• Step 3:Since 0.9% sodium chloride has a freezing point depression of 0.52, one can calculate the percentage concentration of sodium chloride required to lower the difference in freezing points, i.e., the value obtained.in Step 2, _T′f, by the method of proportion.

Step 3 of example

• Find the percentage concentration of sodium chloride required by setting up the proportion as follows:

• It is observed that 1% solution of sodium chloride has a freezing point lowering of 0.58. Therefore, one can also express the proportion as:

solving for X, we get: (0.45/0.58) ×1 = 0.78%

Sample ProbleHow many milligrams each of NaCl and dibucaine

HCl are required to prepare 30 mL of 1% solution of dibucaine HCl isotonic with tears?

• To make the solution isotonic , the freezing point must be lowered to – 0.52oC

• A 1% solution of dibucaine HCl has a freezing point lowering of 0.08oC

• Step 1: A 1% solution of dibucaine HCl has a freezing point lowering of 0.08oC

• Step 2: Thus freezing point must be lowered by an additional of 0.52oC – 0.08oC = 0.44oC by addition of more NaCl

Step 3

1% (NaCl) 0.58oC --------------- = ----------- x % (NaCl) 0.44oC x = 0.76%

To make the 30 mL solution,30 mL x 1% = 0.3 g or 300 mg dibucaine HCl30 mL x 0.76% = 0.228 g or 228 mg NaCl

White-Vincent Method• provided a method for readily finding the

correct volume of water in which to dissolve a drug to produce a solution iso-osmotic with tears.

• followed by the addition of an isotonic vehicle to bring the solution to the final volume.

Formula

• If more than one ingredient is contained in an isotonic preparation, the volume of isotonic solution obtained by mixing each drug with water are additive.

• E value of procaine HCl = 0.21

• The weight of procaine HCl • = (1/100) x 60 = 0.6g

• The volume of isotonic solution that can be prepared with drug

• In order to make the solution isotonic the three ingridients should be dissolved in 54.3mL of water and the preparation is adjusted to 60mL using an isotonic vehicle.

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