Question

1. A 25 mL of commercial vinegar was diluted to 300 mL water; 15 ml of 0.15 M sodium hydroxide was needed for the titration of 20 mL dilute vinegar solution. Determine the % (w/v) of acetic acid in commercial vinegar.

Answer 1

The mass of acetic acid is divided by the volume of vinegar to yield the percentage concentration of acetic acid, which is found to be 8.1075% (w/v).

Step-by-step explanation:

To determine the percentage (w/v) of acetic acid in commercial vinegar using titration, we employ the following steps. First, using the titration data provided, calculate the moles of NaOH used:

Molarity of NaOH (M) × Volume of NaOH (L) = Moles of NaOH

0.15 M × 0.015 L = 0.00225 moles NaOH

Since the reaction between acetic acid (CH3COOH) and NaOH is 1:1, the moles of acetic acid in the 20 mL of diluted vinegar will also be 0.00225. To find the moles in the original 25 mL vinegar, we perform a dilution calculation:

(Moles in diluted vinegar) × (Total volume of diluted vinegar / Volume of aliquot used for titration)

0.00225 moles × (300 mL / 20 mL) = 0.03375 moles of acetic acid in the original 25 mL vinegar.

Using the molar mass of acetic acid (60.05 g/mol), we can find the mass:

0.03375 moles × 60.05 g/mol = 2.026875 g

To calculate the percentage (w/v) of acetic acid, we divide the mass of acetic acid by the volume of vinegar and multiply by 100:

(Mass of acetic acid / Volume of vinegar) × 100 = % (w/v) acetic acid

(2.026875 g / 25 mL) × 100 = 8.1075% acetic acid (w/v)

This indicates that the concentration of acetic acid in the commercial vinegar is 8.1075% (w/v).