Question

an antacid tablet weighs 1.200 g. The active ingredient is CaCO3. This tablet was dissolved in 25.0 mL of 1.0 M HCl and back-titrated with 6.8 mL of 1.0 M NaOH. How much active ingredient is in this tablet?

Answer 1

When CaCO₃ dissolved in HCl, some of the HCl reacts with CaCO₃ and excess unreacted HCl is back titrated with NaOH. So, substract the amount of HCl back titrated from the total amount of HCl. It gives the amount of HCl reacts with CaCO₃.

HCl + NaOH   -----> H₂O + NaCl

Mole ratio of HCl and NaOH is 1:1.

Millimoles of NaOH added = volume NaOH x Molarity of NaOH = 6.8 mL x 1.0 M = 6.8 mmol

Millimoles of HCl back titrated = 6.8 mmol NaOH x ( 1 mol HCl/1 mol NaOH) = 6.8 mmol HCl

Millimoles of total HCl = 25.0 mL x 1.0 M = 25.0 mmol

Moles of HCl reacted with CaCO₃ = 25.0 mmol - 6.8 mmol = 18.2 mmol = 18.2 x 10^-3 mol = 0.0182 mol

CaCO₃ + 2 HCl ------> H₂CO₃ +CaCl₂

Mole ratio of CaCO₃ and HCl is 1:2.

Moles of CaCO₃ = 0.0182 mol HCl x ( 1 mol CaCO₃ / 2 mol HCl) = 0.00910 mol CaCO₃

Mass of CaCO₃ = moles of CaCO₃ x molar mass of CaCO₃ = 0.00910 mol x (100 g/mol) = 0.910 g

So, the mass of active ingredient is 0.910 g.