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Add the following:
Piperazine phosphate (1:1), monohydrate Anhydrous 184.13 » Piperazine Phosphate contains not less than 98.5 percent and not more than 100.5 percent of C4H10N2 · H3PO4, calculated on the anhydrous basis.
Packaging and storage
Preserve in tight containers, and store at room temperature.
Labeling
Label it to indicate that it is for veterinary use only.
USP Reference standards
USP Piperazine Phosphate RS.
Identification
A: Infrared Absorption
Test specimen:
previously dried at 105
B:
In the test for Chromatographic purity, the principal spot in the chromatogram obtained from Test solution 2, observed after spraying with the ninhydrin solutions, corresponds in RF value, color, and size to that in the chromatogram obtained from Standard solution 1.
C:
It meets the requirements of the test for Phosphate
pH
Water, Method I
Chromatographic purity
Solvent
Prepare a mixture of 13.5 N ammonium hydroxide and dehydrated alcohol (3:2).
Test solution 1
Prepare a solution of Piperazine Phosphate in Solvent containing 100 mg per mL.
Test solution 2
Mix 1 mL of Test solution 1 and 9 mL of Solvent.
Standard solution 1
Prepare a solution of USP Piperazine Phosphate RS in Solvent containing 10 mg per mL.
Standard solution 2
Prepare a solution of ethylenediamine in Solvent containing 0.25 mg per mL.
Standard solution 3
Prepare a solution of triethylenediamine in Solvent containing 0.25 mg per mL.
Resolution solution
Prepare a solution in Solvent containing 0.25 mg of triethylenediamine and 10 mg of Piperazine Phosphate per mL.
Procedure
Separately apply 5-µL portions of Test solution 1, Test solution 2, Standard solution 1, Standard solution 2, Standard solution 3, and the Resolution solution to a suitable thin-layer chromatographic plate (see Chromatography
Assay
Dissolve about 200 mg of Piperazine Phosphate in 4 mL of ethylene glycol using a 150-mL beaker. Add 25 mL of glacial acetic acid, rinsing the walls of the beaker with a small amount of the glacial acetic acid. Add crystal violet TS, and titrate with 0.1 N perchloric acid VS. Perform a blank determination, and make any necessary correction. Each mL of 0.1 N perchloric acid is equivalent to 7.953 mg of C4H10N2 · 2HCl.
Auxiliary Information
Please check for your question in the FAQs before contacting USP.
USP32NF27 Page 3316
Pharmacopeial Forum: Volume No. 33(6) Page 1204
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