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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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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

The volume in milliliters (mL) of an HCl solution with a pH of 1.52 that can be neutralized by the given CaCO₃ is 17.87 mL

From the question,

We are to determine the volume of HCl that could be neutralized by the given CaCO₃

First, we will write the balanced chemical equation for the reaction

The balanced chemical equation for the reaction is

2HCl + CaCO₃ → CaCl₂ + CO₂ + H₂O

This means

2 moles of HCl is required to neutralize 1 mole of CaCO₃

Now, we will determine the number of moles of CaCO₃ present

Mass of CaCO₃ = 27.0 mg = 0.027 g

Using the formula

[tex]Number\ of\ moles = \frac{Mass}{Molar\ mass}[/tex]

Molar mass of CaCO₃ = 100.0869 g/mol

∴ Number of moles of CaCO₃ present = [tex]\frac{0.027}{100.0869}[/tex]

Number of moles of CaCO₃ present = 0.00026977 mole

Since

2 moles of HCl is required to neutralize 1 mole of CaCO₃

Then,

0.00053954 mole of HCl will be required to neutralize the 0.00026977 mole of CaCO₃

∴ 0.00053954 mole of HCl is required to neutralize the CaCO₃

Now, for the volume of HCl solution with a pH of 1.52 required

First,

We will determine the concentration of the HCl

From the given information

pH of the HCl = 1.52

Using the formula

pH = -log[H⁺]

Then,

1.52 = -log[H⁺]

∴ [H⁺] = 10^(-1.52)

[H⁺] = 0.0302 M

∴ The concentration of the HCl is 0.0302 M

Now, for the volume

Using the formula,

[tex]Volume = \frac{Number\ of\ moles}{Concentration}[/tex]

∴ Volume of HCl required = [tex]\frac{0.00053954}{0.0302}[/tex]

Volume of HCl required = 0.01787 L

Volume of HCl required = 17.87 mL

Hence, the volume in milliliters (mL) of an HCl solution with a pH of 1.52 that can be neutralized by the given CaCO₃ is 17.87 mL

Learn more here: https://brainly.com/question/14873607

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