Chem 1B
Dr. White
Saddleback College
1
Experiment 13/14: Separation and Identification of Group II Cations
2+
3+
2+
2+
,
3+
(The Acidic Sulfide Group: Cu , Bi , (Pb ), Sn , and Sb )
Objectives
• To understand the chemical reactions involved in
2+
3+
2+
the separation and identification of Cu , Bi , (Pb ),
2+
3+
and Sn , and Sb
• To successfully identify the Group II cation(s)
present in an unknown solution.
Introduction
The cations in Group II and Group III form insoluble
sulfides. Table 1 shows the sulfide solids formed by
the Group II and Group III cations and their Ksp value.
Table 1: Ksp Values at 25°C for the sulfide solids of
the Group II and Group III Cations
Ionic Solid
Ksp
-36
CuS
8.5 x 10
-72
Bi2S3
1.6 x 10
-28
CdS
1.0 x 10
-54
HgS
Group II
1.6 x 10
-27
PbS
1 x 10
-46
SnS2
1 x 10
-93
Sb2S3
1.6 x 10
-21
NiS
2.0 x 10
-21
CoS
1 x 10
-19
FeS
3.7 x 10
-13
MnS
Group III
2.3 x 10
-20
Cr2S3
1 x 10
-14
Al2S3
4.0 x 10
-22
ZnS
2.5 x 10
Referring to the Ksp values of solid sulfides found in
Table 1, one can see that the Group II ions form very
insoluble sulfide solids. Thus, only a small amount of
sulfide ion is needed for precipitation. To see how
this can be, we will look at an example. Copper (II) is
a group II cation, and its sulfide salt is insoluble.
CuS (s) ⇆ Cu
2+
2-
(aq) + S (aq)
A solubility product constant expression can be
written for this reaction and the experimentally
determined value for the solubility product constant of
copper (II) sulfide is given:
2+
2-
Ksp = [Cu ][S ] = 8.5 × 10
–36
During a typical analysis, the cation concentrations
are roughly 0.10 M. To cause copper (II) cations to
precipitate as the sulfide salt, the product of the ion
concentrations must be greater than the Ksp, which is
–36
8.5 × 10 . Solving for the concentration of the
sulfide ions to cause copper (II) sulfide to precipitate:
[S 2− ] =
K sp
2+
[Cu ]
=
8.5 x 10 -36
= 8.5 x 10 -35 M
0.10
The sulfide ion concentration must be just greater
than
€ this to precipitate copper (II) sulfide, and this
value is typical for the other group II cations. A group
III cation that is also precipitated by the sulfide ion is
the nickel (II) ion.
NiS (s) ⇆ Ni
2+
2-
(aq) + S (aq)
However, the solubility product constant for nickel (II)
sulfide is
2+
2-
Ksp = [Ni ][S ] = 2.0 × 10
–21
To precipitate nickel (II) sulfide, the sulfide ion
concentration must be a lot larger:
[S 2− ] =
K sp
2+
[Ni ]
=
2.0 x 10 -21
= 2.0 x 10 -20 M
0.10
This means that even though both of these groups
are€ precipitated by the same reagent, by carefully
controlling the concentration of sulfide ions, the group
II cations may be precipitated out without
precipitating out the group III cations. To precipitate
out the group II cations but not the group III cations,
the sulfide ion concentration must be greater than
–35
–20
8.5 × 10 M but less than 2.0 × 10 M.
The amount of sulfide ion present in solution is
controlled by the pH of the solution. The sulfide
concentration in an acidic solution is extremely small,
-22
due to the small Ka value for H2S (1.0 x 10 ). This is
seen in the equilibrium reaction below:
2-
+
H2S(aq) + 2 H2O(l)  S (aq) + 2 H3O (aq)
An acid ionization constant expression can be written
for it. The value of its acid ionization constant is given
below:
Ka =
[H 3O+ ][S 2− ]
= 1.0 x 10 -22
[H 2 S]
A saturated solution of hydrogen sulfide has a
concentration
of 0.10 M. To precipitate out the
€
cations of group II but not the cations of group III, a
–22
sulfide concentration of about 1.0 × 10
M would
–35
work (larger than 8.5 × 10
M but less than 2.0 ×
–20
10 M). This can be accomplished by adjusting the
pH (the hydronium ion concentration) of the solution.
