To calculate the molarity of Fe2+ and Fe3+ in the original solution, we need to analyze the reactions that occur during each step of the process.
Step 1: Titration of Fe2+ and Fe3+ with KMnO4
We have 25 ml of a solution containing both Fe2+ and Fe3+ ions. When titrated with 23 ml of 0.02 M KMnO4, all the Fe2+ ions are oxidized to Fe3+. This reaction can be represented as follows:
5 Fe2+ + MnO4- + 8 H+ -> 5 Fe3+ + Mn2+ + 4 H2O
From the balanced equation, we can see that 5 moles of Fe2+ ions react with 1 mole of MnO4-. Therefore, the number of moles of Fe2+ in the original solution is given by:
n(Fe2+) = (0.02 mol/L) x (0.023 L) / 5 = 0.000092 mol
Step 2: Treatment with Zn metal to convert Fe3+ to Fe2+
Next, the solution is treated with Zn metal to convert all Fe3+ ions to Fe2+. The reaction can be represented as follows:
Zn + 2 Fe3+ -> Zn2+ + 2 Fe2+
Since all Fe3+ is converted to Fe2+, the number of moles of Fe2+ in the solution remains the same as in Step 1, i.e. 0.000092 mol.
Step 3: Oxidation of Fe2+ to Fe3+ by KMnO4
Finally, the solution containing only Fe2+ ions requires 40 ml of the same 0.02 M KMnO4 solution for oxidation to Fe3+. This reaction can be represented as follows:
5 Fe2+ + MnO4- + 8 H+ -> 5 Fe3+ + Mn2+ + 4 H2O
Using the same approach as in Step 1, we can calculate the number of moles of Fe2+ in the solution:
n(Fe2+) = (0.02 mol/L) x (0.040 L) / 5 = 0.00016 mol
Since all Fe3+ is converted to Fe2+, the number of moles of Fe2+ in the solution remains the same as in Step 2, i.e. 0.000092 mol.
Now, to calculate the molarity of Fe2+ and Fe3+, we divide the number of moles of each species by the total volume of the solution:
Molarity of Fe2+ = 0.000092 mol / (25 ml / 1000) = 3.68 x 10^-3 M
Molarity of Fe3+ = 0.000092 mol / (25 ml / 1000) = 3.68 x 10^-3 M
Therefore, the molarity of Fe2+ and Fe3+ in the original solution is 3.68 x 10^-3 M.