When 25.0 ml of a solution containing both Fe2+and Fe3 +ions is titrated with 23.0 ml of 0.0200M KMnO4(in dilute sulphuric acid), all the Fe2+ are oxidised to Fe3+ ions. Next, the solution is treated with Zinc metal to convert all the Fe3+ ions to Fe2+ ions. Finally 40ml of the same Kmno4 solution is added to the solution oxidise the Fe2+ to Fe3.Calculate the molar concentrations of Fe2+ and Fe3 + in the original solution
23M
To calculate the molar concentrations of Fe2+ and Fe3+ in the original solution, we need to analyze the reactions that occur during the titration process.
Step 1: Titration with KMnO4
During the titration, KMnO4 oxidizes Fe2+ to Fe3+ ions. The balanced equation for this reaction is:
5Fe2+ + MnO4- + 8H+ → 5Fe3+ + Mn2+ + 4H2O
From the balanced equation, we can determine the mole ratio between Fe2+ and KMnO4 as 5:1.
Given:
Volume of KMnO4 solution (V1) = 23.0 mL
Molarity of KMnO4 solution (M1) = 0.0200 M
To calculate the moles of Fe2+ in the original solution, we can use the formula:
Moles of Fe2+ = Moles of KMnO4 × 5
Moles of KMnO4 = M1 × V1
= 0.0200 M × 23.0 mL
= 0.460 mmol (millimoles)
Moles of Fe2+ = 0.460 mmol × 5
= 2.30 mmol
Step 2: Reduction with Zinc
Zinc metal (Zn) is added to convert all Fe3+ ions back to Fe2+ ions. The balanced equation for this reaction is:
Zn + 2Fe3+ → Zn2+ + 2Fe2+
Given:
Volume of the original solution = 25.0 mL
To calculate the moles of Fe3+ ions initially present, we need to know the ratio between Fe2+ and Fe3+ ions in the original solution. However, this ratio is not provided in the question. Therefore, it is not possible to determine the exact moles of Fe3+ ions in the original solution without additional information.
Step 3: Second Titration with KMnO4
After the reduction with zinc, 40 mL of the same KMnO4 solution is added. This results in the oxidation of Fe2+ ions back to Fe3+ ions.
To calculate the moles of Fe2+ oxidized to Fe3+ ions, we can use the same mole ratio between Fe2+ and KMnO4 as before (5:1).
Given:
Volume of KMnO4 solution (V2) = 40 mL
Molarity of KMnO4 solution (M2) = 0.0200 M
Moles of KMnO4 = M2 × V2
= 0.0200 M × 40 mL
= 0.800 mmol (millimoles)
Moles of Fe2+ oxidized to Fe3+ = 0.800 mmol × 5
= 4.00 mmol
However, without knowing the initial moles of Fe3+ or the ratio between Fe2+ and Fe3+ in the original solution, we cannot accurately calculate the molar concentrations of Fe2+ and Fe3+ in the original solution. Additional information is needed to solve the problem completely.