We can solve the acid ionization constant expression
Chem 1B
Dr. White
for the hydronium ion concentration that would give a
–22
sulfide ion concentration of 1.0 × 10 M,
1/2
⎡ K [H S]⎤1 / 2 ⎡ (1.0 x 10 -22 )(0.10) ⎤
[H 3O+ ] = ⎢ a 2−2 ⎥ = ⎢
= 0.32M
⎥
⎣ [S ] ⎦
1.0 x 10 -22
⎣
⎦
A hydronium ion concentration of 0.32 M corresponds
€ to a pH of 0.50, which is very acidic.
+
pH = –log[H3O ] = –log(0.32 M) = 0.50
To get the hydronium ion concentration this high, all
that is needed is to add a strong acid (like
hydrochloric acid) to the solution, and pH paper can
be used to test the solution to insure the proper pH.
This is why the aqueous sample of cations is made
very acidic and then hydrogen sulfide is introduced,
only the group II cations wil precipitate out as
insoluble sulfides under these conditions.
Mercury (II) is a group II cation (one we will not test
for in our lab), and therefore, should precipitate out in
a saturated hydrogen sulfide solution with a pH of
0.50. To verify this, the ion product, Qsp, can be
calculated for mercury (II) sulfide (knowing the
solution is approximately 0.10 M in mercury (II) and
–22
1.0 × 10 M in sulfide when the pH is adjusted to
0.50), and then comparing it to the Ksp for mercury (II)
sulfide:
HgS (s) ⇆ Hg
2+
2+
2-
(aq) + S (aq)
2-
Ksp = [Hg ][S ] = 1.6 × 10
2+
2-
Qsp = [Hg ][S ] = (0.10)(1.0 × 10
–22
–54
) = 1.0 x 10
-23
Because Qsp>Ksp, the reverse reaction is
spontaneous, and the mercury (II) sulfide will
precipitate out from this solution.
To precipitate out the cations of group III, the sulfide
ion concentration would have to be higher than
–19
–6
2.0 × 10 M, so a concentration around 1.0 × 10 M
would work. To determine the pH needed to attain
this concentration, we can again solve for the
hydronium ion concentration in the acid ionization
constant expression for hydrogen sulfide:
1/ 2
⎡ K [H S]⎤1/ 2 ⎡ (1.0 x 10 -22 )(0.10) ⎤
−9
[H 3O+ ] = ⎢ a 2−2 ⎥ = ⎢
⎥ = 3.2x10 M
-6
⎣ [S ] ⎦
1.0 x 10
⎣
⎦
This corresponds to a pH of 8.50, which is basic.
€
+
pH = –log[H3O ] = –log(3.2 × 10
–9
M) = 8.50
So once the group II cations are precipitated out, the
group III cations can be precipitated out by making
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2
the aqueous sample of cations basic and then adding
hydrogen sulfide. To make the solution basic, all that
is needed is to add a base (like ammonia) to the
solution, and pH paper can be used to test the
solution to insure the proper pH.
Manganese (II) is a group III cation, and therefore,
should not precipitate out in a saturated hydrogen
sulfide solution with a pH of 0.50, but should
precipitate out in a saturated hydrogen sulfide
solution with a pH of 8.50. To verify this, the ion
product, Qsp, can be calculated for manganese (II)
sulfide at both pH’s, and then comparing it to the Ksp
for manganese (II) sulfide:
MnS (s) ⇆ Mn
2+
2+
2-
(aq) + S (aq)
2-
Ksp = [Mn ][S ] = 2.3 × 10
2+
2-
–6
–13
Qsp = [Mn ][S ] = (0.10)(1.0 × 10 ) = 1.0 x 10
-7
At pH = 0.50, because Qsp<Ksp, the forward reaction
is spontaneous, and the manganese (II) sulfide will
completely dissolve in this solution. However, at pH =
8.50, because Qsp>Ksp the reverse reaction is
spontaneous, and the manganese (II) sulfide will
precipitate out from this solution. Therefore,
manganese (II) is a Group III cation.
