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.
To calculate the concentrations of Fe2+ and Fe3+ ions in the solution, we can use the information from the titrations with KMnO4.
First, let's determine the moles of KMnO4 used in each titration:
For the first titration, with 15.0 cm^3 of 0.020 M KMnO4:
Moles of KMnO4 = volume (in L) * concentration = 15.0 cm^3 * (1 L / 1000 cm^3) * 0.020 mol/L = 0.003 mol
For the second titration, with 19.0 cm^3 of 0.020 M KMnO4:
Moles of KMnO4 = volume (in L) * concentration = 19.0 cm^3 * (1 L / 1000 cm^3) * 0.020 mol/L = 0.0038 mol
Since KMnO4 acts as an oxidizing agent in these reactions, it reacts with Fe2+ ions to form Fe3+ ions. The balanced equation for this reaction is:
5Fe2+ + MnO4- + 8H+ -> 5Fe3+ + Mn^2+ + 4H2O
Comparing the moles of KMnO4 used with the stoichiometry of the reaction, we can calculate the moles of Fe2+ ions reacted in each titration.
For the first titration:
Using the stoichiometry of the equation, we can see that 1 mole of KMnO4 reacts with 5 moles of Fe2+.
Therefore, Moles of Fe2+ ions in the aliquot = 0.003 mol * (5 mol Fe2+ / 1 mol KMnO4) = 0.015 mol
For the second titration:
Similarly, using the stoichiometry of the equation, we can see that 1 mole of KMnO4 reacts with 5 moles of Fe2+.
Therefore, Moles of Fe2+ ions in the aliquot = 0.0038 mol * (5 mol Fe2+ / 1 mol KMnO4) = 0.019 mol
Since each aliquot used 25.0 cm^3 of solution, the molar concentration of Fe2+ ions can be calculated:
For the first aliquot:
Concentration of Fe2+ ions = Moles of Fe2+ ions / Volume (in L) = 0.015 mol / (25.0 cm^3 * (1 L / 1000 cm^3)) = 0.6 M
For the second aliquot:
Concentration of Fe2+ ions = Moles of Fe2+ ions / Volume (in L) = 0.019 mol / (25.0 cm^3 * (1 L / 1000 cm^3)) = 0.76 M
To calculate the concentration of Fe3+ ions, we can assume that all the Fe2+ ions in the aliquot were converted to Fe3+ ions during the titration. Therefore, the concentration of Fe3+ ions is the same as the concentration of Fe2+ ions.
Therefore, the concentrations of Fe2+ and Fe3+ ions in the solution are:
Fe2+: 0.6 M
Fe3+: 0.76 M
To determine the concentrations of Fe2+ and Fe3+ ions in the solution, we can use the concept of redox reactions and stoichiometry.
1. Let's focus on the first aliquot that was titrated against KMnO4 solution. The balanced redox equation for this reaction is:
5Fe2+ + MnO4- + 8H+ -> 5Fe3+ + Mn2+ + 4H2O
From the stoichiometry of this equation, we can see that 5 moles of Fe2+ react with 1 mole of KMnO4. So, the number of moles of Fe2+ in the solution is given by:
moles of Fe2+ = (0.020 M KMnO4) * (15.0 cm^3) * (1 L / 1000 cm^3) * (5 / 1)
Note that we convert the volume from cm^3 to L and multiply by the stoichiometric ratio of Fe2+ to KMnO4.
2. Now, let's move on to the second aliquot that was reduced using Zn and then titrated. The balanced redox equation for this reaction is:
3Fe2+ + 2MnO4- + 2H2O -> 3Fe3+ + 2Mn2+ + 4OH-
From the stoichiometry of this equation, we can see that 3 moles of Fe2+ react with 2 moles of KMnO4. So, the number of moles of Fe2+ in the solution is given by:
moles of Fe2+ = (0.020 M KMnO4) * (19.0 cm^3) * (1 L / 1000 cm^3) * (3 / 2)
3. To find the number of moles of Fe3+ ions in the solution, we use the difference in the number of moles of Fe2+ before and after the reaction:
moles of Fe3+ = moles of Fe2+ (from step 1) - moles of Fe2+ (from step 2)
4. The total number of moles of iron ions in the solution is the sum of the moles of Fe2+ and Fe3+:
total moles of iron ions = moles of Fe2+ (from step 1) + moles of Fe3+ (from step 3)
5. Finally, to calculate the concentrations of Fe2+ and Fe3+ ions, we divide the number of moles by the volume of the aliquot:
concentration of Fe2+ ion = moles of Fe2+ / (25.0 cm^3 * (1 L / 1000 cm^3))
concentration of Fe3+ ion = moles of Fe3+ / (25.0 cm^3 * (1 L / 1000 cm^3))
By plugging in the calculated values, you should be able to determine the concentrations of Fe2+ and Fe3+ ions in the solution.