1. IC5.8.4 Titration calculations © Oxford University Press Titration calculations

2. IC5.8.4 Titration calculations © Oxford University Press Titrations can be used to find the concentration of an acid or an alkali.

3. IC5.8.4 Titration calculations © Oxford University Press Imagine that a titration is carried out involving 0.1 mol/dm 3 hydrochloric acid and 25.0 cm 3 of an unknown concentration of sodium hydroxide. HCl + NaOH  NaCl + H 2 O If the mean titre of acid is 24.0 cm 3, what is the concentration of the sodium hydroxide?

4. IC5.8.4 Titration calculations © Oxford University Press The first step is to convert all volumes to dm 3.  The titre of HCl is 24.0 ÷ 1000 = 0.024 dm 3  The volume of NaOH is 25.0 ÷ 1000 = 0.025 dm 3 Amount of HCl = concentration × volume = 0.1 × 0.024 = 0.0024 mol

5. IC5.8.4 Titration calculations © Oxford University Press HCl + NaOH  NaCl + H 2 O Using the balanced equation:  1 mol of HCl will react with 1 mol of NaOH  So 0.0024 mol of HCl will react with 0.0024 mol of NaOH Concentration of NaOH = amount ÷ volume = 0.0024 ÷ 0.025 = 0.096 mol/dm 3

6. IC5.8.4 Titration calculations © Oxford University Press Run 1Run 2Run 3Run 4 End reading (cm 3 )21.241.021.241.2 Start reading (cm 3 )0.021.21.021.2 Titre (cm 3 )21.219.820.2 Now work through these calculations using this table of results from a titration. HCl + NaOH  NaCl + H 2 O

7. IC5.8.4 Titration calculations © Oxford University Press 1. Calculate the titre in Run 4.  41.2 – 21.2 = 20.0 cm 3 2. Which is the anomalous titre in Runs 1 to 4?  Run 1 – the result is too high compared to the other three. 3.Calculate the mean titre, ignoring the anomalous titre.  (19.8 + 20.2 + 20.0) ÷ 3 = 20.0 cm 3 Run 1Run 2Run 3Run 4 End reading (cm 3 )21.241.021.241.2 Start reading (cm 3 )0.021.21.021.2 Titre (cm 3 )21.219.820.2

8. IC5.8.4 Titration calculations © Oxford University Press HCl + NaOH  NaCl + H 2 O 4. In this titration 0.2 mol/dm 3 hydrochloric acid was added to 25.0 cm 3 of a sodium hydroxide solution. Calculate the concentration of the alkali.  Titre of HCl was 20.0 ÷ 1000 = 0.02 dm 3 Volume of NaOH is 25.0 ÷ 1000 = 0.025 dm 3  Amount of HCl = concentration × volume = 0.2 × 0.02 = 0.004 mol  Concentration of NaOH = amount ÷ volume = 0.004 ÷ 0.025 = 0.16 mol/dm 3