1. Standardising potassium permanganate solution using oxalic acid

2. The oxalate ion, C 2 O 4 2–, is colourless. It will reduce MnO 4 – to Mn 2+ while itself being oxidised to CO 2. C 2 O 4 2– → 2CO 2 + 2e – MnO 4 – + 8H + + 5e – → Mn 2+ + 4H 2 O 5C 2 O 4 2– + 2MnO 4 – + 16H + → 5CO 2 + 2Mn 2+ + 8H 2 O Oxalic acid is a weak acid, releasing few H+ ions into solution. It is necessary to add extra acid in the form of dilute sulfuric acid. Why is hydrochloric acid unsuitable? Because permanganate oxidises Cl – to Cl 2. The reaction is slow, so we heat the oxalate mixture to around 70 °C before titrating in the permanganate.

3. A standard solution of oxalic acid is prepared. This solution was made by dissolving 3.578 g of (HCOO) 2.2H 2 O in water and making it up to 1.000 L. We know this concentration to 4 sig fig.

4. 20.0 mL of oxalic acid is pipetted into each of 3 conical flasks.

5. About 20 mL of dilute sulfuric acid is added to each flask.

6. The flasks are heated until they are hot (above 70 °C)

7. Permanganate solution is added from the burette. At first the purple solution quickly turns colourless, but as the reaction proceeds it takes longer for the purple colour to completely disappear.

8. Just before the end-point, use a wash bottle to rinse any unreacted permanganate off the sides of the flask. The end-point is the first drop of pink permanganate that is not decolourised. It is very faint!

9. This has been overshot.This is correct.

10. Results Burette reading First titration Second titration Third titration Fourth titration Fifth titration Initial reading 1.2 mL18.8 mL12.4 mL 6.2 mL23.4 mL Final reading 18.8 mL36.1 mL29.6 mL23.4 mL40.7 mL Titre17.6 mL17.3 mL17.2 mL 17.3 mL Our burette and pipette readings are only accurate to 3 sig fig, but we'll carry 4 figures during the calculation.

11. V(Ox)  20.0 mL V(MnO 4 – )  17.25 mL c(Ox)  0.028 38 mol L –1 n(Ox)  5C 2 O 4 2– + 2MnO 4 – + 16H + → 10CO 2 + 2Mn 2+ + 8H 2 O Calculation

12. 5C 2 O 4 2– + 2MnO 4 – + 16H + → 10CO 2 + 2Mn 2+ + 8H 2 O V(Ox)  20.0 mL V(MnO 4 – )  17.25 mL c(Ox)  0.028 38 mol L –1 n(Ox)  5.676 × 10 –4 mol n(MnO 4 – )  ?

13. V(Ox)  20.0 mL V(MnO 4 – )  17.25 mL c(Ox)  0.028 38 mol L –1 c(MnO 4 – )  ? n(Ox)  5.676 × 10 –4 mol n(MnO 4 – )  2.270 × 10 –4 mol The final answer is quoted to 3 sig fig because our pipette and burette are only accurate to 3 sig fig.