How to find cations and anions
In order to analyse cations and anions a series of tests would have to be performed. These can be used to identify the elements found in a specific compounds. There are three sets of tests: flame tests, tests with bases such as NaOH and NH4OH and tests for anions.
Flame tests:
Na+: yellow/orange flame
K+: lilac flame
Mg2+: no coloured flame
Ca2+: red brick flame
Ba2+: green flame
Cu2+: blue-green flame
Pb2+: light blue flame
NH4+: no coloured flame
Al3+: no coloured flame
It is of utter importance to note that it is the ion that produces the coloured flame and not the element.
Hydroxide tests:
| Addition of NaOH | Addition of NH4OH | |
| Na+ | No ppt is formed | No ppt is formed |
| K+ | No ppt is formed | No ppt is formed |
| NH4+ | No ppt is formed | No ppt is formed |
| Mg2+ | A white ppt formed insoluble in excess NaOH | A white ppt formed insoluble in excess NaOH |
| Ca2+ | A white ppt formed insoluble in excess NaOH | A white ppt formed insoluble in excess NaOH |
| Ba2+ | No ppt is formed | No ppt is formed |
| Cu2+ | A blue ppt which dissolves into a blue solution in excess NaOH | A blue ppt which turns into a deep blue solution in excess ammonium hydroxide |
| Fe2+ | A green ppt insoluble in excess | A green ppt insoluble in excess |
| Fe3+ | A brown ppt insoluble in excess | A brown ppt insoluble in excess |
| Pb2+ | A white ppt soluble in excess NaOH | A white ppt insoluble in excess ammonium hydroxide |
| Al3+ | A white ppt soluble in excess NaOH | A white ppt insoluble in excess ammonium hydroxide |
| Zn2+ | A white ppt soluble in excess NaOH | A white ppt soluble in excess ammonium hydroxide |
It must be noted that it is the precipitate that is the most important, and not the colour of the solution even though the solution takes it colour from the ppt. The precipitate is the hydroxide of the cation being analysed.
The reactions taking place are:
M+(aq) + OH–(aq) → MOH(s) for the formation of precipitates
When the ppt dissolves in excess base a complex would be formed, which would have a charge and therefore it would dissolve in an aqueous solution. An example is:
Pb2+ + OH– → Pb(OH)2
Pb(OH)2 + OH– → Pb(OH)4]2-
Although it is important to know that these complexes exist it is a chemistry of its own and students are not expected to know how these complexes form.
Testing for Anions
CO32-: With the addition of an acid a colourless and odourless gas which turns limewater milky is created. This gas is CO2 and the reaction is as following:
CO32-(aq) + 2H+(aq) → H2O + CO2
SO42-: The addition of Barium Chloride would form Barium Sulfate which is insoluble. On addition of acid the white precipitate formed does not dissolve.
SO42-(aq) + Ba2+(aq) → BaSO4(s)
SO32-: The addition of Barium Chloride would form Barium Sulfite which is insoluble. On addition of acid the white precipitate formed disappears. The gas produced when the anion is reacted with an acid turns blue litmus red.
SO32-(aq) + Ba2+(aq) → BaSO3(s)
SO32-(s) + 2H+(aq) → H2O + SO2
S2-: By reacting the anion with an acid a pungent gas is formed.
S2- + 2H+ → H2S
Cl–: Reacting the anion with AgNO3 would form a white ppt.
Ag+(aq) + Cl–(aq) → AgCl(s)
If the chloride ion is reacted with a concentrated acid such concentrated sulfuric acid white fumes would be seen, which would be HCl
Cl– + H+→ HCl
Br–: Reacting the anion with AgNO3 would form a pale yellow ppt.
Ag+(aq) + Br–(aq) → AgBr(s)
If the bromide ion is reacted with a concentrated acid such concentrated sulfuric acid white fumes would be seen, which would be HBr. this might have some brown fumes coming from Bromine.
Br– + H+→ HBr
I–: Reacting the anion with AgNO3 would form a yellow ppt.
Ag+(aq) + l–(aq) → Agl(s)
If the iodide ion is reacted with a concentrated acid such concentrated sulfuric acid white fumes would be seen, which would be Hl. this might have some purple fumes coming from the Iodine vapour.
l– + H+→ Hl
If Pb2+ is reacted with I– a very bright yellow ppt is formed. This would PbI2 and this is the confirmatory test to distinguish Pb2+ from Al3+.
NO3–: By using Devarda’s alloy, a mixture of Zinc, Aluminium and Copper, which is a very strong reducing agent. This would liberate ammonia, which has a pungent smell and turns red litmus blue.
6H+ + NO3– → NH3 + 3H2O +5e–
(This is the half-reaction depicting the reduction of the nitrate)
Another test is the brown ring test, which comprises of the addition of concentrated sulfuric acid and Iron(II) Sulfate. The resultant of this reaction in the presence of a nitrate would be NO, which would form a brown ring in the middle of the solution.