Propylparaben Sodium
» Propylparaben Sodium contains not less than 98.5 percent and not more than 101.5 percent of C10H11NaO3, calculated on the anhydrous basis.
Packaging and storage— Preserve in tight containers.
Completeness of solution 641 One g of it, dissolved in water, meets the requirement.
Identification—
A: Dissolve 0.5 g in 5 mL of water, acidify with hydrochloric acid, and filter the resulting precipitate. Wash the precipitate with water, and dry it over silica gel for 5 hours: the IR absorption spectrum of a mineral oil dispersion of it exhibits maxima only at the same wavelengths as that of a similar preparation of USP Propylparaben RS.
B: Ignite about 0.3 g, cool, and dissolve the residue in about 3 mL of 3 N hydrochloric acid. A platinum wire dipped in this solution imparts an intense, persistent yellow color to a nonluminous flame.
pH 791: between 9.5 and 10.5, in a solution (1 in 1000).
Water, Method I 921: not more than 5.0%.
Chloride 221 A 0.2-g portion shows no more chloride than corresponds to 0.10 mL of 0.020 N hydrochloric acid (0.035%).
Sulfate 221 A 0.25-g portion shows no more sulfate than corresponds to 0.30 mL of 0.020 N sulfuric acid (0.12%).
Assay— Gently reflux about 100 mg of Propylparaben Sodium, accurately weighed, with 30 mL of 1 N sodium hydroxide for 30 minutes. Cool, add 25.0 mL of potassium bromate solution (2.78 in 500), 5 mL of potassium bromide solution (1 in 8), and 10 mL of hydrochloric acid, and immediately insert the stopper into the flask. Cool, shake for 15 minutes, and allow to stand for 15 minutes. Quickly add 15 mL of potassium iodide TS, taking care to avoid the escape of bromine vapor, at once replace the stopper in the flask, and shake vigorously. Rinse the stopper and the neck of the flask with a small quantity of water, and titrate the liberated iodine with 0.1 N sodium thiosulfate VS, adding 3 mL of starch TS as the endpoint is approached. [note—About 15 mL is needed.] Perform a blank determination (see Residual Titrations under Titrimetry 541), and note the difference in volumes required. Each mL of the difference in volume of 0.1 N sodium thiosulfate is equivalent to 3.37 mg of C10H11NaO3.
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Topic/Question Contact Expert Committee
Monograph Robert H. Lafaver, B.A.
Scientist
1-301-816-8335
(EM105) Excipient Monographs 1
Reference Standards Lili Wang, Technical Services Scientist
1-301-816-8129
RSTech@usp.org
USP32–NF27 Page 1330
Chromatographic Column—
Chromatographic columns text is not derived from, and not part of, USP 32 or NF 27.