Chemistry Problem Solving: Oxalic Acid Solution and Potassium Permanganate Reaction
What is the volume of a 0.122 M KMnO4 solution needed to react with a 15.0-mL sample of an oxalic acid solution?
Given data: 15.0-mL sample of oxalic acid solution requires 25.2 mL of 0.149 M NaOH for neutralization.
Calculation and Explanation:
In this problem, we are dealing with the reaction between oxalic acid solution and potassium permanganate solution. The volume of a 0.122 M KMnO4 solution needed to react with a 15.0-mL sample of oxalic acid solution can be determined using the equation:
M₁V₁ = M₂V₂
Where:
M₁ = Molarity of oxalic acid solution
V₁ = Volume of oxalic acid solution (15.0 mL)
M₂ = Molarity of NaOH solution (0.149 M)
V₂ = Volume of NaOH solution (25.2 mL)
First, we need to calculate the molarity of the oxalic acid solution. By using the formula, we find:
M₁ = (M₂V₂) / V₁ = (0.149 * 25.2) / 15 = 0.2503 M
Next, we determine the molarity of KMnO4 required for the reaction. From the balanced net ionic equation, we know that 2 moles of KMnO4 react with 5 moles of oxalic acid, so:
Molarity of KMnO4 = (5/2) * 0.122 = 0.305 M
Finally, we use the formula M₂V₂ = M₃V₃ to find the volume of KMnO4 needed:
V₃ = (M₂V₂) / M₃ = (0.2503 * 15) / 0.305 = 12.3 mL
Further Explanation:
In this chemical reaction, potassium permanganate (KMnO4) acts as an oxidizing agent, while oxalic acid serves as a reducing agent. The reaction proceeds as follows:
8H⁺ + MnO4⁻ + 5e⁻ → Mn²⁺ + 4H2O
C2O4²⁻ → 2CO2 + 2e⁻
16H⁺ + 2MnO4⁻ + 5C2O4²⁻ → 10CO2 + 2Mn²⁺ + 8H2O
This reaction showcases the transfer of electrons between the two substances, leading to the formation of products. By understanding the stoichiometry of the reaction and utilizing the principles of molarity, we can accurately calculate the volume of KMnO4 required for the given scenario.
It is essential to grasp the concept of molarity and its application in reaction stoichiometry to solve such chemistry problems effectively. By mastering these fundamental principles, one can tackle various chemical equations and calculations with confidence and precision.
For additional information on molarity and related topics, you can explore educational resources and references to deepen your understanding of chemical reactions and solution preparations.