**Determine the Percentage Composition of a Mixture of Sodium Oxalate**

** and Oxalic Acid . Provided N/20 KMnO**_{4}

_{4}

Chemistry Lab ManualNCERT Solutions Class 12 Chemistry Sample Papers

**Chemical Equations**

**Theory**

Both oxalic acid and sodium oxalate can be titrated against N/20 KMnO_{4} since both of them are reducing agents.

So normality (N_{2}) of the solution will be due to both of them. From the combined normality (N_{2}), the composition of each can be calculated.

**Indicator**

KMn04 is a self-indicator.

**End Point**

Colourless to permanent pink (KMnO_{4} in burette).

**Procedure**

- Rinse and fill the burette with the N/20 KMnO
_{4}, solution. - Weigh exactly 1.0 g of the given mixture of oxalic acid and sodium oxalate and dissolve in water to prepare exactly 250 ml of solution using a 250 ml measuring flask. Rinse the pipette with the prepared oxalate solution and pipette out 20.0 ml of it in a washed titration flask.
- Add one test-tube (~ 20 ml) full of dilute sulphuric acid (~ 4 N) to the solution in titration flask.
- Note the initial reading of the burette.
- Heat the solution of titration flask to 60-70 °C and run down KMnO
_{4}solution from the burette till a permanent light pink colour is just imparted to the solution in the titration flask. - Note the final reading of the burette.
- Repeat the above steps 4-5 times to get three concordant readings.

**Observations**

Normality of KMnO_{4} solution = 1/20

Volume of oxalate solution taken for each titration = 20.0 ml.

**Calculations**

x ml of N/20 KMnO_{4} solution are equivalent to 20 ml of the given oxalate solution.

Applying normality equation,

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