Arginine Hydrochloride
C6H14N4O2·HCl210.66
l-Arginine monohydrochloride;
l-(+)-Arginine monohydrochloride
[1119-34-2].
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C6H14N4O2·HCl210.66
l-Arginine monohydrochloride;
l-(+)-Arginine monohydrochloride
[1119-34-2].DEFINITION
Arginine Hydrochloride contains NLT 98.5% and NMT 101.5% of arginine hydrochloride (C6H14N4O2·HCl), calculated on the dried basis.
IDENTIFICATION
ASSAY
• Procedure
Sample: 100 mg of Arginine Hydrochloride
Titrimetric system
(See Titrimetry 
541
.)
541.)Mode: Direct titration
Titrant: 0.1 N perchloric acid VS
Endpoint detection: Potentiometric
Blank: 50 mL of glacial acetic acid and 3 mL of 98% formic acid. Add 6 mL of mercuric acetate TS.
Analysis: Dissolve the Sample in 3 mL of 98% formic acid and 50 mL of glacial acetic acid. Add 6 mL of mercuric acetate TS and titrate with the Titrant.
Calculate the percentage of arginine hydrochloride (C6H14N4O2·HCl) in the Sample taken:
Result = [(V 
B) × N × F × 100]/W
B) × N × F × 100]/W| V | = | = Sample titrant volume (mL) |
| B | = | = Blank titrant volume (mL) |
| N | = | = titrant normality (mEq/mL) |
| F | = | = equivalency factor, 105.3 mg/mEq |
| W | = | = weight of Sample (mg) |
Acceptance criteria: 98.5%–101.5% on the dried basis
IMPURITIES
• Residue on Ignition 281: NMT 0.1%
281: NMT 0.1%• Chloride and Sulfate, Sulfate 221: A 1.6-g portion shows no more sulfate than corresponds to 0.50 mL of 0.020 N sulfuric acid (0.03%).
221: A 1.6-g portion shows no more sulfate than corresponds to 0.50 mL of 0.020 N sulfuric acid (0.03%).• Heavy Metals, Method I 231
231 Test preparation: Proceed as directed in the chapter, except to dissolve 1.0 g in 20 mL of water, add 2 mL of 1 N acetic acid, and dilute with water to 25 mL.
Acceptance criteria: NMT 20 ppm
• Chromatographic Purity
System suitability solution: 0.4 mg/mL each of USP Arginine Hydrochloride RS and USP l-Lysine Hydrochloride RS in water
Standard solution: 0.05 mg/mL of USP Arginine Hydrochloride RS in water. [Note—This solution has a concentration equivalent to about 0.5% of that of the Sample solution. ]
Sample solution: 10 mg/mL of Arginine Hydrochloride in water
Chromatographic system
Mode: TLC
Adsorbent: 0.25-mm layer of chromatographic silica gel mixture
Application volume: 5 µL
Developing solvent system: Isopropyl alcohol and ammonium hydroxide (70:30)
Spray reagent: 2 mg/mL of ninhydrin in a mixture of butyl alcohol and 2 N acetic acid (95:5)
Analysis
Samples: System suitability solution, Standard solution, and Sample solution
Proceed as directed in the chapter. Dry the plate between 100
and 105 until the ammonia disappears completely. Spray with Spray reagent, and heat between 100 and 105 for about 15 min. Examine the plate under white light. The System suitability solution exhibits two clearly separated spots.
and 105 until the ammonia disappears completely. Spray with Spray reagent, and heat between 100 and 105 for about 15 min. Examine the plate under white light. The System suitability solution exhibits two clearly separated spots. Acceptance criteria: Any secondary spot from the Sample solution is not larger or more intense than the principal spot from the Standard solution.
Individual impurities: NMT 0.5%
Total impurities: NMT 2.0%
SPECIFIC TESTS
• Optical Rotation, Specific Rotation 781S: +21.4 to +23.6 (t = 20)
781S: +21.4 to +23.6 (t = 20)Sample solution: 80 mg/mL in 6 N hydrochloric acid
• Loss on Drying 731: Dry a sample at 105 for 2 h: it loses NMT 0.2% of its weight.
731: Dry a sample at 105 for 2 h: it loses NMT 0.2% of its weight.• Chloride Content
Sample: 350 mg of Arginine Hydrochloride
Titrimetric system
(See Titrimetry 541.)
541.)Mode: Direct titration
Titrant: 0.1 N silver nitrate VS
Endpoint detection: Colorimetric
Blank: 140 mL of water and 1 mL of dichlorofluorescein TS
Analysis: Transfer the Sample to a porcelain casserole, and add 140 mL of water and 1 mL of dichlorofluorescein TS. Mix and titrate with the Titrant until the silver chloride flocculates and the mixture acquires a faint pink color.
Calculate the percentage of chloride (Cl) in the Sample taken:
Result = [(V B) × N × F × 100]/W
B) × N × F × 100]/W| V | = | = Sample titrant volume (mL) |
| B | = | = Blank titrant volume (mL) |
| N | = | = titrant normality (mEq/mL) |
| F | = | = equivalency factor, 35.45 mg/mEq |
| W | = | = weight of Sample (mg) |
Acceptance criteria: 16.5%–17.1%
ADDITIONAL REQUIREMENTS
• Packaging and Storage: Preserve in well-closed containers.
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Auxiliary Information— Please check for your question in the FAQs before contacting USP.
| Topic/Question | Contact | Expert Committee |
|---|---|---|
| Monograph | Huy T. Dinh, M.S. Scientific Liaison 1-301-816-8594 | (DS2010) Monographs - Dietary Supplements |
| Reference Standards | RS Technical Services 1-301-816-8129 rstech@usp.org |
USP35–NF30 Page 2238
Pharmacopeial Forum: Volume No. 30(2) Page 449Chromatographic Column—
Chromatographic columns text is not derived from, and not part of, USP 35 or NF 30.