Question

A sample containing ${\mathbf{La}}^{\mathbf{3}\mathbf{+}}$ was treated with sodium oxalate to precipitate${\mathbf{La}}_{\mathbf{2}}\mathbf{(}{\mathbf{C}}_{\mathbf{2}}{\mathbf{O}}_{\mathbf{4}}{\mathbf{)}}_{\mathbf{3}}$ which was washed, dissolved in acid and titrated with 18.04mL of$\mathbf{0}\mathbf{.}\mathbf{006363}{\mathbf{MKMnO}}_{\mathbf{4}}$ . Write the titration reaction and find$\mathbf{\left[}{\mathbf{La}}^{\mathbf{3}\mathbf{+}}\mathbf{\right]}$ in the unknown.

Short answer

The titration reaction is $5{\mathrm{C}}_2{\mathrm{O}}_4^{2-}+2{\mathrm{MnO}}_4+16{\mathrm{H}}^{+}\to 10{\mathrm{CO}}_2+2{\mathrm{Mn}}^{2+}+8{\mathrm{H}}_2\mathrm{O}$. The concentration of ${\mathrm{La}}^{3+}$is 2.826mM .

Step-by-step solution

Define redox titration.

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.

Find the final equation.

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

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

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" $5{\mathrm{C}}_2{\mathrm{O}}_4^{2-}+2{\mathrm{MnO}}_4+16{\mathrm{H}}^{+}\to 10{\mathrm{CO}}_2+2{\mathrm{Mn}}^{2+}+8{\mathrm{H}}_2\mathrm{O}$

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

Find the concentration of La3+ .

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

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

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

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

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

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.