Potassium Gluconate
(poe tas' ee um gloo' koe nate).
C6H11KO7·H2O 252.26 d-Gluconic acid, monopotassium salt; Monopotassium d-gluconate. Anhydrous [299-27-4]. Monohydrate [35398-15-3]. DEFINITION
Potassium Gluconate is anhydrous or contains one molecule of water of hydration. It contains NLT 97.0% and NMT 103.0% of anhydrous potassium gluconate (C6H11KO7), calculated on the dried basis.
IDENTIFICATION
• B. Identification TestsGeneral, Potassium 191:
Meets the requirements of the flame test
• C. Thin-Layer Chromatography
Standard solution:
10 mg/mL of USP Potassium Gluconate RS
Sample solution:
10 mg/mL of Potassium Gluconate
Chromatographic system
Mode:
TLC
Adsorbent:
0.25-mm layer of chromatographic silica gel
Application volume:
5 µL
Developing solvent system:
Alcohol, ethyl acetate, ammonium hydroxide, and water (50:10:10:30)
Spray reagent:
Dissolve 2.5 g of ammonium molybdate in 50 mL of 2 N sulfuric acid in a 100-mL volumetric flask, add 1.0 g of ceric sulfate, swirl to dissolve, and dilute with 2 N sulfuric acid to volume.
Analysis
Samples:
Standard solution and Sample solution
Develop the chromatogram until the solvent front has moved about three-fourths of the length of the plate. Remove the plate from the chamber, and dry at 110 for 20 min. Allow to cool, and spray with the Spray reagent. Heat the plate at 110 for about 10 min.
Acceptance criteria:
The principal spot of the Sample solution corresponds in color, size, and RF value to that of the Standard solution.
ASSAY
• Procedure
Standard stock solution:
Transfer 190.7 mg of potassium chloride, previously dried at 105 for 2 h, to a 1000-mL volumetric flask, add sufficient water to dissolve, and dilute with water to volume. Transfer 100.0 mL of this solution to a 1000-mL volumetric flask, and dilute with water to volume. This solution contains 10 µg/mL of potassium (equivalent to 19.07 µg/mL of potassium chloride).
Standard solutions:
Tranfer 10.0, 15.0, and 20.0 mL of Standard stock solution to separate 100-mL volumetric flasks. To each flask add 2.0 mL of 200 mg/mL sodium chloride solution and 1.0 mL of hydrochloric acid. Dilute with water to volume, and mix. The Standard solutions contain 1.0, 1.5, and 2.0 µg/mL of potassium, respectively.
Sample stock solution:
0.18 mg/mL of Potassium Gluconate in water. Filter the solution.
Sample solution:
Transfer 5.0 mL of the filtrate from the Sample stock solution to a 100-mL volumetric flask. Add 2.0 mL of 200 mg/mL sodium chloride solution and 1.0 mL of hydrochloric acid, and dilute with water to volume.
Blank:
Water
Instrumental conditions
Mode:
Atomic absorption spectrophotometry
Analytical wavelength:
766.5 nm
Lamp:
Potassium hollow-cathode
Flame:
Airacetylene
Analysis:
Determine the absorbances of the Standard solutions and the Sample solution. Plot the absorbance of the Standard solutions versus their concentrations, in µg/mL, of potassium, and draw the straight line best fitting the three plotted points. From the graph so obtained, determine the concentration, CK, in µg/mL, of potassium in the Sample solution.
Calculate the percentage of potassium gluconate (C6H11KO7) in the portion of Potassium Gluconate taken:
Result = (CK/CU) × (Mr/Ar) × 100
Acceptance criteria:
97.0%103.0% on the dried basis
IMPURITIES
• Heavy Metals, Method I 231
Test preparation:
1.0 g in 10 mL of water. Add 6 mL of 3 N hydrochloric acid, and dilute with water to 25 mL.
Acceptance criteria:
NMT 20 ppm
• Reducing Substances
Sample:
1.0 g of Potassium Gluconate
Blank:
10 mL of water
Titrimetric system
(See Titrimetry 541.)
Mode:
Residual titration
Titrant:
0.1 N iodine VS
Back titrant:
0.1 N sodium thiosulfate VS
Endpoint detection:
Visual
Analysis:
Transfer the Sample to a 250-mL conical flask, dissolve in 10 mL of water, and add 25 mL of alkaline cupric citrate TS. Cover the flask, boil gently for 5 min, accurately timed, and cool rapidly to room temperature. Add 25 mL of 0.6 N acetic acid, 10.0 mL of Titrant, and 10 mL of 3 N hydrochloric acid, and titrate with Back titrant, adding 3 mL of starch TS as the endpoint is approached. Perform the Blank determination.
Calculate the percentage of reducing substances (as dextrose) in the Sample taken:
Result = {[(VB VS) × N × F]/W} × 100
Acceptance criteria:
NMT 1.0%
SPECIFIC TESTS
• Loss on Drying 731:
Dry a sample in vacuum at 105 for 4 h: it loses NMT 3.0% of its weight for the anhydrous form, and 6.0%7.5% of its weight for the monohydrate form.
ADDITIONAL REQUIREMENTS
• Packaging and Storage:
Preserve in tight containers.
• Labeling:
Label it to indicate whether it is the anhydrous or the monohydrate form.
Auxiliary Information
Please check for your question in the FAQs before contacting USP.
USP35NF30 Page 4368
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