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NaCl needed to make isotonic, Two drugs

8. Buffer and Isotonic Solution 8.2) Isotonic Solution 8.2.3) NaCl needed, Two drugs

8.2.3.1) NaCl needed, Two drugs Easy 1


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Calculate the gram amount of NaCl required to make the following sterile product isotonic.


      Rx:
      Drug PQR: 0.85 %
      Drug XYZ: 0.25 %
      NaCl qs
      Aqua dist q.s. a.d. 20 mL
      M.ft. isotonic solution
      Provided, the E value of the drug PQR is 0.04, and the E value of the drug XYZ is 0.456.


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lb equals 0.150 g kg

    Answer: 0.150 g grams of NaCl will be needed.
    • Volume of the solution: 20 mL
    • Amount of NaCl needed to prepare an isotonic solution of this volume: 0.9/100 × 20 = 0.18 g
    • Concentration of the PQR drug: 0.85 %.
    • Quantity of the PQR drug present in the solution: 0.85 /100 × 20 = 0.17 g
    • E value of the PQR drug (provided): 0.04
    • Now, the NaCl-equivalent tonicity contributed by the PQR drug will be the product of drug's weight in g (0.17) and its E value ( 0.04 ), which is: 0.17 × 0.04 = 0.0068 g
    • Quantity of the XYZ drug present in the solution: 0.25 /100 × 20 = 0.05 g
    • E value of the XYZ drug (provided): 0.456
    • Now, the NaCl-equivalent tonicity contributed by the XYZ drug will be the product of drug's weight in g (0.05) and its E value ( 0.456 ), which is: 0.05 × 0.456 = 0.0228 g
    • Therefore, the quantity of additional amount of NaCl required to make the product isotonic: 0.18 g - 0.0068 g - 0.0228 g = 0.1504 g, rounded to 0.150 g g





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