CSEC Inorganic: Qualitative Analysis
Qualitative analysis is the determination of the chemical composition of a sample. In Inorganic chemistry
we test for the presence of cations and anions by performing various chemical reactions. Products of
these reactions e.g. gases should be identified since they allow us to determine the ions that are present.
You should also include any equation which can support your inference e.g.
Test for Cations
The reagents used are NaOH(aq) and NH3(aq) both of which contain OH- (aq) which will precipitate (ppt)
most metal ions from solution i.e. most metal hydroxides are INSOLUBLE. The metal ion can then be
identified based on
● Whether a ppt forms or not (based on the solubility of the metal hydroxide)
● The colour of the ppt
● Whether the ppt dissolved in excess reagent or not.
Method Observations Inferences Explanation
To a colourless solution These ions all form
A NaOH was added White ppt formed Ca2+, Al3+, Pb2+ or Zn2+ white, insoluble
dropwise may be present hydroxides w/NaOH
Dissolves in excess to Al3+, Pb2+ or Zn2+ may be Ca(OH)2 does NOT
Then in excess give a colourless present dissolve in excess
solution NaOH therefore Ca2+
cannot be present in
A.
NB. Al(OH)3, Pb(OH)2 and Zn(OH)2 are amphoteric hydroxides which means they can react with
bases, this is why the ppt above ‘dissolved in excess NaOH’, it is really reacting with the NaOH and
forming a soluble product.
At this point we are left with 3 possibilities so we can use NH3(aq) on a fresh sample to further narrow
down.
Method Observations Inferences Explanation
To another sample of A These ions all form
NH3(aq) was added White ppt formed Al3+, Pb2+ or Zn2+ may be white, insoluble
dropwise present hydroxides w/NaOH
Then in excess Does not dissolve in Al3+ or Pb2+ may be Zn(OH)2(s) dissolves
excess present in excess NH3(aq)
therefore Zn2+ cannot
be present in A.
NB. Zn(OH)2 and Cu(OH)2 both ‘dissolve in excess’ NH3(aq) but are really reacting to form soluble
complexes. Cu2+ however was not even a possibility because neither solution A nor the ppt were blue.
It can be seen that Pb2+(aq) and Al3+(aq) give identical reactions with NaOH(aq) and NH3(aq) therefore to
distinguish between them KI(aq) (or any solution with iodide ions) is used.
Method Observations Inferences Explanation
To another sample of A A bright, yellow ppt Pb2+ present PbI2 is a bright yellow
add a few drops of KI(aq) is formed solid
Pb2+(aq) + 2I-(aq) → PbI2(s) The equation
supports your
inference
If Al3+ was present in A there would have been no observable reaction with KI(aq)
Observing the colour of the sample can also give an idea of ions present or those that are absent
Method Observations Inferences Explanation
Describe sample K K is a white sold Na+, NH4+, Ca2+, Al3+, Pb2+ These ions are
or Zn2+ may be present colourless and
typically are found in
white compounds
OR Cu2+, Fe2+ or Fe3+ are
absent These ions are
coloured
Describe sample L L is a pale green Fe2+ may be present Fe2+ ions in solution
solution are typically pale
green
Example
Method Observations Inferences
1. To solid J add HCl Effervescence, colourless,
odourless gas evolved which
formed a white ppt in lime water.
Use resulting solution Resulting solution was pale green
in tests 2 and 3 in colour.
2. To solution from test
1 add NaOH dropwise Dirty green ppt formed
Then in excess Does not dissolve in excess
Eqn:
3. To a fresh sample of
solution from test 1,
NH3(aq) was added Dirty green ppt formed
dropwise
Does not dissolve in excess
Then in excess
Identify solid J………………………………………………………………
