Acic base + Redox titration `1^(st)` titration : `M/10` reacts with oxalic acid as well as `H_(2)SO_4` according to equation `H_(2)C_(2)O_4 + NaOH rarrNa_(2)C_(2)O_4 + H_(2)O, H_(2)SO_(4)) + NaOH rarr Na_(2)SO_(4) + H_(2)O`
`2^(nd)` titration : The mixture solution is titrated with `M/10KMnO_4` solution which will reacts with oxalic acid (redox titration) in the presence of `H_(2)SO_(4)`
`H_(2)C_(2)O_(4) + underset(("Purple"))(KMnO_4)+H_(2)SO_(4) rarr K_(2)SO_(4) +underset(("Colourless"))( MnSO_4)+H_2O+CO_2`
The reactionof oxalic acid with `KMnO_4` is very slow therefore the oxalic acid solution is heated to `60-70^@C` initially. Once the reation has started, its rate automatically increases. `MnO_4^(-)` acts as an oxidising agent.
`MnO_(4)^(-)overset(H^(+))rarrunderset("colourless")(Mn^(++),MnO_(4)^(--) ,MnO_(4) underset(H_2O)rarroverset("Brown ppt")(MnO_2) `
If 1.34 gm `Na_(2)C_2O_4` dissolve in 50 ml of water this solution is titrated with `KMnO_4` The volume of `KMnO_4` used is

Options

A. 20 ml
B. 200/3 ml
C. 40 ml
D. 60 ml

Answer

A

Explanation

Use `(V_1M_1)/(n_1) = (V_2M_2)/n_2`