Aluminum Chlorohydrate Solution
»Aluminum Chlorohydrate Solution consists of complex basic aluminum chloride that is polymeric and encompasses a range of aluminum-to-chloride ratios between 1.91:1and 2.10:1.The following solvents may be used:water,propylene glycol,dipropylene glycol,or alcohol.It contains the equivalent of not less than 90.0percent and not more than 110.0percent of the labeled concentration of anhydrous aluminum chlorohydrate.
Packaging and storage— Preserve in well-closed containers.
Labeling— Label Solution to state the solvent used and the claimed concentration of anhydrous aluminum chlorohydrate contained therein.
Identification—
A: Asolution containing the equivalent of about 100mg of anhydrous aluminum chlorohydrate per mLresponds to the tests for Aluminum á191ñand for Chloride á191ñ.
B: Identification of propylene glycol(where stated on the label)—Add about 10mLof isopropyl alcohol to 2g of Solution,mix,and filter.Evaporate the filtrate to about 1mLon a steam bath:the IRspectrum of a film of this solution on a silver chloride disk exhibits maxima only at the same wavelengths as that of a similar preparation of a film of propylene glycol.
C: Identification of dipropylene glycol(where stated on the label)—Add about 10mLof isopropyl alcohol to 2g of Solution,mix,and filter.Evaporate the filtrate to about 1mLon a steam bath:the IRspectrum of a film of this solution on a silver chloride disk exhibits maxima only at the same wavelengths as that of a similar preparation of a film of dipropylene glycol.
D: Identification of alcohol(where stated on the label)—In a small beaker mix 5drops of Solution with 1mLof potassium permanganate solution (1in 100)and 5drops of 2Nsulfuric acid.Immediately cover the beaker with filter paper moistened with a freshly prepared solution of 0.1g of sodium nitroferricyanide and 0.25g of piperazine in 5mLof water:an intense blue color is produced on the filter paper,the color fading after a few minutes.
pHá791ñ: between 3.0and 5.0,in a solution prepared by diluting 3g of the solution with water to obtain 10mL.
Arsenic,Method Iá211ñ Prepare the Test Preparationusing an accurately weighed quantity of the Solution.The limit is 2ppm.
Heavy metals,Method Iá231ñ Prepare the Test Preparationusing an accurately weighed quantity of the Solution.The limit is 0.001%.
Limit of iron—
Standard preparation— Transfer 2.0mLof Standard Iron Solution,prepared as directed under Iron á241ñ,to a 50-mLbeaker.
Test preparation— Transfer 5.3g of the Solution to a 100-mLvolumetric flask,dilute with water to volume,and mix.Transfer 5.0mLof this solution to a 50-mLbeaker.
Procedure— To each of the beakers containing the Standard preparationand the Test preparationadd 5mLof 6Nnitric acid,cover with a watch glass,and boil on a hot plate for 3to 5minutes.Allow to cool,add 5mLof Ammonium Thiocyanate Solution,prepared as directed under Iron á241ñ,transfer to separate 50-mLcolor comparison tubes,dilute with water to volume,and mix:the color of the solution from the Test preparationis not darker than that of the solution from the Standard preparation(75µg 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 400mg of the Solution,accurately weighed,to a 250-mLbeaker,add 20mLof water and 5mLof hydrochloric acid,and boil on a hot plate for not less than 5minutes,and allow to cool.
Procedure— To the Test solutionadd 25.0mLof Edetate disodium titrant,and adjust with 2.5Nammonium hydroxide or 1Nacetic acid to a pHof 4.7±0.1.Add 20mLof acetic acid-ammonium acetate buffer TS,50mLof alcohol,and 5mLof dithizone TS.The pHof this solution should be 4.7±0.1.Titrate with 0.1Mzinc sulfate VSuntil the color changes from green-violet to 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).Calculate the percentage of aluminum (Al)found,and use this value to calculate the Aluminum/chloride atomic ratio.
Content of chloride— Transfer about 1.4g of the Solution,accurately weighed,to a 250-mLbeaker,and add 100mLof water and 10mLof diluted nitric acid with stirring.Titrate with 0.1Nsilver nitrate VSusing a silver-silver chloride glass electrode and a silver billet electrode system,determining the endpoint potentiometrically.Each mLof 0.1Nsilver nitrate is equivalent to 3.545mg of chloride (Cl).Calculate the percentage of chloride (Cl)found,and use this value to calculate the Aluminum/chloride atomic ratio.
Aluminum/chloride atomic ratio— Divide the percentage of aluminum found in the test for Content of aluminumby the percentage of chloride found in the test for Content of chlorideand 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 2.10:1.
Assay— Calculate the percentage of anhydrous aluminum chlorohydrate in the Solution by the formula:
Al{[26.98x +(17.01)(3x-1)+35.453]/26.98x},
in which Al is 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 chloride (Cl).
Auxiliary Information— Staff Liaison:Lawrence Evans,III,Ph.D.,Scientist
Expert Committee:(PA6)Pharmaceutical Analysis 6
USP28–NF23Page 88
Phone Number:1-301-816-8389