Aluminum Chlorohydrex Propylene Glycol
Aly(OH)3y-zClz·nH2O·mC3H8O2 Al2(H2O)y-z(OH)6-n(Cl)n(C3H8O2)z

Aluminum chlorohydroxide,hydrate:propylene glycol complex (1:1).
Aluminum hydroxychloride,hydrate:propylene glycol complex (1:1) [53026-85-0].
»Aluminum Chlorohydrex Propylene Glycol is a complex of aluminum chlorohydrate and propylene glycol in which some of the waters of hydration of the aluminum chlorohydrate have been replaced by propylene glycol.It contains the equivalent of not less than 90.0percent and not more than 110.0percent of the labeled amount of anhydrous aluminum chlorohydrate.
Packaging and storage— Preserve in well-closed containers.
Labeling— The label states the content of anhydrous aluminum chlorohydrate.
Identification—
A: Asolution (1in 10)responds to the tests for Aluminum á191ñand for Chloride á191ñ.
B: Infrared Absorption á197Fñ—
Test specimen— Dissolve 0.5g in about 40mLof water,and while mixing adjust with 2.5Nsodium hydroxide to a pHof 9.55±0.05.Filter the suspension of precipitate thus obtained.Evaporate about 15mLof the filtrate to about 1mLon a hot plate.Deposit this solution on a silver chloride disk.
Standard specimen: a similar preparation of propylene glycol.
pHá791ñ: between 3.0and 5.0,in a solution [15in 100(w/w)].
Limit of iron— Using Aluminum Chlorohydrex Propylene Glycol instead of Aluminum Chlorohydrate,proceed as directed in the test for Limit of ironunder Aluminum Chlorohydrate.The limit is 150µg per g.
Content of aluminum—
Edetate disodium titrant— Prepare and standardize as directed in the Assayunder Ammonium Alum,except to use 37.2g of edetate disodium instead of 18.6g.
Test solution— Transfer about 1.6g of Aluminum Chlorohydrex Propylene Glycol,accurately weighed,to a 100-mLbeaker,add 15to 20mLof water and 5to 6mLof hydrochloric acid,and boil on a hot plate for 15to 20minutes.Cool the solution,and with the aid of water transfer to a 100-mLvolumetric flask.Dilute with water to volume,and mix.
Procedure— Transfer 5.0mLof the Test solutionto a 250-mLbeaker,add 10to 15mLof water,and adjust with 1Nsodium hydroxide to a pHof 1.5±0.5.Add 10.0mLof Edetate disodium titrant,and heat to boiling.Cool the solution and carefully introduce a magnetic stirring bar into the beaker.Add 10to 15mLof acetic acid–ammonium acetate buffer TS,40to 50mLof alcohol,and while stirring adjust with glacial acetic acid to a pHof 4.6±0.1.Add 1to 2mLof dithizone TSand 40to 50mLof alcohol,and titrate with 0.1Mzinc sulfate VSuntil the color changes from a green-violet to a rose-pink.Perform a blank titration,and make any necessary correction.Each mLof 0.1MEdetate disodium titrantconsumed is equivalent to 2.698mg of aluminum (Al).Use the aluminum content thus obtained to calculate the Aluminum/chloride atomic ratio.
Content of chloride— Using Aluminum Chlorohydrex Propylene Glycol instead of Aluminum Chlorohydrate,proceed as directed in the test for Content of chlorideunder Aluminum Chlorohydrate.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 Assayby the percentage of chloride found in the test for Content of chloride,and multiply by 35.453/26.98,in which 35.453and 26.98are the atomic weights of chlorine and aluminum,respectively:the ratio is between 1.91:1and 2.1:1.
Assay— Calculate the percentage of anhydrous aluminum chlorohydrate in the Aluminum Chlorohydrex Propylene Glycol by the formula:
Al({26.98x +[17.01(3x -1)]+35.453}/26.98x),
in which Alis the percentage of aluminum found in the test for Content of aluminum,xis the aluminum/chloride atomic ratio,26.98is the atomic weight of aluminum,17.01is the molecular weight of the hydroxide ion (OH),and 35.453is the atomic weight of chlorine (Cl).
Auxiliary Information— Staff Liaison:Lawrence Evans,III,Ph.D.,Scientist
Expert Committee:(PA6)Pharmaceutical Analysis 6
USP28–NF23Page 90
Phone Number:1-301-816-8389
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