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Identification
of metallic and non-metallic ions
(Pt. 2)
Francine
Taylor Campbell, Contributor
LAST
WEEK, we looked at the reactions of
some anions and cations, which are mentioned
in the CXC chemistry syllabus.This week,
we hope to use the knowledge gained
to answer questions, which require the
identification of these ions. Attempt
the question given above using the guidelines
shown.
COMMENTS
*
Whenever the reagent silver nitrate
is used this implies that we are searching
for halide ions. Cl-, Br- and I- form
white, cream and yellow precipitates
with silver nitrate solution, respectively.
Once this solution is added and no
precipitate is seen, this suggests
that no halide ion is present.
*
Whenever barium ions are added to
a solution, one can assume that we
are searching for either CO32-, SO42-
or SO32- ions. The addition of acid
allows us to determine which of these
ions is present. CO32- and SO32- will
react with the acid and give off CO2
and SO2 gases, respectively. SO42-
ion does not react and is in fact
insoluble in the acid.
*
There are three possible ions (Zn2+,
Al3+ and Pb2+) that produce a white
precipitate, which is soluble when
excess sodium hydroxide is added.
These
three ions must be listed and further
tests and observations done to determine
which ion is present. Alkaline substances
turn red litmus paper to blue. Ammonia
is therefore the gas evolved, as it
is alkaline in nature and would be
formed from an ammonium ion (NH4+).
*
The ions that form a white precipitate
with ammonia and which is insoluble
in excess are Pb2+ and Al3+. Zn2+
is, therefore, not the cation present,
as it is now eliminated from the list
of possibilities.
*
To identify which metal ion is really
present further tests must be carried
out to differentiate between Al3+
and Pb2+. In an earlier lesson, it
was seen that aluminium and lead ions
exhibited similar reactions with aqueous
ammonia and sodium hydroxide.
Thus
if one hopes to differentiate between
them then solutions of iodide, chloride
or sulphate ions can be used.
Remember
that Pb2+ forms precipitates with
Cl-, I- and SO42- to form PbCl2 (white),
PbI2 (yellow) and PbSO4 (white), respectively.
In the question above, no yellow precipitate
was formed when potassium iodide was
added and this suggests that lead
ions were not present. The metal ion
present was therefore Al3+.
| TEST |
OBSERVATIONS |
INTERFERENCES |
| i.
To a sample of Solution Y, dilute
nitric acid was added, followed
by a few drops of silver nitrate
solution. |
No
precipitate formed. |
No
halide ion is present (CI, I or
BR) |
| ii.
To a sample of Solution Y, dilute
hydrochloric acid was added, followed
by a few drops of barium chloride
solution. |
White
precipitate formed. |
SO4²-
ion is present.
Ba²+
(aq) + SO4²-
(aq) = BaSO4 (s)
White precipitate is barium sulphate.
|
| iii.
To a sample of Solution Y, sodium
hydroxide was added until in excess.
The mixture was warmed and gas
tested with blue and red litmus. |
White
precipitate
soluble in excess.
No
effect on blue litmus, red litmus
turned blue
|
A13+,
Pb²+
or Zn²+
ion may be present.
Alkaline
gas is given off. Gas is NH3
Cation is NH4+
|
| iv.
To a sample of Solution Y, aqueous
ammonia was added until in excess. |
White
precipitate
insoluble in excess. |
A13+,
Pb²+
may
be present |
| v.
To a sample of Solution Y, some
potassium iodide was added. |
No
yellow precipitate. |
Pb²+
is
not present Cation is A13+ |
| vi.
Which metal ion was present? The
aluminium metal ion (A13+)
was present. |
| vii.
Give a reason for your answer.
The aluminium ion would not
form a precipitate with iodide
ions. |
*
Francine Taylor-Campbell is
an independent contributor.
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