Answer:
This is a Redox Titration reaction and the overall ionic equation is given as follows;
2 MnO⁴⁻ + 5C₂O₄
²⁻+ 16H+ → 2Mn²⁺ + 10CO₂ + 8H₂O
From here, we could see that the stoichiometry ratio of KMnO₄ and oxalic acid = 2:5
To calculate the molarity of KMnO₄, we will use the following equation.
x₁M₁V₁= x₂M₂V₂ (1)
Where,
x₁ = 2, stoichiometry number of KMnO₄ in the balanced reaction
x₂ = 5, stoichiometry number of C₂O₄ in the balanced reaction
M₁ and M₂ are the molarities of oxalic acid and KMnO₄ solutions
V1 and V2 are the volumes of oxalic acid and KMnO₄ solutions.
Making M₂, the subject of the formula and substituting the given values, we have,
2M₁V₁ = 5M₂V₂, M₂=2M₁V₁ / 5V₂
= = 2x0.02Mx25mL/
5x42.32mL
M₂=0.0047M
Hence, the molarity of the KMnO4 at endpoint =0.0047M
