Boric acid needed to make isotonic, One drug
8. Buffer and Isotonic Solution 8.2) Isotonic Solution 8.2.4) Boric acid needed, One drug
8.2.4.1) Boric acid needed, One drug Easy 1
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Calculate the gram amount of boric acid required to make the following sterile product isotonic.
Rx:
Drug ABC: 1.25 %
Boric acid qs
Aqua dist q.s. a.d. 90 mL
M.ft. isotonic solution
Provided, the E value of the drug is 0.7 , and the E value of boric acid is 0.52
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lb equals 0.043 g kg
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Answer: 0.043 g grams of boric acid will be needed.
- Volume of the product: 90 mL
- Concentration of the drug: 1.25 %.
- Amount of NaCl needed to prepare an isotonic solution of this volume: 0.9/100 × 90 = 0.81 g
- Quantity of the drug present in the solution: 1.25 /100 × 90 = 1.125 g
- E value of the drug (provided): 0.7
- Now, the NaCl-equivalent tonicity contributed by the drug will be the product of drug's weight in g (1.125) and its E value ( 0.7 ), which is: 1.125 × 0.7 = 0.7875 g
- Therefore, the quantity of additional amount of NaCl required to make the product isotonic: 0.81 g - 0.7875 g = 0.0225 g
- Since the product has to be made isotonic with the help of boric acid, we need to divide the gram amount of sodium chloride with the E value of boric acid (0.52)
- Therefore, the quantity of boric acid needed is 0.0225 / 0.52 = 0.043 g