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Identification
of metallic and non-metallic ions
- revision 2
Francine
Taylor-Campbell, Contributor
This
week, we will continue to look at
questions having to do with qualitative
analysis. We will first examine the
question that was given at the end
of last week's lesson. I hope you
attempted it.
ANSWERS
TO LAST WEEK'S QUESTION
Q3.
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.
|
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 litus, 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+
|
Comments
1.
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.
2.
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.
3.
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+).
4.
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.
5.
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, 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+.
Francine
Taylor-Campbell is an independent
contributor. Send questions and comments
to kerry-ann.hepburn@gleanerjm.com
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