Question #a8451

1 Answer
Oct 25, 2015


#"0.284 M"#


The idea here is that the iron(II) ions will reduce the manganese from a #color(blue)(+7)# oxidation state in the permanganate ion, to a #color(blue)(+2)# oxidation state.

In the process, the iron(II) ions will be oxidized to iron(III) ions.

I assume that you're familiar with the redox reaction, so I won't go into that here.

The reaction takes place in acidic solution, so you can expect some water molecules and some protons to pop up in the balanced chemical equation.

#8"H"_text((aq])^(+) + "MnO"_text(4(aq])^(-) + color(red)(5)"Fe"_text((aq])^(2+) -> "Mn"_text((aq])^(2+) + 5"Fe"_text((aq])^(3+) + 4"H"_2"O"_text((l])#

Notice that you have a #1:color(red)(5)# mole ratio between the permanganate ions and the iron(II) ions. Keep this in mind.

Now, you know that you dissolve a #"1.08-g"# sample of iron in acidic solution, producing iron(II) cations. Use iron's molar mass to determine how many moles of iron(II) cations you have

#1.08color(red)(cancel(color(black)("g"))) * "1 mole Fe"^(2+)/(55.845color(red)(cancel(color(black)("g")))) = "0.01934 moles Fe"^(2+)#

The important thing to remember here is that you have this many moles of iron(II) cations in #"250.0 mL"# of the acidic solution!

This means that if you take a #"10.0-mL"# sample, you will have

#10.0color(red)(cancel(color(black)("mL"))) * "0.01934 moles Fe"^(2+)/(250.0color(red)(cancel(color(black)("mL")))) = "0.0007736 moles Fe"^(2+)#

Now the mole ratio comes in handy. You know that the equivalence point of the titration is recorded when you have

#0.0007736color(red)(cancel(color(black)("moles Fe"^(2+)))) * (color(red)(5)" moles MnO"_4^(-))/(1color(red)(cancel(color(black)("mole Fe"^(2+))))) = "0.003868 moles MnO"_4^(-)#

Now simply use the average equivalence point volume to get the concentration of the permanganate ions

#c = n/V#

#c = "0.003868 moles"/(13.6 * 10^(-3)"L") = color(green)("0.284 M")#

Check out this very cool video of the entire iron(II) - permanganate titration