After careful adjustment of the pH, the sulfide ion
needed for precipitation within the solution is
generated with the addition of thioacetamide
(CH3CSNH2) and heat. Then, the solid is treated with
1 M NaOH solution, which dissolves the sulfides of tin
and antimony and allows for their separation and
confirmation. The sulfides unaffected by NaOH, with
2+
will dissolve in hot 6 M HNO3. Then, Cu is
3+
2+
separated from Bi and Pb by the reaction with
2+
ammonia to form ammine complex ions of Cu .
These are the main ways the ions are separated in
the Group II analysis. Of course, confirmation tests
will be used to make a definitive conclusion of each
of the cations in Group II. The left column in the
experimental procedure will give more information
about the reactions in each step.
Since five different metals form cations which fall into
this group, the procedure is relatively complicated.
Extreme care must be taken to carefully follow the
procedure and make careful observations.
Note that since, PbCl2 tends to not precipitate
completely in the Group I separation, it is also listed
as one of the Group II ions. You will only concern
yourself with the lead part of this lab in the general
unknown at the end of the semester.
Chem 1B
Dr. White
Reagents Available
0.3 M HCl
1M thioacetimide
6 M HCl
6 M NaOH
6 M NH3
1 M NH4Cl
1M NaOH
0.5 M NaC2H3O2
Al (s)
0.1 M HgCl2
12 M HCl
6 M HC2H3O2 Na2S2O3 (s)
6 M HNO3
0.1 M SnCl2
Na2S2O4 (s)
Dilute, Known Solution that contains all the Group II
2+
cations (NOT Pb )
Safety
All of the Group II cations are toxic and 12 M HCl, 6
M HCl, NH3 and NaOH, and HNO3 are irritants.
Avoid contact and wash immediately if any is spilled
or splashed on you. Wear eye protection at all times.
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3
As you perform the experiment, collect all waste
solutions in a waste beaker. This mixture should then
be discarded in the appropriate waste container. DO
NOT POUR ANY OF THE SOLUTIONS DOWN THE
DRAIN.
Unknowns and Knowns
A known solution containing four Group II cations
2+
(excluding Pb ) is provided for your use. Testing a
known sample is helpful in this analysis since doing
so will allow you to observe what a positive test looks
like. It is usually convenient to test a known sample
simultaneously with your unknown. Note that the
experimental conditions, such as pH, for the known
test must be the same as that for the unknown.
The H2S (g) produced is foul smelling and highly
toxic. Work under a fume hood at all times and avoid
smelling the gas.
Outline of Procedure: Record all observations for each step in the procedure.
Experimental Procedure:
1. (a) As a control, use 0.3 M HCI (it's on the
reagent shelf) to see what a pH 0.5 solution should
look like on 0-3 pH paper. Put a drop of this 0.3 M
HCl on the pH paper and note the color. This is a
pH of 0.5.
Chemistry, Comments, and Background Information:
1. Precipitation of the Group II cations at a pH of 0.5
2-22
A pH of 0.5 gives a [S ] of 1 x 10 M, which will precipitate
the Group II cations without precipitating the Group III
cations. So, before precipitation, the pH must be adjusted to
0.5.
(b) Take 2 mL of the solution to be tested (known
or unknown) and add 6 M NaOH dropwise with
mixing until you see some additional precipitate
that remains after mixing or until you've added 20
drops of the 6 M NaOH, whichever comes first.
Then add 2 drops of 6 M HCl. Next, test a drop of
the solution on the pH paper. Note the color you
first see (not a later color resulting from spreading,
separating or changing); if it matches the color from
the 0.3 M HCl, proceed to step 1d, if not go onto
the next step (1c).
(c) Adjust with 6 M NaOH or 6 M HCI to get a pH
close to 0.5. If it is too acidic add a drop of NaOH.
If it is too basic add a drop HCl. (Use the color
scale on the pH paper box to judge this.)
(d) Once the 0.5 pH is established, add 1 mL 1 M
thioacetamide and stir. Heat the centrifuge tube in
a boiling water bath for 5 minutes. A precipitate will
form that will darken over time. Heat an additional
2 minutes once the color stops changing.
