calculating the sodium chloride equivalent of a substance.
remember that the sodium chloride equivalent of a substance may be calculated as follows:
molecular weight of sodium chloride x i factor of sub.
i factor of sodium chloride x molecular weight of sub.
example:
pap.hydrochloride (m.w. 376) is a 2-ion electrolyte, dissociating 80%. in a given concentration. calculate its sodium chloride equivalent. because papaverine hydrochloride is a 2-ion electrolyte, dissociating 80%-, its i factor is 1.8.
58.5/1.8 x 1.8/376 =0.156
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.
ex/
p. n 0.3
nacl q.s
p.w ad 30 ml
make isotonic solution
0.23 x 0.3 g = 0.069 g of sodium chloride represented by the pilocarpine nitrate
2. 30 x 0.009 = 0.270 g of sodium chloride in 30 ml of an isotonic sodium chloride solution
0.270 g (from step 2)
- 0.069 g (from step 1)
=0.201 g of sodium chloride to be used, answer.
ex: how many grams of boric acid should be used in compounding the following prescription?
ph hcl 1%
chlo. ½ %
the prescription calls for 0.6 g of phenacaine hydrochloride and 0.3 g of chlorobutanol.
step1. 0.20 x 0.6g = 0.120 g of sodium chloride represented by phenacaine hydrochloride
0.24 x 0.3g = 0.072 g of sodium chloride represented by chlorobutanol
total:= 0.192 g of sodium chloride represented by both ingredients
step 2. 60 x 0.009 = 0.540 g of sodium chloride in 60 ml of an isotonic sodium
chloride solution step 3. 0.540 g (from step 2) - 0.192 g (from step 1)
0.348 g of sodium chloride required to make the solution isotonic
but because the prescription calls for boric acid:
step 4. 0.348 g -r- 0.52 (sodium chloride equivalent of boric acid) = 0.669 g of boric acid to be used, answer.
how many grams of potassium nitrate could be used to make the following prescription isotonic?
i£ sol. silver nitrate 60.0
1 : 500 w/v
make isoron. sol.
sig. for eye use.
the prescription contains 0.120 g of silver nitrate.
boric acid s.q.
p.w ad 60 ml
60 x 0.009= 0.54 g of nacl in 60 ml
0.54 – 0.192 =0.348 g of nacl required
0.348/ 0.52 = 0.669 g of boric acid
ex:
solution silver n. 60
1:5000w/v
make isotonic solution
sig for eye use
step 1. 0.33 x 0.120 g = 0.040 g of sodium chloride represented by silver nitrate
step 2. 60 x 0.009 = 0.540 g of sodium chloride in 60 ml of an isotonic sodium chloride solution
step 3. 0.540 g (from step 2)
- 0.040 g (from step 1)
=0.500 g of sodium chloride required to make solution isotonic
because, in this solution, sodium chloride is incompatible with silver nitrate, the tonic agent of choice is potassium nitrate.
therefore,
step 4. 0.500 g / 0.58 (sodium chloride equivalent of potassium nitrate) = 0.862 g of potassium nitrate to be used, answer.
ingredient x 0.5
sodium chloride s.q.
purified water ad 50
make isoton. sol.
sig. eye dropings.
how many grams of sod. chloride should be used in compounding the following prescription?
let us assume that ingredient x is a new substance for which no sodium chloride equivalent is to be found in table 9.1, and that its molecular weight is 295 and its i factor is 2. 4. the sodium chloride equivalent of ingredient x may be calculated as follows