A 47.26 mL sample of solution containing Fe2+ ions is titrated with a 0.0112 M KMnO4 solution. It required ...

Question

17.78 mL of KMnO4 solution to oxidize all the Fe2+ ions to Fe3+ by the following reaction.

MnO4-(aq) + Fe2+(aq) Mn2+(aq) + Fe3+(aq) (unbalanced)

(a) What was the concentration of Fe2+ ions in the sample solution? in M

What volume of 0.0150 M K2Cr2O7 solution would it take to do the same titration? The reaction is shown below. in mL

Cr2O72-(aq) + Fe2+(aq) Cr3+(aq) + Fe3+(aq) (unbalanced)

Answer 1

The essential step here is to balence the reaction equation:
MnO4- + 5Fe(2+) + 8H+ <==> Mn(2+) + 5Fe(3+) + 4H2O
17.78 mL of 0.0112 M KMnO4 <=> 0.199 mMol KMnO4 ==> Ozidize 5x0.199 = 0.996 (mMol) Fe(2+)
0.996 mMol Fe(2+) in 47.26 mL <=> 0.0211 M

Also balence the new reaction equation:
Cr2O7(2-) + 6Fe(2+) + 14H+ <==> 2Cr(3+) + 6Fe(3+) +7H2O
Starting from 0.996 mMol Fe(2+) ==> Need (0.996/6) or 0.166 mMol K2Cr2O7 to oxidize ==> The volume of 0.0150 M K2Cr2O7 solution is:
0.166 mMol /0.0150 M = 11.1 mL