# Question a8451

Oct 25, 2015

$\text{0.284 M}$

#### Explanation:

The idea here is that the iron(II) ions will reduce the manganese from a $\textcolor{b l u e}{+ 7}$ oxidation state in the permanganate ion, to a $\textcolor{b l u e}{+ 2}$ 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.

$8 {\text{H"_text((aq])^(+) + "MnO"_text(4(aq])^(-) + color(red)(5)"Fe"_text((aq])^(2+) -> "Mn"_text((aq])^(2+) + 5"Fe"_text((aq])^(3+) + 4"H"_2"O}}_{\textrm{\left(l\right]}}$

Notice that you have a $1 : \textcolor{red}{5}$ mole ratio between the permanganate ions and the iron(II) ions. Keep this in mind.

Now, you know that you dissolve a $\text{1.08-g}$ 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 $\text{250.0 mL}$ of the acidic solution!

This means that if you take a $\text{10.0-mL}$ 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

$c = \frac{n}{V}$

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