Aqueous glycerol solution weighing 100.0m gwas treated with 50.0 mL of 0.083 7 M ${\mathbf{Ce}}^{\mathbf{4}\mathbf{+}}$in 4 MHCI${\mathbf{O}}_{\mathbf{4}}$at $\mathbf{60}\mathbf{°}$for15minto oxidize glycerol to formic acid.

$$\begin{gathered} {\mathbf{CH}}_{\mathbf{2}}\mathbf{-}\mathbf{CH}\mathbf{-}{\mathbf{CH}}_{\mathbf{2}} \\ \mathbf{}\mathbf{|}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{|}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{|} \\ \mathbf{OH}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{OH}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{OH} \end{gathered}$$ ${\mathbf{HCO}}_{\mathbf{2}}\mathbf{H}$

Glycerol Formic acid

FM92.095

The excess ${\mathbf{Ce}}^{\mathbf{4}\mathbf{+}}$ required 12.11mL of 0.044 8 MF${\mathbf{e}}^{\mathbf{2}\mathbf{+}}$to reach a ferroin end point. Find wt%glycerol in the unknown.

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 ${\mathrm{Ce}}^{4+}$is reduced to ${\mathrm{Ce}}^{3+}$and glycerol is oxidized to formic acid.

Reduction:${\mathrm{Ce}}^{4+}+{\mathrm{e}}^{-}\rightleftharpoons {\mathrm{Ce}}^{3+}$

Oxidation:${\mathrm{C}}_3{\mathrm{H}}_8{\mathrm{O}}_3+3{\mathrm{H}}_2\mathrm{O}\rightleftharpoons 3\mathrm{HCOOH}+8{\mathrm{e}}^{-}+8{\mathrm{H}}^{+}$

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

Reduction:$8\mathrm{Ce}{4}^{+}+8{\mathrm{e}}^{-}\rightleftharpoons 8\mathrm{Ce}{3}^{+}$

Oxidation:${\mathrm{C}}_3{\mathrm{H}}_8{\mathrm{O}}_3+3{\mathrm{H}}_2\mathrm{O}\rightleftharpoons 3\mathrm{HCOOH}+8{\mathrm{e}}^{-}+8{\mathrm{H}}^{+}$

By adding these two reactions we get,

role="math" localid="1663604753334" $8{\mathrm{Ce}}^{4+}+{\mathrm{C}}_3{\mathrm{H}}_8{\mathrm{O}}_3+3{\mathrm{H}}_2\mathrm{O}\rightleftharpoons 8{\mathrm{Ce}}^{3+}+\mathrm{HCOOH}+8{\mathrm{e}}^{-}+8{\mathrm{H}}^{+}$

The net reaction is :

role="math" $8{\mathrm{Ce}}^{4+}+{\mathrm{C}}_3{\mathrm{H}}_8{\mathrm{O}}_3+3{\mathrm{H}}_2\mathrm{O}\rightleftharpoons 8{\mathrm{Ce}}^{3+}+\mathrm{HCOOH}+8{\mathrm{H}}^{+}$.

To determine the number of moles of ${\mathrm{Ce}}^{4+}$that was used in the reaction with glycerol, we need to determine the number of moles of excess ${\mathrm{Ce}}^{4+}$that was used in a reaction with ${\mathrm{Fe}}^{2+}$. The reaction of excess role="math" localid="1663604894428" ${\mathrm{Ce}}^{4+}$and ${\mathrm{Fe}}^{2+}$is:

${\mathrm{Ce}}^{4+}+{\mathrm{Fe}}^{2+}\rightleftharpoons {\mathrm{Ce}}^{3+}+{\mathrm{Fe}}^{3+}$

Since the number of moles of and are equal, we can write,

$$\begin{gathered} \mathrm{n}\left(\mathrm{excess}{\mathrm{Ce}}^{4+}\right)=\mathrm{n}\left({\mathrm{Fe}}^{2+}\right) \\ \mathrm{n}\left(\mathrm{excess}{\mathrm{Ce}}^{4+}\right)=\mathrm{n}\left({\mathrm{Fe}}^{2+}\right)\times \left({\mathrm{Fe}}^{2+}\right) \\ \mathrm{n}\left(\mathrm{excess}{\mathrm{Ce}}^{4+}\right)=0.0448\mathrm{M}\times 12.11\mathrm{mL} \\ \mathrm{n}\left(\mathrm{excess}{\mathrm{Ce}}^{4+}\right)=0.543\mathrm{mmol} \end{gathered}$$

The total number of moles of used in a reaction with glycerol is:

$$\begin{gathered} \mathrm{n}\left(\mathrm{total}{\mathrm{Ce}}^{4+}\right)=\mathrm{c}\left({\mathrm{Ce}}^{4+}\right)\times \mathrm{V}\left({\mathrm{Ce}}^{4+}\right) \\ \mathrm{n}\left(\mathrm{total}{\mathrm{Ce}}^{4+}\right)=0.0837\mathrm{M}\times 50.0\mathrm{mL} \\ \mathrm{n}\left(\mathrm{total}{\mathrm{Ce}}^{4+}\right)=4.185\mathrm{mmol} \end{gathered}$$

Hence the total number of moles of ${\mathrm{Ce}}^{4+}$is 4.185 mmol .

To calculate the number of moles that reacted with glycerol, we have to subtract the total moles with excess moles:

$$\begin{gathered} \mathrm{n}\left({\mathrm{Ce}}^{4+}\right)=\mathrm{n}\left(\mathrm{total}{\mathrm{Ce}}^{4+}\right)-\mathrm{n}\left(\mathrm{excess}{\mathrm{Ce}}^{4+}\right) \\ \mathrm{n}\left({\mathrm{Ce}}^{4+}\right)=\left(4.185-0.543\right)\mathrm{mmol} \\ \mathrm{n}\left({\mathrm{Ce}}^{4+}\right)=3.642\mathrm{mmol} \end{gathered}$$

Now we can put moles of glycerol and ${\mathrm{Ce}}^{4+}$in ratio. As we can see in net reaction, the ratio is so we can write

$$\begin{gathered} \mathrm{n}\left(\mathrm{glycerol}\right)=\frac{1}{8}\mathrm{n}\left({\mathrm{Ce}}^{4+}\right) \\ \mathrm{n}\left(\mathrm{glycerol}\right)=\frac{1}{8}\times 3.642\mathrm{mmol} \\ \mathrm{n}\left(\mathrm{glycerol}\right)=0.455\mathrm{mmol} \end{gathered}$$

Now we can calculate the wt % of glycerol in 100.0 m g of unknown sample:

$$\begin{gathered} \mathrm{w}\left(\mathrm{glycerol},\mathrm{sample}\right)=\frac{\mathrm{m}\left(\mathrm{glycerol}\right)}{\mathrm{m}\left(\mathrm{sample}\right)}\times 100\% \\ \mathrm{w}\left(\mathrm{glycerol},\mathrm{sample}\right)=\frac{\mathrm{n}\left(\mathrm{glycerol}\right)\times \mathrm{M}\left(\mathrm{glycerol}\right)}{\mathrm{m}\mathrm{sample}}\times 100\% \\ \mathrm{w}\left(\mathrm{glycerol},\mathrm{sample}\right)=\frac{0.455\mathrm{mmol}\times 92.095\mathrm{g}/\mathrm{mol}}{100.0\mathrm{mg}}\times 100\% \\ \mathrm{w}\left(\mathrm{glycerol},\mathrm{sample}\right)=41.9\% \end{gathered}$$