Question

What is the equivalent mass of ${\mathrm{KMnO}}_4$ in an acidic and alkaline medium?

Answer 1

Step $1$: Finding molecular mass:

• "The number of moles of portion by mass of a substance that combines with or displaces $1\mathrm{g}$ of hydrogen, $8\mathrm{g}$of oxygen, or $35.5\mathrm{g}$ of chlorine is known as its equivalent mass."

$\mathrm{Equivalent}\mathrm{mass}=\frac{\mathrm{Molecular}\mathrm{mass}}{\mathrm{n}\mathrm{factor}}$

• In the case of ${\mathrm{KMnO}}_4$, the n factor will be the number of electrons lost or gained.
• So, the molecular mass of ${\mathrm{KMnO}}_4$should be first calculated.

$$\begin{gathered} \mathrm{Molecular}\mathrm{mass}\mathrm{of}{\mathrm{KMnO}}_4=39.1+54.93+4\left(16\right) \\ =158.034\mathrm{g}{\mathrm{mol}}^{-1} \end{gathered}$$

Step $2$: Finding n- factor of ${\mathrm{KMnO}}_4$in acidic medium:

The n factor will be different in the case of the acidic and basic medium.

• In an acidic medium, the ${\mathrm{KMnO}}_4$ will exist as:

$$\begin{gathered} {{\mathrm{MnO}}_4}^{-}+8{\mathrm{H}}^{+}+5{\mathrm{e}}^{-}\to {\mathrm{Mn}}^{2+}+4{\mathrm{H}}_2\mathrm{O}(\mathrm{n}=5) \\ {\mathrm{KMnO}}_4\to {\mathrm{Mn}}^{2+} \\ \mathrm{Oxidation}\mathrm{Number}\mathrm{Oxidation}\mathrm{Number} \\ \mathrm{of}\mathrm{Mn}=+7\mathrm{of}\mathrm{Mn}=+2 \\ \mathrm{Electrons}\mathrm{used},\mathrm{n}-\mathrm{factor}=7-2=5 \end{gathered}$$

Thus, the equivalent mass will be:

$$\begin{gathered} \mathrm{Equivalent}\mathrm{mass}=\frac{158.034}{5} \\ =31.61\mathrm{g}{\mathrm{eq}}^{-1} \end{gathered}$$

Step $3$: Finding n factor value of ${\mathrm{KMnO}}_4$ in basic medium:

• The ${\mathrm{KMnO}}_4$acts as an oxidizer which means the loss of electrons in acidic media.
• In a basic medium, the ${\mathrm{KMnO}}_4$will exist as

$$\begin{gathered} {{\mathrm{MnO}}_4}^{-}+2{\mathrm{H}}_2\mathrm{O}+3{\mathrm{e}}^{-}\to {\mathrm{MnO}}_2+4{\mathrm{OH}}^{-}(\mathrm{n}=3) \\ +7+4 \\ \mathrm{Electrons}\mathrm{used},\mathrm{n}-\mathrm{factor}=7-4=3 \end{gathered}$$

Thus, the equivalent mass will be:

$$\begin{gathered} \mathrm{Equivalent}\mathrm{mass}=\frac{158.034}{3} \\ =52.68\mathrm{g}{\mathrm{eq}}^{-1} \end{gathered}$$