ISOTONIC ISOTONIC SOLUTIONSSOLUTIONS

Author: ass. Yu.Yu.

Plaskonis

Learning ObjectivesLearning Objectives1. Aseptic conditions.2. Algorithm of preparation injection solutions.3. Classification of solutions for injections.4. Methods of calculating the isotonic

concentration: based on equation Mendeleev-Klapeyron or law

Van't Hoff based on a law Raul (for cryoscopy constants) using isotonic equivalents by sodium chloride.

Terms to RememberTerms to Remember

osmotic pressure

isotonic solutionhypertonic soluti

onhypotonic soluti

on

Injection dosage forms – specific group of drugs

that require special conditions of

preparation, the strictest adherence

to aseptic, technological

discipline, full responsibility for

the preparation, quality control and

design to dispensing drugs.

Injection solutions prepares in aseptic conditions

Aseptic conditions - defined conditions, and complex institutional arrangements required to enable to save the drugs from getting into these microorganisms.

Technology process of preparation solutions for Technology process of preparation solutions for injection consists of the following stages:injection consists of the following stages:

Preparation of aseptic unit and the organization of work in aseptic conditions.

Preparation of table-wear and auxiliary materials.

Preparation of solvents and drugs. Dissolution of drugs. Stabilization or isotones solutions Quality control of solutions. Filtering solutions into bottles for dispense,

checking the absence of mechanical inclusions. Closing, leak check, preparation for sterilization

(marking). Sterilization. Quality control and design of drugs to dispense.

Preparation of aseptic unit and the Preparation of aseptic unit and the organization of work in aseptic organization of work in aseptic conditionsconditions

Preparation of table-wear Preparation of table-wear and auxiliary materialsand auxiliary materials

Preparation of solvents and Preparation of solvents and drugsdrugs

Dissolution of drugsDissolution of drugs

Stabilization or isotones Stabilization or isotones solutions solutions

Quality control of Quality control of solutionssolutions

Filtering solutions into Filtering solutions into bottles for dispensebottles for dispense

Corking bottles

SterilizationSterilization

CLASSIFICATION OF CLASSIFICATION OF SOLUTIONS SOLUTIONS FOR INJECTIONSFOR INJECTIONS

Isotonic solutions are solutions, which have an osmolality, equal to the osmolality of liquids of organism (blood, plasma, lymph, tear liquid).

hypertonic solution solution plasmolysis

hypotonic solution solution hemolysis

LET’S REPEAT

B. Amsden CHEE 440

Methods of Adjusting Methods of Adjusting TonicityTonicity

Isotonic equivalent (E) of sodium chloride shows the amount of sodium chloride, which creates conditions identical osmotic pressure, osmotic pressure equal to 1.0 g of drug.

The isotonic concentration of solution sodium chloride is 0,9%.

Solutions of medicinal substances in concentrations, which create osmotic pressure, even such to 0,9% solution of sodium chloride, also are isotonic.

THE CALCULATION OF ISOTONIC THE CALCULATION OF ISOTONIC CONCENTRATIONS OF SOLUTIONSCONCENTRATIONS OF SOLUTIONS USING THE SODIUM CHLORIDE ISOTONIC USING THE SODIUM CHLORIDE ISOTONIC EQUIVALENTS EQUIVALENTS

The name of

medicinal substances

The equivalent of

substances by NaCl

Isotonic

concentration, %

Sodium chloride

(NaCl)- 0.9

Sodium nitrite

(NaNО2)0.66 1.3

Sodium sulphate

(Na2SО4)0.23 3.9

Glucose (anhydrous)

(C6H12O6)

0.18 5.2

Boric acid

(H3BO3)0.53 1.7

Rp.: Sol. Glucosi 200 ml isotonicae

Sterilisa!D.S. For intravenous

introduction

Equivalent glucose/sodium chloride - 0,18; depression of temperature of

freezing 1% solution - 0,1; М - 180,0

NaCl Equivalent MethodE = amount of NaCl equivalent in P to 1 g of drug

The sodium chloride equivalent is also known as “tonicic equivalent”. The sodium chloride equivalent of a drug is the amount of sodium chloride that is equivalent to (i.e., has the same osmotic effect as) 1 gram, or other weight unit, of the drug.

Example:Equivalent glucose/sodium chloride - 0,181,0 Glucose = 0,18 Sodium chloride

The sodium chloride equivalent values of many drugs are listed in tables.

