Problem 64 - Chemistry

Question

The oxidation of \(25.0 \mathrm{~mL}\) of a solution containing \(\mathrm{Fe}^{2+}\) requires \(26.0 \mathrm{~mL}\) of \(0.0250 \mathrm{M} \mathrm{K}_{2} \mathrm{Cr}_{2} \mathrm{O}_{7}\) in acidic solution. Balance the following equation, and calculate the molar concentration of \(\mathrm{Fe}^{2+}\) : $$ \mathrm{Cr}_{2} \mathrm{O}_{7}^{2-}+\mathrm{Fe}^{2+}+\mathrm{H}^{+} \longrightarrow \mathrm{Cr}^{3+}+\mathrm{Fe}^{3+} $$

Step-by-Step Solution

Answer

The molar concentration of \(\mathrm{Fe}^{2+}\) is 0.156 M.

Step 1: Determine the balanced chemical equation

To balance the redox reaction, first balance the atoms other than O and H. Then, balance the oxygen atoms using water molecules, and hydrogen atoms using H⁺ ions. Afterward, balance the charges using electrons. The balanced equation is: \[ \mathrm{Cr}_{2} \mathrm{O}_{7}^{2-} + 6\,\mathrm{Fe}^{2+} + 14\,\mathrm{H}^{+} \rightarrow 2\,\mathrm{Cr}^{3+} + 6\,\mathrm{Fe}^{3+} + 7\,\mathrm{H}_{2}\mathrm{O} \]

Step 2: Calculate moles of K₂Cr₂O₇

The volume and molarity of \(\mathrm{K}_{2} \mathrm{Cr}_{2} \mathrm{O}_{7}\) are given: 26.0 mL and 0.0250 M. Convert the volume from mL to L by dividing by 1000, then calculate the moles of \(\mathrm{K}_{2} \mathrm{Cr}_{2} \mathrm{O}_{7}\):\[ \text{Moles of } \mathrm{K}_{2} \mathrm{Cr}_{2} \mathrm{O}_{7} = 0.0250\,\mathrm{M} \times 0.0260\,\mathrm{L} = 6.50 \times 10^{-4} \text{ mol} \, \mathrm{K}_{2} \mathrm{Cr}_{2} \mathrm{O}_{7} \]

Step 3: Relate moles of Cr₂O₇²⁻ to Fe²⁺

According to the balanced equation, 1 mole of \(\mathrm{Cr}_{2} \mathrm{O}_{7}^{2-}\) reacts with 6 moles of \(\mathrm{Fe}^{2+}\). Therefore, the moles of \(\mathrm{Fe}^{2+}\) are: \[ \text{Moles of } \mathrm{Fe}^{2+} = 6 \times (6.50 \times 10^{-4}) = 3.90 \times 10^{-3} \text{ mol} \, \mathrm{Fe}^{2+} \]

Step 4: Calculate the molarity of Fe²⁺

Use the moles of \(\mathrm{Fe}^{2+}\) calculated in Step 3 and the volume of the \(\mathrm{Fe}^{2+}\) solution to find its molarity. Convert the given volume from mL to L:\[ \text{Volume of } \mathrm{Fe}^{2+} = 25.0\,\mathrm{mL} = 0.0250\, \mathrm{L} \]The molarity of \(\mathrm{Fe}^{2+}\) is:\[ \mathrm{Molarity\ of\ } \mathrm{Fe}^{2+} = \frac{3.90 \times 10^{-3} }{0.0250} = 0.156 \text{ M} \]

Key Concepts

Molarity Calculation

Molarity is an essential concept in chemistry that helps us understand the concentration of a solution. It is defined as the number of moles of a solute divided by the volume of the solution in liters. In this exercise, we calculated the molarity of \( \mathrm{Fe}^{2+} \) ions after determining the moles present in the solution.

To calculate molarity, follow these steps:

• Calculate the moles of solute using the relation: \( \text{Moles} = \text{molarity} \times \text{volume (in L)} \).
• Determine the volume of the solution in liters by converting mL to L. Remember: 1 mL = 0.001 L.
• Calculate molarity using \( \text{Molarity} = \frac{\text{moles of solute}}{\text{volume of solution in L}} \).

Using these steps ensures precise calculations in chemistry problems and helps predict how substances react in a solution. For example, in our problem, the molarity of \( \mathrm{Fe}^{2+} \) was found to be 0.156 M.

Balancing Chemical Equations

Balancing chemical equations is crucial for understanding how substances react with each other. It involves making sure that the number of each type of atom on the reactant side equals the number on the product side. For this, one often uses the redox method, especially with redox reactions like ours.

Here are steps to balance oxidation-reduction reactions:

• Separate the equation into two half-reactions: one for oxidation and one for reduction.
• For each half-reaction, balance the atoms other than O and H first.
• Then, balance oxygen atoms by adding water (\( \mathrm{H}_2\mathrm{O} \)).
• Balance hydrogen atoms by adding protons (\( \mathrm{H}^+ \)).
• Finally, balance the electrons by ensuring equal charge on both sides of the half-reactions.

Once each half-reaction is balanced, combine them, ensuring that electrons lost and gained are equal. Our original exercise resulted in a balanced equation by following these steps.

Stoichiometry

Stoichiometry involves calculating the quantities of reactants and products in a chemical reaction. It is pivotal for linking molarity to real-world reactions, as it relies on balancing chemical equations to find the mole ratio of reactants and products.

To apply stoichiometry effectively:

• Use the balanced chemical equation to understand the mole ratio of each reactant and product.
• Convert all given information to moles (e.g., using molarity and volume).
• Apply the mole ratio from the balanced equation to find unknown quantities.

In our exercise, \( \mathrm{Cr}_2\mathrm{O}_7^{2-} \) reacted with \( \mathrm{Fe}^{2+} \) in a 1:6 ratio. Using stoichiometry, we determined the moles of \( \mathrm{Fe}^{2+} \) reacted given the moles of \( \mathrm{Cr}_2\mathrm{O}_7^{2-} \) provided. Understanding stoichiometry is crucial for accurately predicting how much of each substance is needed or produced in a reaction.

Acidic Solution

In chemistry, an acidic solution refers to any solution that has a higher concentration of hydrogen ions (\( \mathrm{H}^+ \)) than pure water, which contributes to its acidic properties. This is often necessary for certain reactions, including many redox processes, as in our exercise.

Key aspects of acidic solutions in reactions:

• The presence of \( \mathrm{H}^+ \) ions can help balance the reaction, especially for oxygen and hydrogen.
• Acidic environments can alter the rate and equilibrium of reactions, making certain reductions or oxidations possible.
• Understanding the role of \( \mathrm{H}^+ \) assists in balancing redox equations and predicting product formation in reactions.

In our redox reaction, the acidic solution facilitated the oxidation of \( \mathrm{Fe}^{2+} \) ions, as shown by the inclusion of \( \mathrm{H}^+ \) in the balanced equation. Recognizing the conditions under which a reaction occurs can be vital for mastering chemistry concepts.