91Ó°ÊÓ

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.

To determine the number of moles of $C{e}^{4+}$that was used in the reaction with nitrate, we need to determine the number of moles of excess $C{e}^{4+}$and subtract it with excess moles of $C{e}^{4+}$that were back-titrated with ${\mathrm{Fe}}^{2+}$.

The total number of mmoles of $C{e}^{4+}$used in a reaction with nitrate is:

$$\begin{gathered} \mathrm{n}\left({\mathrm{totalCe}}^{4+}\right)=\mathrm{c}\left({\mathrm{Ce}}^{4+}\right)×\mathrm{V}\left({\mathrm{Ce}}^{4+}\right) \\ \mathrm{n}\left({\mathrm{totalCe}}^{4+}\right)=0.1186\mathrm{M}×50.0\mathrm{mL} \\ \mathrm{n}\left({\mathrm{totalCe}}^{4+}\right)=5.93\mathrm{mmol} \end{gathered}$$

Since we can see in the second reaction that the number of moles of $C{e}^{4+}$androle="math" localid="1663606500487" $F{e}^{2+}$are equal we can write:

$$\begin{gathered} \mathrm{n}\left({\mathrm{excessCe}}^{4+}\right)=\mathrm{n}\left({\mathrm{Fe}}^{2+}\right) \\ \mathrm{n}\left({\mathrm{excessCe}}^{4+}\right)=\mathrm{c}\left({\mathrm{Fe}}^{2+}\right)×\mathrm{V}\left({\mathrm{Fe}}^{2+}\right) \\ \mathrm{n}\left({\mathrm{excessCe}}^{4+}\right)=0.04289\mathrm{M}×31.13\mathrm{mL} \\ \mathrm{n}\left({\mathrm{excessCe}}^{4+}\right)=1.335\mathrm{mmol} \end{gathered}$$

To calculate the number of milimoles that reacted with nitrite, we have to subtract excess milimoles from the total milimoles:

$$\begin{gathered} \mathrm{n}\left({\mathrm{Ce}}^{4+}\right)=\mathrm{n}\left({\mathrm{totalCe}}^{4+}\right)-\mathrm{n}\left({\mathrm{excessCe}}^{4+}\right) \\ \mathrm{n}\left({\mathrm{Ce}}^{4+}\right)=(5.93-1.335)\mathrm{mmol} \\ \mathrm{n}\left({\mathrm{Ce}}^{4+}\right)=4.595\mathrm{mmol} \end{gathered}$$

Hence the total number of moles of $C{e}^{4+}$is 4.595mmol .

Now we can put moles of nitrite and ${\mathrm{Ce}}^{4+}$in ratio. As we can see in first reaction, the ratio of nitrite and $C{e}^{4+}$ is 1 : 2 so we can write,

$$\begin{gathered} \mathrm{n}\left(\mathrm{nitrite}\right)=\frac{1}{2}\mathrm{n}\left({\mathrm{Ce}}^{4+}\right) \\ \mathrm{n}\left(\mathrm{nitrite}\right)=\frac{1}{2}×4.595\mathrm{mmol} \\ \mathrm{n}\left(\mathrm{nitrite}\right)=2.298\mathrm{mmol}=2.298×{10}^{-3}\mathrm{mol} \end{gathered}$$

Now we can calculate the mass of nitrite in 25mL aliquot.

$$\begin{gathered} \mathrm{m}\left({\mathrm{NaNO}}_2\right)=\mathrm{n}\left({\mathrm{NaNO}}_2\right)×\mathrm{M}\left({\mathrm{NaNO}}_2\right) \\ \mathrm{m}\left({\mathrm{NaNO}}_2\right)=2.298×{10}^{-3}\mathrm{mol}×68.995\mathrm{g}/\mathrm{mol} \\ \mathrm{m}\left({\mathrm{NaNO}}_2\right)=0.1586\mathrm{g}) \end{gathered}$$

For 500mL :

$$\begin{gathered} \frac{500\mathrm{mL}}{25\mathrm{mL}}=\frac{\mathrm{x}}{0.159\mathrm{g}} \\ 25\mathrm{mL}×\mathrm{x}=500\mathrm{mL}×0.1586\mathrm{g} \\ \mathrm{x}=\frac{500\mathrm{mL}×0.1586\mathrm{g}}{25\mathrm{ml}} \\ \mathrm{x}=3.172\mathrm{g} \end{gathered}$$

Now we calculate the wt% of ${\mathrm{NaNO}}_2$of unknown sample:

$$\begin{gathered} w(NaN{O}_2,sample)=\frac{m\left(NaN{O}_2\right)}{m\left(sample\right)}×100\% \\ w(NaN{O}_2,sample)=\frac{3.172g}{4.030g} \\ w(NaN{O}_2,sample)=78.71\% \end{gathered}$$

The formula for ferrous ammonium sulfate is $\left({\mathrm{NH}}_4{)}_2\mathrm{Fe}\right({\mathrm{SO}}_4{)}_2$.

Hence the wt% of $NaN{O}_2$is78.71%