Equivalent glucose/sodium chloride - 0,18I variant of calculation

0,18 sodium chloride – 1,0 glucose0,9 sodium chloride – ХХ = (0,9 · 1,0) : 0,18 = 5,0By the prescription glucose for 200 ml:5,0 - 100 mlХ - 200 ml Х = 10,0 Glucose

II variant of calculation100 ml isoton. sol. – 0,9 sodium chloride 200 ml – ХX = 1,8 NaCl0,18 sodium chloride – 1,0 glucose1,8 sodium chloride – ХX = 10,0 Glucose

Sample calculation: Calculate the amount of NaCl

required to make the following ophthalmic solution isotonic.

Atropine Sulfate 2% NaCl q.s. Aqua. pur. q.s. ad. 30 mlM.ft. isotonic solution

1. Determine the amount of NaCl to make 30 ml of an isotonic solution

2. Calculate the contribution of atropine sulfate to the NaCl equivalent

3. Determine the amount of NaCl to add to make the solution isotonic by subtracting (2) from (1)

Rp. Sol. Dicaini 0,3% 100 ml Rp. Sol. Dicaini 0,3% 100 ml Natrii chloridi q. s. ut. f. sol. isotonica Natrii chloridi q. s. ut. f. sol. isotonica D.S. D.S.

Е Dicaini /sodium chloride = 0,18

0,18 sodium chloride - 1,0 Dicaini Х sodium of chloride – 0,3 Dicaini Х = (0,18 · 0,3) : 1,0 = 0,054 NaCl

100 ml isotonic sol. NaCl C 0,9 NaCl

0,9 - 0,054 = 0,846 ≈ 0,85 NaCl

Cryoscopy Method Cryoscopy Method Tf blood and tears = - 0.52 ˚CSo for any solution to be isotonic with blood, it must also have a depression of 0.52 C. For a number of drugs the freezing point depression caused by a 1% solution is given in tables in literature.  Steps: We find the freezing point depression caused by the given amount of the drug In the prescription in the given volume of water. We subtract it from 0.52. For the remaining depression in freezing point, we add sufficient sodium chloride, knowing that 1% sodium chloride has a freezing point lowering of 0.58 C.

Calculation after depression of freezing temperature of solution:

Δt - depression of freezing temperature of solution shows how many degrees Celsius reduces the freezing temperature of 1% solution compared to the freezing point of pure solvent.

Freezing point depression of serum 0,52. 1% - ΔtX - 0,52 o C

X= (0,52/ Δt) 1%

For Glucose (Δt 1% solution 0,1 °C):

х=0,52/0,1=5,2%.

For Sodium chloride (Δt 1% solution 0,576 °C):

х=0,52/0,576=0,903%

For Magnesia sulphate (Δt 1% solution 0,08 °C):

х=0,52/0,08=6,5%

Calculation after equation of Calculation after equation of Mendeleyev- Mendeleyev- KlapeyronKlapeyron::

Calculation after equation of Calculation after equation of Van’t Van’t HoffHoff::

When calculating isotonic concentrations of electrolytes as the law of Van't Hoff and by Mendeleev-Clapeyron equation, should make corrections, i.e. the value (0.29 x M) must be divided into isotonic coefficient "and," which shows how many times an increasing number of particle dissociation (compared with nedysotsiyuyuchoyu matter), and numerically equal to:

i = 1 + a (n-1), wherei- isotonic coefficient;a - the degree of electrolytic dissociation;n - number of particles formed from one molecule of the

substance at dissociation.

Substance Isotonic coefficient

Acid boric 1,06

Glucose 1,00

Novocain 1,57

Sodium chloride, Sodium iodide, Sodium nitrate, Potassium chloride, Potassium iodide, Potassium nitrate

1,87

Zinc sulphate 1,12

Rp.: Sol. Glucosi 200 ml isotonicaeSterilisa!D.S. For intravenous introduction

m = 0,29 · 180,0 = 52,2 (1 l of solution) 100 ml 5,22 200 ml X X = 10,44 Thus, for 200 ml of solution it is necessary to take

a 10,44 glucose (waterless).

m = 0,29 · 180,0 / 1,00 = 52,2 (1 l of solution), 100 ml 5,22 200 ml X X = 10,44

Calculation after equation of Calculation after equation of Van’t HoffVan’t Hoff::

Calculation after equation of Calculation after equation of Mendeleyev- Mendeleyev- KlapeyronKlapeyron::

For preparation of injection solutions and eye drops use glucose taking into account it actual humidity.

A calculation is conducted after a formula:

Х = (100 · а) : (100 - в), whereand is an amount of glucose after the

sample of writing;in is humidity of glucose.If humidity of glucose of 10%, then:Х = (100 · 10,44) : (100 - 10) = 11,6

THANKTHANK YOU FOR YOU FOR ATTENTIOATTENTIO

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