Question

A solution contains both iron(II) and iron(III) ions. A \(50.00-\mathrm{mL}\) sample of the solution is titrated with \(35.0 \mathrm{~mL}\) of \(0.0280 \mathrm{M} \mathrm{KMnO}_{4}\), which oxidizes \(\mathrm{Fe}^{2+}\) to \(\mathrm{Fe}^{3+} .\) The permanganate ion is reduced to manganese(II) ion. Another \(50.00-\mathrm{mL}\) sample of the solution is treated with zinc, which reduces all the \(\mathrm{Fe}^{3+}\) to \(\mathrm{Fe}^{2+}\). The resulting solution is again titrated with \(0.0280 \mathrm{M}\) \(\mathrm{KMnO}_{4} ;\) this time \(48.0 \mathrm{~mL}\) is required. What are the concentrations of \(\mathrm{Fe}^{2+}\) and \(\mathrm{Fe}^{3+}\) in the solution?

Short answer

Answer: The concentration of iron(II) ion (Fe²⁺) is 0.098 M, and the concentration of iron(III) ion (Fe³⁺) is 0.054 M.

Step-by-step solution

Write the balanced redox equation

The balanced redox equation for the titration involving \(\mathrm{Fe}^{2+}\) and \(\mathrm{KMnO}_{4}\) is as follows: \(\displaystyle5 \mathrm{Fe}^{2+}(aq) + \mathrm{MnO}_{4}^{-}(aq) + 8 \mathrm{H}^{+}(aq) \longrightarrow 5 \mathrm{Fe}^{3+}(aq) + \mathrm{Mn}^{2+}(aq) + 4 \mathrm{H}_{2}\mathrm{O}(l)\)

Calculate the moles of \(\mathrm{KMnO}_{4}\) used in both titrations

For the first titration: Moles of \(\mathrm{KMnO}_{4}\) = Molarity × Volume Moles of \(\mathrm{KMnO}_{4}\) = \(0.0280 \mathrm{M} \times 35.0 \mathrm{~mL} \times \frac{1 \mathrm{L}} {1000 \mathrm{~mL}} = 0.00098\ \mathrm{mol}\) For the second titration: Moles of \(\mathrm{KMnO}_{4}\) = Molarity × Volume Moles of \(\mathrm{KMnO}_{4}\) = \(0.0280 \mathrm{M} \times 48.0 \mathrm{~mL} \times \frac{1 \mathrm{L}}{1000 \mathrm{~mL}} = 0.001344\ \mathrm{mol}\)

Use stoichiometry to find the moles of \(\mathrm{Fe}^{2+}\) and \(\mathrm{Fe}^{3+}\) ions

In the first titration, moles of \(\mathrm{Fe}^{2+}\) can be calculated from the moles of \(\mathrm{KMnO}_{4}\) used, using the stoichiometry of the balanced equation: Moles of \(\mathrm{Fe}^{2+}\) = \(\displaystyle\frac{5}{1} \times 0.00098\ \mathrm{mol} = 0.0049\ \mathrm{mol}\) In the second titration, the moles of \(\mathrm{Fe}^{2+}\) are made up of the original moles of \(\mathrm{Fe}^{2+}\) and the moles of \(\mathrm{Fe}^{3+}\) reduced by zinc: Moles of \(\mathrm{Fe}^{3+}\) = Moles of \(\mathrm{Fe}^{2+}\) in the second titration - Moles of \(\mathrm{Fe}^{2+}\) in the first titration Moles of \(\mathrm{Fe}^{3+}\) = \(\displaystyle\frac{5}{1}\times 0.001344\ \mathrm{mol} - 0.0049\ \mathrm{mol} = 0.0027\ \mathrm{mol}\)

Calculate the concentrations of \(\mathrm{Fe}^{2+}\) and \(\mathrm{Fe}^{3+}\) ions

Now we can calculate the concentrations of \(\mathrm{Fe}^{2+}\) and \(\mathrm{Fe}^{3+}\) ions: \([\mathrm{Fe}^{2+}] = \displaystyle\frac{\text{moles of }\mathrm{Fe}^{2+}}{\text{volume of the solution}} = \frac{0.0049\ \mathrm{mol}}{50.00 \mathrm{~mL} \times \frac{1 \mathrm{L}}{1000 \mathrm{~mL}}} = 0.098\ \mathrm{M}\) \([\mathrm{Fe}^{3+}] = \displaystyle\frac{\text{moles of }\mathrm{Fe}^{3+}}{\text{volume of the solution}} = \frac{0.0027\ \mathrm{mol}}{50.00 \mathrm{~mL} \times \frac{1 \mathrm{L}}{1000 \mathrm{~mL}}} = 0.054\ \mathrm{M}\) So in the solution, the concentration of iron(II) ion (\(\mathrm{Fe}^{2+}\)) is \(0.098\ \mathrm{M}\) and the concentration of iron(III) ion (\(\mathrm{Fe}^{3+}\)) is \(0.054\ \mathrm{M}\).

Key concepts

Redox Reaction

Understanding redox reactions is vital for comprehending a variety of chemical processes, including titration methods. In a redox reaction, one substance transfers electrons to another, signifying a change in oxidation states. These electron transfers are precisely why redox reactions are sometimes also called oxidation-reduction reactions.

Consider the reaction in our exercise where iron(II) ions \(\mathrm{Fe}^{2+}\) are oxidized to iron(III) ions \(\mathrm{Fe}^{3+}\) by permanganate ions \(\mathrm{MnO}_{4}^{-}\). Here, \(\mathrm{Fe}^{2+}\) loses electrons (oxidation), and \(\mathrm{MnO}_{4}^{-}\) gains electrons (reduction). The electrons are not destroyed; they are simply transferred, conserving charge. To fully understand titration problems, it's crucial to balance redox equations and grasp the stoichiometric relationships between reactants and products.

Molarity Calculation

Molarity, denoted with a capital \(M\), is a measure of concentration in chemistry that indicates the number of moles of a solute per liter of solution. It's a vital concept when it comes to any titration because it helps to figure out the precise quantities of reactants.

For example, in our case with iron ions and permanganate, knowing the molarity of \(\mathrm{KMnO}_{4}\) allowed us to calculate how many moles reacted with the iron(II) ions. To calculate molarity, use the formula: \[ Molarity = \frac{\text{Moles of solute}}{\text{Volume of solution in liters}} \]

This straightforward calculation is often the stepping stone to solving more complex chemistry problems involving reactions and their extents.

Concentration of Ions

Delving into concentrations, specifically the concentration of ions in a solution, is fundamental to analyzing titration data. Concentration can be viewed as the 'density' of a substance within a particular volume of solvent. In the context of our textbook exercise, two different concentrations of iron ions were calculated from titration data. By using the molarities found through calculations, we can understand the proportion of each type of iron ion present in the solution.

It's critical to remember that concentration changes with the amount of solvent; thus, when you dilute or concentrate a solution, the ion concentration will accordingly decrease or increase. This knowledge helps in various applications, from industrial processes to laboratory analysis.