
home	test	drill	game	contact

Back Titration (Indirect Titration) Calculations

Key Concepts

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
direct titration 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 Substance

A 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 25.00mL of the cloudy ammonia solution into a 250.0mL conical flask.
50.00mL of 0.100M HCl_(aq) was immediately added to the conical flask which reacted with the ammonia in solution.
The excess (unreacted) HCl was then titrated with 0.050M Na₂CO_3(aq).
21.50mL of Na₂CO_3(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) + Na₂CO_3(aq) → 2NaCl_(aq) + CO_2(g) + H₂O_(l)

acid + carbonate → salt + carbon
dioxide + water
Calculate the moles, n, of Na₂CO_3(aq) that reacted in the titration:
n = M x V
M = 0.050 molL^-1
V = 21.50mL = 21.50 x 10^-3L
n(Na₂CO_3(aq)) = 0.050 x 21.50 x 10^-3 = 1.075 x 10^-3 mol
Use 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(HCl_titrated) = 2 x 1.075 x 10^-3 = 2.150 x 10^-3 mol
The 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(HCl_{total added}) = M x V
M = 0.100 molL^-1
V = 50.00mL = 50.00 x 10^-3L
n(HCl_{total added}) = 0.100 x 50.00 x 10^-3 = 5.00 x 10^-3 mol
Calculate the moles of HCl that reacted with the ammonia in the diluted cloudy ammonia solution
n(HCl_titrated) + n(HCl_{reacted with ammonia}) = n(HCl_{total added})
n(HCl_{total added}) = 5.00 x 10^-3 mol
n(HCl_titrated) = 2.150 x 10^-3 mol
2.150 x 10^-3 + n(HCl_{reacted with ammonia}) = 5.00 x 10^-3
n(HCl_{reacted with ammonia}) = 5.00 x 10^-3 - 2.150 x 10^-3 = 2.85 x 10^-3 mol
Write the balanced chemical equation for the reaction between ammonia in the cloudy ammonia solution and the HCl_(aq).
NH_3(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.
M = n ÷ V
n = 2.85 x 10^-3 mol (moles of NH₃ that reacted with HCl)
V = 25.00mL = 25.00 x 10^-3L (volume of ammonia solution that reacted with HCl)
M = 2.85 x 10^-3 ÷ 25.00 x 10^-3 = 0.114 M
The concentration of ammonia in the cloudy ammonia solution was 0.114M

Example : Back (Indirect) Titration to Determine the Amount of an Insoluble Salt

A student was asked to determine the mass, in grams, of calcium carbonate present in a 0.125g sample of chalk.
The student placed the chalk sample in a 250mL conical flask and added 50.00mL 0.200M HCl using a pipette.
The excess HCl was then titrated with 0.250M NaOH.
The average NaOH titre was 32.12mL
Calculate the mass of calcium carbonate, in grams, present in the chalk sample.

Step 1: Determine the amount of HCl in excess from the titration results

Write the equation for the titration:
HCl_(aq) + NaOH_(aq) → NaCl_(aq) + H₂O_(l)

acid + base → salt + water
Calculate the moles, n, of NaOH_(aq) that reacted in the titration:
n = M x V
M = 0.250 molL^-1
V = 32.12mL = 32.12 x 10^-3L
n(NaOH_(aq)) = 0.250 x 32.12 x 10^-3 = 8.03 x 10^-3 mol
Use 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 NaOH reacts with 1 mole of HCl
So, 8.03 x 10^-3 mole NaOH reacted with 8.03 x 10^-3 moles HCl
The amount of HCl that was added to the chalk in excess was
8.03 x 10^-3 mol

Step 2: Determine the amount of calcium carbonate in chalk

Calculate the total moles of HCl originally added to the chalk:
n(HCl_{total added}) = M x V
M = 0.200 molL^-1
V = 50.00mL = 50.00 x 10^-3L
n(HCl_{total added}) = 0.200 x 50.00 x 10^-3 = 0.010 mol
Calculate the moles of HCl that reacted with the calcium carbonate in the chalk
n(HCl_titrated) + n(HCl_{reacted with calcium carbonate}) = n(HCl_{total added})
n(HCl_{total added}) = 0.010 mol
n(HCl_titrated) = 8.03 x 10^-3 mol
8.03 x 10^-3 + n(HCl_{reacted with calcium carbonate}) = 0.010
n(HCl_{reacted with calcium carbonate}) = 0.010 - 8.03 x 10^-3 = 1.97 x 10^-3 mol
Write the balanced chemical equation for the reaction between calcium carbonate in the chalk and the HCl_(aq).
CaCO₃_(s) + 2HCl_(aq) → CaCl_2(aq) + CO_2(g) + H₂O_(l)
From the balanced chemical equation, calculate the moles of CaCO₃ that reacted with HCl.
From the equation, 1 mol CaCO₃ reacts with 2 mol HCl so, 1 mol HCl reacts with ½ mol CaCO₃
So, 1.97 x 10^-3 mol HCl had reacted with ½ x 1.97 x 10^-3 = 9.85 x 10^-4 mol CaCO₃ in the chalk.
Calculate the mass of calcium carbonate in the chalk.
n = mass ÷ MM
n = 9.85 x 10^-4 mol (moles of CaCO₃ that reacted with HCl)
MM(CaCO₃) = 40.08 + 12.01 + (3 x 16.00) = 100.09 g/mol
mass = n x MM = 9.85 x 10^-4 x 100.09 = 0.099g
The mass of calcium carbonate in the chalk was 0.099g