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A redox titration happens when the analyte and the titrant undergo an oxidation–reduction process. The endpoint is frequently detected using an indicator, much as it is in acid–base titrations. Oxalic acid titrated against potassium permanganate in acid medium is an example of redox titration.

First we need to write the redox reactions. In acidic conditions ${\mathrm{KMnO}}_4$is reduced to ${\mathrm{Mn}}^{2+}$and $C_2{O}_4^{2-}$ is oxidized to .

Reduction: ${\mathrm{MnO}}_4^{-}+8{\mathrm{H}}^{+}+5{\mathrm{e}}^{-}→{\mathrm{Mn}}^{2+}+4{\mathrm{H}}_2\mathrm{O}×2$

Oxidation: ${\mathrm{C}}_2{\mathrm{O}}_4^{2-}→2{\mathrm{CO}}_2+2{\mathrm{e}}^{-}×5$

To balance the electrons on the left and right side, we multiply the first reaction with and the second reaction with and we get the final equation:

role="math" localid="1663604875571" $5{\mathrm{C}}_2{\mathrm{O}}_4^{2-}+2{\mathrm{MnO}}_4+16{\mathrm{H}}^{+}→10{\mathrm{CO}}_2+2{\mathrm{Mn}}^{2+}+8{\mathrm{H}}_2\mathrm{O}$

Hence the final equation is $5{\mathrm{C}}_2{\mathrm{O}}_4^{2-}+2{\mathrm{MnO}}_4+16{\mathrm{H}}^{+}→10{\mathrm{CO}}_2+2{\mathrm{Mn}}^{2+}+8{\mathrm{H}}_2\mathrm{O}$.

$$\begin{gathered} \mathrm{V}\left({\mathrm{KMnO}}_4\right)=18.04\mathrm{ml} \\ \mathrm{c}\left({\mathrm{KMnO}}_4\right)=0.006363\mathrm{M} \end{gathered}$$

Now we can calculate the moles of oxalate used, by using the following relation:

$$\begin{gathered} \mathrm{n}\left({\mathrm{C}}_2{\mathrm{O}}_4^{2-}\right)=\frac{5}{2}·\mathrm{n}\left({\mathrm{KMnO}}_4\right) \\ =\frac{5}{2}×\mathrm{V}\left({\mathrm{KMnO}}_4\right)×\mathrm{c}\left({\mathrm{KMnO}}_4\right) \\ =\frac{5}{2}×0.006363\mathrm{mol}/\mathrm{L}×18.04\mathrm{mL} \\ =0.287\mathrm{mmol} \end{gathered}$$.

The moles of ${\mathrm{La}}^{3+}$:

$$\begin{gathered} \mathrm{n}\left({\mathrm{La}}^{3+}\right)=\frac{2}{3}×\mathrm{n}\left({\mathrm{C}}_2{\mathrm{O}}_4^{2-}\right) \\ =\frac{2}{3}×0.287\mathrm{mmol} \\ =0.1913\mathrm{mmol} \end{gathered}$$

Finally we can calculate the concentration of ${\mathrm{La}}^{3+}$using the following relation:

$\mathrm{c}\left({\mathrm{La}}^{3+}\right)=\frac{\mathrm{n}\left({\mathrm{La}}^{3+}\right)}{\mathrm{V}\left(\mathrm{solution}\right)}=\frac{0.1913\mathrm{mmol}}{50.00\mathrm{mL}}=3.826\mathrm{mM}$

Hence the concentration of ${\mathrm{La}}^{3+}$is 3.826mM.