Packaging and storage—
Preserve in well-closed containers.
Identification—
A:
A solution (1 in 10) responds to the tests for
Magnesium 
191
.
pH 791:
between 6.0 and 7.8, in a solution (1 in 20).
Water, Method Ib 921:
between 3.0% and 12.0%, 30 minutes being allowed for solubilization of the specimen and for the reaction to reach completion, and a blank determination being performed with the same volume of
Reagent but without the specimen. Calculate the water content of the specimen, in mg, taken by the formula:
F(
X b 
X)
R,
in which
Xb is the volume, in mL, of standardized
Water-Methanol Solution required to neutralize the unconsumed
Reagent in the blank determination; and the other terms are as defined therein.
Chloride 221—
A 1.0-g portion shows no more chloride than corresponds to 0.70 mL of 0.020 N hydrochloric acid (0.05%).
Sulfate 221—
A 2.0-g portion shows no more sulfate than corresponds to 1.0 mL of 0.020 N sulfuric acid (0.05%).
Heavy metals 231—
Dissolve 1.0 g in 10 mL of water, add 6 mL of 3 N hydrochloric acid, and dilute with water to 25 mL: the limit is 0.002%.
Reducing substances—
Transfer 1.0 g 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 minutes, accurately timed, and cool rapidly to room temperature. Add 25 mL of 0.6 N acetic acid, 10.0 mL of 0.1 N iodine VS, and 10 mL of 3 N hydrochloric acid, and titrate with 0.1 N sodium thiosulfate VS, adding 3 mL of
starch TS as the endpoint is approached. Perform a blank determination, omitting the specimen, and note the difference in volumes required. Each mL of the difference in volume of 0.1 N sodium thiosulfate consumed is equivalent to 2.7 mg of reducing substances (as dextrose): the limit is 1.0%.
Organic volatile impurities, Method I 467:
meets the requirements.
Assay—
Weigh accurately about 800 mg of Magnesium Gluconate, dissolve in 20 mL of water, add 5 mL of ammonia-ammonium chloride buffer TS and 0.1 mL of
eriochrome black TS, and titrate with 0.05
M edetate disodium VS to a blue endpoint. Each mL of 0.05
M edetate disodium is equivalent to 20.73 mg of C
12H
22MgO
14.
;)
