1. Tests for cations in solution

2. Cations are positive ions.
The cations you need to be able to identify are:
copper
iron(III)
silver
iron(II)
magnesium
zinc
aluminium
lead
barium
sodium

3. Copper Solutions containing Cu2+(aq) will be blue/green in colour.
Pour a little of the test solution into a clean test tube.

4. Cu2+(aq) + 2OH–(aq) → Cu(OH)2(s)
Add a few drops of dilute sodium hydroxide solution.
A pale blue precipitate indicates the presence of copper ions.
Cu2+(aq) + 2OH–(aq) → Cu(OH)2(s)

5. To confirm the presence of Cu2+(aq), add about 5 mL of ammonia solution.
The pale blue precipitate redissolves to form a clear, royal blue solution.
Cu(OH)2(s) + 4NH3(aq) → [Cu(NH3)4]2+(aq) + 2OH–(aq)

6. Iron(III)
Solutions containing Fe3+(aq) will be orange in colour (or yellow if very dilute).
Pour a little of the test solution into a clean test tube.

7. Fe3+(aq) + 3OH–(aq) → Fe(OH)3(s)
Add a few drops of dilute sodium hydroxide solution.
An orange or dark brown precipitate forms.
Fe3+(aq) + 3OH–(aq) → Fe(OH)3(s)

8. 2Ag+(aq) + 2OH–(aq) → Ag2O(s) + H2O(l)
Silver
Silver ions also react with sodium hydroxide solution to form a brown precipitate
Iron(III) solutions are usually coloured and the precipitate is dark brown or orange.
Silver solutions are colourless and the precipitate is mud-brown or the colour of milky coffee.
2Ag+(aq) + 2OH–(aq) → Ag2O(s) + H2O(l)

9. To confirm Fe3+
Potassium thiocyanate, KSCN, can be used to detect the presence of Fe3+.
This test can be used on very dilute solutions or in the presence of other metal ions (especially Fe2+).

10. Fe3+(aq) + SCN–(aq) → [FeSCN]2+(aq)
To a fresh sample of the test solution, add a few drops of potassium thiocyanate solution.
If Fe3+ is present the solution will turn blood-red.
Fe3+(aq) + SCN–(aq) → [FeSCN]2+(aq)

11. Iron(II) Solutions of Fe2+ are colourless or pale green.
Pour a little of the test solution into a clean test tube.

12. Fe2+(aq) + 2OH–(aq) → Fe(OH)2(s)
Add a few drops of sodium hydroxide solution.
If Fe2+ is present you will see an olive-green precipitate or gel form.
Fe2+(aq) + 2OH–(aq) → Fe(OH)2(s)

13. Magnesium, zinc, aluminium and lead
These four cations all form white precipitates with sodium hydroxide solution. Three form soluble complexes with excess hydroxide, one also forms a soluble complex with ammonia solution.
It is very important when adding sodium hydroxide to colourless solutions that you start with one or two drops only – otherwise you might miss the formation of the precipitate when it redissolves in excess hydroxide.

14. Mg2+(aq) + 2OH–(aq) → Mg(OH)2(s)
Magnesium
Pour a little of the test solution into a clean test tube.
Add one or two drops only of sodium hydroxide solution.
If Mg2+ is present a white precipitate forms.
Mg2+(aq) + 2OH–(aq) → Mg(OH)2(s)

15. When excess sodium hydroxide is added, more precipitate is formed: the precipitate does NOT redissolve.

16. Zn2+(aq) + 2OH–(aq) → Zn(OH)2(s)
Zinc
Pour a little of the test solution into a clean test tube.
Add one or two drops only of sodium hydroxide solution.
If Zn2+ is present a white precipitate forms.
Zn2+(aq) + 2OH–(aq) → Zn(OH)2(s)

17. Zn(OH)2(s) + 2OH–(aq) → [Zn(OH)4]2–(aq)
Add excess sodium hydroxide solution (about 5 mL).
The white precipitate slowly redissolves to form a clear, colourless solution.
Zn(OH)2(s) + 2OH–(aq) → [Zn(OH)4]2–(aq)
Al3+ and Pb2+ also form complexes with OH–, but only Zn2+ also forms a complex with ammonia.

18. To a fresh sample of the test solution add a few drops of sodium hydroxide to form a white precipitate as before.

19. Add excess ammonia solution (about 5 mL).
If Zn2+ is present the white precipitate will slowly redissolve to form a clear, colourless solution.
Zn(OH)2(s) + 4NH3(aq) → [Zn(NH3)4]2+(aq) + 2OH–(aq)

20. Al3+(aq) + 3OH–(aq) → Al(OH)3(s)
Aluminium
Pour a little of the test solution into a clean test tube.
Add one or two drops only of sodium hydroxide solution.
If Al3+ is present a white precipitate forms.
Al3+(aq) + 3OH–(aq) → Al(OH)3(s)

21. Add excess hydroxide (about 5 mL).
The precipitate redissolves to form a clear, colourless solution.
Al(OH)3(s) + OH–(aq) → [Al(OH)4]–(aq)

22. To a fresh sample of the test solution, add a few drops of sodium hydroxide solution to form a white precipitate as before.

23. Add excess (5 mL) ammonia solution.
The white precipitate does NOT redissolve.

24. Both Al3+ and Pb2+ form white precipitates with OH– which redissolve in excess OH– but not in excess NH3. However, aluminium sulfate is soluble in water, while lead sulfate is not.
To a fresh sample of the test solution add a few drops of sulfuric acid.
If Al3+ is present there will be NO precipitate.

25. Pb2+(aq) + 2OH–(aq) → Pb(OH)2(s)
Lead
Pour a little of the test solution into a clean test tube.
Add one or two drops only of sodium hydroxide solution.
A white precipitate forms.
Pb2+(aq) + 2OH–(aq) → Pb(OH)2(s)

26. Pb(OH)2(s) + 2OH–(aq) → [Pb(OH)4]–(aq)
Add excess (5 mL) sodium hydroxide solution. If Pb2+ is present the white precipitate redissolves to form a clear, colourless solution.
Pb(OH)2(s) + 2OH–(aq) → [Pb(OH)4]–(aq)

27. To a fresh sample of the test solution add a few drops of sodium hydroxide solution as before.

28. Add excess (5 mL) ammonia solution.
If Pb2+ is present the white precipitate does NOT redissolve.

29. Pb2+(aq) + SO42–(aq) → PbSO4(s)
To a fresh sample of the test solution add a few drops of dilute sulfuric acid.
If Pb2+ is present a white precipitate will form.
Pb2+(aq) + SO42–(aq) → PbSO4(s)

30. Barium and sodium Solutions containing Ba2+ or Na+ will be colourless.
Barium hydroxide is moderately soluble, while sodium hydroxide is very soluble.

31. Ba2+(aq) + 2OH–(aq) → Ba(OH)2(s)
Add a few drops of sodium hydroxide.
No precipitate suggests barium or sodium. Add more hydroxide (5 mL).
If barium is present there may be a slight cloudiness after excess hydroxide is added.
If sodium is present the solution will remain clear and colourless.
Ba2+(aq) + 2OH–(aq) → Ba(OH)2(s)

32. Ba2+(aq) + SO42–(aq) → BaSO4(s)
Barium salts form a white precipitate with sulfuric acid.
Sodium salts do not form a precipitate with sulfuric acid.
Ba2+(aq) + SO42–(aq) → BaSO4(s)