Question

A 0.2005 g sample of a chalk containing CaCO3 and other nonreactive materials (impurities) to HCl was mixed with 30.0 mL of 0.229 M HCl. After boiling the solution to eliminate the CO2 of the reaction the excess acid was titrated with 17.06 mL of 0.1889 M NaOH. What is the weight percent (wt%) of impurities in the chalk?

Answer 1

The sample contained CaCO₃ and other materials non-reactive to HCl. After the reaction between CaCO₃ and HCl, the excess acid was titrated with NaOH. The two reactions are represented below.

CaCO₃ + 2 HCl -------> CaCl₂ + CO₂ + H₂O …….(1)

NaOH + HCl --------> NaCl + H₂O ……..(2)

As per stoichiometric equations (1) and (2),

1 mole CaCO₃ = 2 moles HCl.

1 mole NaOH = 1 mole HCl.

Millimoles of HCl added originally = (volume of HCl added in mL)*(molarity of HCl) = (30.0 mL)*(0.229 M) = (30.0 mL)*(0.229 mol/L) [1 M = 1 mol/L] = 6.87 mmol.

Millimoles of NaOH added = (17.06 mL)*(0.1889 mol/L) = 3.222634 mmol.

Therefore, millimoles of HCl neutralized = millimoles of NaOH added = 3.222634 mmol; therefore, millimoles of HCl reacted with CaCO₃ = (6.87 – 3.222634) mmol = 3.647366 mmol.

Millimoles of CaCO₃ present = (3.647366 mmol HCl)*(1 mole CaCO₃/2 mole HCl) = 1.823683 mmol.

Molar mass of CaCO₃ = (1*40.078 + 1*12.01 + 3*15.9994) g/mol = 100.0862 g/mol.

Mass of CaCO₃ present in the sample = (1.823683 mmol)*(1 mole/1000 mmol)*(100.0862 g/1 mol) = 0.1825 g.

Mass of impurities in the sample = (0.2005 – 0.1825) g = 0.018 g.

Percentage of impurities in the sample = (0.018 g)/(0.2005 g)*100 = 8.977% (ans).