1. Author: ass. Yu.Yu. Plaskonis
ISOTONIC SOLUTIONS
Author:
ass. Yu.Yu. Plaskonis

2. Learning Objectives Aseptic conditions.
Algorithm of preparation injection solutions.
Classification of solutions for injections.
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.

3. Terms to Remember osmotic pressure isotonic solution
hypertonic solution
hypotonic solution

4. 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.

5. Injection solutions prepares in aseptic conditions

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

7. Technology process of preparation solutions for 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.

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

9. Preparation of table-wear and auxiliary materials

10. Preparation of solvents and drugs

11. Dissolution of drugs

12. Stabilization or isotones solutions

13. Quality control of solutions

14. Filtering solutions into bottles for dispense

15. Corking bottles

16. Sterilization

17. CLASSIFICATION OF SOLUTIONS FOR INJECTIONS

19. LET’S REPEAT hypertonic solution plasmolysis
Isotonic solutions are solutions, which have an osmolality, equal to the osmolality of liquids of organism (blood, plasma, lymph, tear liquid).
hypertonic solution plasmolysis
hypotonic solution hemolysis

22. Methods of Adjusting Tonicity
B. Amsden
CHEE 440

23. 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.

24. THE CALCULATION OF ISOTONIC CONCENTRATIONS OF SOLUTIONS USING THE SODIUM CHLORIDE ISOTONIC 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

25. 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

26. Equivalent glucose/sodium chloride - 0,18
NaCl Equivalent Method
E = 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 - 0,18
1,0 Glucose = 0,18 Sodium chloride
The sodium chloride equivalent values of many drugs are listed in tables.

27. Equivalent glucose/sodium chloride - 0,18 I variant of calculation
0,18 sodium chloride – 1,0 glucose
0,9 sodium chloride – Х
Х = (0,9 · 1,0) : 0,18 = 5,0
By the prescription glucose for 200 ml:
5, ml
Х ml Х = 10,0 Glucose
II variant of calculation
100 ml isoton. sol. – 0,9 sodium chloride
200 ml – Х
X = 1,8 NaCl
1,8 sodium chloride – Х
X = 10,0 Glucose

28. 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 ml M.ft. isotonic solution

29. 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)

30. Rp. Sol. Dicaini 0,3% 100 ml Natrii chloridi q. s. ut. f. sol
Rp. Sol. Dicaini 0,3% 100 ml Natrii chloridi q. s. ut. f. sol. isotonica 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 ̶ 0,9 NaCl 0,9 - 0,054 = 0,846 ≈ 0,85 NaCl

31. Cryoscopy Method DTf blood and tears = - 0.52 ˚C
So 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 C.

32. 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% Δt
X - 0,52 o C
X= (0,52/ Δt) 1%

33. 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%

34. Calculation after equation of Van’t Hoff:
Calculation after equation of Mendeleyev- Klapeyron:

35. 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:

36. i = 1 + a (n-1), where i- 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

37. Rp.: Sol. Glucosi 200 ml isotonicae
Sterilisa!
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),
Calculation after equation of Van’t Hoff:
Calculation after equation of Mendeleyev- Klapeyron:

38. 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), where
and 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) : ( ) = 11,6

39. THANK YOU FOR ATTENTION