1. Vinegar is a solution of ethanoic acid (acetic acid) - Leaving Cert Chemistry - Question 1 - 2002

The conversion of ethanol to ethanoic acid is an oxidation process. This involves the addition of oxygen or the removal of hydrogen, resulting in the formation of acetic acid.

To measure the 50 cm³ sample of vinegar, the following steps should be taken:

1. Use a pipette to draw up exactly 50 cm³ of vinegar from the bottle.
2. Transfer the 50 cm³ of vinegar into a clean volumetric flask that can hold at least 500 cm³.
3. Add deionized water slowly into the volumetric flask until the bottom of the meniscus aligns with the 500 cm³ mark.
4. Stopper the flask and invert it several times to ensure thorough mixing of the vinegar and deionized water.

The piece of equipment used to measure the ethanoic acid solution during the titration is a burette.

To prepare the burette for titration:

1. Rinse the burette with deionized water to remove any residues from previous uses.
2. Then, rinse the burette with the ethanolic acid solution to ensure that the inner surface is coated with the solution that will be used in the titration.
3. Fill the burette with the ethanoic acid solution, ensuring that there are no air bubbles in the nozzle. Use a funnel to help fill the burette if necessary, ensuring that you remove it before starting the titration.

A suitable indicator for this titration is phenolphthalein. It changes color from colorless to pink as the solution transitions from acidic to slightly alkaline, which is ideal for this type of titration.

To calculate the concentration of ethanoic acid:

• Given the titration mean volume is 20.5 cm³, or 0.0205 L.
• The concentration of NaOH is 0.12 M.

Using the reaction stoichiometry (1:1 mole ratio):

Number of moles of NaOH used = concentration × volume = 0.12 imes 0.0205 = 0.00246 moles

Thus, the concentration of ethanoic acid (since it's a 1:1 reaction) in the original 0.05 L of vinegar solution:

Concentration in moles/L = rac{0.00246 ext{ moles}}{0.050 ext{ L}} = 0.0492 M.

To convert to % (w/w):

Using the formula for concentration:

ext{mass of solute (g)} = 	ext{molar concentration (mol/L)} 	imes 	ext{molar mass (g/mol)} 	imes 	ext{volume (L)} \n
ext{mass of ethanoic acid} = 0.0492 	imes 60 	ext{ g/mol} 	imes 0.05 = 0.1476 	ext{ g}

Assuming the density of the vinegar is roughly 1 g/cm³, for 500 cm³:

ext{total mass of vinegar} 	ext{(approx)} = 500 	ext{ g}

Thus, the % w/w = rac{0.1476 ext{ g}}{500 ext{ g}} imes 100 = 0.02952 ext{ extit{or approximately }} 0.03 ext{ extit{%.}}