# How can I do redox titration calculations?

Apr 18, 2015

You do them the same way as you do acid-base titration calculations.

The only difference is how you balance the equation for the reaction.

EXAMPLE

A student dissolved 0.3101 g of ${\text{Na"_2"C"_2"O}}_{4}$ in 30 mL of water and 15 mL of 3 mol/L ${\text{H"_2"SO}}_{4}$. She titrated this solution with a ${\text{KMnO}}_{4}$ solution. What was the molarity of the ${\text{KMnO}}_{4}$ if it took 24.90 mL of titrant to reach a permanent pale pink equivalence point?

Solution:

First, you would write the balanced equation for the reaction.

The skeleton ionic equation is

${\text{MnO"_4^(-) + "C"_2"O"_4^(2-) → "Mn"^(2+) + "CO}}_{2}$

Then you would probably balance this by the ion electron method to get the balanced net ionic equation

$\text{2MnO"_4^(-) + "5C"_2"O"_4^(2-) + "16H"^+ → "2Mn"^(2+) + "10CO"_2 + "8H"_2"O}$

and perhaps put the spectator ions back in to get the molecular equation

$\text{2KMnO"_4 + "5Na"_2"C"_2"O"_4 + "8H"_2"SO"_4→ "2MnSO"_4 + "10CO"_2 +"K"_2"SO"_4 + "5Na"_2"SO"_4 + "8H"_2"O}$

From here on, the calculation is the same as for an acid-base titration.

Step 1. Calculate the moles of titrant.

${\text{Moles of KMnO"_4 = 0.3101 cancel("g Na₂C₂O₄") × (1 cancel("mol Na₂C₂O₄"))/(134.00 cancel("g Na₂C₂O₄")) × ("2 mol KMnO"_4)/(5 cancel("mol Na₂C₂O₄")) = 9.2567 × 10^-4"mol KMnO}}_{4}$

Step 2. Calculate the molarity of the titrant.

$\text{Molarity" = (9.2567 × 10^-4"mol")/"0.024 90 L" = "0.037 17 mol/L}$