Question

Equivalent mass of $KMn{O}_4$ in acidic, basic and neutral are in the ratio of :
(Given, molecular mass of $KMn{O}_4=158g$)

• A. $3:5:15$
• B. $5:3:1$
• C. $5:1:3$
• D. $3:15:5$

Answer 1

The correct option is D $3:15:5$

In acidic medium: $Mn{O}_4^{-}+8H^{+}+5e^{-}\to M{n}^{2+}+4H_{2}O$
In basic medium; $Mn{O}_4^{-}+e^{-}\to Mn{O}_4^{2-}$
In neutral medium : $Mn{O}_4^{-}+4H^{+}+3e^{-}\to Mn{O}_2+2H_{2}O$
Equivalent mass is molecular mass divided by number of electrons gained.
In acidic medium, equivalent weight $=\frac{\text{Molecular weight}}{5}=\frac{\text{158}}{5}$
In basic medium, equivalent weight $=\frac{\text{Molecular weight}}{1}=\frac{\text{158}}{1}$
In neutral medium, equivalent weight $=\frac{\text{Molecular weight}}{3}=\frac{\text{158}}{3}$
Ratio of equivalent mass $=\frac{\text{158}}{5}:\frac{\text{158}}{1}:\frac{\text{158}}{3}$
Ratio of equivalent mass $3:15:5$