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

ISOTONIC
SOLUTIONS
Author:
ass.Yu.Yu. Plaskonis
Learning Objectives
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.
1.
Terms to Remember
osmotic pressure
 isotonic solution
 hypertonic solution
 hypotonic solution

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
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 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
Corking bottles
Sterilization
CLASSIFICATION OF
SOLUTIONS
FOR INJECTIONS
LET’S REPEAT
Isotonic solutions are
solutions, which have an
osmolality, equal to the
osmolality of liquids of
organism (blood, plasma,
lymph, tear liquid).
 hypertonic
 hypotonic
solution
plasmolysis
solution
hemolysis
Methods of Adjusting Tonicity
B. Amsden
CHEE 440
 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
CONCENTRATIONS OF SOLUTIONS
USING THE SODIUM CHLORIDE ISOTONIC
EQUIVALENTS
The name of
The equivalent of
Isotonic
medicinal
substances
substances by NaCl
concentration, %
Sodium chloride
-
0.9
0.66
1.3
0.23
3.9
0.18
5.2
0.53
1.7
(NaCl)
Sodium nitrite
(NaNО2)
Sodium sulphate
(Na2SО4)
Glucose (anhydrous)
(C6H12O6)
Boric acid
(H3BO3)
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 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.
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:
5,0 - 100 ml
Х - 200 ml Х = 10,0 Glucose
II variant of calculation
100 ml isoton. sol. – 0,9 sodium chloride
200 ml – Х
X = 1,8 NaCl
0,18 sodium chloride – 1,0 glucose
1,8 sodium chloride – Х
X = 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 ml
M.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
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
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
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% - Δt
X - 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 Van’t Hoff:
Calculation after equation of MendeleyevKlapeyron:

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),
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
Rp.: Sol. Glucosi 200 ml isotonicae
Sterilisa!
D.S. For intravenous introduction
Calculation after equation of Van’t Hoff:
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).
Calculation after equation of MendeleyevKlapeyron:
m = 0,29 · 180,0 / 1,00 = 52,2 (1 l of solution),
100 ml  5,22
200 ml  X
X = 10,44
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) : (100 - 10) = 11,6
THANK
YOU FOR
ATTENTIO
N