Question

Calcium carbonate is a common ingredient in antacids that reduces the discomfort associated with acidic stomach or heartburn. Stomach acid is hydrocholoric acid, HCl. What volume in milliliters (mL) of an HCl solution with a pH of 1.52 can be neutralized by 27.0 mg of CaCO3

Answer 1

Answer:

17.86mL of the HCl solution

Explanation:

The reaction of CaCO₃ with HCl is:

CaCO₃ + 2HCl → CaCl₂ + CO₂ + H₂O

The concentration of HCl with a pH of 1.52 is:

pH = 1.52 = -log [H⁺]

[H⁺] = 0.0302M = [HCl]

27.0mg = 0.0270g of CaCO₃ (Molar mass: 100.09g/mol) are:

0.0270g of CaCO₃ ₓ (1mol / 100.09g) = 2.70x10⁻⁴ moles of CaCO₃

Moles of HCl to react completely with these moles of CaCO₃ are:

2.70x10⁻⁴ moles of CaCO₃ ₓ (2 mol HCl / 1 mol CaCO₃) =

5.40x10⁻⁴ moles of HCl

As the concentration of HCl is 0.0302M, volume in 5.40x10⁻⁴ moles is:

5.40x10⁻⁴ moles of HCl * (1L / 0.0302mol) = 0.01786L =

17.86mL of the HCl solution