Smt. Dr. Jayanti Vijaya Ratna
There are four methods of adjusting isotonicity. For preparations meant for use as intra
venous injections or for use in the eye or nasal tract or ear the liquid must
be isotonic with the body fluids. This can be done by any one of four methods.
Blood has a freezing point of –0.52 degreeC. So for any solution to be isotonic with blood, it must also have a depression of 0.52 0C. For a number of drugs the freezing point depression caused by a 1% solution is given in tables in literature.
Steps:
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. The sodium chloride equivalent values of many drugs are
listed in tables.
In this method we find out the E value of the drug; either from tables or from the
formula,
Liso
E = 17 –––––
M
Where E is sodium chloride equivalent value
M is Molecular Weight
L_{ iso}is a factor which depends on ionic state of the salt.
For Non electrolytes L_{ iso} is 1.9
Weak electrolytes L _{ iso } is 2.0
Divalent electrolytes L _{ iso } is 2.0
Uniuni valent electrolytes L _{ iso }is 3.4
Unidi valent electrolytes L _{iso } is 4.3
Diuni valent electrolytes L_{iso}is4.8
Unitrivalent electrolytes L _{ iso } is 5.2
Tri Univalent electrolytes L _{ iso } is 6.0
Tetraborate electrolytes L _{ iso } is 7.6
The steps are
1.We find the E value of the drug.
2.We multiplythe quantity of the drug with its E value. We get the weight (x ) that is equivalent to sodium chloride with respect to osmotic pressure.
3. Since, for every 100ml of
solution, 0.9g of sodium chloride is required for isotonicity, we subtract the
amount obtained in step 2 (x) from 0.9g; let this be y.
4. We add y of NaCl, to every 100ml of solution.
In this method we add enough water to the drug to make the solution isotonic and then
we add an isotonic sodium chloride solution to it to bring up the volume to the
required level.
Steps involved are
1. Find the weight of the drug prescribed (w), the volume
prescribed (v) and its sodium chloride equivalent value (E).
2. Multiply the weight (w) with the sodium chloride equivalent value (E).
W X E = X
So x is the weight of sodium chloride osmotically
equivalent to the given weight W of the drug.
3.The volume V of isotonic solution that can be prepared
from W g of drug is obtained by solving the equation
9.0 /100 =x/v
V= ( Xx 100 )/0.9
V = X x 111.1
OrV = W x E x 111.1 =
4.So V is the volume of solution that is isotonic with
blood. Dissolve Wg drug in Y ml of water. This solution is isotonic.
5.Now, make up the volume of this solution to the
required volume with an isotonic solution, such as 0.9% sodium chloride
solution.
In this method we make use of the V values which were defined and calculated for
many drugs by Sprowls. Fixing the W as 0.3g for many drugs, and knowing their E
values he calculated the V values for many drugs.
Steps :
1.Find the V value from the table. V is the volume of solution that is isotonic with
blood for 0.3.
2.For the prescribed amount of drug, calculate the
volume. Suppose, prescribed weight is X g.
For 0.3g, volume of water for isotonicity is v ml.
For Xg, volume of water is ?
V x ( X/0.3)=y
3.Now dissolve x g in y ml of water.
4.Make up this solution to the required volume with 0.9%
sodium chloride solution.
1.How much sodium chloride is required to render 100ml of a 1% solution of
apomorphine hydrochloride isotonic with blood serum?
Depression in Freezing Point available 0.08^{}
Further depression in Freezing Point required is0.44
0.44^{}C depression in Freezing Point is caused by ? NaCl solution ?
1 x0.44= 0.7586 or 0.76%
0.58
So 0.76 g of Nacl in 100 ml will give a lowering in Freezing Point of 0.44^{}.
So to make the required drug
solution isotonic, we dissolve 1g of apomorphine hydrochloride and 0.76 g of
sodium chloride in 100ml of water.
Method 2 :
This is the amount of sodium chloride equivalent to 1g of apomorphine hydrochloride
Method 3 :
Volume of solution = 100ml
Sodium Chloride Equivalent E = 0.14
1 x 0.14 = 0.14
= 15.55 ml
Method 4 :