The concentration of oxalic acid is ' x ' mol L ⁻¹ . 40 ~mL of this solution reacts with 16 ~mL of 0.05 M acidified KMnO ₄ . What is the pH of ' x ' M oxalic acid solution ? (Assume that oxalic acid dissociates completely)

Step-by-step solution

Oxalic acid =x ~mol / L Oxalic acid KMnO ₄M₁ V₁=M₂ V₂40 ~mL x=16 ~mL 0.05x= 16 0.05 40 = 1 50 x= 1 50 M Now convert molarity into normality N eq. wt. =M mol. wt. of oxalic acid N 45= 1 50 90N= 1 25 This normality represents the hydrogen ion concentration. So, [ H ⁺ ]= 1 25 pH = 1 [ H ⁺ ] = 25=1.3