Question

For a product to be called "vinegar," it must contain at least \(5.0 \%\) acetic acid, \(\mathrm{HC}_{2} \mathrm{H}_{3} \mathrm{O}_{2},\) by mass. A 10.00 -g sample of a "raspberry vinegar" is titrated with \(0.1250 \mathrm{M} \mathrm{Ba}(\mathrm{OH})_{2}\) and required \(37.50 \mathrm{~mL}\) for complete neutralization. Can the product be called a "vinegar"?

Short answer

Answer: No, the raspberry vinegar sample cannot be called "vinegar" according to the required minimum acetic acid content of 5.0 % because its mass percentage of acetic acid is only 2.815 %.

Step-by-step solution

Determine the moles of the titrant used for neutralization.

To do this, multiply the volume of Ba(OH)2 by its molarity: \( moles \ of \ Ba(OH)_{2} = Volume \ of \ Ba(OH)_{2} \times Molarity \) moles of Ba(OH)2 = 0.0375 L × 0.1250 M moles of Ba(OH)2 = 0.0046875 mol

Determine the moles of acetic acid in the sample.

Since the stoichiometric ratio between acetic acid and Ba(OH)2 in the neutralization reaction is 1:1, the moles of acetic acid in the sample are equal to the moles of Ba(OH)2 used: moles of acetic acid = 0.0046875 mol

Calculate the mass of acetic acid in the sample.

Use the molar mass of acetic acid (60.05 g/mol) to calculate its mass in the sample: mass_acetic_acid = moles_acetic_acid × molar_mass mass_acetic_acid = 0.0046875 mol × 60.05 g/mol mass_acetic_acid ≈ 0.2815 g

Calculate the mass percentage of acetic acid in the sample.

Divide the mass of acetic acid by the total mass of the sample, and multiply by 100: mass_percentage_acetic_acid = (mass_acetic_acid / mass_sample) × 100 mass_percentage_acetic_acid ≈ (0.2815 / 10) × 100 mass_percentage_acetic_acid ≈ 2.815 %

Compare with the required minimum percentage.

Since the mass percentage of acetic acid in the sample is only 2.815 %, which is less than the required minimum of 5.0 %, the product cannot be called "vinegar."