A 0.204 g sample of a CO3 2- antacid is dissolved with 25.0ml of 0.0981 M HCL. The hydrochloric acid that is not neutralized by the antacid is titrated to a bromophenol blue endpoint with 5.83 ml of 0.104 M NaOH. Assuming the active ingredient in the antsacid sample is CaCO3, calculate the mass of CaCO3 in the sample.

0.0922 g

Explanation:

Number of moles of acid present = 25/1000 × 0.0981

= 0.00245 moles

Number of moles of base = 5.83/1000 × 0.104

= 0.000606 moles

Since the reaction of HCl and NaOH is 1:1

Number of moles of HCl that reacted with antacid = 0.00245 moles - 0.000606 moles

= 0.001844 moles

From the reaction;

CaCO3 + 2HCl > CaCl2 + H2O + CO2

1 mole of CaCO3 reacts with 2 moles of HCl

x moles of CaCO3 reacts with 0.001844 moles ofHCl

x = 1 × 0.001844/2

= 0.000922 moles

Mass of CaCO3 = 0.000922 moles × 100 g/mol

= 0.0922 g