Add the following:
Packaging and storage
Preserve in well-closed, light-resistant containers. Store at room temperature.
NF24
Labeling
Label it to indicate whether it is anhydrous or the monohydrate.
Specific rotation 781S:
between +33.0
and +36.5
, measured at 20
.
Test solution:
10 mg per mL in 6 N hydrochloric acid.
Microbial limits 61
The total aerobic microbial count does not exceed 1000 cfu per g, and the total combined molds and yeasts count does not exceed 100 cfu per g.
Loss on drying 731
Dry it at 130
for 3 hours: the anhydrous form loses not more than 1.0% of its weight, and the monohydrate loses between 11.5% and 12.5% of its weight.
Lead 251
Prepare a
Test Preparation using a 1-g portion of Asparagine, and use 5 mL of
Diluted Standard Lead Solution (5 µg of Pb) for the test: the limit is 5 µg per g.
Change to read:
Chromatographic purity
Adsorbent:
0.25-mm layer of chromatographic silica gel mixture.
Test solution:
10 mg per mL.
Standard solution
Prepare a solution of
USP Asparagine Anhydrous RS or USP Asparagine Monohydrate RS
NF24 in water having a known concentration of about 0.05 mg per mL.
[NoteUse USP Asparagine Anhydrous RS and USP Asparagine Monohydrate RS for the evaluation of the anhydrous and monohydrate forms of Asparagine, respectively.
]NF24
Application volume:
5 µL.
Developing solvent system:
a mixture of butyl alcohol, water, and glacial acetic acid (3:1:1).
Spray reagent
Dissolve 0.2 g of ninhydrin in 100 mL of a mixture of butyl alcohol and glacial acetic acid (95:5).
Procedure
Proceed as directed for
Thin-Layer Chromatography under
Chromatography 621, and then dry the plate at 80
for 30 minutes. Spray the plate with the
Spray reagent, heat at 80
for 10 minutes, and examine under white light: no secondary spot in the chromatogram obtained from the
Test solution is larger or more intense than the principal spot in the chromatogram obtained from the
Standard solution (0.5%); and not more than 1.0% of total impurities is found.
Assay
Dissolve about 130 mg of Asparagine, accurately weighed, in 3 mL of formic acid and 50 mL of glacial acetic acid, and titrate with 0.1 N perchloric acid VS (see
Titrimetry 541), determining the endpoint potentiometrically. Perform a blank determination, and make any necessary corrections. Each mL of 0.1 N perchloric acid is equivalent to 13.21 mg of C
4H
8N
2O
3, calculated on the dried basis.
Auxiliary Information
Staff Liaison :
Catherine Sheehan, B.Sc., Scientist
Expert Committee : (EM105) Excipient Monographs 1
USP29NF24 Page 3276
Pharmacopeial Forum : Volume No. 31(1) Page 87
Phone Number : 1-301-816-8262