Question icon
Mechanics

The equivalent weight of KMnO4, the acidic medium is

Profile image of Regina Phalange
7 Years agoGrade
Answers icon

1 Answer

Profile image of Arun
7 Years ago
In Acidic medium, This Mn+7 goes to Mn+2 state and hence there is a net gain of 5 electrons. 
Now, equivalent weight = [molar mass] /[number of electrons gained or lost]
so, eq wt = 158/5 = 31.6 g.
Now, for alkaline medium, there are two possibilities
1) faintly alkaline/neutral:
Mn+7 changes into Mn+4 therefore, gain of 3 electrons
Hence, eq wt = 158/3 = 52.67g.
2) highly alkaline:
Mn+7 changes into Mn+6 therefore, gain of 1 electron
Hence, eq wt = 158/1 = 158g