Aly(OH)3y-zClz·H2O
Aluminum chlorohydroxide.
Aluminum hydroxychloride.
Dihydrate
[
12042-91-0].
Aluminum chlorohydroxide, dihydrate.
Aluminum hydroxychloride, dihydrate.
Dihydrate
210.48
[
12042-91-0].
Anhydrous
174.45
[
1327-41-9].
Packaging and storage—
Preserve in well-closed containers.
Labeling—
The label states the content of anhydrous aluminum chlorohydrate.
Identification—
A solution (1 in 10) responds to the tests for
Aluminum 
191
and for
Chloride 191.
pH 791:
between 3.0 and 5.0, in a solution [15 in 100 (w/w)].
Limit of iron—
Standard preparation—
Transfer 2.0 mL of
Standard Iron Solution, prepared as directed under
Iron 241, to a 50-mL beaker.
Test preparation—
Transfer 2.7 g of Aluminum Chlorohydrate to a 100-mL volumetric flask, dilute with water to volume, and mix. Transfer 5.0 mL of this solution to a 50-mL beaker.
Procedure—
To each of the beakers containing the
Standard preparation and the
Test preparation add 5 mL of 6 N nitric acid, cover with a watch glass, and boil on a hot plate for 3 to 5 minutes. Allow to cool, add 5 mL of
Ammonium Thiocyanate Solution, prepared as directed under
Iron 241, transfer to separate 50-mL color comparison tubes, dilute with water to volume, and mix: the color of the solution from the
Test preparation is not darker than that of the solution from the
Standard preparation (150 µg per g).
Content of aluminum—
Edetate disodium titrant—
Prepare and standardize as directed in the
Assay under
Ammonium Alum, except to use 37.2 g of edetate disodium instead of 18.6 g.
Test solution—
Transfer about 200 mg of Aluminum Chlorohydrate, accurately weighed, to a 250-mL beaker, add 20 mL of water and 5 mL of hydrochloric acid, boil on a hot plate for not less than 5 minutes, and allow to cool.
Procedure—
To the
Test solution add 25.0 mL of
Edetate disodium titrant, and adjust with 2.5 N ammonium hydroxide or 1 N acetic acid to a pH of 4.7 ± 0.1. Add 20 mL of acetic acid-ammonium acetate buffer TS, 50 mL of alcohol, and 5 mL of
dithizone TS. The pH of this solution should be 4.7 ± 0.1. Titrate with 0.1
M zinc sulfate VS until the color changes from green-violet to rose-pink. Perform a blank titration, and make any necessary correction. Each mL of 0.1 M
Edetate disodium titrant consumed is equivalent to 2.698 mg of aluminum (Al). Use the aluminum content thus obtained to calculate the
Aluminum/chloride atomic ratio.
Content of chloride—
Transfer about 700 mg of Aluminum Chlorohydrate, accurately weighed, to a 250-mL beaker, and add 100 mL of water and 10 mL of diluted nitric acid with stirring. Titrate with 0.1 N silver nitrate VS using a glass silver-silver chloride electrode and a silver billet electrode system, determining the endpoint potentiometrically. Each mL of 0.1 N silver nitrate is equivalent to 3.545 mg of chloride (Cl). Use the chloride content thus obtained to calculate the
Aluminum/chloride atomic ratio.
Aluminum/chloride atomic ratio—
Divide the percentage of aluminum found in the test for
Content of aluminum by the percentage of chloride found in the test for
Content of chloride, and multiply by 35.453/26.98, in which 35.453 and 26.98 are the atomic weights of chlorine and aluminum, respectively: the ratio is between 1.91:1 and 2.10:1.
Assay—
Calculate the percentage of anhydrous aluminum chlorohydrate in the Aluminum Chlorohydrate by the formula:
Al({26.98
x + [17.01(3
x 
1)] + 35.453} / 26.98
x),
in which
Al is the percentage of aluminum as obtained in the test for
Content of aluminum,
x is the aluminum/chloride atomic ratio, 26.98 is the atomic weight of aluminum, 17.01 is the molecular weight of the hydroxide anion (OH), and 35.453 is the atomic weight of chlorine (Cl).
