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15 April, 01:37

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

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  1. 15 April, 05:02
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    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
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