Figure 19.16
The general procedure for separating ions in qualitative analysis
19-2
Add
precipitating
ion
Centrifuge
Centrifuge
Add
precipitating
ion
Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display.
A qualitative analysis scheme for separating cations into five ion groups
19-3
Add
(NH4)2HPO4
Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display.
Centrifuge
Add
NH3/NH4+
buffer(pH 8)
Centrifuge
Centrifuge
Centrifuge
Add
6M HCl
Acidify to
pH 0.5;
add H2S
A qualitative analysis scheme for separating cations
into five ion groups
19-4
Add
NH3/NH4+
buffer(pH 8)
Add
(NH4)2HPO4
Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display.
Centrifuge
Centrifuge
Centrifuge
Add
6M HCl
Acidify to
pH 0.5;
add H2S
Centrifuge
Figure 19.17
Using pH and complexation
to Separate Ions
For Qualitative Analysis
19-5
Hot water
AgCl(s); Hg2Cl2(s)
Pb2+
CrO42,aqueous
ppt
2
Pbaqueous
 CrO42,aqueous 
 PbCrO4 , solid
19-6
Hot water
AgCl(s); Hg2Cl2(s)
Pb2+
CrO42,aqueous
NH3
HgNH2Cl(s)
ppt
Ag(NH3)6+


Ag aqueous
 NH3aqueous 
 AgNH 3aqueous
AgNH3aqueous  NH3aqueous 
 Ag NH 3  2 aqueous

19-7
Centrifuge
Centrifuge
Figure 19.18 A qualitative analysis scheme for Ag+,Al3+,Cu2+, and Fe3+
Step 1
Step 2
Add
Add HCl
NH3(aq)
Step 3 Add
NaOH
Centrifuge
Step 4
Add HCl,
Na2HPO4
19-8
Step 5
Dissolve in
HCl and
add KSCN
Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display.
Background
General Unknown
HCl
Group I
Insoluble Metal Chlorides
and Ammonia
HCl / H2S
Group II
Acid Insoluble Metal Sulfides
NH3 / H2S
Group III
Alkaline Insoluble Metal
Sulfides and Hydroxides
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Group IV
Soluble Metal
Ions
Background
• For this experiment, the group III ions are Fe3+, Ni2+, Mn2+,
Al3+ and Zn2+.
• These ions initially precipitate as either metal sulfides (in
an alkaline environment) or metal hydroxides.
• This requires the chemist to generate a small quantity of
sulfide ion to precipitate the metals.
• A convenient source of S2- is thioacetamide, which
decomposes when heated to give hydrogen sulfide
(H2S) which yields S2- in chemical reactions.
• A reagent that is made and consumed in the same
flask is said to be produced in situ.
Valdosta State University
Background – Hydrogen Sulfide
S
H
H
C
C
H
O
N
H
H
+ 2 H2O +
H
H+(aq)
H
C
C
H
O
+ NH4+(aq) + H2S(g)
H
Thioacetamide
H2S(aq) + 2 H2O(l) D 2 H3O+(aq) + S2-(aq)
• The addition of base to the second reaction consumes the hydronium ion
and drives the reaction to the right, increasing the concentration of S2-(aq).
Valdosta State University
Background – Group III Separation Scheme
Group III Unknown
NH3, H2S,
heat
NiS, FeS, MnS, ZnS,
Fe(OH)3, Al(OH)3
Group IV ions
HCl, HNO3, heat
Ni2+, Fe3+, Mn2+, Zn2+, Al3+
Waste
6 M NaOH
Fe(OH)3, Ni(OH)2,
Mn(OH)2
Al(OH)4-, Zn(OH)42-
HNO3
Divide
sample
HNO3
NaBiO3
NH3
MnO4purple
Conc.
NH3
Ni(NH3)62
Fe(OH)3
+
HCl /
NH4SCN
H2DMG
Fe(SCN)63blood red
Ni(DMG)2
strawberry red
ppt.
