Qualitative Analysis of Metallic Elements
Ag+, Pb2+, Bi3+
Cu2+, Al3+, Cr3+
Ni2+, Co2+, Zn2+
Given a solution that contains
one or more of these cations
how can we use chemistry to
identify which ions are
present and which are not?
Differences in solubility
• Soluble vs. insoluble
Sb3+/Sb5+
• Solubility with temperature
Sn2+/Sn4+,
• Solubility and pH
Fe2+/Fe3+
• Complex ion formation
• Amphoteric behavior
• Changes in solubility with change
in oxidation state
Chemistry 123 – Dr. Woodward
Solubility Rules
Soluble vs. Insoluble is not very helpful. It
can really only be used to separate Ag+ and
Pb2+ from the other ions.
Chemistry 123 – Dr. Woodward
Ag+, Pb2+, Bi3+, Cu2+, Sb3+,
Sb5+, Sn2+, Sn4+, Fe2+, Fe3+,
Al3+, Cr3+, Ni2+, Co2+, Zn2+
Separation into
Groups
cold dilute HCl
AgCl, PbCl2
Remaining Cations & Pb2+
Group I –
Insoluble chlorides
H2S (+ HNO3)
PbS, Bi2S3, CuS,
Sb2S5, SnS2
Group II – Acid
insoluble sulfides
Group III– Base
insoluble sulfides
& hydroxides
Remaining Cations
1. NH4Cl, NH3
2. H2S
NiS, CoS, ZnS,
Fe(OH)3, Al(OH)3, Cr(OH)3
Groups 4 & 5
Chemistry 123 – Dr. Woodward
Group I – Insoluble chlorides
AgCl(s) ↔ Ag+(aq) + Cl−(aq)
Ksp = 1.8  10−10
PbCl2(s) ↔ Pb2+(aq) + 2Cl−(aq)
Ksp = 1.7  10−5
• AgCl & PbCl2 are both white solids.
• Dilute HCl is used rather than a chloride salt (i.e. NaCl)
to avoid precipitating SbOCl and BiOCl (which are soluble
in acidic solutions).
• If the chloride concentration is too high (conc. HCl)
complex ion formation, AgCl2− & PbCl42−, causes the
precipitates to dissolve.
• Because PbCl2 is only moderately insoluble adding HCl
does not completely remove Pb2+ from the mixture. So we
need to consider Pb2+ once again with the
group
cations.
Chemistry
123II
– Dr.
Woodward
Separating and Confirming Pb2+
• Because the solubility of PbCl2 increases considerably
upon heating (6.73 g/L at 0 °C vs. 33.4 g/L at 100 °C)
we can dissolve PbCl2 from AgCl by heating.
• To confirm the presence of lead we add potassium
dichromate to precipitate PbCrO4 which is a distinctive
orange-yellow solid
Favored for low pH (acidic)
Favored for high pH (basic)
Cr2O72−(aq) + H2O(l) ↔ 2CrO42−(aq) + 2H+(aq)
orange
yellow
Pb2+(aq) + CrO42−(aq) ↔ PbCrO4(s)
Ksp = 1.8  10−14
orange-yellow
Chemistry 123 – Dr. Woodward
Confirming Ag+
After separating Pb2+ we should have a white precipitate
of AgCl. How can we make sure the precipitate is AgCl
and not undissolved PbCl2? Concentrated NH3 should
dissolve the precipitate due to complex ion formation
AgCl(s) + 2NH3(aq) ↔ Ag(NH3)+(aq) + Cl−(aq)
If Pb2+ remains in solution the white precipitate Pb(OH)2
will form on making the solution basic.
