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: 0.25 %
Drug XYZ: 0.45 %
NaCl qs
Aqua dist q.s. a.d. 10 mL
M.ft. isotonic solution
Provided, the E value of the drug PQR is 0.03, and the E value of the drug XYZ is 0.231.
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lb equals 0.079 g kg
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Answer: 0.079 g grams of NaCl will be needed.
- Volume of the solution: 10 mL
- Amount of NaCl needed to prepare an isotonic solution of this volume: 0.9/100 × 10 = 0.09 g
- Concentration of the PQR drug: 0.25 %.
- Quantity of the PQR drug present in the solution: 0.25 /100 × 10 = 0.025 g
- E value of the PQR drug (provided): 0.03
- Now, the NaCl-equivalent tonicity contributed by the PQR drug will be the product of drug's weight in g (0.025) and its E value ( 0.03 ), which is: 0.025 × 0.03 = 0.00075 g
- Quantity of the XYZ drug present in the solution: 0.45 /100 × 10 = 0.045 g
- E value of the XYZ drug (provided): 0.231
- Now, the NaCl-equivalent tonicity contributed by the XYZ drug will be the product of drug's weight in g (0.045) and its E value ( 0.231 ), which is: 0.045 × 0.231 = 0.010395 g
- Therefore, the quantity of additional amount of NaCl required to make the product isotonic: 0.09 g - 0.00075 g - 0.010395 g = 0.078855 g, rounded to 0.079 g g