Al(OH)3
aluminon,
NH3
Al(OH)3aluminon
cherry red ppt.
Valdosta State University
Zn(NH3)42+
K4Fe(CN)6
K2Zn3[Fe(CN)6]2
white ppt.
Background – Group III Separation Scheme
A – Preparation of Group III cations
Group III unknown
NH3, H2S, Heat
NiS, FeS, MnS, Fe(OH)3,
Al(OH)3, ZnS
Group IV ions
• The group III ions are initially separated
from the bulk solution by precipitation as
either insoluble metal sulfides or
hydroxides.
Ni2+(aq) + S2(aq) D NiS(s) (black)
Fe2+(aq) + S2(aq) D FeS(s) (black)
Zn2+(aq) + S2(aq) D ZnS(s) (white)
Mn2+(aq) + S2(aq) D MnS(s) (pink)
Al3+(aq) + 3 OH(aq) D Al(OH)3(s) (white, gel)
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Background – Group III Separation Scheme
A – Preparation of Group III cations
Group III unknown
NH3, H2S, Heat
NiS, FeS, MnS, Fe(OH)3,
Al(OH)3, ZnS
Group IV ions
• Since iron has two common oxidation
states, its chemistry in this step is more
complex.
• If iron(III) is present it is reduced to iron(II)
and elemental sulfur in produced.
2 Fe3+(aq) + H2S (aq)  2 Fe2+(aq) + S(s) + 2 H+(aq)
Valdosta State University
Background – Group III Separation Scheme
A – Preparation of Group III cations
Group III unknown
• Alternately, the iron(III) can combine with the
hydroxide ion and precipitate as iron(III)
hydroxide.
NH3, H2S, Heat
NiS, FeS, MnS, Fe(OH)3,
Al(OH)3, ZnS
Group IV ions
Fe3+(aq) + 3 OH-(aq) D Fe(OH)3 (rust color)
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Background – Group III Separation Scheme
B1 – Separation of Group III cations
NiS, FeS, MnS, Fe(OH)3,
Al(OH)3
HCl, HNO3, Heat
Waste
• Following the precipitation, the metal
ions are combined with acid to form the
free (and soluble) metal ions.
Ni2+, Fe3+, Mn2+, Zn2+, Al3+
3NiS(s) +8H+(aq) + 2NO3(aq)  3Ni2+(aq) + 2NO(g) + 3S(s) + 4H2O(l)
FeS(s) + 2 H+ (aq)  Fe2+(aq) + H2S(aq)
3Fe2+(aq) + 4H+(aq) + NO3(aq)  3Fe3+(aq) + NO(g) + 2H2O(l)
MnS(s) + 2 H+(aq)  Mn2+(aq) + H2S(aq)
ZnS(s) + 2 H+(aq)  Zn2+(aq) + H2S(aq)
Al(OH)3(s) + 3 H+(aq)  Al3+(aq) + H2O(l)
Valdosta State University
Background – Group III Separation Scheme
B2 – Separation of Group III cations
Ni2+, Fe3+, Mn2+, Zn2+, Al3+
6 M NaOH
Fe(OH)3, Ni(OH)2, Mn(OH)2
Al(OH)4-, Zn(OH)42-
• Aluminum and zinc ions are amphoteric.
• This means that at high acid or base
concentrations, these metals form soluble
complexes, but precipitate at moderate pH.
• Iron, manganese and nickel form insoluble
hydroxides at high pH.
Fe3+(aq) + 3 OH(aq) D Fe(OH)3(s) (rust-color)
Ni2+(aq) + 2 OH(aq) D Ni(OH)2(s) (green)
Mn2+(aq) + 2 OH(aq) D Mn(OH)2(s) (light brown)
Valdosta State University
Background – Group III Separation Scheme
B2 – Separation of Group III cations
Ni2+, Fe3+, Mn2+, Zn2+, Al3+
6 M NaOH
Fe(OH)3, Ni(OH)2, Mn(OH)2
Al(OH)4-, Zn(OH)42-
• Aluminum and zinc ions are amphoteric.