To confirm Ag+ we reverse this by making the sol’n acidic
Ag(NH3)+(aq) + Cl−(aq) + 2H+(aq) ↔ AgCl(s) + 2NH4+(aq)
Chemistry 123 – Dr. Woodward
Separation into
Groups
Ag+, Pb2+, Bi3+, Cu2+, Sb3+,
Sb5+, Sn2+, Sn4+, Fe2+, Fe3+,
Al3+, Cr3+, Ni2+, Co2+, Zn2+
cold dilute HCl
AgCl, PbCl2
Remaining Cations & Pb2+
Group I –
Insoluble chlorides
H2S (+ HNO3)
PbS, Bi2S3, CuS,
Sb2S5, SnS2
Group II – Acid
insoluble sulfides
Group III– Base
insoluble sulfides
& hydroxides
Acidic solution
Remaining Cations
Basic (buffered)
solution
1. NH4Cl, NH3
2. H2S
NiS, CoS, ZnS,
Fe(OH)3, Al(OH)3, Cr(OH)3
Groups 4 & 5
Chemistry 123 – Dr. Woodward
Aqueous Chemistry of S2H2S(aq) ↔ H+(aq) + HS−(aq)
Ka1 = 9.5  10−8
HS−(aq) ↔ H+(aq) + S2−(aq)
Ka2 = 1  10−19
Because Ka2 is so small there are almost no S2− ions in solution.
We can neglect the second reaction.
Formation of metal sulfides occurs through reaction of HS− with
metal cations, and the concentration of HS− is dependent on the
pH. Consider the precipitation of CoS.
H2S(aq) ↔ H+(aq) + HS−(aq)
Ka1 = 9.5  10−8
Co2+(aq) + HS−(aq) ↔ H+(aq) + CoS(s)
1/Ksp = 2  10+21
Co2+(aq) + H2S(aq) ↔ 2H+(aq) + CoS(s) K = 1.9  10+14
Chemistry 123
Dr. Woodward
Increasing [H+] (lowering pH) drives the equilibrium
to –the
left.
Separation of Group II from Group III
Group II Cations
Pb2+(aq) + 2HS−(aq) ↔ PbS(s) + H+(aq)
Ksp = 3  10−28
Cu2+(aq) + 2HS−(aq) ↔ CuS(s) + H+(aq)
Ksp = 6  10−37
Group III Cations
Ni2+(aq) + 2HS−(aq) ↔ NiS(s) + H+(aq)
Ksp = 3  10−20
Co2+(aq) + 2HS−(aq) ↔ CoS(s) + H+(aq)
Ksp = 6  10−22
By making the solution acidic we shift the equilibrium to
the left (favors reactants) which makes the Group III
Chemistry 123 – Dr. Woodward
sulfides dissolve, but not the Group
II sulfides.
Group II – Acid Insoluble Sulfides
Pb2+(aq) + 2HS−(aq) ↔ PbS(s) + H+(aq)
Black ppt
2Bi3+(aq) + 3HS−(aq) ↔ Bi2S3(s) + 3H+(aq)
Dk. Brown ppt
Cu2+(aq) + HS−(aq) ↔ CuS(s) + H+(aq)
Black ppt
SnCl62−(aq) + 2HS−(aq) ↔ SnS2(s) + 4H+(aq) + 6Cl−(aq)
Yellow ppt
2SbCl6−(aq) + 5HS−(aq) ↔ Sb2S5(s) + 5H+(aq) + 12Cl−(aq)
Orange ppt
Chemistry 123 – Dr. Woodward
Pb2+, Bi3+, Cu2+, Sb3+, Sb5+, Sn2+, Sn4+
HNO3 acts as an
oxidizing agent
Cl− acts as a
complexing agent
HNO3 + HCl
(Aqua regia)
Pb2+, Bi3+, Cu2+, SbCl61−, SnCl62−
CH3CSNH2
(thioacetamide)
decomposes on
heating to give
~0.10 M H2S(aq)
Removes excess
acid, be careful
not to overdo it.
Evaporate to
a paste
HNO3
CH3CSNH2, heat
PbS (black), Bi2S3 (dark brown), CuS (black),
Sb2S5 (orange), SnS2 (yellow)
NaOH
Copper subgroup
PbS, Bi2S3, CuS
SnS2 & Sb2S5 are
amphoteric
Antimony subgroup
3− SnO 3−
SbS43−,Chemistry
SbO43−,123
SnS
4
– 4Dr. ,Woodward