Question

In setting up your 3 kinetics experiment, you will be mixing different volumes the two reactants (oxalic acid (H2C2O4) and potassium permanganate (KMnO4)). If you mix 10 ml of 0.02 M with 20 ml of 0.5 M oxalic acid, what is the concentration of H2C2O4 in the mixture? Hint: Use M1V1 = M2 V2.

Answer 1

First calculate initial moles of both reactants .

Moles of KMnO ₄ = Molarity Volume of solution in L = 0.02 mol / L 0.01 L = 2 10 ^-04 mol

Moles of H₂C₂O₄ = Molarity Volume of solution in L = 0.5 mol / L 0.02 L = 0.01 mol

From above values , it is clear that KMnO ₄ is limiting reactant. H₂C₂O₄ is excess reactant.

Consider a reaction, 2 MnO ₄^- + 5 C₂O₄^2- + 16 H ⁺ 2 Mn ²⁺ + 10 CO ₂ + 8 H₂O

From reaction, Stoichiometric ratio = 2 mol MnO ₄^- / 5 mol C₂O₄^2- = 2/5

We have correlation, No. of moles of MnO ₄^- = No. of moles of C₂O₄^2- Stoichiometric ratio

No. of moles of C₂O₄^2- consumed in the reaction = No. of moles of MnO ₄^- / Stoichiometric ratio

= 2 10 ^-04 mol / (2/5)

= 5 10 ^-04 mol

No. of moles C₂O₄^2- remained non reacted = initial moles - No. of moles of C₂O₄^2- consumed in the reaction

= 0.01 mol - 5 10 ^-04 mol

= 9.5 10 ^-03 mol

Volume of mixture = Volume of oxalic acid + volume of potassium permangnate

=10 ml + 20 ml = 30 ml

= 0.03 L

[C₂O₄^2- ] = [oxalic acid] = No. of moles / Volume of solution in L

= 9.5 10 ^-03 mol / 0.03 L

= 0.3167 M

ANSWER : Concentration of oxalic acid in the mixture = 0.3167 M