The source of H2S is thioacetimide (CH3CSNH2). The
compound decomposes when heated to produce H2S:
(e) Cool the centrifuge tube under the water tap
and let stand for a minute or so. Centrifuge out the
precipitate and with a capillary pipet, carefully
decant out as much of the clear supernatant
(absolutely no traces of precipitate) into a clean
centrifuge tube. Set the precipitate aside for step
The H2S then creates S needed for precipitation of the
Group II ions by the following equilibrium reaction:
+
CH3CSNH2(aq) + 2H2O(l) + H (aq) →
+
CH3COOH(aq) + NH4 (aq) + H2S(aq)
2-
2-
+
H2S (aq) + 2H2O (l)  S (aq) + 2H3O (aq)
Heating in step 1d increases the particle size of the
Chem 1B
Dr. White
Perform steps 1f-h ONLY in the general
unknown. Otherwise discard the supernatant
from step 1e and proceed to step 2.
(f) Check the pH of the supernatant. If it is too low
(below pH 0.5), add 0.5 M NaC2H3O2 drop by drop
with stirring until the pH is again 0.5. A brown or
yellow precipitate may form which indicates that not
st
all the Group II ions precipitated the 1 time.
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4
precipitate and this makes it easier to separate from the
solution.
CuS, and PbS are black precipitates; Bi2S3 is brown; SnS2 is
yellow; and Sb2S5 is an orange-red precipitate. A precipitate
of unpredictable color will result from a mixture containing
various sulfide salts.
(g) Add 0.5 mL 1 M thioacetimide and heat for 3
minutes in the boiling water bath. Cool the
centrifuge tube under the water tap and let sit for 1
minute. Centrifuge out any precipitate and decant
the supernatant into a tube (save this for the Group
III analysis). Add 1 mL 1 M NH4Cl and 1 mL DI
water to the precipitate, stir, and add this mixture to
the centrifuge tube containing the precipitate from
step (1e). Make sure to get all the precipitate
transferred. If necessary, use two small portions of
NH4Cl to get it all transferred.
(h) Centrifuge the combined precipitates and
decant, discarding the supernatant.
2.* (a) Take the precipitate from step 1 and wash it
with 1 mL 1 M NH4Cl and 1 mL DI water.
Centrifuge and then discard the wash. Repeat this
washing once more.
(b) Add 2 mL 1 M NaOH. Heat in the water bath
with stirring for 2 minutes.
(c) Centrifuge and decant the supernatant into a
clean centrifuge tube. BE SURE TO LABEL THE
TUBE. Wash the precipitate twice with 2 mL DI
water and 1 mL 1 M NaOH. Stir, centrifuge, and
decant, discarding the wash each time. Set the
precipitate aside for further testing. LABEL THE
TUBE!
3. To the supernatant from step (2c) add 6 M HCl
until you start to get a precipitate, then check with
litmus paper to get it acidic (turns blue paper red).
Then, check the pH on pH paper to get it to a pH of
0.5. Stir well for 30 seconds, centrifuge and then
discard the supernatant.
2. Separation of Sn
complex ions:
4+
and Sb
3+
by the formation of
Washing the precipitate with NH4Cl enhances the production
of large particles in the precipitate. A precipitate with very
small-sized particles is difficult to wash and separate from a
solution. Washing the precipitate with water releases any
contaminating ions trapped inside the particles of the
precipitate.
The tin (IV) and antimony (III) ions form soluble complex ions
2with NaOH. Sn(OH)6 (aq) and Sb(OH)4 (aq) form.
*Do this step carefully because insufficient extraction
will cause the Sn and Sb tests to fail.
4+
3+
3. Re-precipitation of Sn and Sb :
2The Sn(OH)6 (aq) and Sb(OH)4 (aq) are precipitated by
+
adding an acid. The H3O reacts with the OH in the complex
and the sulfide solid is formed again.
The balanced net ionic chemical equation, for the reprecipitation of SnS2 and Sb2S3 follow:
2-
+
2-
Sn(OH)6 (aq) + 6H3O (aq) + 2S (aq) → SnS2 (s) + 12 H2O (l)
2-
+
2-
2Sb(OH)4 (aq) + 8H3O (aq) + 3S (aq) → Sb2S3 (s) + 16 H2O(l)
4. (a) To the precipitate from step (3), add 2 mL 6
M HCl.
(b) Stir and transfer the mixture to a 50-mL beaker.
Boil the liquid gently for one minute to drive out the
H2S and dissolve the sulfides. (Don’t boil to
dryness).
