a quantity of 25 ml of a solution containing both Fe2+ and Fe3+ ions is titrated with 23 ml of .02 M KMnO4 solution. as a result all the Fe2+ ions are oxidised to Fe3+ ions. Next the solution is treated with Zn metal to convert all Fe3+ ions to Fe2+ ions. Finally, the solution containing only the Fe2+ ions requires 40ml of the same KMnO4 solution for oxidation to Fe3+. Calculate the molarity of Fe2+ and Fe3+ in the original solution.
To solve this problem, we need to follow the steps given and use the information provided to calculate the molarity of Fe2+ and Fe3+ in the original solution.
Step 1: Calculate the moles of KMnO4 used in the titration with Fe2+ and Fe3+ ions.
- The volume of the KMnO4 solution used in the titration is 23 mL.
- The molarity of KMnO4 solution is 0.02 M.
- Using the equation Molarity (M) = Moles (mol) / Volume (L), we can rearrange the equation to find the moles.
- Moles = Molarity * Volume = 0.02 M * 0.023 L = 0.00046 mol KMnO4
Step 2: Determine the number of moles of Fe2+ oxidized to Fe3+ ions.
- The balanced chemical equation for the titration is:
5Fe2+ + MnO4- + 8H+ -> 5Fe3+ + Mn2+ + 4H2O
- From the balanced equation, we can see that 5 moles of Fe2+ are required to react with 1 mole of KMnO4.
- Therefore, the moles of Fe2+ oxidized would be 5 * 0.00046 mol = 0.0023 mol
Step 3: Calculate the moles of Fe2+ and Fe3+ in the original solution.
- Since all the Fe2+ ions are oxidized to Fe3+ ions, the moles of Fe3+ in the original solution is equal to the moles of Fe2+ oxidized in the titration, which is 0.0023 mol.
- The volume of the original solution is 25 mL = 0.025 L.
- Therefore, the molarity of Fe3+ in the original solution is 0.0023 mol / 0.025 L = 0.092 M.
Step 4: Determine the moles of Fe2+ in the original solution after reduction with Zn metal.
- The balanced chemical equation for the reduction of Fe3+ to Fe2+ is:
Fe3+ + e- -> Fe2+
- From the equation, we can see that each Fe3+ ion is reduced to one Fe2+ ion.
- Since all the Fe3+ ions are converted to Fe2+ ions, the moles of Fe2+ in the original solution are also 0.0023 mol.
Step 5: Calculate the molarity of Fe2+ in the original solution.
- The volume of the original solution is still 25 mL = 0.025 L.
- Therefore, the molarity of Fe2+ in the original solution is 0.0023 mol / 0.025 L = 0.092 M.
So, the molarity of Fe2+ and Fe3+ in the original solution is both 0.092 M.
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.
Write the equation for Fe^2+ and MnO4^-
The pertinent information is this.
5Fe^2+ + MnO4^- ==> Mn^2+ + 5Fe^3+
From the first titration you have
mols MnO4^- = M x L = ?
Convert mols MnO4^- to mols Fe^2+ using the coefficients in the balanced equation. That's mols Fe^2+ = 5x mols MnO4^-
M Fe^2+ = mols Fe^2+/0.025L = ?
Do the second part as above. This time you have total mols Fe^2+ + Fe^3+. Subtract from this number mols Fe^2+ from the first titration. This give you Fe^3+, then M Fe^3+ = mols/0.025L = ?