0.004 M Benzethonium Chloride VS

This article is compiled based on the United States Pharmacopeia (USP) – 2025 Edition

Issued and maintained by the United States Pharmacopeial Convention (USP)

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Dissolve 1.792 g of benzethonium chloride, previously dried at 100-105° to constant weight, in water to make 1000 mL. ^▲[NOTE-This solution is commercially available ready to be used. Use a suitable grade.]

1 Standardization

See Volumetric Solutions. 1. Definitions.

See Titrimetry (541).

Standardize by one of the following procedures. [NOTE-Other standardization procedures may be used. See Volumetric Solutions. 2. Preparation and Standardization, 2.3 Standardization.]

1.1 Standardization with visual endpoint

_{▲(USP 1-Aug-2020)} Dissolve 0.350 g of the dried benzethonium chloride in 30 mL of glacial acetic acid and add 6 ml. of mercuric acetate TS. Titrate with 0.1 N perchloric acid ^▲in glacial acetic acid _{▲(USP-1-Aug-2020)} VS, using 0.05 mL of crystal violet TS as an indicator. Carry out a blank titration. One mL of 0.1 N perchloric acid VS is equivalent to 44.81 mg of benzethonium chloride (C₂₇H₄₂CINO₂). ^▲Calculate the molarity of the volumetric solution from the content of benzethonium chloride in the dried benzethonium chloride, as determined through the titration above.

1.2 Standardization with potentiometric endpoint

Accurately measure 2.0 mL of standardized 0.02 M sodium tetraphenylboron VS into a titration vessel, add 50 mL of water and 1 mL of 1 N sodium hydroxide VS. Titrate with the benzethonium chloride solution using a surfactant electrode suitable for non-ionic surfactants.

M = [mL NaB(C₆H₅)₄ × M NaB(C₆H₅)₄ ]/ mL benzethonium chloride

[NOTE-If this volumetric solution is used in a qualitative application such as pH adjustment, dissolution medium, or diluent, its standardization is not required.)_{▲(USP 1-Aug-2020)}