• This means that at high acid or base
concentrations, these metals form soluble
complexes, but precipitate at moderate pH.
• Iron, manganese and nickel form insoluble
hydroxides at high pH.
Al3+(aq) + 3 OH(aq) D Al(OH)3(s) (white, gelatinous)
Zn2+(aq) + 2 OH(aq) D Zn(OH)2(s) (white)
Valdosta State University
Background – Group III Separation Scheme
B2 – Separation of Group III cations
Ni2+, Fe3+, Mn2+, Zn2+, Al3+
6 M NaOH
Fe(OH)3, Ni(OH)2, Mn(OH)2
Al(OH)4-, Zn(OH)42-
Excess Acid
Al(OH)3(s) + 3H+(aq) D Al3+ + 3 H2O(aq)
Zn(OH)2(s) + 2H+(aq) D Zn2+ + 2 H2O(aq)
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Background – Group III Separation Scheme
B2 – Separation of Group III cations
Ni2+, Fe3+, Mn2+, Zn2+, Al3+
6 M NaOH
Fe(OH)3, Ni(OH)2, Mn(OH)2
Al(OH)4-, Zn(OH)42-
Excess Base
Al(OH)3(s) + OH(aq) D Al(OH)4-(aq)
Zn(OH)2(s) + 2OH(aq) D Zn(OH)42-(aq)
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Background – Group III Separation Scheme
Fe(OH)3, Ni(OH)2, Mn(OH)2
HNO3, KNO2
Fe3+, Ni2+, Mn2+
Divide
Sample
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C1 – Test for Mn2+, Fe3+, Ni2+
• The precipitate is redissolved by adding
acid to the precipitate.
• The addition of nitric acid neutralizes the
sodium hydroxide and regenerates the
free cations.
• There is no easy method which will allow
Mn2+, Fe3+ and Ni2+ to be separated;
therefore, the sample is divided.
Background – Group III Separation Scheme
C2 – Test for Mn2+
Fe3+,
Ni2+,
Mn2+
Divide
Sample
NaBiO3
• If sodium bismuthate is added to a
solution containing manganese(II), a
redox reaction occurs resulting in the
formation of the purple permanganate
ion.
MnO4purple
14H+(aq) + 2Mn2+(aq) + 5BiO3-(s)  2 MnO4-(aq) + 5Bi3+(aq) + 7H2O(l)
Valdosta State University
Background – Group III Separation Scheme
D1 – Separation of Fe3+ and Ni2+
Fe3+,
Ni2+,
Mn2+
Divide
Sample
Conc. NH3
Fe(OH)3
Ni(NH3)62+
• The nickel and iron ions can be
separated by the addition of ammonia.
• The increased pH causes the formation
of the insoluble iron(III) hydroxide.
• The nickel ion combines with ammonia
to form a soluble complex ion,
hexaamminenickel(II).
Fe3+(aq) + 3NH3(aq) + 3H2O(l)  3NH4+(aq) + Fe(OH)3(s) (brown)
Ni2+(aq) + 6NH3(aq) D Ni(NH3)62+(aq) (blue)
Valdosta State University
Background – Group III Separation Scheme
D2 – Test for Fe3+
Fe3+,
Ni2+,
Mn2+
• The presence of the iron(III) ion is
confirmed by the addition of ammonium
thiocyanate.
• If iron(III) is present, a blood red solution
forms.
Divide
Sample
Conc. NH3
Fe(OH)3
Ni(NH3)62+
HCl / NH4SCN
Fe(SCN)63blood red
Fe3+(aq) + 6SCN-(aq) D Fe(SCN)63-(aq) blood red
Valdosta State University
Background – Group III Separation Scheme
E – Test for Ni2+
Fe3+,
Ni2+,
Mn2+
• The presence of the nickel ion is
confirmed by the addition of
dimethylglyoxime.
