Benzethonium Chloride Tincture
» Benzethonium Chloride Tincture contains, in each 100 mL, not less than 190 mg and not more than 210 mg of C27H42ClNO2.
Benzethonium Chloride 2 g
Alcohol 685 mL
Acetone 100 mL
Purified Water, a sufficient quantity to make 1000 mL
Dissolve the Benzethonium Chloride in a mixture of the Alcohol and the Acetone. Add sufficient Purified Water to make 1000 mL.
note—Benzethonium Chloride Tincture may be colored by the addition of any suitable color or combination of colors certified by the FDA for use in drugs.
Packaging and storage— Preserve in tight, light-resistant containers.
Identification— The residue obtained by evaporating 50 mL on a steam bath responds to Identification tests A and B under Benzethonium Chloride.
Specific gravity 841: between 0.868 and 0.876.
Alcohol and acetone content— To a 100-mL volumetric flask transfer 20.0 mL of Benzethonium Chloride Tincture and 5.0 mL of methanol as the internal standard, dilute with water to volume, and mix. Similarly prepare four 100-mL standard solutions in water, each containing 5.0 mL of methanol as the internal standard, and individually containing, respectively, 11.0 mL of dehydrated alcohol, 14.0 mL of dehydrated alcohol, 1.7 mL of acetone, and 2.2 mL of acetone. Inject 0.8 µL of the solution containing the substance under test into a suitable gas chromatograph equipped with a flame-ionization detector, and record the chromatogram. Similarly and successively record the chromatograms for 0.8-µL injected volumes of the four standard solutions. Under typical conditions, the instrument contains a 120-cm × 4-mm column packed with a suitable type of support, such as 80- to 100-mesh S3; the column is maintained at about 120; the injection port and detector block temperatures are maintained at about 240; and dry helium is used as the carrier gas at a flow rate of about 90 mL per minute. From the respective chromatograms obtained as described previously, calculate the ratios of peak areas for alcohol to internal standard and for acetone to internal standard.
Calculate the percentage of alcohol and of acetone in the Tincture by the formula:
[A(Y Z) + B(Z X)] / (Y X)
in which A and B are the percentage of alcohol, or of acetone, in the lower and higher standards, respectively; and X, Y, and Z are the ratios of the alcohol peak areas, or the acetone peak areas, to the internal standard peak areas for the lower standard, higher standard, and the material under test, respectively: the content of C2H5OH is between 62.0% and 68.0%, and the content of acetone (C3H6O) is between 9.0% and 11.0%.
Assay— Transfer 50.0 mL of Tincture to a 150-mL beaker, and add, with continuous stirring, 10 mL of sodium tetraphenylboron solution (1 in 40). Cover, and allow to stand for 16 hours. Decant the supernatant into a tared sintered-glass crucible, applying vacuum filtration. Suspend the precipitate in 20 mL of water, and transfer the precipitate to the crucible, washing well with water. Dry the precipitate and the crucible at 105 for 1 hour, cool, and weigh. The weight of the precipitate so obtained, multiplied by 0.6122, represents its equivalent of C27H42ClNO2.
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Topic/Question Contact Expert Committee
Monograph Behnam Davani, Ph.D., M.B.A.
Senior Scientist
(MDAA05) Monograph Development-Antivirals and Antimicrobials
USP32–NF27 Page 1639