You titrated a 25.00 ml solution of 0.02 m oxalic acid with a freshly prepared solution of kmno4. if it took 42.32 ml of this solution to reach the endpoint, what was the molarity of the kmno4?

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

wher is the rest of the question