Chromatographic purity
0.005 M Octane sulfonic acid
Transfer 1.08 g of sodium 1-octane sulfonate to a 1-liter volumetric flask. Dilute with 1.5% (v/v) acetic acid solution to volume, add 5.0 mL of triethylamine, mix, and filter.
Mobile phase
Prepare a filtered and degassed mixture of
0.005 M Octane sulfonic acid and acetonitrile (39:11). Make adjustments if necessary (see
System Suitability under
Chromatography 621).
System suitability solution
Dissolve suitable quantities of phenylethyl alcohol and
USP Pheniramine Maleate RS in water to obtain a solution containing about 3.6 and 0.24 mg per mL, respectively.
Test solution
Transfer about 24 mg of Pheniramine Maleate, accurately weighed, to a 100-mL volumetric flask, dissolve in and dilute with water to volume, and mix.
Chromatographic system
(see
Chromatography 621)The liquid chromatograph is equipped with a 265-nm detector and a 3.9-mm × 30-cm column that contains packing L1. The flow rate is about 2 mL per minute. Chromatograph the
System suitability solution, and record the peak responses as directed for
Procedure: the relative retention times are about 0.5 phenylethyl alcohol and 1.0 for pheniramine maleate, and the resolution,
R, between phenylethyl alcohol and pheniramine maleate is not less than 2.0, the tailing factor is not more than 2.5, and the relative standard deviation for replicate injections is not more than 2.0%.
Procedure
Inject a volume (about 10 µL) of the
Test solution into the chromatograph, record the chromatogram, and measure the peak responses. Calculate the percentage of each impurity (not including the solvent peak and maleic acid, if observed) in the portion of Pheniramine Maleate taken by the formula:
100(ri / rs),
in which
ri is the peak response for each impurity, and
rs is the sum of the responses of all of the peaks: not more than 0.5% of any individual impurity is found, and not more than 2.0% of total impurities is found.
Assay
Dissolve about 500 mg of Pheniramine Maleate, accurately weighed, in 25 mL of glacial acetic acid. Add 2 drops of
crystal violet TS, and titrate with 0.1 N perchloric acid VS. Perform a blank determination, and make any necessary corrections. Each mL of 0.1 N perchloric acid is equivalent to 17.82 mg of C
16H
20N
2·C
4H
4O
4.