Question

Nitrite $\left({\mathrm{NO}}_2^{-}\right)$can be determined by oxidation with excess localid="1663607686215" ${\mathbf{Ce}}^{\mathbf{4}\mathbf{+}}$ , followed by back titration of unreacted . A sample of solid containing only ${\mathbf{NaNO}}_{\mathbf{2}}\mathbf{\left(}{\mathbf{FM}}_{\mathbf{68}\mathbf{.}\mathbf{995}}\mathbf{\right)}$ and ${\mathbf{NaNO}}_{\mathbf{3}}\mathbf{}$was dissolved in 500.0mL . A sample of this solution was treated with 50.00mL of$\mathbf{0}\mathbf{.}\mathbf{118}\mathbf{}\mathbf{}\mathbf{6}\mathbf{M}\mathbf{}{\mathbf{Ce}}^{\mathbf{4}\mathbf{+}}$ in strong acid for 5min , and excess ${\mathbf{Ce}}^{\mathbf{4}\mathbf{+}}$ was back-titrated with 31.13mL of ferrous ammonium sulfate.

localid="1663606208971" $$\begin{gathered} \mathbf{2}\mathit{C}{\mathit{e}}^{\mathbf{4}\mathbf{+}}\mathbf{+}\mathit{N}{\mathit{O}}_{\mathbf{2}}^{\mathbf{-}}\mathbf{+}{\mathbf{H}}_{\mathbf{2}}\mathbf{O}\mathbf{\to }\mathbf{2}\mathit{C}{\mathit{e}}^{\mathbf{3}\mathbf{+}}\mathbf{+}\mathit{N}{\mathit{O}}_{\mathbf{3}}^{\mathbf{-}}\mathbf{+}\mathbf{2}{\mathbf{H}}^{\mathbf{+}} \\ \mathit{C}{\mathit{e}}^{\mathbf{4}\mathbf{+}}\mathbf{+}\mathit{F}{\mathit{e}}^{\mathbf{2}\mathbf{+}}\mathbf{\to }\mathit{C}{\mathit{e}}^{\mathbf{3}\mathbf{+}}\mathbf{+}\mathit{F}{\mathit{e}}^{\mathbf{3}\mathbf{+}} \end{gathered}$$

What is the formula for ferrous ammonium sulfate? Calculate wt in the solid.

Short answer

The wt% of ${\mathrm{NaNO}}_2$ is 78.71% . The formula for ferrous ammonium sulfate is $\left({\mathrm{NH}}_4{)}_2\mathrm{Fe}\right({\mathrm{SO}}_4{)}_2$ .

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 number of mmoles that reacted with nitrite.

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:

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

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:

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

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 .

Find the wt% of  NaNO2.

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,

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

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

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

For 500mL :

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

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

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

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%