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Identification
of metallic and non-metallic ions
Francine
Taylor-Campbell, Contributor
Last
week, we looked at the reactions of
some anions and cations. This week
we hope to use the knowledge gained
to answer questions which require
the identification of these ions.
Q1.
A student conducted a number of tests
on an aqueous solution of Compound
Y. The observations that were made
are recorded in the table below. You
are required to fill in the interferences
that could be made based on the observations
recorded.
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 (NH3+).
- The
ions that form a white precipitate
with ammonia and which are 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),
Pbl2 (yellow) and PbSO4
(white) respectively.
| 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
(Cl-, 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. |
Al3+,
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.
|
Al3+,
or 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 Al3+
|
| vi.
Which metal ion was present?
The aluminium metal ion
(Al3+)
was present. |
| vii.
Give a reason for your answer.
The aluminium ion would
not form a precipitate with
iodide ions. |
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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+.
Francine
Taylor-Campbell is an independent
contributor.
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