(c) Add 1 mL 6 M HCl and then pour the solution
4. Re-dissolving the SnS2(s) and Sb2S3(s):
The SnS2(s) and Sb2S3(s) readily dissolve in excess HCl at
2100°C with the formation of the chloro complexes SnCl6 and
3SbCl6 .
Chem 1B
Dr. White
into a centrifuge tube. Centrifuge out any solid
residue and transfer the liquid to a centrifuge tube.
Discard the solid.
5. (You must finish this step completely – don’t
stop half way through)
(a) Pour half the solution from step (4c) in a
centrifuge tube and add a 1-cm length of filed 24guage Al wire. Heat the test tube in the hot water
for three minutes or until all of the aluminum has
reacted away. Once the aluminum has reacted
away, heat for two more minutes. If the
aluminum wire is still present after the three
minutes, remove it, and heat the solution for two
additional minutes. Do not let the solution
evaporate to dryness. If the volume of the
solution gets low, replenish it with drops of 6 M
hydrochloric acid. Separate the solution from any
black solid residue (which would be antimony
metal, indicating that antimony is present) using
a capillary pipet, and transfer the solution into a
separate, clean glass test tube.
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5
5. Confirmation of the presence of tin.
4+
2+
The aluminum wire causes Sn to be reduced to Sn :
4+
2+
3+
3Sn (aq) + 2Al(s) → 3Sn (aq) + 2Al (aq)
The confirmation test for tin involves the following redox
reaction:
2+
2+
4+
2Hg (aq) + Sn (aq) + 2Cl (aq) → Hg2Cl2 (s) + Sn (aq)
The Hg2Cl2 (s) formed in the confirmation of tin does not
dissolve in 12 M HCl.
(b) To the liquid, IMMEADIATELY add a drop or
two of 0.1 M HgCl2. A white or gray cloudiness that
forms establishes the presence of tin.
(c) To check that this is not a false postive, add 12
M HCl to the precipitate. If it dissolves, you do not
have tin.
6. (a) To the other half of the solution from step
(4c), add 6 M NH3 until the solution becomes basic
to litmus. Any precipitate that forms can be
ignored; it will not interfere, but does indicate that
tin or antimony must be present.
7. Confirmation of the presence of antimony:
Addition of ammonia (NH4OH) causes antimony to
precipitate:
3SbCl6 (aq) + 4NH4OH (aq) 
+
SbOOH (s) + 3NH4 (aq) + 6Cl (aq) + H2O (l)
(b) Now add 6 M HC2H3O2 until the mixture
becomes acidic to litmus. Then add 0.5 mL more.
(c) Add a little (spatula tip full) of solid Na2S2O3 and
put the centrifuge tube in the boiling water bath for
a few minutes. Watch the color of the mixture for
the first couple of minutes. A peach color confirms
the presence of antimony, even if turns darker later.
7.(a) To the precipitate from step (2b), add 2 mL 6
M HNO3. Heat in the boiling water bath as some of
the sulfides dissolve and sulfur is formed. Continue
heating until no more reaction occurs.
(b) If you have a precipitate, centrifuge and decant
2+
3+
the supernatant which may contain Cu , Bi [and
2+
Pb ] into a centrifuge tube. LABEL IT. You may
discard any precipitate
8.(a) To the supernatant from step (7b), add 6M
NH3 until the solution is basic to litmus. Then add
0.5 mL more and stir. If copper is present the
solution will turn blue. A white precipitate is
indicative of bismuth (or possibly lead).
2-
The thiosulfate ion, S2O3 reacts sparingly to yield H2S:
22S2O3 (aq) + H2O (l)  SO4 (aq) + H2S (aq)
The H2S reacts with the SbOOH (s) to create a peach
precipitate of Sb2OS2:
2SbOOH (s) + 2H2S (aq)  Sb2OS2 (s) + 3H2O
7. Dissolving CuS (s), Bi2S3 (s) [and PbS (s)]
CuS, Bi2S3, and PbS readily dissolve in hot 6 M HNO3 in a
redox reaction. An example reaction is:
+
Bi2S3 (s) + 8H (aq) + 2NO3 (aq) →
3+
2Bi (aq) + 3S(s) + 2NO(g) + 4 H2O (l)
Elemental sulfur (yellow or tan) will be produced as a side
product, but will be removed.
