When \(25.0 \mathrm{~mL}\) of a solution containing both \(\mathrm{Fe}^{2+}\) and \(\mathrm{Fe}^{3+}\) ions is titrated with \(23.0 \mathrm{~mL}\) of \(0.0200 \mathrm{M}\) \(\mathrm{KMnO}_{4}\) (in dilute sulfuric acid), all of the \(\mathrm{Fe}^{2+}\) ions are oxidized to \(\mathrm{Fe}^{3+}\) ions. Next, the solution is treated with \(Z\) n metal to convert all of the \(\mathrm{Fe}^{3+}\) ions to \(\mathrm{Fe}^{2+}\) ions. Finally, \(40.0 \mathrm{~mL}\) of the same \(\mathrm{KMnO}_{4}\) solution are added to the solution in order to oxidize the \(\mathrm{Fe}^{2+}\) ions to \(\mathrm{Fe}^{3+}\). Calculate the molar concentrations of \(\mathrm{Fe}^{2+}\) and \(\mathrm{Fe}^{3+}\) in the original solution.

Short Answer

Expert verified
The molar concentrations of \(\mathrm{Fe}^{2+}\) and \(\mathrm{Fe}^{3+}\) in the original solution were 0.092 M and 0.16 M, respectively.

Step by step solution

01

Determine Moles of \(\mathrm{Fe}^{2+}\) Consumed in First Titration

Calculate the moles of \(\mathrm{KMnO}_{4}\) used in the first titration using the volume and molarity. This is done by multiplying the volume in liters (23.0 mL = 0.023 L) by the molarity (0.0200 M) to get moles of \(\mathrm{KMnO}_{4}\). We get 0.0200 M * 0.023 L = 0.00046 moles of \(\mathrm{KMnO}_{4}\). From the balanced redox equation, 1 mol of \(\mathrm{KMnO}_{4}\) reacts with 5 mol of \(\mathrm{Fe}^{2+}\), therefore, moles of \(\mathrm{Fe}^{2+}\) = 5 * moles of \(\mathrm{KMnO}_{4}\) = 5 * 0.00046 mol = 0.0023 mol.
02

Calculate Concentration of \(\mathrm{Fe}^{2+}\)

To obtain the concentration of \(\mathrm{Fe}^{2+}\) ions in the original solution, divide the moles of \(\mathrm{Fe}^{2+}\) by the volume of the original solution in liters. Concentration = 0.0023 mol / 0.025 L = 0.092 M.
03

Determine Moles of \(\mathrm{Fe}^{2+}\) Formed from \(\mathrm{Fe}^{3+}\) in Second Titration

Calculate the moles of \(\mathrm{KMnO}_{4}\) used in the second titration using the given volume and molarity. Multiply the volume in liters (40.0 mL = 0.040 L) by the molarity (0.0200 M) to get moles of \(\mathrm{KMnO}_{4}\). We get 0.0200 M * 0.040 L = 0.0008 moles of \(\mathrm{KMnO}_{4}\). As in the first step, 1 mol of \(\mathrm{KMnO}_{4}\) reacts with 5 moles of \(\mathrm{Fe}^{2+}\), therefore, moles of \(\mathrm{Fe}^{2+}\) = 5 * moles of \(\mathrm{KMnO}_{4}\) = 5 * 0.0008 mol = 0.004 mol.
04

Calculation of Concentration of \(\mathrm{Fe}^{3+}\)

The moles of \(\mathrm{Fe}^{2+}\) formed in the second reaction are equal to the moles of \(\mathrm{Fe}^{3+}\) present in the original solution since all \(\mathrm{Fe}^{3+}\) ions were converted to \(\mathrm{Fe}^{2+}\). Thus, to get the concentration of \(\mathrm{Fe}^{3+}\), divide the moles of \(\mathrm{Fe}^{3+}\) by the original solution volume. Concentration = 0.004 mol / 0.025 L = 0.16 M.

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