Question

You titrated a 25.00 mL solution of .0350 M oxalic acid with a freshly prepared solution of KMnO4. If it took 37.55 mL of this solution, what was the molarity of KMnO4?

Answer 1

Approx. `0.1*mol*L^-1`..........with respect to `MnO_4^-`

Explanation:

We need a stoichiometric equation that represents the oxidation of oxalic acid by permanganate........

`2MnO_4^−+"5HO(O=)CC(=O)OH"+6H^+⟶2Mn^(2+)+10CO_2+8H_2O`

You should be able to know how to get this reaction by the method of half-equations.......we will address this later...

We have a molar quantity of `25.00xx10^-3Lxx0.350*mol*L^-1` with respect to oxalic acid, i.e. `8.75xx10^-3*mol`.

Given the stoichiometry expressed above, we need `2/5*"equiv"` precisely with respect to permanganate, and thus we can express the concentration of the `MnO_4^-` as........

`(2/5xx8.75xx10^-3*mol)/(37.55xx10^-3*L)=0.0932*mol*L^-1`.

As to how to access the redox equation we set up the individual redox equations. Deep purple permanganate ion, `Mn(VII+)`, is reduced to colourless `Mn^(2+)`:

`MnO_4^(-)+8H^+ + 5e^(-) rarr Mn^(2+) + 4H_2O` `(i)`

Charge and mass are balanced as required......

Meanwhile oxalic acid is oxidized to carbon dioxide.......`C(+III)rarrC(+IV)`.........

`C_2O_4^(2-) rarr 2CO_2(g)uarr+2e^-` `(ii)`

We takes `2xx(i)+5xx(ii)` to eliminate the electrons.......

`2MnO_4^(-)+6H^++5HO(O=)C-C(=O)OH(aq)rarr 2Mn^(2+) + 10CO_2(g)uarr+ 8H_2O(l)`

When you perform this reaction, the endpoint is signalled by the disappearance of the deep purple colour of the permanganate ion to give the ALMOST colourless `Mn^(2+)`. The reaction thus has a built-in indicator............