Add the oxidation and reduction half equations together to give a balanced redox reaction equation.

Extract all the relevant information from the question.

Check the data for consistency, for example, concentrations are often given in M or mol L^{-1} but volumes are often given in mL. You will need to convert the mL to L for consistency. The easiest way to do this is to multiply the volume in mL x 10^{-3} (enter the volume in mL into your calculator, then click "EXP", then enter -3 and the conversion is done!).

Calculate the moles of reactant (titrant) (n) for which you have both volume (V) and concentration (molarity, c): n = c x V

From the redox reaction equation find the mole ratio of known reactant (titrant) : unkown reactant (analyte)

Calculate moles of unknown reactant (analyte) using this mole ratio.

From the volume (V) of the unknown reactant (analyte) and its calculated moles (n), calculate its concentration ( molarity, c): c = n ÷ V

Example

A 0.0484 mol L^{-1} standard solution of potassium permanganate was titrated against 25.00 mL of an iron(II) sulfate solution.

The equivalence point, as indicated by a faint pink colour, was reached when 15.50 mL of potassium permanganate solution had been added.

Calculate the concentration of the iron(II) sulfate solution.

reduction half equation:

MnO_{4}^{-} + 8H^{+} + 5e → Mn^{2+} + 4H_{2}O

oxidation half equation:

Fe^{2+} → Fe^{3+} + e

redox equation:

MnO_{4}^{-} + 8H^{+} + 5e

→

Mn^{2+} + 4H_{2}O

5Fe^{2+}

→

5Fe^{3+} + 5e

MnO_{4}^{-} + 8H^{+} + 5Fe^{2+}

→

Mn^{2+} + 4H_{2}O + 5Fe^{3+}

Extract all the relevant information:

MnO_{4}^{-}

Fe^{2+}

[MnO_{4}^{-}] = 0.0484 mol/L

[Fe^{2+}] = ? mol/L

V (MnO_{4}^{-}) = 15.50 mL

V (Fe^{2+}) = 25.00 mL

Check the data for consistency: convert volumes in mL to L

MnO_{4}^{-}

Fe^{2+}

[MnO_{4}^{-}] = 0.0484 mol/L

[Fe^{2+}] = ? mol/L

V (MnO_{4}^{-}) = 15.50 x 10^{-3} L

V (Fe^{2+}) = 25.00 x 10^{-3} L

Calculate moles of known reactant (titrant):

MnO_{4}^{-}

[MnO_{4}^{-}] = 0.0484 mol/L

V (MnO_{4}^{-}) = 15.50 x 10^{-3} L

n (MnO_{4}^{-}) = c x V = 0.0484 x 15.50 x 10^{-3} = 7.502 x 10^{-4} mol

Find the mole ratio from balanced redox equation
MnO_{4}^{-} : Fe^{2+} 1 : 5

Calculate moles (n) of Fe^{2+} n (Fe^{2+}) = 5 x n (MnO_{4}^{-}) = 5 x 7.502 x 10^{-4} = 0.003751 mol

Calculate the concentration of Fe^{2+} [Fe^{2+}] = n ÷ V = 0.003751 ÷ 25.00 x 10^{-3} = 0.1500 mol/L