A 49.45 mL sample of solution containing Fe2+ ions is titrated with a 0.0103 M KMnO4 solution. It required 23.76 mL of KMnO4 solution to oxidize all the Fe2+ ions to Fe3+ by the following reaction.
MnO4-(aq) + Fe2+(aq) Mn2+(aq) + Fe3+(aq) (unbalanced)
(a) What was the concentration of Fe2+ ions in the sample solution?
(b) What volume of 0.0150 M K2Cr2O7 solution would it take to do the same titration? The reaction is shown below.
Cr2O72-(aq) + Fe2+(aq) Cr3+(aq) + Fe3+(aq) (unbalanced)
To find the concentration of Fe2+ ions in the solution and the volume of K2Cr2O7 solution required for the titration, we can follow these steps:
(a) Finding the concentration of Fe2+ ions in the sample solution:
1. First, we need to balance the given equation: MnO4-(aq) + Fe2+(aq) -> Mn2+(aq) + Fe3+(aq).
The balanced equation is: 5Fe2+(aq) + MnO4-(aq) + 8H+(aq) -> 5Fe3+(aq) + Mn2+(aq) + 4H2O(l).
2. We can write the balanced equation in terms of moles and use the given volume and concentration of KMnO4 to calculate the moles of MnO4- used:
Moles of MnO4- = (Volume of KMnO4 used in liters) x (Molarity of KMnO4).
Moles of MnO4- = 23.76 mL / 1000 mL/L x 0.0103 M = 0.002447 moles.
3. From the balanced equation, we can see that the stoichiometric ratio between MnO4- and Fe2+ is 1:5.
Therefore, the moles of Fe2+ ions present in the sample solution can be calculated by multiplying the moles of MnO4- by the stoichiometric ratio:
Moles of Fe2+ = 0.002447 moles x 5 = 0.01224 moles.
4. Finally, we can find the concentration of Fe2+ ions in the sample solution by dividing the moles of Fe2+ by the volume of the sample solution (in liters):
Concentration of Fe2+ = Moles of Fe2+ / Volume of sample solution in liters.
Concentration of Fe2+ = 0.01224 moles / 49.45 mL / 1000 mL/L = 0.2473 M.
Therefore, the concentration of Fe2+ ions in the sample solution is 0.2473 M.
(b) Finding the volume of 0.0150 M K2Cr2O7 solution required for the titration:
1. Again, we need to balance the given equation: Cr2O72-(aq) + Fe2+(aq) -> Cr3+(aq) + Fe3+(aq).
The balanced equation is: 6Fe2+(aq) + Cr2O72-(aq) +14H+(aq) -> 6Fe3+(aq) + 2Cr3+(aq) + 7H2O(l).
2. We can use the stoichiometry of the balanced equation to find the moles of K2Cr2O7 required to oxidize all the Fe2+ ions.
The stoichiometric ratio between Cr2O72- and Fe2+ is 1:6.
3. Moles of K2Cr2O7 required = Moles of Fe2+ / 6.
Moles of K2Cr2O7 required = 0.01224 moles / 6 = 0.00204 moles.
4. Now, we can use the molarity and the equation Moles = Molarity x Volume to find the volume of K2Cr2O7 required.
Volume of K2Cr2O7 required = Moles of K2Cr2O7 required / Molarity of K2Cr2O7.
Volume of K2Cr2O7 required = 0.00204 moles / 0.0150 M = 0.136 L or 136 mL.
Therefore, it would take 136 mL of 0.0150 M K2Cr2O7 solution to titrate the given solution.