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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. 100 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 0.048 g kg

    Answer: 0.048 g grams of boric acid will be needed.
    • Volume of the product: 100 mL
    • Concentration of the drug: 1.25 %.
    • Amount of NaCl needed to prepare an isotonic solution of this volume: 0.9/100 × 100 = 0.9 g
    • Quantity of the drug present in the solution: 1.25 /100 × 100 = 1.25 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 (1.25) and its E value ( 0.7 ), which is: 1.25 × 0.7 = 0.875 g
    • Therefore, the quantity of additional amount of NaCl required to make the product isotonic: 0.9 g - 0.875 g = 0.025 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 0.025 / 0.52 = 0.048 g



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