1. 3.4.3 Titrations Starter: Calculate the concentrations of the following solutions a) in g/dm 3 and b) in mol/dm 3. 5g of NaOH dissolved in 1dm 3 of water 15g of MgCl 2 dissolved in 0.5dm 3 of water 1.3kg of K 2 Cr 2 O 7 dissolved in 150dm 3 of water 250g Ba(NO 3 ) 2 dissolved in 1300cm 3 of water.

2. 3.4.3 Titrations 5g of NaOH dissolved in 1dm 3 of water a)mass/volume = 5/1 = 5g/dm 3 b)n = m/Mr = 5/(23+16+1) = 0.125 mol c = n/v = 0.125/1 = 0.125mol/dm 3 15g of MgCl 2 dissolved in 0.5dm 3 of water a)mass/volume = 15/0.5 = 30g/dm 3 b)n = m/Mr = 15/(24+71) = 0.158 mol c = n/v = 0.158/0.5 = 0.316 mol/dm 3

3. 3.4.3 Titrations 1.3kg of K 2 Cr 2 O 7 dissolved in 150dm 3 of water a)mass/volume = 1300/150 = 8.67g/dm 3 b)n = m/Mr = 1300/294 = 4.42 mol c = n/v = 4.42/150 = 0.0295 mol/dm 3 250g Ba(NO 3 ) 2 dissolved in 1300cm 3 of water. a)mass/volume = 250/1.3 = 192g/dm 3 b)n = m/Mr = 250/261 = 0.958 mol c = n/v = 0.958/1.3 = 0.737 mol/dm 3

4. Acid-base titrations Suppose you had two bottles of hydrochloric acid, one with a concentration of 0.1mol/dm 3 the other 1mol/dm 3, neither of them with labels, how could you work out which was which?

5. Acid-base titrations The most effective way is to add an alkali to a set volume of each of the acids. a)What else would you need to add? b)How could you tell which acid was which? a) Add an indicator to the acid to show when it had been neutralised. b) The acid which required more alkali to neutralise it is the more concentrated (1mol/dm 3 ).

6. Acid-base titrations In actual fact this technique can be used not simply to find out which acids are more concentrated, but to find the actual concentration (in mol/dm 3 ) of a particular acid or alkali. In order to do this you carry out a technique called titration. Titration works by reacting an alkali for which you know the volume and concentration with an acid for which you know the volume only… …or vice versa.

7. Example You have been given a solution of sodium hydroxide (alkali) but you don’t know the concentration. You have also been given some hydrochloric acid solution with a concentration of 0.1mol/dm 3.

8. A known volume of the alkali (NaOH) is transferred into the conical flask, using a volumetric pipette. In this example 25.0cm 3 is transferred An indicator called methyl orange is added. This turns yellow in alkaline conditions Conical flask Pipette

10. The acid (0.1mol/dm 3 HCl in this case) is added to the conical flask from a burette until all of the alkali has been neutralised. When one drop too much of acid has been added, the indicator turns red. The volume of acid added can be worked out using the graduations on the burette. Burette

11. The Formula Concentration = number of moles / volume (in dm 3 ) or C = n / v

12. NaOH+ HCl  NaCl + H 2 O C Write the balanced equation for the reaction. Underneath the reactants, create a table as shown. n V

13. NaOH+ HCl  NaCl + H 2 O C 0.1mol/dm 3 Fill in the information you know for each of the reactants. Convert the volumes to dm 3 if necessary. n V = 25/1000 = 0.025dm 3 = 23.4/1000 = 0.0234dm 3

14. NaOH+ HCl  NaCl + H 2 O C 0.1mol/dm 3 You have the concentration and volume of HCl so you can calculate the number of moles of HCl. Do so. n V = 25/1000 = 0.025dm 3 = 23.4/1000 = 0.0234dm 3

15. NaOH+ HCl  NaCl + H 2 O C 0.1mol/dm 3 You have the concentration and volume of HCl so you can calculate the number of moles of HCl. Do so. n n = C V = 0.1 x 0.0234 = 2.34 x 10 -3 mol V = 25/1000 = 0.025dm 3 = 23.4/1000 = 0.0234dm 3

16. NaOH+ HCl  NaCl + H 2 O C 0.1mol/dm 3 The balancing numbers in equation above tells us that 1 mole of NaOH reacts with 1 mole of HCl. n n = C V = 0.1 x 0.0234 = 2.34 x 10 -3 mol V = 25/1000 = 0.025dm 3 = 23.4/1000 = 0.0234dm 3

17. NaOH+ HCl  NaCl + H 2 O C 0.1mol/dm 3 The balancing numbers in equation above tells us that 1 mole of NaOH reacts with 1 mole of HCl. Therefore the number of moles of NaOH = number of moles of HCl n Reaction ratio 1:1 n = 2.34 x 10 -3 mol n = C V = 0.1 x 0.0234 = 2.34 x 10 -3 mol V = 25/1000 = 0.025dm 3 = 23.4/1000 = 0.0234dm 3

18. NaOH+ HCl  NaCl + H 2 O C 0.1mol/dm 3 We now have enough information to calculate the concentration of the NaOH. Do so. n Reaction ratio 1:1 n = 2.34 x 10 -3 mol n = C V = 0.1 x 0.0234 = 2.34 x 10 -3 mol V = 25/1000 = 0.025dm 3 = 23.4/1000 = 0.0234dm 3

19. NaOH+ HCl  NaCl + H 2 O C C = n / V = 2.34 x 10 -3 /0.025 = 0.0936mol/dm 3 0.1mol/dm 3 n Reaction ratio 1:1 n = 2.34 x 10 -3 mol n = C V = 0.1 x 0.0234 = 2.34 x 10 -3 mol V = 25/1000 = 0.025dm 3 = 23.4/1000 = 0.0234dm 3

20. Exam technique Typically, GCSE (and AS) questions will break the calculation into a series of steps, for example: i.Calculate the number of moles of HCl; ii.deduce the number of moles of NaOH; iii.calculate the concentration of NaOH.

21. Monitoring the “end point” The “end point” of an acid-base titration is when the acid or alkali has been neutralised. This can be monitored using an indicator or a pH meter. Although universal indicator, which is a mixture of a number of different indicators, would work in theory, it is more usual to use single indicators. Single indicators have one colour change when the pH passes a certain value. For example, methyl orange changes colour between pH 4.4 (yellow) and 3.1 (red). Phenolphthalein changes colour between pH 8.2 (colourless) and 10 (pink)

22. By the end of this session you should know that: The volumes of acid and alkali solutions that react with each other can be measured by titration using a suitable indicator. If the concentration of one of the reactants is known, the results of a titration can be used to find the concentration of the other reactant.

23. Extension: try this for sulphuric acid