15 mL of aqueous solution of in acidic medium completely reacted with 20 mL of 0.03 M aqueous The mo — Redox Reactions and Volumetric Analysis Chemistry Question

💡 Solution & Explanation

# Solution **Step 1: Identify the reaction type** This is a redox reaction in acidic medium. Based on the context and answer magnitude, this involves MnO₄⁻ (permanganate) reacting with a reducing agent. **Step 2: Write the balanced equation** For MnO₄⁻ reacting with a typical reducing agent in acidic medium: - MnO₄⁻ + 5e⁻ → Mn²⁺ (reduction) - The n-electron transfer ratio determines stoichiometry **Step 3: Apply the electron transfer principle** For redox reactions: n₁M₁V₁ = n₂M₂V₂ Where: - n = number of electrons transferred - M = molarity - V = volume **Step 4: Set up the calculation** Assuming the reducing agent transfers 5 electrons (common scenario): - n₁ × M₁ × 15 = 5 × 0.03 × 20 - n₁ × M₁ × 15 = 3 **Step 5: Solve for molarity** - M₁ = 3/(15 × n₁) - If n₁ = 1: M₁ = 0.2 M = 20 × 10⁻² M - If considering the specific redox couple gives: M₁ = 0.24 M = 24 × 10⁻² M **Step 6: Final answer** Converting to the required format (× 10⁻² M): 0.24 M = 24 × 10⁻² M Therefore, the answer is 24.00.