1. Making Salts Insoluble salts – precipitation Mix two solutions together one with the cation, one with the anion. Filter, wash and dry. Soluble Na +, K + and NH 4+ salts – titration Add acid to sodium, potassium or ammonium hydroxide (or carbonate). Evaporate off the water. Other soluble salts Add acid to solid metal / metal oxide / metal carbonate / metal hydroxide. Filter off excess solid, evaporate off the water.

2. Titration Burette contains acid of known concentration. Conical flask contains alkali of unknown concentration but known volume. Titrate and record exactly how much acid is required to neutralise the acid. Titrations

4. Procedure  Do titration once quite quickly to get an approximate volume.  Always read the burette to two decimal places but the last decimal place is 0 or 5.  Repeat the titration but this time add the last 2-4cm 3 very slowly to get an accurate titrate.  Repeat the titration until you get 2 concordant results.  Concordant results are those volumes that differ from each other by 0.20 cm 3 or less.

5. Which titration results to use?

6. Titration Calculation  1 mol dm -3 (1M) contains 1 mole in 1dm 3  1dm 3 = 1000cm 3  Eg 1 mol dm -3 HCl contains 36.5g in 1000cm 3 water.

7. What is the concentration in mol dm -3 1.96g of sulphuric acid in 1000cm3 2g sodium hydroxide in 1000cm3

8. What is the concentration in mol dm -3 1.96g of sulphuric acid in 1000cm 3 (RMM=98) 1.96/98 = 0.003 moles 0.003mol dm -3 = 3mmol dm -3 2g sodium hydroxide in 1000cm 3 (RMM=40) 2/40 = 0.05 moles 0.05mol dm -3 Now do exercise 2 in the titrations booklet

9. What is the concentration in mol dm -3  19.6g sulphuric acid in 200cm3  1g sodium hydroxide in 25cm3

10. What is the concentration in mol dm -3 19.6g sulphuric acid in 200cm 3 (RMM=98) 19.6/98 = 0.03 moles 0.03 moles in 200cm 3 0.15 mol dm -3 1g sodium hydroxide in 25cm 3 (RMM=40) 1/40 = 0.025 moles 0.025 moles in 25cm 3 1 mol dm -3

11. Number of moles = conc (mol dm -3 ) x volume (dm 3 ) How many moles? a) In 20cm 3 of 0.5 mol dm -3 HCl? b) In 150cm 3 of 0.25 mol dm -3 NaOH?

12. Number of moles = conc (mol dm -3 ) x volume (dm 3 ) How many moles? a) In 20cm 3 of 0.5 mol dm -3 HCl? 0.5 x 0.02 = 0.01 moles b) In 150cm 3 of 0.25 mol dm -3 NaOH? 0.25 x 0.15 = 0.0375 moles Now do exercise 3 in the “Titrations” booklet

13. Titrations In a titration, 20 cm 3 of NaOH was neutralised by 23 cm 3 of 0.24 mol dm -3 HCl. What concentration was the NaOH?

14. Step 1 Use the equation to work out the ratio of HCl to NaOH in the reaction. Step 2 Work out the number of moles of HCl Step 3 This equals the number of moles of NaOH in 20cm 3, so convert this to concentration in mol dm -3.

15. Step 1 Use the equation to work out the ratio of HCl to NaOH in the reaction. HCl + NaOH NaCl + H 2 O Step 2 Work out the number of moles of HCl 0.023 x 0.24 = 0.0052 moles Step 3 This equals the number of moles of NaOH in 20cm 3, so convert this to concentration in mol dm -3. 0.0052 moles in 20cm 3 = 0.276 mol dm -3

16. Calculations 1. 50cm 3 of 2 mol dm -3 NaOH was neutralised exactly by 30cm 3 of HCl. What was the concentration of the HCl? 1. 20cm 3 of 0.5 mol dm -3 NaOH was neutralised exactly by 40cm 3 of H 2 SO 4. What was the concentration of the H 2 SO 4 ?