Identification of metallic and non-metallic ions I
IMPORTANT POINTS
Metallic ions or cations are identified based on the colour and solubility of the hydroxides formed from
their reaction with aqueous ammonia and sodium hydroxide.
Non-metallic ions or anions are generally identified based on the evolution and identification of gases
produced when the compounds are heated or reacted with acids. The colour and solubility of precipitates
formed on reaction with barium ions and ammonia (for the halide
ions) are also used.
2+
3+
Let us summarise the reactions of the metallic ions: Pb , Al ,
2+
2+
3+
2+
4+
2+
Ca , Zn , Fe , Fe , NH and Cu with aqueous ammonia and
sodium hydroxide.



2+
Fe produces a dirty-green precipitate, which is insoluble
in excess aq NH3 and NaOH
3+
Fe produces a rusty-brown precipitate, which is insoluble
2+
in excess aq NH and NaOH
2+
Cu forms a blue precipitate, which is insoluble in excess
aq NaOH, but soluble in excess aq NH3.
These cations can clearly be identified based on their colours
alone.
3+
2+
3 Al and Pb both form white precipitates when reacted with aq
NaOH and NH3.
In aq NaOH, the precipitates are soluble in excess, while in aq NH3 they are insoluble in excess. As can
3+
2+
be seen, Al and Pb give the same results. So, to differentiate between them, other reactions must be
2+
2taken into consideration. Pb will form precipitates with I , Cl and SO4 ions (PbI2 (yellow), PbC2 (white)
3+
and PbSO4 (white)) while Al will not.
2+
Two other cations form white precipitates. Zn forms a white precipitate, which is soluble both in excess
3
2+
aq NaOH and NH Ca forms a white precipitate which is insoluble in excess NaOH but does not form a
precipitate when treated with ammonia.
+
The NH4 ion can be identified based on the evolution of ammonia (NH 3) gas when warmed with aq
NaOH.
2-
The identification of the anions is a lot more complicated. The ones we are concerned with are CO 3 ,
22SO4 , SO3 , NO3 , Br , Cl and I .
One way to do this is to heat the compound and identify the gas produced.
2-
CO3 =(heat) produces the gas CO2
NO3- ===== (heat) gives off O2 and/or NO2
2-
SO3 ===== (heat) gives off SO2
These gases can be identified by their characteristic tests.





CO2 is weakly acidic and turns lime water (Ca(OH) 2) milky
O2 is neutral and relights a glowing splint
NO2 is a red-brown/ yellow-brown acidic gas
SO2 is an acidic gas which turns acidified potassium manganate (VII) solution from purple to
colourless and changes acidified potassium dichromate (VI) from orange to green.
NH3 is an alkaline gas which forms dense white fumes with hydrogen chloride gas.
Some anions react with dilute and concentrated acids (if they are added to the solid).
2-
CO3 reacts with dilute and concentrated acids to release CO 2.
2-
SO3 forms SO2 gas on reaction with dilute and concentrated acids.
-
-
-
If conc H2SO4 acid is added to a solid containing the halide ions (I , Cl , and Br ), the following gases can
be formed: HI(g), which decomposes to I2, HCl(g) and HBr, which decomposes to Br2.
The halide ions, in solution, can be identified differently, based on their reaction with silver nitrate
solution (AgNO3) and the solubility of the precipitate formed in ammonia.
-
Cl forms a white precipitate with AgNO3 solution, which is soluble in aq NH3.
-
Br forms a cream precipitate with AgNO3, which is partially soluble in aq NH3.
-
I forms a light-yellow precipitate, which is insoluble in aq NH3.
Finally, some anions form precipitates with a solution of Ba
2+
ions
2-
CO3 forms a white precipitate, which dissolves on addition of acid to form CO2.
2-
SO3 forms a white precipitate, which dissolves on addition of acid to give SO2
2-
SO4 forms a white precipitate, which is insoluble in dilute acid
Identification of metallic and non-metallic ions II
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
222CO3 , SO4 or SO3 ions. The addition of acid allows us to determine which of these ions is
22present. CO3 and SO3 will react with the acid and give off CO2 and SO2 gases, respectively.
2SO4 ion does not react and is in fact insoluble in the acid.
2+
3+
2+
There are three possible ions (Zn , Al and Pb ) 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 ).
