Electron-Transfer Equivalents
323
E-T equivalents of KMnO 4 = E-T equivalents of Na 2 C 2 O 4
(0.04045 liter) f/V ?SHIX\ = °'
2814 g
= 0.004200 equiv
V
1'ter /
6? OQ _JL_
equiv
= 0.1038 N KMnO 4
0.04045 liter
Because there are 5 equiv/mole in KMnO 4 ,
0.1038
molarity = - - = 0.02077 M KMnO 4
equiv
mole
Note that it is not necessary to have a balanced chemical equation to work this
problem. You need know only the valence changes that are involved.
PROBLEM:
A 45.34 ml sample of FeSO 4 requires 35.76 ml of 0.1047 N Na 2 Cr 2 O 7 in acid
solution for titration. Calculate the normality of the FeSO 4 solution.
SOLUTION:
Using equiv to mean E-T equiv, we see that (at the endpoint)
equiv of Cr 2 O|~ = equiv of Fe
2+
( 0.1047 £9H!X) (0.03576 liter) = (0.04534 liter) (N £9H1X\
\
liter /
\
liter /
N = 0.008258 N FeSO 4
Note that, in a problem like this, you need not know the equation for the
reaction or the valence changes involved with the reactants. If you wanted to
know the molarities of the two solutions, you would have to know that Cr 2 O?~
has 6 equiv/mole (because Cr 2 O?~ goes to 2Cr
3+ ) and that Fe
2+ has 1 equiv/mole
(because Fe
2+ goes to Fe
3+ ). Thus,
molarity of Na 2 Cr 2 O 7 = ( 0. ,047
\
liter / \ 6 equiv/
= 0.01745 M Na 2 Cr 2 O 7
molarity of FeSO 4 = ( 0.08258 £SHLl) (
\
liter / \ equiv/
= 0.08258 M FeSO 4