5fe2+ + mno4− + 8 h+ ⇄ 5 fe3+ + mn2+ + 4 h2o in a titration experiment based on the equation above, 25.0 milliliters of an acidified fe2+ solution requires 14.0 milliliters of standard 0.050-molar mno4− solution to reach the equivalence point. the concentration of fe2+ in the original solution is…

(a) 0.0010 m
(b) 0.0056 m
(c) 0.028 m
(d) 0.090 m
(e) 0.14 m

0.14 M

Explanation:

Volume of Fe2+ required for titration = 25 mL

Volume of MnO4- required for titration = 14 mL

Concentration of MnO4- = 0.05 M

We have to find the concentration of Fe2+.

Let it be x.

In a redox reaction:

Number of equivalence of oxidising agent = Number of equivalence of reducing agent.

Formula for number of equivalence: n*M*V

n = n-factor (number of electrons gained or lost by 1 mole of reducing or oxidising agent)

M = molarity

V = volume

n-factor of Fe2+ is 1 as it changes to Fe3+ by loosing 1 electron

n-factor of MnO4- is 5 as it changes to Mn2+ by gaining 5 electrons.

Number of equivalence of Fe2+ = Number of equivalence of MnO4-

1*(x)*25 = 5*0.05*14

x = 0.14 M

i think its 100

i hope this !