Titration problem: 5Fe+2 + MnO4- + 8H+ --> 5Fe+3 + Mn+2 + 4H20 ?

Question

100 ml of acidified Fe+2 requires 12ml of standard .050M MnO2- solution to reach the equiv point. What is the concentration of Fe+2 in the ORIGINAL solution??

Thanks!

Answer 1

Moles MnO4- = 0.050 M x 0.012 L = 0.00060

Moles Fe2+ = 5 x 0.00060 =0.0030 in 100 mL

[Fe2+] = 0.030 M