Question

2.4.0 mL of 3.00 M HCl reacts completely with the CaCO₃ (active ingredient) in an antacid tablet. The remaining solution (excess HCl) is then titrated with 0.0981 M NaOH to the bromothymol blue endpoint.

1. Write the balanced chemical equation for the reaction of CaCO₃ (s) with HCl (aq).

2. If the volume of NaOH delivered at the endpoint is 22.20 mL, determine the amount (in milligrams) of CaCO₃ that was contained in the antacid tablet.

Answer 1

The balanced chemical equation between Cacoz (s) and Hell (as) can be given as below - Ca CO₂ (s) +2Hd (ar) Cach (as) + H2O (1) t coz (g) Now in 24 mL of 3.00 M HU solution the no. of moles of Hu = 24 x x 3.moliki = 0.072 mol At the end point 22.20 mL of 0.0981M of NaOH is required : No. mol of NaOH in the above solution 22.20 - K x 0.0981 molt = 2.18 x 10-3 mol Now , NaOH + HU- Nau + H₂O Thus, 1 mol NaOH neutralize 1 mol Hu

Thus, 2.18 x 10-3 mol NaOH will neutralize exactly 2.18 x 10-3 mol of HU. Thus HU reacted in the equation ④ =10.072 - (2-18810-8)] moli = 0.0698 mol Now according to the equation ② 2 mol of He will react with 1 mol of Caloz This 0.0698 mol of Hu will neact with 0.0698 mol CaCO3 = 0.0349 mol Ca Loz - Molar mass of Calog = 100 g/mol i mass of 0.0349 mol of CaCO3 = 100 x 0.0349 g = 3.49 g = 3.49 X 103 mg Amount of Ca coz contained in the tablet = 3.49 X 103 mg

I have given the answer in three significant figures....if the answer do not match due to significant figures then please comment below......