A quantity of \(25.0 \mathrm{~mL}\) of a solution containing both \(\mathrm{Fe}^{2+}\) and \(\mathrm{Fe}^{3+}\) ions is titrated with \(23.0 \mathrm{~mL}\) of \(0.0200 M \mathrm{KMnO}_{4}\) (in dilute sulfuric acid). As a result, all of the \(\mathrm{Fe}^{2+}\) ions are oxidized to \(\mathrm{Fe}^{3+}\) ions. Next, the solution is treated with Zn metal to convert all of the \(\mathrm{Fe}^{3+}\) ions to \(\mathrm{Fe}^{2+}\) ions. Finally, the solution containing only the \(\mathrm{Fe}^{2+}\) ions requires \(40.0 \mathrm{~mL}\) of the same \(\mathrm{KMnO}_{4}\) solution for oxidation to \(\mathrm{Fe}^{3+}\). Calculate the molar concentrations of \(\mathrm{Fe}^{2+}\) and \(\mathrm{Fe}^{3+}\) in the original solution. The net ionic equation is \(\mathrm{MnO}_{4}^{-}+5 \mathrm{Fe}^{2+}+8 \mathrm{H}^{+} \longrightarrow\) \(\mathrm{Mn}^{2+}+5 \mathrm{Fe}^{3+}+4 \mathrm{H}_{2} \mathrm{O}\)

Step by step solution

01

Calculate the moles of \(KMnO_{4}\) in each titration.

To find the amount of \(KMnO_{4}\) in each titration, use the formula moles = Molarity x Volume. For the first titration, moles = \(23.0 mL * 0.0200 M = 0.00046 moles\) of \(KMnO_{4}\). For the second titration, moles = \(40.0 mL * 0.0200 M = 0.0008 moles\) of \(KMnO_{4}\).

02

Calculate the moles of \(Fe^{2+}\) and \(Fe^{3+}\) in the original solution.

Use the stoichiometric ratio between \(MnO_{4}^{-}\) and \(Fe^{2+}\) in the balanced equation to calculate the moles of \(Fe^{2+}\) and \(Fe^{3+}\). There is a 1:5 ratio between \(MnO_{4}^{-}\) and \(Fe^{2+}\), so for each mole of \(KMnO_{4}\) used in the titration, 5 moles of \(Fe^{2+}\) were present in the solution. Thus, in the first titration, \(5 * 0.00046 moles = 0.0023 moles\) of \(Fe^{2+}\). The second titration gives the total moles of \(Fe\) in the original solution: \(5 * 0.0008 moles = 0.004 moles\). Therefore, through subtraction, the amount of \(Fe^{3+}\) in the original solution is \(0.004 moles - 0.0023 moles = 0.0017 moles\).

03

Find the molar concentrations in the original solution.

The molar concentrations of \(Fe^{2+}\) and \(Fe^{3+}\) can be calculated by dividing the number of moles by the original volume in liters (0.025 L). Thus, for \(Fe^{2+}\), the concentration is \(0.0023 moles/0.025 L = 0.092 M\). For \(Fe^{3+}\), the concentration is \(0.0017 moles/0.025 L = 0.068 M\).

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