Back titration: purity of calcium carbonate

Worked example

A 1.221 g sample of impure calcium carbonate was added to 50.0 cm³ of 1.00 mol dm⁻³ hydrochloric acid (an excess). When the reaction was complete, the remaining acid needed 26.73 cm³ of 1.00 mol dm⁻³ sodium hydroxide for neutralisation. Calculate the percentage purity of the calcium carbonate. M(CaCO₃) = 100.1 g mol⁻¹.

1. 1. Concentration from moles and volume: solve for n(HCl) added

   $n = cV$

   = 0.05000 mol

2. 2. Concentration from moles and volume: solve for n(HCl) left over = n(NaOH)

   $n = cV$

   = 0.02673 mol

3. 3. Subtract n(HCl) added − n(HCl) left over = n(NaOH) → n(HCl) that reacted with CaCO₃

   = 0.02327 mol

4. 4. Mole ratio from the balanced equation: solve for n(CaCO₃)

   $n_\mathrm{B} = \dfrac{b}{a}\,n_\mathrm{A}$

   = 0.01163 mol

5. 5. Moles from mass: solve for m(CaCO₃)

   $m = nM$

   = 1.165 g

6. 6. Divide m(CaCO₃) ÷ m(sample) → Purity

   = 0.9539

Answer: Percentage purity = 0.954

Method

n(HCl) added = cV. The NaOH titre gives the HCl left over (1 : 1). The difference is the HCl that reacted, and CaCO₃ + 2HCl → CaCl₂ + H₂O + CO₂ halves it to n(CaCO₃). Multiply by M for the mass, then divide by the sample mass.

Common mistakes

• CaCO₃ + 2HCl → CaCl₂ + H₂O + CO₂: halve the moles of HCl that reacted to get moles of CaCO₃.
• The NaOH measures the HCl left over. The HCl that reacted with the carbonate is the amount added minus the amount left over.

Formulas used

• Concentration from moles and volume

  $c = \dfrac{n}{V}$

• Mole ratio from the balanced equation

  $n_\mathrm{B} = \dfrac{b}{a}\,n_\mathrm{A}$

• Moles from mass

  $n = \dfrac{m}{M}$

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