1. To determine the concentration of ethanoic acid, CH₃COOH, in a sample of vinegar, the vinegar was first diluted and then titrated against 25.0 cm³ portions of a previously standardised 0.10 M solution of sodium hydroxide, NaOH - Leaving Cert Chemistry - Question 1 - 2008

1. Rinse the pipette (using a pipette filler) with deionised water, followed by a rinse with the vinegar solution to avoid contamination.

2. Using the pipette, carefully draw up 25.0 cm³ of the vinegar solution, ensuring that the bottom of the meniscus is at the 25.0 cm³ mark on the pipette.

3. Transfer the 25.0 cm³ of vinegar into a clean 250 cm³ volumetric flask, ensuring that no liquid is lost during transfer.

4. Rinse a graduated cylinder or volumetric flask with deionised water before adding distilled water to the flask containing vinegar.

5. Add deionised distilled water to the volumetric flask until the level reaches exactly the 250 cm³ mark. Use a dropper or pipette for the last few drops to achieve precision.

6. Stopper the volumetric flask and invert it several times to ensure the solution is homogenous.

The suitable indicator for this titration would be phenolphthalein. This indicator is colorless in acidic solutions and turns pink in basic solutions, which corresponds to the expected pH range of the endpoint of the reaction between ethanoic acid and NaOH. At the endpoint, the solution transitions through neutral pH where this indicator provides a clear visual cue of the completion of the titration.

At the endpoint of the titration using phenolphthalein, the solution changes from colorless (in the acid) to pink (in the base).

The mean titre volume is calculated as follows: $\text{Mean Titre} = \frac{22.9 + 22.6 + 22.7}{3} = 22.76 \text{ cm}^3$

First, convert the mean titre from cm³ to dm³: $22.76 \text{ cm}^3 = 0.02276 \text{ dm}^3$

Using the molarity of NaOH (0.10 M) and the reaction stoichiometry, the number of moles of NaOH used is: $0.10 \text{ mol dm}^{-3} \times 0.02276 \text{ dm}^3 = 0.002276 \text{ mol NaOH}$

Since the ratio of CH₃COOH to NaOH is 1:1, the moles of CH₃COOH is also 0.002276 mol. Therefore, the concentration of the diluted solution of ethanoic acid is: $\text{Concentration} = \frac{0.002276 \text{ mol}}{0.025 \text{ dm}^3} = 0.09104\text{ mol dm}^{-3}$

Using the concentration obtained, we can find the concentration of the ethanoic acid in grams per litre: $\text{Concentration (g/L)} = 0.09104\text{ mol dm}^{-3} \times 60.05 \text{ g/mol} = 5.47 \text{ g/L}$

To express the concentration in % (w/v), where w/v is grams of solute per 100 mL of solution:

We have 5.47 g in 1000 mL, thus: \text{Concentration (%) = } \frac{5.47}{10} = 0.547 \% (w/v)

The carboxylic acid found in nettles and stinging ants is formic acid (HCOOH).