Question

A 0.6747 gram tablet containing CaCO₃ (100.087 g/mol) is added to a flask along with 15.00 mL of 2.0 M HCl. Enough water is added to make 250.0 mL of solution A. A 20.00 mL of aliquot of solution A is taken and titrated with 15.03 mL of 0.1160 M NaOH. What is the wt% of pure CaCO₃ in the sample?

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

Confused on this whole problem, please help

Answer 1

15.00 mL of 2.0 M HCl = 0.015 L * 2.0 mole / L = 0.03 mole HCl

and

15.03 mL of 0.1160 M NaOH = 0.01503 L * 0.1160 mole / L = 1.7435 * 10^-3 mole.

for 20.00 ml solution, 1.7435 * 10^-3 mole NaOH needed.

for 250.0 ml solution, 250.0 ml * 1.7435 * 10^-3 mole / 20.00 ml = 0.0218 mole NaOH.

NaOH + HCl .................> NaCl + H2O

1 mole NaOH is neutralized by one mole HCl.

thus

mole of HCl used for tablet = (0.03 - 0.0218) = 8.2 * 10^-3 mole.

1 mole CaCO3 is neutralized by 2 mole HCl.

so

mole of CaCO3 = 8.2 * 10^-3 / 2 = 4.1 * 10^-3 mole

and

mass of CaCO3 = mole * molar mass = 4.1 * 10^-3 mole * 100.087 g/mole = 0.4104 g

thus

wt% of pure CaCO₃ in the sample = 0.4104 * 100 / 0.6747 = 60.8 %