Question

PART B. Determination of Calcium Carbonate in Antacid Tablets Brand name Tablet 1 Tablet 2 Walgreens antiacid carbonak Sopo mg ) Active ingredient Mass of active ingredient printed on label Mass of tablet Molarity of HCl solution Volume of HCI added to tablet 1.2939 1.30g 3.0M 3.0M 4m2 4mL "0.1M~O.IM 0 13.50 3.50 8.00 Molarity of NaOH titrant Initial burette reading for NaOH (mL) Final burette reading for NaOH (mL) Volume of NaOH used (mL) Moles of HCI added to tablet Moles of NaOH required to neutralize excess acid Moles of HCl consumed by reaction with tablet Moles of Caco, contained in tablet Milligrams of Caco, contained in tablet % difference between measured and manufacturers' reported value of Caco, in tablet

Answer 1

   NaOH + HCl ---> NaCl + H₂O

Trial 1:

Volume = 3.5- 0 = 3.5 ml = 3.5/1000 = 0.0035 lts

Moles of NaOH = Molarity x Vol in lts = 0.1 M x 0.0035 lts = 0.00035 moles

So, moles of excess HCl titrated = moles of NaOH (molar ratio is 1:1) = 0.00035 moles

Initial Moles of HCl added to the antacid = Molarity x vol = 3 M x 4 /1000 = 0.012 moles

   CaCO₃ + 2HCl ----> CaCl₂ + CO₂ + H₂O

Moles of HCl neutralised by the tablet = Initial Moles of HCl - moles of HCl titrated = 0.012 - 0.00035 = 0.01165 moles

Molar ratio of CaCO₃ and HCl = 1:2

Hence moles of CaCO₃ = 0.01165/2 = 0.005825 moles

Molar mass of CaCO₃ = 100.09 g/mol

Mass of CaCO₃ = moles x molar mass = 0.005825 x 100.09 = 0.583 g = 583 mg

Percent difference = ([583-500)]500) x 100 = 16.6 %

Trial 2 :

Volume = 8- 3.5 = 4.5 ml = 4.5/1000 = 0.0045 lts

Moles of NaOH = Molarity x Vol in lts = 0.1 M x 0.0045 lts = 0.00045 moles

So, moles of excess HCl titrated = moles of NaOH (molar ratio is 1:1) = 0.00045 moles

Initial Moles of HCl added to the antacid = Molarity x vol = 3 M x 4 /1000 = 0.012 moles

   CaCO₃ + 2HCl ----> CaCl₂ + CO₂ + H₂O

Moles of HCl neutralised by the tablet = Initial Moles of HCl - moles of HCl titrated = 0.012 - 0.00045 = 0.01155 moles

Molar ratio of CaCO₃ and HCl = 1:2

Hence moles of CaCO₃ = 0.01155/2 = 0.005775 moles

Molar mass of CaCO₃ = 100.09 g/mol

Mass of CaCO₃ = moles x molar mass = 0.005775 x 100.09 = 0.578 g = 578 mg

Percent difference = ([578-500)]500) x 100 = 15.6 %