Question

\(.500 \mathrm{ml}\) of \(0.150 \mathrm{M} \mathrm{AgNO}_{3}\) solution is mixed with 500 \(\mathrm{ml}\) of \(1.09 \mathrm{M} \mathrm{Fe}^{2+}\) solution and the reaction is allowed to reach equilibrium at \(25^{\circ} \mathrm{C}\). \(\mathrm{Ag}^{+}(\mathrm{aq})+\mathrm{Fe}^{2+}(\mathrm{aq})=\mathrm{Fe}^{3+}(\mathrm{aq})+\mathrm{Ag}(\mathrm{s})\) For \(25 \mathrm{ml}\) of the equilibrium solution, \(30 \mathrm{ml}\) of \(0.0833\) \(\mathrm{M} \mathrm{KMnO}_{4}\) were required for oxidation. Calculate the approximate equilibrium constant for the reaction at \(25^{\circ} \mathrm{C}\).

Short answer

The equilibrium constant \( K \) is approximately 147.5.

Step-by-step solution

Understand the Reaction Scheme

The given reaction is \( \mathrm{Ag}^{+} + \mathrm{Fe}^{2+} \rightarrow \mathrm{Fe}^{3+} + \mathrm{Ag} \). This is a redox reaction where Ag+ is reduced and Fe2+ is oxidized.

Calculate Initial Concentrations After Mixing

The initial concentration of \( \mathrm{Ag}^{+} \) is: \[ \text{Ag}^+ = \frac{0.150 \, \text{M} \times 0.500 \text{ L}}{1.000 \text{ L}} = 0.075 \, \text{M} \] The initial concentration of \( \mathrm{Fe}^{2+} \) is: \[ \text{Fe}^{2+} = \frac{1.09 \, \text{M} \times 0.500 \text{ L}}{1.000 \text{ L}} = 0.545 \, \text{M} \] These concentrations assume complete mixing of the solutions.

Calculate the Amount of \( \mathrm{MnO}_4^- \) Used

In a titration, we used \( 30 \, \text{ml} \) of \( 0.0833 \, \text{M} \) \( \mathrm{KMnO}_4 \), hence: \[ n_{\mathrm{MnO}_4^-} = 0.030 \, \text{L} \times 0.0833 \, \text{M} = 0.002499 \text{ moles} \]

Relate \( \mathrm{MnO}_4^- \) to \( \mathrm{Fe}^{2+} \) Reaction

\( \mathrm{MnO}_4^- \) reacts with \( \mathrm{Fe}^{2+} \) at a ratio of 1:5 per the reaction \( \mathrm{MnO}_4^- + 5 \mathrm{Fe}^{2+} \rightarrow \mathrm{Mn}^{2+} + 5 \mathrm{Fe}^{3+} \). Using stoichiometry, the moles of \( \mathrm{Fe}^{2+} \) that reacted is: \[ 0.002499 \text{ mol} \times 5 = 0.012495 \text{ mol} \]

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

This moles of \( \mathrm{Fe}^{2+} \) are in 25 ml, therefore, their concentration decrease is: \[ \Delta \text{Fe}^{2+} = \frac{0.012495}{0.025} = 0.4998 \text{ M} \]

Calculate Equilibrium Concentrations

Prom initial \( 0.545 \text{ M} \), subtracting reacted \( \text{Fe}^{2+} \) gives: \( [\mathrm{Fe}^{2+}]_{eq} = 0.545 - 0.4998 = 0.0452 \text{ M} \). Adding it's equal to \( [\mathrm{Fe}^{3+}]_{eq} \). The \( \mathrm{Ag}^+ \) concentration is approximately unchanged \( 0.075 \text{ M} \) as reaction extent was major on \( \mathrm{Fe}^{2+} \).

Calculate the Equilibrium Constant

The equilibrium constant \( K \) is \( \frac{[\mathrm{Fe}^{3+}]}{[\mathrm{Fe}^{2+}][\mathrm{Ag}^{+}]} \). Thus: \[ K = \frac{0.4998}{(0.0452)(0.075)} \approx 147.5 \]

Key concepts

Redox Reaction

A redox reaction involves the transfer of electrons between two species. One substance undergoes oxidation, which means it loses electrons, while the other undergoes reduction, gaining electrons. In this exercise, the redox reaction entails

• Silver ions (Ag^+) being reduced to metallic silver (Ag), and
• Ferrous ions (Fe^{2+}) being oxidized to ferric ions (Fe^{3+}).

To identify the redox nature, locate the changes in oxidation states. This process helps in understanding which species gain or lose electrons and is crucial for balancing charge in chemical equations. A simple way to memorize is: \( \text{LEO} \text{says} \text{GER} \) — Loss of Electrons is Oxidation, Gain of Electrons is Reduction.
This understanding is crucial for performing accurate stoichiometry and calculating equilibrium constants.

Stoichiometry

Stoichiometry is the quantitative relationship between reactants and products in a chemical reaction. In this exercise, stoichiometry helps relate the different substances' amounts, ensuring the balanced reaction reflects actual change.
Understanding the stoichiometric coefficients from balanced chemical equations allows for the precise calculation of how much product will form or how much reactant is needed. For example, in this case:
The reaction of MnO_4^- with Fe^{2+} uses a 1:5 stoichiometric ratio. This means one mole of permanganate ion reacts with five moles of ferrous ions. By using stoichiometry, you can determine the Fe^{2+} consumed when given the MnO_4^- amount used. This step is crucial for calculating how much Fe^{3+} forms and understanding its contribution to reaching equilibrium.

Concentration Calculations

Calculating concentrations accurately is essential for understanding the changes in a reaction. In this example, the initial concentration of species is calculated assuming complete mixing of the solutions:

• The concentration of Ag^+ after mixing is determined by dividing its total moles by the total volume of the solution.
• The same method is applied for Fe^{2+}.

The change in concentration (9 ext{C}) occurs when the reaction proceeds and depends on the amount of reactant consumed. For the Fe^{2+} to Fe^{3+} conversion:
The concentration decrease stems from the titration data and stoichiometric ratio, linking directly to the equilibrium calculation. Accurate concentration calculations involve understanding molarity, volume, and the chemical changes through the reaction pathway.

Titration Analysis

Titration is an analytical technique to determine the concentration of an unknown solution. By using a solution of known concentration (titrant), the unknown solution's concentration is calculated based on the stoichiometrical relationship. In the presented problem:

• KMnO_4 is a titrant that oxidizes Fe^{2+}, allowing the calculation of oxidized Fe^{3+} concentration.
• The volume and concentration of the titrant give the moles of MnO_4^- used.

By applying stoichiometry (1:5 ratio with Fe^{2+}), you can calculate the moles of Fe^{2+} that reacted. Understanding these results contributes to

• calculating the equilibrium constant (K), and
• providing insight into the reaction's efficiency.

Overall, titration is a practical method for assessing reaction progress and verifying stoichiometry-based calculations.