Sodium Chloride Equivalents
A special problem arises when a prescription directs us to make a solution isotonic by adding the proper amount of some substance other than the active ingredient or ingredients. Given a 0.5% (w/v) solution of sodium chloride, we may easily calculate that 0.9 g — 0.5 g = 0.4 g of additional sodium chloride that should be contained in each 100 mL if the solution is to be made isotonic with a body fluid. But how much sodium chloride should be used in preparing 100 mL of a 1% (w/v) solution of atropine sulfate, which is to be made isotonic with lacrimal fluid? The answer depends on how much sodium chloride is in effect represented by the atropine sulfate.
The relative tonic effect of two substances, i.e., the quantity of one that is the equivalent in tonic effects to a given quantity of the other, may be calculated if the quantity of one having a certain effect in a specified quantity of solvent is divided by the quantity of the other having the same effect in the same quantity of solvent.
For example, we calculated that 17.3 g of boric acid per 1000 g of water and 9.09 g of sodium chloride per 1000 g of water are both instrumental in making an aqueous solution isotonic with lacrimal fluid. If, however, 17.3 g of boric acid are equivalent in tonicity to 9.09 g of sodium chloride, then 1 g of boric acid must be the equivalent of 9.09 g / 17.3 g or 0.52 g of sodium chloride.
Similarly, 1 g of sodium chloride must be the "tonicic equivalent" of 17.3 g + 9.09 g or 1.90 g of boric acid.
We have seen that one quantity of any substance should in theory have a constant tonic effect if dissolved in 1000 g of water: 1 g molecular weight of the substance divided by its / or dissociation value.
Hence, the relative quantity of sodium chloride that is the tonicic equivalent of a quantity of boric acid may be calculated by these ratios:
58.5 / 1.8 or 38.3 X 1.0
61.8 / 1.0 61.8 X 1.8
and we formulate a convenient rule: quantities of substances that are tonicic equivalents are proportional to the molecular weights of each multiplied by the i value of the other.
To return to the problem involving 1 g of atropine sulfate in 100 mL of solution:
Molecular weight of sodium chloride = 58.5; i = 1.8 Molecular weight of atropine sulfate = 695; i = 2.6
693 X 1.8 = 1 (g)
58.5 X 2.6 x (g)
x = 0.12 g of sodium chloride represented by I g of atropine sulfate
Because a solution isotonic with lacrimal fluid should contain the equivalent of 0.90 g of sodium chloride in each 100 mL of solution, the difference to be added must be 0.90 g — 0.12 g = 0.78 g of sodium chloride.
Table 9.1 gives the sodium chloride equivalents (E values) of each of the substances listed. These values were calculated according to the rule stated previously. If the number of grams of a substance included in a prescription is multiplied by its sodium chloride equivalent, the amount of sodium chloride represented by that substance is determined.
The procedure for the calculation of isotonic solutions with sodium chloride equivalents may be outlined as follows:
Step 1. Calculate the amount (in grams) of sodium chloride represented by the ingredients in the prescription. Multiply the amount (in grams) of each substance by its sodium chloride equivalent.