Posted by tam on .
A 25.0cm^3 aliquot of a solution containing Fe2+ ions and Fe3+ ions was acidified and titrated against KMnO4 solution. 15.0cm^3 of 0.020M KMnO4 was needed. A second 25.0cm^3 aliquot was reduced using Zn, then titrated. This time, 19.0cm^3 of the 0.020M KMnO4 was needed. Calculate the concentrations of:
Fe2+ ions and Fe3+ ions in the solution
Write the equation and balance it for
Fe^+2 + MnO4^- ==> Fe^+3 + Mn^+2
moles MnO4- = M x L = ??
Using the coefficients in the balanced equation, convert moles MnO4^- to moles iron(II). [This first step titrates only iron(II) and no iron(III).]
M of the 25cc is moles/L.
The second titration is carried out in which the iron(III) is reduced to iron(II) with the Zn (called a Jones reductor) so it titrates all of the iron(II) that it titrated in the first titration plus all of the iron(III) that was reduced to iron(II) with Zn.
Total moles = total iron = M x L in second step = ??
moles second titn 0 moles first titn = moles iron(III) and M = moles/L.
A comment here. The problem asks for concn of iron(II) and iron(III) IN THE SOLUTION and the 25 cc is just an aliquot. My instructions provide concn in the 25 cc BECAUSE the total volume of the solution is not given. Without knowing that, you can calculate only the amount of iron in the 25 cc and not in the original solution. If you wish to know the concn in the original, then
moles to titrate 25 cc x (volume original/25 cc) = moles in the original and moles/volume original = M original.