You have £20.00 in your pocket. You go to the shops and buy a few things. When you get home, you have £3.73 left. How much did you spend? You hopefully said £16.27. A back titration is essentially the same as the calculation you’ve just done – by determining how much of something was left at the end, we can work out how much of it was used up!
A classic situation where a back-titration is used is to determine the amount of an insoluble base (e.g., Mg(OH)2, CaCO3 etc). Here’s an example:
| Question 1: 10 mol of HCl are added to a sample of solid Mg(OH)2. The resulting solution is then titrated with 2 mol NaOH. Reaction 1: Mg(OH)2 + 2 HCl –> MgCl2 + 2H2O Reaction 2: NaOH + HCl –> NaCl + H2O Determine the initial moles of Mg(OH)2 The key to answering these questions is to work backwards! Step 1 If 2 mol NaOH were required for Reaction 2, then there must have been 2 mol HCl at the start of Reaction 2 since we know that HCl and NaOH react in a 1:1 ratio. Step 2 If there were 10 mol HCl at the start of Reaction 1 and 2 mol HCl at the start of Reaction 2, then 8 mol HCl were used in Reaction 1 since 10 – 2 = 8 mol. Step 3 If 8 mol HCl were used up in Reaction 1, there must have been 4 mol Mg(OH)2 since HCl and Mg(OH)2 react in a 2:1 ratio. |
Here are another example, using the same set of reactions, to check your understanding:
- 50 mol of HCl are added to a sample of solid Mg(OH)2
- The resulting solution is titrated with 10 mol NaOH
- Determine the initial moles of Mg(OH)2
- 50 – 10 = 40 mol HCl used up, so 40 / 2 = 20 mol Mg(OH)2
and another:
- 0.24 mol of HCl are added to a sample of solid Mg(OH)2
- The resulting solution is titrated with 0.04 mol NaOH
- Determine the initial moles of Mg(OH)2
- 0.24 – 0.04 = 0.20 HCl used up, so 0.20 / 2 = 0.10 mol Mg(OH)2
We’ll now look at a more complicated example:
| Question 2: 1.50 g of impure CaCO3 is added to 100 cm3 of 0.5 mol dm-3 HCl. A 10 cm3 sample of this solution is titrated with 25 cm3 of 0.1 mol dm-3 NaOH. Reaction 1: CaCO3 + 2 HCl –> CaCl2 + H2O + CO2 Reaction 2: NaOH + HCl –> NaCl + H2O Calculate the % by mass purity of CaCO3. Step 1 Calculate the number of moles of substances from the data given in the question. There will always be 1 or 2 marks for this so it’s a good first step to do whilst you’re working out what the question actually wants. n NaOH = cv = 0.1 x (25/1000) = 2.5 x 10-3 mol n HCl = cv = 0.5 x (100/1000) = 5.0 x 10-2 mol
Step 2 There is a 1:1 ratio between NaOH and HCl moles in Reaction 2 so we can work out the HCl moles in the 10 cm3 sample: n HCl = 2.5 x 10-3 mol (in 10 cm3 sample)
Step 3 The 100 cm3 total is 10x bigger than the 10 cm3 sample so we can work out the total HCl moles at the end of Reaction 1: n HCl = 10 x 2.5 x 10-3 = 2.5 x 10-2 mol (total)
Step 4 We know that there was 5.0 x 10-2 mol HCl at the start and 2.5 x 10-2 mol HCl at the end of Reaction 1. By subtraction, we can work out the HCl moles used up in Reaction 1: 5.0 x 10-2 – 2.5 x 10-2 = 2.5 x 10-2 mol
Step 5 There is a 1:2 ratio between CaCO3 and HCl in Reaction 1: n CaCO3 = 0.5 x 2.5 x 10-2 = 1.25 x 10-2 mol
Step 6 The relative formula mass (Mr) of CaCO3 is 100 so the mass of CaCO3 is: m CaCO3 = 100 x 1.25 x 10-2 = 1.25 g
Step 7 The % by mass purity of CaCO3 is therefore: % CaCO3 = 1.25 / 1.50 x 100 = 83% |