Calcium carbonate is a common ingredient in antacids that reduces the discomfort associated with acidic stomach or heartburn. Stomach acid is hydrochloric acid, HCl.

Question

What volume in milliliters of an HCl solution with a pH of 1.81 can be neutralized by 17.0mg of CaCO3? What percentage of the acid is neutralized? If all of the acid is neutralized enter 100%

Answer 1

Calcium carbonate = CaCO3

2HCl + CaCO3 ==> CaCl2 + CO2 + H2O ... balance equation for the reaction taking place

note it takes 1 mole CaCO3 to neutralize 2 moles HCl

How many moles of CaCO3 are present?

molar mass CaCO3 = 100 g / mole

17.0 mg x 1 g / 1000 mg x 1 mol / 100 g = 1.70x10^-4 moles

moles HCl needed = 1.70x10^-4 moles CaCO3 x 2 mols HCl / mol CaCO3 = 3.40x10^-4 mols HCl

Since pH of HCl = 1.81, we can determine the molarity of the HCl:

[H+] = 1x10^-1.81 M = 0.0155 M = 0.0155 mols HCl / L

Volume HCl needed to provide 3.40x10^-4 mols:

3.40x10^-4 mols x 1 L / 0.0155 mols = 0.0219 L = 21.9 mls HCl needed