Question

Suppose that 200.0mL of a 0.1220 M HCL solution was added to a full antacid tabet containing CaCO3. The HCl fully reacted with the CO3^2- and the excess HCL was then back titrated with a 0.1050 M NaOH solution. The back titratin required 32.20 mL based on this data, calculate the moles of CaCO3 in the balet and the weight of CaCo3 IN THE TABLET (the molecular weight of CaCO3 is 100.09 g/mol)

What would these calculations look like?

Answer 1

initial moles of HCl = 200 x 0.1220 / 1000 = 0.0244

moles of NaOH = 0.1050 x 32.20 / 1000 = 3.38 x 10^-3

moles of HCl = moles of NaOH

so moles of HCl excess reacted with NaOH = 3.38 x 10^-3

moles of HCl reacted with antacid = 0.0244 - 3.38 x 10^-3 = 0.02102

CaCO3 + 2HCl -------------------> CaCl2 + H2O + CO2

1                 2

x               0.02102

moles of CaCO3 = 0.02102 / 2 = 0.01051

moles of CaCO3 = 0.01051

mass of CaCO3 = moles x molar mass

                          = 0.01051 x 100.09

                           = 1.052 g

mass of CaCO3 =1.052 g