• Dimethylglyoxime combines with the
nickel ion to form a complex which forms
a strawberry red precipiate.
Divide
Sample
Conc. NH3
Ni(NH3)62+
Fe(OH)3
HCl / NH4SCN
H2DMG
Fe(SCN)63blood red
Ni(DMG)2
strawberry red ppt.
Ni(NH3)62+(aq) + 2 HC4H7N2O2(aq)  4NH3(aq) + 2NH4+(aq) + Ni(C4H7N2O2)2(s) (red)
Valdosta State University
Background – Group III Separation Scheme
Al(OH)4-,
Zn(OH)42-
HNO3
NH3
Al(OH)3
Zn(NH3)42+
F1 – Separation of Al3+ and Zn2+
• Careful control of pH allows for the
separation of aluminum and zinc ions.
• The solution is made very slightly basic.
• At these conditions, the aluminum ion
precipitates as aluminum hydroxide.
• The zinc ion remains in solution.
Al3+(aq) + 3 NH3(aq) + 3 H2O(l) D 3 NH4+(aq) + Al(OH)3(s)
Zn2+(aq) + 4 NH3(aq) D Zn(NH3)42+(aq)
Valdosta State University
Background – Group III Separation Scheme
Al(OH)4-, Zn(OH)42-
NH3
HNO3
Al(OH)3
Zn(NH3)42+
NH3,
aluminon
F2 – Test for Al3+
• A successful test for aluminum requires
that the previous reactions and their pH
control were properly performed.
• If not, false positive tests result.
• The test for aluminum requires the free
aluminum ion to react with ammonia in the
presence of a reagent called aluminon and
form a red precipitate.
• Be careful, if there is iron or zinc left in the
sample, a red precipitate will form resulting
in a false positive.
Al(OH)3 aluminon
cherry red ppt.
Al3+(aq) + 3 NH3(aq) + 3 H2O + aluminon(aq) D 3 NH4+(aq) + Al(OH)3aluminon(s) (red)
Valdosta State University
Background – Group III Separation Scheme
F2 – Test for Al3+
Al(OH)4-, Zn(OH)42-
• To confirm that the red precipitate is the
aluminum complex, ammonium carbonate
is added.
• If the red color does not fade, aluminum is
present.
NH3
HNO3
Al(OH)3
Zn(NH3)42+
NH3,
aluminon
Al(OH)3 aluminon
cherry red ppt.
Valdosta State University
Background – Group III Separation Scheme
G – Test for Zn2+
Al(OH)4-, Zn(OH)42-
NH3
HNO3
Zn(NH3)42+
Al(OH)3
NH3,
aluminon
K4Fe(CN)6
Al(OH)3 aluminon
cherry red ppt.
• To test for the zinc ion, a solution of
potassium hexacyanoferrate(II) is
added to the test solution.
• If zinc is present a white precipitate
forms.
• The exact color of the precipitate can
vary depending on the presence of
other ions.
• If iron is present the color can change
to yellow, green or blue.
K2Zn3[Fe(CN)6]2
white ppt.
3 Zn2+(aq) + 2 K+(aq) + 2 Fe(CN)64 (aq) D K2Zn3[Fe(CN)6]2 (s)
Valdosta State University
Background – Group III Separation Scheme
Zn2+
Al3+
Valdosta State University
Ni2+
Fe3+
Mn2+
Background
Group IV Unknown
Flame Test K+
lavender flame
Flame Test Na+
orange-yellow flame
HC2H3O2,
K2CrO4
BaCrO4
yellow ppt.
6M HCl
Ba2+
Flame Test
apple - green
6M H2SO4
(NH4)2C2O4
BaSO4
white ppt.
CaC2O4
white ppt.
Mg2+
NH3(aq), NaH2PO4,
heat
6M HCl
Flame Test
red-orange
Valdosta State University
Ca2+, Mg2+
MgNH4PO46H2O
white ppt.
