Question

Calculate the mass in milligrams of CaCO3 in the antacid tablets (500mg) which were tested.

Concentration of 25mL HCl 0.1310M

Mass of tablet 1.3177g

Concentration of NaOH Solution 0.1055M

Volume of Solute used to titrate 24.10mL

Answer 1

Let us write the balanced reaction between CaCO3 and HCl  

2 HCl + CaCO₃ ====> CaCl₂ + H₂O + CO₂

Volume of HCl = 25 ml

Concentration of HCl = 0.1310 M

Moles of HCl = 25 x 0.1310 / 1000 =   0.003275 M

Volume of NaOH = 24.1 ml

Concentration of NaOH = 0.1055 M

Moles of NaOH = 24.1 x 0.1055 / 1000 =  0.00254255 M

Moles of HCl reacted with CaCO3 = 0.0073245 Moles

Moles CaCO3 reacted =  0.003662 Moles

The ratio of CaCO3 to HCl is 1:2.

The molar mass of CaCO3 is 100 g/mol

The mass of CaCO3 presents =  0.003662 Moles x  100 g/mol =  0.3665 gm

1.311 gm of tablet contains 0.3665 g of CaCO3

Mass of CaCO3 present in 500 mg of antacid tablet = 0.3665 g x 500 mg / 1.311 g = 139.78 mg