Question

Balance the following reaction in acidic conditions. If a 74.8 mL sample of a solution of H2C2O4 (aq) requires 27.5 mL of a 0.037 M KMnO4 solution for a complete reaction, what is the concentration of H2C2O4 in the original solution?

MnO4- (aq) + H2C2O4(aq) -> Mn2+(aq) + CO2(g)

Answer 1

Write the unbalanced equation first:

(MnO₄)^-(aq) + H₂C₂O₄ Mn²⁺(aq) + CO₂(g)

Now, write the half-reactions and balanced them based on redox rules:

H₂C₂O₄ 2CO₂(g) + 2 H⁺(aq) + 2 e^- -----------------------------------------------------(1)

(MnO₄)^-(aq) + 8 H⁺(aq) + 5 e^- Mn²⁺(aq) + 4 H₂O(l) ----------------------(2)

Multiplying equation (1) by 5 and equation (2) by 2, we get;

5H₂C₂O₄ 10CO₂(g) + 10 H⁺(aq) + 10 e^- -----------------------------------------------------------(3)

2(MnO₄)^-(aq) +   16 H⁺(aq) + 10 e^- 2Mn²⁺(aq) + 8 H₂O(l) ----------------------------(4)

Adding equation (3) and (4), we get;

2(MnO₄)^-(aq)    + 10 e^- + 5H₂C₂O₄ + 16 H⁺(aq) 2Mn²⁺(aq) + 8 H₂O(l) + 10CO₂(g) + 10 H⁺(aq) + 10 e^-

2(MnO₄)^-(aq) + 5H₂C₂O₄ + 6 H⁺(aq)    2Mn²⁺(aq) + 10CO₂(g) + 8 H₂O(l)

Molarity = Moles / Liter

Moles = Molarity x Liter

27.5 mL of a 0.037 M KMnO4 solution for a complete reaction

Moles of KMnO4 = 0.037 M x 0.0275 L = 0.0010175 moles

Now, the balanced equation is

2(MnO₄)^-(aq) + 5H₂C₂O₄ + 6 H⁺(aq)    2Mn²⁺(aq)   + 10CO₂(g) + 8 H₂O(l)

In the above reaction equation

2 moles of KMnO₄ reacts with 5 moles of H₂C₂O₄.

1 mole of KMnO₄ reacts with 5/2 moles of H₂C₂O₄.

0.0010175 moles of KMnO₄ reacts with (5/2) (0.0010175) moles of H₂C₂O₄.

0.0010175 moles of KMnO₄ reacts with 0.00254375 moles of H₂C₂O₄.

Moles of H₂C₂O₄ = 0.00254375 moles

Volume of  H₂C₂O₄ = 74.8 mL = 0.0748 L

[H₂C₂O₄] = 0.00254375 moles / 0.0748 L = 0.034 M