If 11.5 ml of vinegar sample (d=1g/ml) is titrated with 18.5 ml of standardized Sodium hydroxide
solution. What is the percentage of acetic acid (by mass) in the vinegar

4.83% of acetic acid in the vinegar

Explanation:

The neutralization reaction of acetic acid (CH₃COOH) with sodium hydroxide (NaOH) is:

CH₃COOH + NaOH → CH₃COO⁻Na⁺ + H₂O

Assuming the sodium hydroxide solution has a concentration of 0.500M, moles used in the neutralization reaction are:

0.0185L × (0.500mol / L) = 0.00925 moles of NaOH = 0.00925 moles CH₃COOH

Because 1 mole of acid reacts per mole of NaOH

Using molar mass of acetic acid (60g/mol):

0.00925moles CH₃COOH ₓ (60g / mol) = 0.555g of CH₃COOH

As mass of vinegar sample is 11.5g (d = 1g/mL), percentage of acetic acid by mass is:

0.555g CH₃COOH / 11.5g vinegar × 100 = 4.83% of acetic acid in the vinegar