posted by Albert on .
when 25.0 mL of a solution containing both Fe2+ and Fe3+ ions is titrated with 23.0 mL of 0.0200 M KMnO4 (in dilute sulfuric acid). As a result, all of the Fe2+ ions are oxidized to Fe3+ ions. Next, the solution is treated with Zn metal to convert all of the Fe3+ ions to Fe2+ ions. Finally, the solution containing only the Fe2+ ions requires 40.0 mL of the same KMnO4 solution for oxidation to Fe3+. Calculate the molar concentrations of Fe2+ and Fe3+ in the original solution
I had tremendous problems with this question as well, and since I now have the answer, I'll spread the love.
First, you figure out how many moles of KMnO4 is used for both titrations using the equation
---- M=(mol)/(L) ---- (M=Molarity)
Doing this, you would get the first amount of KMnO4 to be
(.02M)(.023L)= .00046 mol Fe2+
and the second to be
(.02M)(.040L)= .00080 mol Fe2+
Then, using stoichiometry* and the provided equation
- 2+ + 2+ 3+
MnO4 +5Fe +8H -->Mn+ 5Fe + 4H20
You get .0024mol Fe2+ to begin with and .0040mol Fe2+ to end with.
To get how many moles of Fe3+, you just subtract .0040-.0023 = .0017mol Fe3+
Molar Concentration is the same as Molarity, so...
----Here's the answer!----
M Fe2+ = .0023mol/.025L = .092M
M Fe3+ = .0017mol/.025L = .068M
*I assume you know how to set up the ratios, but I'll just say that you would use 1mol KMnO4=1mol MnO4 and
1mol MnO4=5mol Fe2+
Oops the charges in the equation got messed up so instead:
MnO4- + 5Fe2+ + 8H+ -> 5Fe3+ + 4H2O