3+
2+
2+
8. Separating the Bi [and the Pb ] from the Cu
2+
Upon addition of NH3, the complex ions Cu(NH3)4 (aq, blue)
2+
and Cd(NH3)4 (aq, colorless) form. The NH3 also raises the
pH and the [OH ] in solution so that Bi(OH)3 (s) and Pb(OH)2
(s) form.
Chem 1B
Dr. White
(b) Centrifuge and decant the solution into a
centrifuge tube. LABEL IT!
(c) Wash the precipitate with 1 mL DI water and 0.5
mL 6 M NH3. Stir, centrifuge and discard the wash
solution.
9. (a) To the precipitate from step (8c), add 0.5 mL
6 M HCl and 0.5 mL DI water. Stir to dissolve any
Bi(OH)3 that is present. A white insoluble solid may
2+
contain lead. (If there is no Pb in the sample,
then obviously there is no ppt here and you can
proceed to step 10)
(b) If this is the general unknown and there is a
precipitate, centrifuge and decant the supernatant
into a centrifuge tube. Wash the residue with 1 mL
DI water and 0.5 mL 6 M HCl. Centrifuge and
discard the wash. LABEL YOUR TUBES.
10. (a) Add 2 or 3 drops of the supernatant from
step (9) to 300 mL water in a beaker. A white
cloudiness caused by the precipitation of BiOCl
appears if the sample contains bismuth.
(b) To the rest of the supernatant from step (9), add
6 M NaOH until it is definitely basic. A white
precipitate of Bi(OH)3 will appear. To the mixture
add 2 drops 0.1 M SnCl2 and stir. A black metallic
residue indicates the presence of bismuth.*
*If you don’t get a black ppt, add an extra 2-3 drops
6 M NaOH and mix.
11. If the solution from step (8b) was blue complete
this step.
(a) Take the solution from step (8b) and start by
adding 5-6 microspatula tipfuls of Na2S2O4 (not
Na2S2O3). If the blue color is not gone, add more
Na2S2O4. Heat the centrifuge tube and centrifuge
to separate any dark ppt. If copper was present
(blue color previously) a dark ppt should have
formed on heating - if it didn't, add more Na2S2O4
and reheat.
12. Skip this step unless you are doing the
general unknown and you know your sample
may contain lead.
(a) To the white precipitate from step (9b), add 1
mL 6 M HC2H3O2. Heat gently if necessary to
dissolve any lead containing salts.
(b) Add 1 mL DI water and 1 mL 6 M H2SO4. A
white precipitate confirms the presence of lead.
Saddleback College
Cu
2+
6
+
(aq) + 4NH3 (aq) → Cu(NH3)4 (aq)
The net ionic equation for the reaction that occurs when the
bismuth hydroxide precipitate forms:
3+
+
Bi (aq) + 3H2O(l) + 3NH3(aq) → Bi(OH)3(s) + 3NH4 (aq)
9. Dissolving Bi(OH)3
If you are doing the general unknown, this step separates the
lead and the bismuth. If not, this step just serves to dissolve
the Bi(OH)3 (s). Adding HCl dissolves the Bi(OH)3 (s):
+
3+
Bi(OH)3 (s) + 3H (aq)  Bi (aq) + 3H2O (l)
The HCl reacts with Pb(OH)2 to form PbCl2 (s, white).
3+
10. Confirmation of Bi
There are 2 confirmation tests for bismuth.
3+
(a) The reaction of H2O with Bi in HCl gives BiOCl – a white
solid.
3+
+
Cl (aq) + H2O (l) + Bi (aq) → BiOCl (s) + 2H (aq)
3+
(b) In another test for the presence of Bi , the solution is
3+
made basic by adding NaOH. This precipitates Bi as
Bi(OH)3. SnCl2 is then added. This causes Bi(OH)3 to be
0
reduce to elemental Bi , a black solid. The unbalanced net
ionic chemical equation for this reaction is:
0
2Bi(OH)3(s) + Sn(OH)3 (aq) + OH (aq)→ Bi (s) + Sn(OH)6 (aq)
2+
11. Confirmation of Cu
2+
To test for the presence of Cu , Na2S2O4 is added and a
2+
redox reaction occurs. The Cu is reduced to metallic
0
copper (Cu ) in a basic solution and the reaction is shown
below.
