# 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?

Jul 10, 2017

Approx. $0.1 \cdot m o l \cdot {L}^{-} 1$..........with respect to $M n {O}_{4}^{-}$

#### 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 $25.00 \times {10}^{-} 3 L \times 0.350 \cdot m o l \cdot {L}^{-} 1$ with respect to oxalic acid, i.e. $8.75 \times {10}^{-} 3 \cdot m o l$.

Given the stoichiometry expressed above, we need $\frac{2}{5} \cdot \text{equiv}$ precisely with respect to permanganate, and thus we can express the concentration of the $M n {O}_{4}^{-}$ as........

$\frac{\frac{2}{5} \times 8.75 \times {10}^{-} 3 \cdot m o l}{37.55 \times {10}^{-} 3 \cdot L} = 0.0932 \cdot m o l \cdot {L}^{-} 1$.

As to how to access the redox equation we set up the individual redox equations. Deep purple permanganate ion, $M n \left(V I I +\right)$, is reduced to colourless $M {n}^{2 +}$:

$M n {O}_{4}^{-} + 8 {H}^{+} + 5 {e}^{-} \rightarrow M {n}^{2 +} + 4 {H}_{2} O$ $\left(i\right)$

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

Meanwhile oxalic acid is oxidized to carbon dioxide.......$C \left(+ I I I\right) \rightarrow C \left(+ I V\right)$.........

${C}_{2} {O}_{4}^{2 -} \rightarrow 2 C {O}_{2} \left(g\right) \uparrow + 2 {e}^{-}$ $\left(i i\right)$

We takes $2 \times \left(i\right) + 5 \times \left(i i\right)$ to eliminate the electrons.......

$2 M n {O}_{4}^{-} + 6 {H}^{+} + 5 H O \left(O =\right) C - C \left(= O\right) O H \left(a q\right) \rightarrow 2 M {n}^{2 +} + 10 C {O}_{2} \left(g\right) \uparrow + 8 {H}_{2} O \left(l\right)$

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 $M {n}^{2 +}$. The reaction thus has a built-in indicator............