2+
The ions that form a white precipitate with ammonia and which are insoluble in excess are Pb
3+
2+
and Al . Zn 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
3+
2+
between Al and Pb . 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.
2+
- 2Remember that Pb forms precipitates with Cl , I and SO4 to form PbCl2 (white), Pbl2 (yellow)
and PbSO4 (white) respectively.
TEST
i. To a sample of Solution Y, dilute
nitric acid was added, followed by a
few drops of silver nitrate solution.
OBSERVATIONS
INTERFERENCES
No Precipitate formed.
No halide ion is present
- (Cl , I or Br
2-
ii. To a sample of Solution Y, dilute
hydrochloric acid was added,
followed by a few drops of barium
chloride solution.
White precipitate formed.
SO4 ion is present.
2+
2Ba (aq)+ SO4 (aq) =
BaSO4(s)
White precipitate is barium
sulphate
3+
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.
iv. To a sample of Solution Y,
aqueous ammonia was added until in
excess.
v. To a sample of Solution Y, some
potassium iodide was added.
White precipitate, soluble in
excess.
No effect on blue litmus, red
litmus turned blue.
White precipitate, insoluble in
excess.
No yellow precipitate.
2+
Al , Pb or Zn
be present.
2+
ion may
Alkaline gas is given off
Gas is Nh3
+
Cation is NH4
3+
Al , or Pb
present
2+
may be
2+
Pb is not present
3+
Cation is Al
3+
vi. Which metal ion was present? The aluminium metal ion (Al ) was present.
vii. Give a reason for your answer. The aluminium ion would not form a precipitate with iodide
ions.
In the question above, no yellow precipitate was formed when potassium iodide was added and this
3+
suggests that lead ions were not present. The metal ion present was, therefore, Al .
Identification of metallic and non-metallic ions III
Question 1
Identify each lettered substance below by giving its name and formula (with the aid of equations).
Q is a sodium salt. When a solution of Q was added to aqueous silver nitrate, a white precipitate, R, was
formed. This precipitate did not dissolve in dilute nitric acid.
S is a sodium salt. When a solution of S was added to aqueous barium nitrate, a white precipitate, T, was
formed. When diluted nitric acid was added to T, it dissolved and a gas, U, which turned lime water milky.
When aqueous sodium hydroxide was added to aqueous zinc chloride, a white precipitate, V, was
formed. This precipitate dissolved in excess of sodium hydroxide to form a colourless solution in which
zinc was present as a salt, W.
ANSWERS
-
(a) The ion present is Cl , which means that R is AgCl (silver chloride) and Q is NaCl (sodium chloride).
+
Ag (aq) + Cl (aq) = AgCl (white precipitate).
2-
(b) The gas produced, U, is CO2, which means that the ion is CO3 . T is BaCO3 (barium carbonate),
hence the sodium salt S is Na2CO3 (sodium carbonate).
2
2-
Ba +(aq) + CO3 (aq) == BaCO3 (s) white
BaCO3(s) + HNO3 (aq) == Ba(NO3)2(aq) + H2O (l) + CO2(g)
(c) V is zinc hydroxide (Zn(OH)2) and in excess NaOH, W, which is sodium zincate (Na2ZnO2), is formed.
2+
Zn (aq) + 2OH (aq) == Zn(OH)2 (s)
Zn(OH)2(s) + 2NaOH (aq) == Na2ZnO2 (aq) + 2H2O (l)
Question 2
Describe what is observed (with the aid of equations) in each of the following reactions.
(i) Aqueous NaOH is added to aq iron(III) sulphate
(ii) Dilute HCl is added to solid sodium carbonate
(iii) Aqueous barium chloride is added to dilute sulphuric acid
(iv) Aqueous silver nitrate is added to aq sodium chloride.
(8 marks)
ANSWERS
(a) A rusty-brown precipitate is formed, which is insoluble in excess NaOH.
3+
Fe (aq) + 3OH-(aq) == Fe(OH)3 (s)
(b) Sodium carbonate would dissolve and a gas would be given off
Na2CO3(s) + 2HCl (aq) == 2NaCl (aq) + H2O(l) + CO2 (g)
(c) A white precipitate will be seen, which will remain insoluble.
2+
2-
Ba (aq) + SO4 (aq) == BaSO4(s)
(d) The silver nitrate would react with the chloride ions to form a white precipitate of silver chloride
+
-
Ag (aq) + Cl (aq) == AgCl (s)