A 0.204-g sample of CO3

Question

A 0.204-g sample of CO₃^2- antacid is dissolved with 25.0 mL of 0.0981 M HCl. The Hydrochloric acid that is NOT neutralized by the antacid is is titrated to a bromophenol blue endpoint with 5.83 mL of 0.104 M NaOH

1. Calculate the number of moles of base in the antacid.

I need some clarification. How does one find the amount of a base in the antacid, if the antacid is just a simple ion? Is it just moles of CO₃? But then why does the other information exist?

Definitely missing something here.

Thanks!

Answer 1

All of the information provided is needed to answer this question. This is a technique known as "back titration". In this method, you add more titrant (HCl) than is needed to completely react with the CO₃^2-, and then use NaOH to "back titrate" the excess HCl.

The reaction in the first step is ...

CO₃^2- + 2HCl ==> H₂O + CO₂ + 2Cl^-

moles HCl added = 25.0 ml x 1 L / 1000 ml x 0.0981 mol / L = 0.002453 mols HCl

The reaction in the 2nd step is ...

HCl + NaOH ==> NaCl + H2O

mols NaOH needed = 5.83 ml x 1 L / 1000 ml x 0.104 mol / L = 0.0006063 mols

Moles HCl used to neutralize the CO₃^2- = 0.002453 mol - 0.0006063 mol = 0.001847 mols

Moles CO₃^2- present = 0.001847 mols HCl x 1 mol CO32- / 2 mols HCl = 0.000923 mols CO₃^2-