A student carries out a titration to find the concentration of a solution of oxalic acid (H2C2O4\text{H}_2\text{C}_2\text{O}_4H2C2O4).
She is given:
She uses this method to do the titration:
Name the piece of apparatus that the student should use to add the oxalic acid solution in step 1.
What is the colour change of the phenolphthalein indicator in step 4?
Why is it better to use phenolphthalein indicator rather than universal indicator in this titration?
The student repeats the experiment four times to find the volume of NaOH\text{NaOH}NaOH needed for neutralisation. The table shows her results:
| Trial | 1 | 2 | 3 | 4 |
|---|---|---|---|---|
| Burette reading after adding sodium hydroxide solution (cm3\text{cm}^3cm3) | 24.85 | 24.30 | 24.55 | 24.20 |
| Burette reading before adding sodium hydroxide solution (cm3\text{cm}^3cm3) | 0.50 | 0.40 | 0.50 | 0.30 |
| Volume of sodium hydroxide solution added (cm3\text{cm}^3cm3) | 24.35 | 23.90 | 24.05 | 23.90 |
The average (mean) volume of sodium hydroxide solution should be calculated using only concordant results. Concordant results are those volumes that differ from each other by 0.20 cm3 or less.
Calculate the average volume of sodium hydroxide solution added.
In a different experiment under identical conditions, the student records the following titration results with another sample of oxalic acid:
The equation for the reaction is:
2NaOH+H2C2O4→Na2C2O4+2H2O 2\text{NaOH} + \text{H}_2\text{C}_2\text{O}_4 \rightarrow \text{Na}_2\text{C}_2\text{O}_4 + 2\text{H}_2\text{O} 2NaOH+H2C2O4→Na2C2O4+2H2OCalculate the amount, in moles, of NaOH\text{NaOH}NaOH in 22.50 cm3 of the sodium hydroxide solution.
Calculate the amount, in moles, of H2C2O4\text{H}_2\text{C}_2\text{O}_4H2C2O4 in the oxalic acid solution.
Calculate the concentration, in mol/dm3\text{mol/dm}^3mol/dm3, of the oxalic acid.
