1. These are necessary when the use of a particular reactant: gives a slow chemical reaction, unsuitable for titration gives a poor or unclear end point colour change does not give a solution that is readily titratable.

2. To overcome this, a third reactant is used. The reaction between the second and third reactant is fast, suitable for titration and the end point can clearly be seen when it self-indicates or uses another indicator e.g. starch or acid/base indicator.

3. In a back titration, all of reactant one is reacted with reactant two, because reactant two is added in excess. The amount of reactant two left over (that didn’t react with reactant one) is then titrated against reactant three. Mole amounts can be calculated from known concentrations and known volumes.

4. Example: The reaction of unknown concentration calcium carbonate CaCO 3 (aq) and known concentration hydrochloric acid HCl(aq) may be slow. Therefore, we do a back titration. Excess acid is used to neutralise all the CaCO 3, with leftover HCl able to be reacted with known concentration NaOH, using a phenolphthalein indicator.

5. 20.0mL of CaCO 3 solution was pipetted into a flask. 25.0mL of 0.550mol L -1 HCl solution was pipetted into the flask too. All CaCO 3 reacted after 3 minutes. A titre of 23.45mL of 0.425molL -1 NaOH was added from the burette to get a pink end point.

6. Total excess amount of HCl n = cV = 0.550molL -1 x 0.025L = 0.0138mol

7. Amount of NaOH needed to neutralise excess HCl n = cV = 0.425molL -1 x 0.02345L = 0.00997mol

8. Balanced equation: NaOH + HCl  NaCl + H 2 O Therefore n(NaOH) : n(HCl) = 1:1 Leftover amount of HCl = 0.00997mol also.

9. Amount of HCl used to neutralise CaCO 3 is n(excess) – n(leftover) = 0.0138mol – 0.00997mol = 0.00378mol

10. Balanced equation: CaCO 3 + 2HCl  CaCl 2 + H 2 O + CO 2 Therefore n(CaCO 3 ) : n(HCl) = 1:2

11. Amount of CaCO 3 reacted = 0.00378mol x ½ = 0.00189mol

12. This amount was found in 20.0mL. c(CaCO 3 ) = n / V = 0.00189mol / 0.020L = 0.0946molL -1

13. Back titrations can also be used in redox reactions where it is difficult to directly measure one reactant against another. An intermediate product, often iodine, is used. This reaction can be tracked using starch as an indicator.

14. Example: 20mL of 0.21 mol L -1 acidified potassium permanganate solution was reacted with excess potassium iodide solution in a flask. The liberated iodine was titrated against an average volume of 22.8mL of potassium thiosulfate solution. Calculate the concentration of the thiosulfate solution.

15. Equation 1 2MnO 4 - + 16H + + 10I - 2Mn 2+ + 4H 2 O + 5I 2 Equation 2 I 2 + 2S 2 O 3 2- 2I - + S 4 O 6 2-

16. Number of moles of MnO 4 - n= cV = 0.21 mol L-1 x 0.02L = 4.2 x 10 -3 mol

17. Mole ratios: From equation 1: 2 mol MnO 4 - produces 5 mol I 2 From equation 2: 2 mol S 2 O 3 2- uses 1 mol I 2 Therefore 1 mol MnO 4 - is equivalent to 5 mol S 2 O 3 2-

18. Number of moles of S 2 O 3 2- n(S 2 O 3 2- ) = 5 x 4.2 x 10 -3 = 0.021 mol

19. This amount was found in 22.8mL c(S 2 O 3 2- ) = n / V = 0.021 / 0.0228 = 0.92 mol L -1