Question

What is the equivalent weight of ${\mathrm{KMnO}}_4$ in acidic, basic, and neutral medium?

Answer 1

Step 1. Equivalent weight

• Equivalent weight is equal to the molecular/atomic mass divided by the number of electrons involved in the reaction per molecule/atom/ion.

Step 2. Formula for the equivalent weight of an oxidising agent

• $\text{Equivalent weight}=\frac{\mathrm{Molecular}\mathrm{weight}}{\text{Number of electron gained}}$

Step3. Equivalent weight of ${\mathrm{KMnO}}_4$ in acidic medium

• The reaction in an acidic medium is as follows:-
• ${\mathrm{MnO}}_4^{-}+8{\mathrm{H}}^{+}+5{\mathrm{e}}^{-}\to {\mathrm{Mn}}^{2+}+4{\mathrm{H}}_2\mathrm{O}$
• $$\begin{gathered} {\text{Equivalent weight of KMnO}}_{\text{4}}=\frac{158}{5} \\ =31.6{\text{g eq}}^{-1} \end{gathered}$$

Step4. Equivalent weight of ${\mathrm{KMnO}}_4$ in basic medium

• The reaction in basic medium is as follows:-

${\mathrm{MnO}}_4^{-}+2{\mathrm{H}}_2\mathrm{O}+3{\mathrm{e}}^{-}\to {\mathrm{MnO}}_2+4{\mathrm{OH}}^{–}$

• $$\begin{gathered} {\text{Equivalent weight of KMnO}}_{\text{4}}=\frac{158}{3} \\ =52.6{\text{g eq}}^{-1} \end{gathered}$$

Step5. Equivalent weight of ${\mathrm{KMnO}}_4$ in neutral medium

• In a neutral medium as well, the value of equivalent weight is $52.6{\text{g eq}}^{-1}$.

Hence, the equivalent weight of ${\mathrm{KMnO}}_4$ in acidic, basic, and neutral medium is $31.6{\text{g eq}}^{-1}$, and $52.6{\text{g eq}}^{-1}$.