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 "Na"_2"C"_2"O"_4Na2C2O4 in 30 mL of water and 15 mL of 3 mol/L "H"_2"SO"_4H2SO4. She titrated this solution with a "KMnO"_4KMnO4 solution. What was the molarity of the "KMnO"_4KMnO4 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
"MnO"_4^(-) + "C"_2"O"_4^(2-) → "Mn"^(2+) + "CO"_2MnO−4+C2O2−4→Mn2++CO2
Then you would probably balance this by the ion electron method to get the balanced net ionic equation
"2MnO"_4^(-) + "5C"_2"O"_4^(2-) + "16H"^+ → "2Mn"^(2+) + "10CO"_2 + "8H"_2"O"2MnO−4+5C2O2−4+16H+→2Mn2++10CO2+8H2O
and perhaps put the spectator ions back in to get the molecular equation
"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" 2KMnO4+5Na2C2O4+8H2SO4→2MnSO4+10CO2+K2SO4+5Na2SO4+8H2O
From here on, the calculation is the same as for an acid-base titration.
Step 1. Calculate the moles of titrant.
"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.
"Molarity" = (9.2567 × 10^-4"mol")/"0.024 90 L" = "0.037 17 mol/L"
