A 0.450 g sample of impure CaCO3 is dissolved in 50.0 mL of 0.150 M HCL. The excess HCL is titrated by 8.55 mL of 0.125 M NaOH. Calculate the mass percentage of CaCO3 in the sample

Answer 1

The reaction of CaCO3 with HCl is CaCO3 + 2HCl ====> CaCl2 + CO2 + H2O.

The moles of HCl added to the impure CaCO3 is 0.150 m/L/ 1000 mL x 50.0 mL = 0.0075 moles HCl added.

There are moles of HCl in excess which is titrated with 0.125 M NaOH. Since NaOH reacts on a 1:1 mole ratio with HCL. The number of moles of NaOH will be equal to the moles of HCl in excess.

Moles of HCl in excess = 0.125 m/L / 1000 mL x 8.55 mL NaOH = 0.00107 moles HCl left over.

The difference between the moles HCl initially added minus the moles left over = moles of HCl reacted with CaCO3. 0.0075 - 0.00107 = 0.00643 moles HCl reacted.

Since the reactions stoichiometry shows that for every mole of HCl reacted, 0.5 moles of CaCO3 are consumed.

Grams CaCO3 = Molecular mass of CaCO3 in grams / 0.00643 moles / 2 = 100.086 g/mole x 0.00322 moles = 0.322 g CaCO3 in the impure sample. The % purity is then = 0.322 g / 0.450 x 100 = 71.6%