Background – Group IV Separation Scheme
A – Flame test for Na+ and K+
Group IV Unknown
Flame Test K+
lavender flame
Flame Test Na+
orange-yellow flame
Na
Valdosta State University
• Insoluble salts of sodium and potassium
are not known.
• One method of determining the presence
of these ions is the flame test.
K
Background – Group IV Separation Scheme
B – Test for Ba2+
HC2H3O2,
K2CrO4
• The formation of a yellow precipitate on
the addition of potassium chromate
indicates the presence of the barium ion.
BaCrO4
yellow ppt.
6M HCl
Ba2+
Flame Test
apple - green
6M H2SO4
BaSO4
white ppt.
Valdosta State University
Ba2+(aq) + K2CrO4(aq)  BaCrO4(s) + 2K+(aq)
Background – Group IV Separation Scheme
B – Test for Ba2+
HC2H3O2,
K2CrO4
• A flame test (apple green) is used to
confirm the presence of the ion.
BaCrO4
yellow ppt.
6M HCl
Ba2+
Flame Test
apple - green
6M H2SO4
BaSO4
white ppt.
Valdosta State University
Background – Group IV Separation Scheme
B – Test for Ba2+
HC2H3O2,
K2CrO4
• The final test for barium is the formation of
a white precipitate on the addition of a
sulfate to the solution.
BaCrO4
yellow ppt.
6M HCl
Ba2+
Flame Test
apple - green
6M H2SO4
BaSO4
white ppt.
Valdosta State University
Ba2+(aq) + H2SO4(aq)  BaSO4(s) + 2H+(aq)
Background – Group IV Separation Scheme
Ca2+,
C – Test for Ca2+
Mg2+
• The calcium ion is separated from
magnesium by precipitating calcium
oxalate.
(NH4)2C2O4
CaC2O4
6M HCl
Flame test
red-orange
Valdosta State University
Mg2+
Ca2+(aq) + (NH4)2C2O4(aq) 
CaC2O4(s) + 2NH4+(aq)
Background – Group IV Separation Scheme
Ca2+,
C – Test for Ca2+
Mg2+
• A flame test (red-orange flame) is used to
confirm the presence of the calcium ion.
(NH4)2C2O4
CaC2O4
6M HCl
Flame test
red-orange
Valdosta State University
Mg2+
Background – Group IV Separation Scheme
Ca2+,
C – Test for Mg2+
Mg2+
• The magnesium ion is precipitated as a
white solid (ammonium phosphate) in an
alkaline solution.
K2C2O4
Mg2+
CaC2O4
Mg2+(aq) + NH3(aq) + HPO42-(aq)  MgNH4PO4(s)
NH3(aq)
Na2HPO4(aq)
6M HCl
Flame test
red-orange
MgNH4PO46H2O
white ppt.
Valdosta State University
Sample Problem 19.12
Separating Ions by Selective Precipitation
PROBLEM: A solution consists of 0.20M MgCl2 and 0.10M CuCl2. Calculate
the [OH-] that would separate the metal ions as their hydroxides.
Ksp of Mg(OH)2= is 6.3x10-10; Ksp of Cu(OH)2 is 2.2x10-20.
PLAN: Both precipitates are of the same ion ratio, 1:2, so we can compare
their Ksp values to determine which has the greater solubility.
It is obvious that Cu(OH)2 will precipitate first so we calculate the
[OH-] needed for a saturated solution of Mg(OH)2. This should
ensure that we do not precipitate Mg(OH)2. Then we can check
how much Cu2+ remains in solution.
SOLUTION:
Mg(OH)2(s)
Mg2+(aq) + 2OH-(aq) Ksp = 6.3x10-10
Cu(OH)2(s)
Cu2+(aq) + 2OH-(aq)
[OH-] needed for a saturated Mg(OH)2 solution =
= 5.6x10-5M
Ksp = 2.2x10-20
K sp
[Mg2 ]
6.3x1010

0.20