Universal indicator must not be used in titration experiments because it changes color over a broad pH range. This makes it difficult to determine the precise endpoint of the titration. Instead, a single indicator such as phenolphthalein or methyl orange should be used for clearer results.

Titration Experiment

1. Gather Materials: Prepare a burette, pipette, conical flask, and a suitable indicator (e.g., phenolphthalein).
2. Rinse Equipment: Rinse the pipette with sodium hydroxide solution and the burette with hydrochloric acid to avoid contamination.
3. Measure Alkali: Use the pipette to measure 25.0 cm³ of sodium hydroxide solution into a clean conical flask.
4. Add Indicator: Add a few drops of the chosen indicator to the sodium hydroxide solution in the flask.
5. Fill Burette: Fill the burette with hydrochloric acid, ensuring there are no air bubbles.
6. Perform Titration: Slowly add hydrochloric acid from the burette to the sodium hydroxide solution while swirling the flask until a permanent color change indicates neutralisation.
7. Record Volume: Note the volume of hydrochloric acid used at the endpoint.
8. Repeat: Perform the experiment two more times to obtain concordant results.
9. Crystallisation: Once you have determined the exact volume, evaporate the remaining liquid in an evaporating dish to obtain dry crystals of sodium chloride.

The volume of hydrochloric acid that must be used to calculate the concentration of sodium hydroxide solution is the average of the titration results:

$\frac{22.6 + 22.8}{2} = 22.7 \text{ cm}^3$

To calculate the concentration of the sodium hydroxide solution, we can use the following steps:

1. Find Moles of HCl Used:

   • Volume of HCl = 23.2 cm³ = 0.0232 dm³
   • Concentration of HCl = 0.100 mol dm⁻³
   • Moles of HCl: $ext{moles}_{ ext{HCl}} = ext{concentration} \times ext{volume} = 0.100 \times 0.0232 = 0.00232 \text{ mol}$

2. Use Stoichiometry: The reaction ratio is 1:1, so the moles of NaOH are equal to the moles of HCl: $ext{moles}_{ ext{NaOH}} = 0.00232 \text{ mol}$

3. Calculate Concentration:

   • Volume of NaOH = 25.0 cm³ = 0.0250 dm³
   • Concentration of NaOH: $ext{concentration}_{ ext{NaOH}} = \frac{ ext{moles}}{ ext{volume}} = \frac{0.00232}{0.0250} = 0.0928 \text{ mol dm}^{-3}$