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: 1.25 %
Drug XYZ: 1.15 %
NaCl qs
Aqua dist q.s. a.d. 70 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 0.462.
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lb equals 0.232 g kg
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Answer: 0.232 g grams of NaCl will be needed.
- Volume of the solution: 70 mL
- Amount of NaCl needed to prepare an isotonic solution of this volume: 0.9/100 × 70 = 0.63 g
- Concentration of the PQR drug: 1.25 %.
- Quantity of the PQR drug present in the solution: 1.25 /100 × 70 = 0.875 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 (0.875) and its E value ( 0.03 ), which is: 0.875 × 0.03 = 0.02625 g
- Quantity of the XYZ drug present in the solution: 1.15 /100 × 70 = 0.805 g
- E value of the XYZ drug (provided): 0.462
- Now, the NaCl-equivalent tonicity contributed by the XYZ drug will be the product of drug's weight in g (0.805) and its E value ( 0.462 ), which is: 0.805 × 0.462 = 0.37191 g
- Therefore, the quantity of additional amount of NaCl required to make the product isotonic: 0.63 g - 0.02625 g - 0.37191 g = 0.23184 g, rounded to 0.232 g g