Chemical reaction problem

Question

A piece of CaCO3 is put into a 1 litre (l) of 1.1 mol/l HCl solution. After whole CaCO3 dissolved, 10 ml of the resulting solution was being titrated with 1.08 mol/l NaOH solution. 1.69 ml of NaOH solution was used for the titration. What was the mass of CaCO3?

Answer 1

So, this problem involves what is referred to as a "back titration". The original CaCO₃ is placed in HCl and reacts with it as follows: CaCO₃ + 2HCl ==> CaCl2 + H2O + CO2. However, there is excess HCl added, so then in the second reaction, the xs HCl is back titrated with NaOH as follows: HCl + NaOH => NaCl + H₂O.

(1). moles HCl originally present = 1 L x 1.1 mol / L = 1.1 moles HCl

moles HCl taken for titration = 1.1 moles/ L x 10 ml x 1 L / 1000 ml = 0.011 moles HCl

(2). moles NaOH needed = 1.69 ml x 1 L / 1000 ml x 1.08 mol / L = 0.00183 mols NaOH.

This is the amount of excess HCl, i.e. 0.00183 moles. So, now we can find moles HCl that reacted:

0.011 mols HCl - 0.00183 mols = 0.108 mols HCl reacted in reaction #1

finding moles CaCO₃: 0.108 mols HCl x 1 mol CaCO₃ / 2 mol HCl = 0.0541 mols CaCO₃

mass CaCO₃ = 0.0541 mols CaCO₃ x 100. g / mol = 5.41 g CaCO₃ in the 10 ml samples that was used.

To convert this to the original mass in the 1 liter, we need to scale it up as follows:

5.41 g / 10 ml x 1000 ml / L = 541 g CaCO₃ in original sample.