Question #a8451

1 Answer
Oct 25, 2015

"0.284 M"0.284 M

Explanation:

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

http://socratic.org/questions/how-do-you-balance-this-redox-reaction-using-the-oxidation-number-method-fe2-aq-

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])8H+(aq]+MnO4(aq]+5Fe2+(aq]Mn2+(aq]+5Fe3+(aq]+4H2O(l]

Notice that you have a 1:color(red)(5)1:5 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"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