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# Back Titration (Indirect Titration) Calculations

## Key Concepts

A back titration, or indirect titration, is generally a two-stage analytical technique:
1. Reactant A of unknown concentration is reacted with excess reactant B of known concentration.

2. 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 25.00 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

1. Write the equation for the titration:
 2HCl(aq) + Na2CO3(aq) → 2NaCl(aq) + CO2(g) + H2O(l) acid + carbonate → salt + carbondioxide + water

2. 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 mol

3. 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(HCltitrated) = 2 x 1.075 x 10-3 = 2.150 x 10-3 mol

4. 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

1. 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 mol

2. Calculate 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 mol

3. Write the balanced chemical equation for the reaction between ammonia in the cloudy ammonia solution and the HCl(aq).
NH3(aq) + HCl(aq) → NH4Cl(aq)

4. 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.

5. 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-1

6. The concentration of ammonia in the cloudy ammonia solution was 0.114 mol L-1

## 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 0.125 g sample of chalk.
The student placed the chalk sample in a 250 mL conical flask and added 50.00 mL 0.200 mol L-1 HCl using a pipette.
The excess HCl was then titrated with 0.250 mol L-1 NaOH.
The average NaOH titre was 32.12 mL
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

1. Write the equation for the titration:
 HCl(aq) + NaOH(aq) → NaCl(aq) + H2O(l) acid + base → salt + water

2. Calculate the moles, n, of NaOH(aq) that reacted in the titration:
n = c x V
c = concentration = 0.250 mol L-1
V = volume = 32.12 mL = 32.12 x 10-3 L
n(NaOH(aq)) = 0.250 x 32.12 x 10-3 = 8.03 x 10-3 mol

3. 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

4. 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

1. Calculate the total moles of HCl, n(HCl), originally added to the chalk:
n(HCltotal added) = c x V
c = concentration (molarity) = 0.200 mol L-1
V = volume = 50.00 mL = 50.00 x 10-3 L
n(HCltotal added) = 0.200 x 50.00 x 10-3 = 0.010 mol

2. Calculate the moles of HCl that reacted with the calcium carbonate in the chalk
n(HCltitrated) + n(HClreacted with calcium carbonate) = n(HCltotal added)
n(HCltitrated) = 8.03 x 10-3 mol
8.03 x 10-3 + n(HClreacted with calcium carbonate) = 0.010
n(HClreacted with calcium carbonate) = 0.010 - 8.03 x 10-3 = 1.97 x 10-3 mol

3. Write the balanced chemical equation for the reaction between calcium carbonate, CaCO3(s), in the chalk and the HCl(aq).
CaCO3(s) + 2HCl(aq) → CaCl2(aq) + CO2(g) + H2O(l)

4. 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.

5. Calculate the mass of calcium carbonate in the chalk.
moles = mass ÷ molar mass
moles = 9.85 x 10-4 mol (moles of CaCO3 that reacted with HCl)
molar mass(CaCO3) = 40.08 + 12.01 + (3 x 16.00) = 100.09 g mol-1
mass = moles x molar mass = 9.85 x 10-4 x 100.09 = 0.099 g

6. The mass of calcium carbonate in the chalk was 0.099 g
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