Isotonic Solutions Lab 10

Definitions Isotonic Solution: is a solution having the same osmotic pressure as a body fluid. Ophthalmic (eye), nasal(nose), and parenteral (injection) solutions should be isotonic. Hypotonic solution: is a solution of lower osmotic pressure than that of a body fluid. Hypertonic solution: is a solution having a higher osmotic pressure than that of a body fluid.

Calculation of Isotonic Solutions Using Sodium Chloride Equivalents (E values) Rx Pilocarpine nitrate 0.3 g Sod. Chloride Q.S. Purified water ad 30 ml Make isotonic solution.

Calculation of Isotonic Solutions Using Sodium Chloride Equivalents (E values) Calculate amount in grams of NaCl represented by the ingredients in prescription. Multiply the amount in grams of each substance by its NaCl equivalent “either from tables or by calculation”. Calculate amount in grams of NaCl, alone, that would be contained in an isotonic solution (0.9%) of the total volume specified in prescription.

Calculation of Isotonic Solutions Using Sodium Chloride Equivalents (E values) Subtract the amount of NaCl represented by the ingredients in the prescription (step#1) from the amount of NaCl needed to prepare isotonic solution (step#2) If an agent other than sodium chloride such as boric acid, dextrose, sodium or potassium nitrate is to be used to make a solution isotonic, divide the amount of NaCl (step 3) by NaCl equivalent of the other substance.

Calculation of Sodium Chloride equivalent (E values) i factor is the dissociation factor.

Calculation of Sodium Chloride equivalent (E values) Problems on calculating the dissociation factor (i) of an electrolyte: Zinc Sulfate ZnSO4 is a 2-ion electrolyte, dissociation 40% in a certain concentration. Calculate its dissociation factor. If we have 100 particles of ZnSo4, 40% dissociation: 40 Zinc 40 SO4 60 undissociated particles. 140 particles So the dissociation factor (i) of ZSO4 = = 1.4 where 100 is the starting number of particles.

Calculation of Sodium Chloride equivalent (E values) In general, we may use the following tabulated values in case of solutions of 80% or higher concentrations: (i) Nonelectrolytes, slightly dissociated subs. 1 Substances that dissociate into two ions 1.8 Substances that dissociate into three ions 2.6 Substances that dissociate into four ions 3.4 Substances that dissociate into five ions 4.2

Calculation of Sodium Chloride equivalent (E values) Papavarine HCl (M.wt = 376) is a 2-ion electrolyte dissociating 80%. Calculate its E-value, where its dissociation factor (i) = 1.8 M.Wt of NaCl = 58.5 By substituting the values in the equation: E = 58.5/1.8 X 1.8/376 = 0.156 or 0.16

Calculation of Isotonic Solutions Using Sodium Chloride Equivalents (E values) How many grams of NaCl should be used in compounding the following prescription: Rx Pilocarpine nitrate 0.3 g E =0.23 Sod. Chloride Q.S. Purified water ad 30 ml Make isotonic solution. The amount of NaCl represented by Pilocarpine = 0.3 x 0.23 = 0.069g. The amount of NaCl required to make 30 ml of isotonic (0.9%): 0.9  100 X  30 ml X = 0.27 g of NaCl. Amount of NaCl to be added to prepare 30 ml of isotonic solution = 0.27 – 0.069 = 0.201 g

Calculating Isotonic Solutions Using Freezing Point Data The freezing point of blood and lachrymal fluid is – 0.52 oC. The dissolved substances in plasma or tear depress the solution freezing point below 0.52°C. Any solution that freeze at T=-0.52°C is isotonic with blood and tear

Calculating Isotonic Solutions Using Freezing Point Data How many grams each of NaCl and dibucaine HCl are required to prepare 30 ml of a 1% solution of dibucaine HCl isotonic with tears? From tables: 1% NaCl Tf = 0.58 1% Dibucaine HCl Tf = 0.08 Freezing point of blood and lachrymal fluid is – 0.52 oC. We need additional freezing point depression in the amount = 0.52 – 0.08 = 0.44 oC. Concentration of NaCl needed to lower the freezing point by 0.44 oC to make the solution isotonic: 1% NaCl Tf = 0.58 X 0.44 X= 0.76% Actual amount of NaCl to be added: 0.76 100 X 30 X= 0.228 g NaCl Actual amount of dibucaine to be added: 1 100 X 30 X = 0.3 g

Practice problems Zinc Chloride (ZnCl2) is a 3-ion electrolyte; dissociation 80% in a certain concentration, calculate (i). How many grams of boric acid should be used in compounding the following: RX Phenacaine HCl 1% E (0.2) Chlorobutanol 0.5% E (0.24) Boric acid Q.S. E (0.52) Purified water ad 60 ml Make isotonic solution

Practice problems Rx Oxytetracyline HCl 0.5% E = 0.12 Tetracaine HCl 2% solution 15 ml Sodium Chloride Q.S. Purified water ad 30 ml Make isotonic solution 2% solution of Tetracaine HCl is isotonic. How many mls of 0.9% solution of Sodium Chloride should be used? Rx Tetracaine HCl 0.5% E(0.18) Epinephrine bitartarate 1:1000 solution 10 ml Boric acid Q.S E(0.52) Purified water ad 30 Make isotonic solution The solution of Epinephrine bitartarate (1:1000) is already isotonic. How many grams of boric acid should be used in compounding the prescription?

Practice problems How many grams of NaCl and Naphazoline HCl should be used in compounding the following prescription: Rx Naphazoline HCl 1% NaCl Q.S. Purified water ad 30ml Make isotonic solution. Use the freezing point depression method. Tf blood = 0.52 Tf 1% NaCl = 0.58 Tf 1% Naphazoline HCl = 0.16

Practice problems For agent having 0.32 sodium chloride equivalent, calculate the percentage concentration of an isotonic solution. Calculate the E value for holocaine hydrochloride, if 6.7 mL of water will produce an isotonic solution from 0.3 g of drug substance.

Home work The freezing point of 5% solution of boric acid is -1.55 oC. How many grams of boric acid should be used in preparing 1500 mL of an isotonic solution? Rx Dextrose, anhydrous 2.5% Sodium chloride q.s Sterile water for injection ad 1200mL Label: Isotonic Dextrose and Saline Solution. How many grams of sodium chloride should be used in preparing the solution?