A back titration, or indirect titration, is generally a two-stage analytical technique:Reactant A of unknown concentration is reacted with excess reactant B of known concentration.A titration is then performed to determine the amount of reactant B in excess.

Back titrations are used when:

one of the reactants is volatile, for example ammonia.
an acid or a base is an insoluble salt, for example calcium carbonate
a particular reaction is too slow
directtitration would involve a weak acid - weak base titration
(the end-point of this type of direct titration is very difficult to observe)
Example : Back (Indirect) Titration to Determine the Concentration of a Volatile SubstanceA student was asked to determine the concentration of ammonia, a volatile substance, in a commercially available cloudy ammonia solution used for cleaning.
First the student pipetted 25.00 mL of the cloudy ammonia solution into a 250.0 mL conical flask.
50.00 mL of 0.100 mol L-1 HCl(aq) was immediately added to the conical flask which reacted with the ammonia in solution.
The excess (unreacted) HCl was then titrated with 0.050 mol L-1 Na2CO3(aq).
21.50 mL of Na2CO3(aq) was required.
Calculate the concentration of the ammonia in the cloudy ammonia solution.Step 1: Determine the amount of HCl in excess from the titration results

Write the equation for the titration:


2HCl(aq) + Na2CO3(aq) → 2NaCl(aq) + CO2(g) + H2O(l)
acid + carbonate → salt + carbon
dioxide + water

Calculate the moles, n, of Na2CO3(aq) that reacted in the titration:
n = c x V
c = concentration (molarity) = 0.050 mol L-1
V = volume = 21.50 mL = 21.50 x 10-3 L
n(Na2CO3(aq)) = moles of Na2CO3(aq)= 0.050 x 21.50 x 10-3 = 1.075 x 10-3 molUse the balanced chemical reaction for the titration to determine the moles of HCl that reacted in the titration.
From the balanced chemical equation, 1 mole Na2CO3 react with 2 moles of HCl
So, 1.075 x 10-3 mole Na2CO3 reacted with 2 x 1.075 x 10-3 moles HCl
n(HCltitrated) = 2 x 1.075 x 10-3 = 2.150 x 10-3 molThe amount of HCl that was added to the cloudy ammonia solution in excess was
2.150 x 10-3 mol

Step 2: Determine the amount of ammonia in the cloudy ammonia solution

Calculate the total moles of HCl originally added to the diluted cloudy ammonia solution:
n(HCltotal added) = c x V
c = concentration (molarity) = 0.100 mol L-1
V = volume = 50.00 mL = 50.00 x 10-3 L
n(HCltotal added) = 0.100 x 50.00 x 10-3 = 5.00 x 10-3 molCalculate the moles of HCl that reacted with the ammonia in the diluted cloudy ammonia solution
n(HCltitrated) + n(HClreacted with ammonia) = n(HCltotal added)
n(HCltotal added) = 5.00 x 10-3 mol
n(HCltitrated) = 2.150 x 10-3 mol
2.150 x 10-3 + n(HClreacted with ammonia) = 5.00 x 10-3
n(HClreacted with ammonia) = 5.00 x 10-3 - 2.150 x 10-3 = 2.85 x 10-3 molWrite the balanced chemical equation for the reaction between ammonia in the cloudy ammonia solution and the HCl(aq).
NH3(aq) + HCl(aq) → NH4Cl(aq)From the balanced chemical equation, calculate the moles of NH3 that reacted with HCl.
From the equation, 1 mol HCl reacts with 1 mol NH3
So, 2.85 x 10-3 mol HCl had reacted with 2.85 x 10-3 mol NH3 in the cloudy ammonia solution.Calculate the ammonia concentration in the cloudy ammonia solution.
c = n ÷ V
n = moles (NH3) = 2.85 x 10-3 mol (moles of NH3 that reacted with HCl)
V = volume (NH3(aq)) = 25.00 mL = 25.00 x 10-3 L (volume of ammonia solution that reacted with HCl)
c = concentration (molarity) = 2.85 x 10-3 ÷ 25.00 x 10-3 = 0.114 mol L-1The concentration of ammonia in the cloudy ammonia solution was 0.114 mol L-1