Asked by Marina
An analysis of generic antacid tablets labeled to contain 250mg of CaCO3 (MM= 100.09g/mol) per tablet of active ingredient was performed. The analysis was performed by dissolving a 0.550g sample of the antacid tablet in 75.00mL of 0.1055M HCl. The excess acid was back-titrated with exactly 44.97mL of 0.125M NaOH. The average weight of a tablet is 1.045g. The tablet came from a bottle of 125 tablets that cost $3.49.
Calculate the actual mg per tablet of CaCO3 in the generic antacid.
Calculate the actual mg per tablet of CaCO3 in the generic antacid.
Answers
Answered by
DrBob222
I will estimate all of the numbers which means you must recalculate all of them for better accuracy.
millimols HCl added initially = mL x M = about 7.9 millimols HCl.
millimols NaOH to backtitrate = mL x M = about 5.6 millimoles.
millimoles HCl used by the tablet when disolved = 7.9-5.6 = about 2.3.
CaCO3 + 2HCl ==> CaCl2 + CO2 + H2O
Convert 2.3 millimoles HCl to millimoles CaCO3 so 2.3/2 = 1.2 millimoles CaCO3 in the sample or about 0.0012 mols.
g CaCO3 = mols CaCO3 x molar mass CaCO3 and that x 1000 = about 120mg CaCO3 in the 0.550 g sample. Convert to mg in a single table with the average weight of 1.045 by about 120 mg x (1.045/0.550) = ?
Post your work if you get stuck.
millimols HCl added initially = mL x M = about 7.9 millimols HCl.
millimols NaOH to backtitrate = mL x M = about 5.6 millimoles.
millimoles HCl used by the tablet when disolved = 7.9-5.6 = about 2.3.
CaCO3 + 2HCl ==> CaCl2 + CO2 + H2O
Convert 2.3 millimoles HCl to millimoles CaCO3 so 2.3/2 = 1.2 millimoles CaCO3 in the sample or about 0.0012 mols.
g CaCO3 = mols CaCO3 x molar mass CaCO3 and that x 1000 = about 120mg CaCO3 in the 0.550 g sample. Convert to mg in a single table with the average weight of 1.045 by about 120 mg x (1.045/0.550) = ?
Post your work if you get stuck.
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