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

To measure a 50 cm³ sample of vinegar, follow these steps:

1. Use a pipette or a burette to draw up 50 cm³ of vinegar from the bottle.
2. Transfer the 50 cm³ of vinegar into a clean volumetric flask.
3. Add deionized water to the flask gradually until the total volume reaches the 500 cm³ mark. Ensure that all of the vinegar is transferred and mixed well with the water for a homogeneous solution.

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

To prepare the burette for the titration, wash it with deionized water first to remove any impurities. Then, rinse the burette with the ethanoic acid solution to ensure that the solution does not get contaminated. This involves filling the burette with the solution and allowing it to drain out before filling it again to the desired mark.

A suitable indicator for this titration is phenolphthalein, which changes color when the solution transitions from acidic to neutral.

To calculate the concentration of ethanoic acid in moles per liter, we need to use the titration data. Given:

• The mean titration volume of NaOH used is 20.5 cm³ = 0.0205 L
• The concentration of NaOH is 0.12 M

Using the reaction stoichiometry:

1. Calculate the moles of NaOH used:

   $ext{Moles of NaOH} = ext{Concentration} imes ext{Volume} = 0.12 imes 0.0205 = 0.00246 ext{ moles}$

2. Since the reaction ratio of ethanolic acid to NaOH is 1:1, the moles of ethanoic acid present in 20.5 cm³ of the solution are also 0.00246 moles.

3. To find the concentration in the diluted solution:

   ext{Concentration in mol/L} = rac{ ext{Moles}}{ ext{Volume}} = rac{0.00246 ext{ moles}}{0.025 ext{ L}} = 0.0984 ext{ mol/L}

4. To convert to % (w/w), knowing that the molar mass of ethanoic acid (C₂H₄O₂) is approximately 60 g/mol:

   $ext{Mass of ethanoic acid} = 0.0984 ext{ mol/L} imes 60 ext{ g/mol} = 5.904 ext{ g}$

5. To express this in terms of % (w/w), considering 500 g of solution:

   ext{Percentage} = rac{5.904 ext{ g}}{500 ext{ g}} imes 100 = 1.18 ext{ % (w/w)}