Titration can be used to determine the exact amount of hydrochloric acid that reacts with a given amount of sodium hydroxide solution - Edexcel - GCSE Chemistry - Question 5 - 2013 - Paper 1

Universal indicator is not suitable for titration experiments because it provides a broad range of colors instead of a sharp color change at the endpoint. A sharper indicator, such as phenolphthalein or methyl orange, is preferred because it allows for more precise determination of the endpoint.

To conduct a titration experiment:

1. Rinse a pipette with sodium hydroxide solution and use it to fill a clean conical flask with exactly 25.0 cm³ of the NaOH solution.
2. Add a few drops of a suitable indicator, such as phenolphthalein, to the flask containing the NaOH solution.
3. Fill a burette with dilute hydrochloric acid and record the initial volume.
4. Begin titrating by slowly adding the hydrochloric acid to the sodium hydroxide solution while continuously swirling the flask.
5. Observe the color change of the solution. The endpoint is reached when the solution changes from pink (due to the NaOH) to colorless (due to neutralisation).
6. Record the final volume of hydrochloric acid in the burette and calculate the volume used by subtracting the initial volume from the final volume.
7. Repeat the titration to obtain concordant results.
8. To obtain pure sodium chloride, mix the acid and alkali in the correct proportions, evaporate the solution to leave sodium chloride crystals.

The volume of hydrochloric acid to be used is the average of the titration results:

$ext{Volume of hydrochloric acid} = \frac{22.6 + 22.8}{2} = 22.7 \text{ cm}^3$

To find the concentration of sodium hydroxide:

1. Use the formula for concentration: $c_1v_1 = c_2v_2$ where:

   • $c_1$ is the concentration of HCl (0.100 mol dm⁻³)
   • $v_1$ is the volume of HCl used (23.2 cm³ = 0.0232 dm³)
   • $c_2$ is the concentration of NaOH (unknown)
   • $v_2$ is the volume of NaOH used (25.0 cm³ = 0.0250 dm³)

2. Rearrange the equation to solve for $c_2$: $c_2 = \frac{c_1 v_1}{v_2}$

3. Substituting the values: $c_2 = \frac{0.100 \times 0.0232}{0.0250} = 0.0928 \text{ mol dm}⁻³$

4. Therefore, the concentration of the sodium hydroxide solution is 0.0928 mol dm⁻³.