Question

Experiment 13 - Acids and Bases: Titration Techniques

Data:

Titration 1:

2HCl (aq) + CaCO3 (aq) ---> CaCl2 (aq) + H2O (l) + CO2 (aq)

Titration 2:

HCl (aq) + NaOH (aq) ---> H2O (l) + NaCl (aq)  

Acid: 0.097 M HCl

Base: 0.1331 M NaOH

Trial 1

Piece of antacid tablet used (Tums): CaCO3 = 0.6535 g

Mass of 1 tablet of antacid: 2.5866 g

Total volume of HCl (Titration 1): 52.6mL HCl

Moles of HCl (acid) added to the sample:  

.0526L HCl x .097 mol HCl/ 1L = 0.005102 mol HCl

Total volume of NaOH (Titration 2): 3.6mL  

Moles of NaOH (base) added to neutralize excess acid:   

.0036L NaOH x .1331 mol NaOH/1L = 0.0004792 mol NaOH

Data Analysis:

Calculate the moles of acid that were neutralized by the piece of antacid tablet.  

Use the moles of acid neutralized by the piece of tablet to calculate the moles of acid that could be neutralized by the entire antacid tablet.  

Answer 1

The antacid tablet contains CaCO3 which neutralises the acid HCl.

The scheme of the reaction between them is

mass of the piece of actacid tablet/ mass of CaCO3 in the piece of antacid = 0.6535 g

Molar mass of CaCO3 = 100 g/mol

Hence, the number of moles of CaCO3 in the piece of antacid tablet is

According to the balanced reaction, 1 mole of CaCO3 neutralises 2 moles of HCl.

Hence, the number of moles of HCl neutralised by the piece of antacid tablet is

Mass of entire tablet = 2.5866 g

Number of pieces of antacid tablet in whole tablet is

if 1 piece of tablet neutralises 0.01307 mol of HCl, 1 antacid tablet will neutralise

Hence, number of moles of HCl neutralised by

Piece of antacid = 0.01307 mol

Entire tablet = 0.05175 mol