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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.
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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 ChlorideEquivalents (E values) RxPilocarpine nitrate 0.3 g Sod. Chloride Q.S.Purified water ad 30 mlMake isotonic solution.
Calculation of Isotonic Solutions Using Sodium ChlorideEquivalents (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 ChlorideEquivalents (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:
Calculation of Sodium Chloride equivalent (E values) • PapavarineHCl (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 = 58.5By 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 ChlorideEquivalents (E values) • How many grams of NaCl should be used in compounding the following prescription:RxPilocarpine nitrate 0.3 g E =0.23Sod. Chloride Q.S.Purified water ad 30 mlMake 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 dibucaineHCl are required to prepare 30 ml of a 1% solution of dibucaineHCl isotonic with tears?From tables:1% NaClTf = 0.581% DibucaineHClTf= 0.08Freezing 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% NaClTf = 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 PhenacaineHCl 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 • RxOxytetracylineHCl 0.5% E = 0.12TetracaineHCl 2% solution 15 mlSodium Chloride Q.S.Purified water ad 30 mlMake isotonic solution2% solution of TetracaineHCl is isotonic. How many mls of 0.9% solution of Sodium Chloride should be used? • RxTetracaineHCl 0.5% E(0.18)Epinephrine bitartarate 1:1000 solution 10 mlBoric acid Q.S E(0.52)Purified water ad 30Make isotonic solutionThe 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 NaphazolineHClshould be used in compounding the following prescription:RxNaphazolineHCl 1%NaCl Q.S. Purified water ad 30mlMake isotonic solution. Use the freezing point depression method.Tf blood = 0.52Tf 1% NaCl = 0.58Tf 1% NaphazolineHCl = 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? • RxDextrose, anhydrous 2.5%Sodium chloride q.sSterile water for injection ad 1200mLLabel: Isotonic Dextrose and Saline Solution.How many grams of sodium chloride should be used in preparing the solution?