2+
2-
-
5Cu (aq) + S2O4 (aq) + 10OH (aq) →
25Cu (s) + 2SO4 (aq) + 5H2O (l)
2+
11. Confirmation of Pb
Lead is dissolved in acetic acid:
PbCl2 (s) + 2HC2H3O2 (aq) 
2+
+
Pb (aq) + 2C2H3O2 (aq) + 2Cl (aq) + 2H (aq)
Then it is confirmed by adding sufate:
2+
2Pb (aq) + SO4 (aq) → PbSO4 (s)
If it is present, a white precipitate will form.
Chem 1B
Dr. White
Saddleback College
Outline of Procedure for Group II Cations (Group II Flow Chart)
7
Chem 1B
Dr. White
Saddleback College
Pre-Lab Questions
Exp. 13/14 Qualitative Analysis of the Group II Ions
For Numerical Problems, you must show all work for credit! Please box your final answer.
1. Using the Ksp values given in Table 1 of the Introduction to this lab, calculate the molar solubility of
a) CuS
b) Al2S3
2+
2-
2. If the [Cu ] in solution is 0.10 M, what must the [S ] be greater than in order for CuS to precipitate?
3+
2-
3. If the [Al ] in solution is 0.10 M, what must the [S ] be greater than in order for Al2S3 to precipitate?
8
Chem 1B
Name: ___________________________
Dr. White
Saddleback College
Lab Day/Time: ______________
Data and Results for Exp. 13/14: Group II Qualitative Analysis
Draw a flow chart indicating the behavior of your unknown:
Conclusion: The cations present in unknown number_____________are ________________.
9
Chem 1B
Dr. White
Saddleback College
10
Post Lab Questions
Exp. 13/14 Qualitative Analysis of the Group II Ions
1. Explain how changing the pH changes the concentration of sulfide in solution. Then, explain how this separates
the group II and group III ions. Be as specific as possible.
2+
(HINT: think Le Chatlier’s Principle for this reaction: H2S(aq) + 2 H2O(l)  S (aq) + 2 H3O (aq)
2-
-
2. In step 2, the tin (IV) and antimony (III) ions form soluble complex ions with NaOH. Sn(OH)6 (aq) and Sb(OH)4
(aq) form.
a. Write the balanced net ionic chemical equation, including phase labels, for the reaction that occurs
when NaOH(aq) is added to SnS2(s):
b. Write the balanced net ionic chemical equation, including phase labels, for the reaction that occurs
when NaOH(aq) is added to Sb2S3(s):
3. In step 4, the SnS2(s) and Sb2S3(s) dissolve in excess HCl at 100°C with the formation of the chloro- complexes
23SnCl6 and SbCl6 .
a. Write the balanced net ionic chemical equation, including phase labels, for the reaction that
2occurs when HCl(aq) is added to SnS2(s) to form SnCl6 :
b. Write the balanced net ionic chemical equation, including phase labels, for the reaction that
3occurs when HCl(aq) is added to Sb2S3(s) to form SbCl6 :
Chem 1B
Dr. White
Saddleback College
11
4. In step 7, CuS, Bi2S3, and PbS are dissolved in hot nitric acid. Balance the following redox reaction that occurs
in acid below:
2+
CuS (s) + NO3 (aq) → Cu (aq) + S(s) + NO(g)
5. The following reaction occurs in step 10b. This redox reaction is not balanced. Balance it in a basic solution.
0
2Bi(OH)3(s) + Sn(OH)3 (aq) → Bi (s) + Sn(OH)6 (aq)
6. For a Cation Group II unknown solution, determine from the following observations whether each of the ions in
2+
3+
2+
3+
our Group II analysis (Cu , Bi , Sn , and Sb ) is present, absent or in doubt. The sulfide precipitate is not
affected by treatment with 1 M NaOH (i.e. none of it dissolves). It does dissolve completely in 6 M HNO3. Addition
of an excess of 6 M NH3 to the acid solution produces only a blue solution. Explain your answer. (It is helpful to
use the flowchart on these types of problems)
Cation(s) Present:_________________
Cation(s) Absent:__________________
Cation(s) in Doubt:_________________