Back Titration or Indirect Titration Tutorial

Key Concepts

1. Reactant A of unknown concentration is reacted with excess reactant B of known concentration.
2. A direct 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)

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Example : Back (Indirect) Titration to Determine the Concentration of a Volatile Substance

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	+	carbon dioxide	+	water

2. Calculate the moles, n, of Na₂CO_3(aq) that reacted in the titration:

   moles = concentration (mol L^-1) × volume (L)
   n(Na₂CO_3(aq)) = c × V
   c(Na₂CO_3(aq)) = 0.050 mol L^-1
   V(Na₂CO_3(aq)) = 21.50 mL = 21.50 × 10^-3 L
   n(Na₂CO_3(aq)) = 0.050 × 21.50 × 10^-3 = 1.075 × 10^-3 mol

3. Use the balanced chemical reaction for the titration to determine the moles of HCl that reacted in the titration (mole ratio (stoichiometric ratio)).

   From the balanced chemical equation, 1 mole Na₂CO₃ react with 2 moles of HCl
   So, 1.075 × 10^-3 mole Na₂CO₃ reacted with 2 × 1.075 × 10^-3 moles HCl
   n(HCl_titrated) = 2 × 1.075 x 10^-3 = 2.150 × 10^-3 mol

4. The amount of HCl that was added to the cloudy ammonia solution in excess was :

   n(HCl_exccess) = 2.150 × 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:

   moles = concentration (mol L^-1) × volume (L)
   n(HCl_{total added}) = c × V
   c(HCl_{total added}) = 0.100 mol L^-1
   V(HCl_{total added}) = 50.00 mL = 50.00 × 10^-3 L
   n(HCl_{total added}) = 0.100 × 50.00 × 10^-3 = 5.00 × 10^-3 mol

2. 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 × 10^-3 mol
   n(HCl_titrated) = 2.150 × 10^-3 mol (calculated in Step 1 above)
   2.150 × 10^-3 + n(HCl_{reacted with ammonia}) = 5.00 × 10^-3
   n(HCl_{reacted with ammonia}) = 5.00 × 10^-3 - 2.150 × 10^-3 = 2.85 × 10^-3 mol

3. 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)

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 × 10^-3 mol HCl had reacted with 2.85 × 10^-3 mol NH₃ in the cloudy ammonia solution.

5. Calculate the ammonia concentration in the cloudy ammonia solution.

   concentration (mol L^-1) = moles ÷ volume (L)
   c(NH_3(aq)) = n(NH_3(aq)) ÷ V(NH_3(aq))
   n(NH_3(aq)) = 2.85 × 10^-3 mol (moles of NH₃ that reacted with HCl)
   V(NH_3(aq)) = 25.00 mL = 25.00 × 10^-3 L (volume of ammonia solution that reacted with HCl)
   c(NH_3(aq)) = 2.85 × 10^-3 ÷ 25.00 × 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 of 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:

   moles = concentration (mol L^-1) × Volume (L)
   n(NaOH_(aq)) = c(NaOH_(aq)) × V(NaOH_(aq))
   c(NaOH_(aq)) = 0.250 mol L^-1
   V(NaOH_(aq)) = 32.12 mL = 32.12 × 10^-3 L
   n(NaOH_(aq)) = 0.250 × 32.12 × 10^-3 = 8.03 × 10^-3 mol

3. Use the balanced chemical reaction for the titration to determine the moles of HCl that reacted in the titration (mole ratio (stoichiometric ratio)).

   From the balanced chemical equation, 1 mole NaOH reacts with 1 mole of HCl
   So, 8.03 × 10^-3 mole NaOH reacted with 8.03 × 10^-3 moles HCl

4. The amount of HCl that was added to the chalk in excess was 8.03 × 10^-3 mol

Step 2: Determine the amount of calcium carbonate in chalk

1. Calculate the total moles of HCl originally added to the chalk:

   moles = concentration (mol L^-1) x Volume (L)
   n(HCl_{total added}) = c(HCl_{total added}) × V(HCl_{total added})
   c(HCl_{total added}) = 0.200 mol L^-1
   V(HCl_{total added}) = 50.00 mL = 50.00 × 10^-3 L
   n(HCl_{total added}) = 0.200 × 50.00 × 10^-3 = 0.010 mol

2. 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 × 10^-3 mol (calculated in Step 1 above)
   8.03 × 10^-3 + n(HCl_{reacted with calcium carbonate}) = 0.010 mol
   n(HCl_{reacted with calcium carbonate}) = 0.010 - 8.03 × 10^-3 = 1.97 × 10^-3 mol

3. Write the balanced chemical equation for the reaction between calcium carbonate in the chalk and the HCl_(aq).

   CaCO_3(s) + 2HCl_(aq) → CaCl_2(aq) + CO_2(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 × 10^-3 mol HCl had reacted with ½ × 1.97 × 10^-3 = 9.85 × 10^-4 mol CaCO₃ in the chalk.

5. Calculate the mass of calcium carbonate in the chalk.

   moles = mass (g) ÷ molar mass (g mol^-1)
   n(CaCO₃) = mass(CaCO₃) ÷ M(CaCO₃)
   n(CaCO₃) = 9.85 × 10^-4 mol (moles of CaCO₃ that reacted with HCl)
   M(CaCO₃) = 40.08 + 12.01 + (3 × 16.00) = 100.09 g mol^-1 (using the Periodic Table to find molar mass of each element)
   mass(CaCO₃) = n(CaCO₃) × M(CaCO₃) = 9.85 × 10^-4 × 100.09 = 0.099 g

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