LABORATORY EXERCISE 7: Qualitative Analysis II
  Group II Cations: Pb2+, Sn4+, (Sn2+), Cu2+, Bi3+

As you can see from the flowchart in the General Laboratory Directions, Group II is comprised of more cations than we will address here. Analysis of many Group II ions requires the use of extremely hazardous chemicals, and disposal also requires special measures. For safety, as well as time, we will confine our research to lead, tin, copper, and bismuth.

To begin our analysis, we must first precipitate the sulfide salts of Group II ions. These salts are insoluble in highly acidic solutions (pH of 0.5), but we must be careful. A pH greater than 0.5 can result in precipitation of Group III ions as well.

After careful adjustment of the pH, we can generate the sulfide ion needed for precipitation within our solution with the addition of thioacetamide (CH3 CSNH2) and heat:

                 (Heat added)
CH3CSNH2 + H20 --> CH3COO- (aq) + NH4+(aq) + H2S(aq)

H2S is a weak diprotic acid that ionizes slightly in aqueous solution. This ionization is sufficient to provide the sulfide ion necessary for precipitation:

H2S (aq) + 2H20 <-- --> 2H30+ (aq) + S2- (aq)

With an understanding of LeChatelier's principle, it is simple to see from the above equation the direct relationship between pH and sulfide ion concentration. Since we have taken great care to adjust the pH of the test solution to 0.5 or less, we can deduce that only small amounts of sulfide ions are necessary to achieve complete precipitation and that Group II ions have low Ksp values compared to Group III ions.

The only other matter of concern in separating the Group II ions is the fact that stannous sulfide, SnS, is a gelatinous precipitate and therefore difficult to separate from the solution. So, in commencing the analysis, we must oxidize all Sn2+ to Sn4+ through the addition of hot HNO3.

A. Precipitation of the Cations for Analysis:

1. Oxidation of Tin and pH Adjustgment:
Place 2mL of test solution from Qual l, part B.2 or a solution containing the four test cations in an evaporating dish and add 15 drops 6M HNO3. Heat gently until a moist residue remains, then cool and add 2mL deionized water. Now add drops of 6M NH3 until the solution is basic to red litmus. Finally, add drops of 6M HCl until the solution is acidic to litmus. Once this is so, add 2 more drops of 6M HCl. This should adjust the pH of the solution to about 0.5. Transfer the solution to a centrifugable test tube.

2. Precipitation of Cations as Sulfide:
Add 10 - 15 drops of CH3CSNH2 (thioacetamide) to the solution. Heat in a hot water bath for 5 minutes, then cool and centrifuge. Repeat this process once more to ensure complete precipitation. Decant the supernatant and save it for Qual III.

The Tin Ion:
After completing the oxidation and precipitation, we have SnS2 precipitate, which we can now separate from the other sulfide salts by adding additional thioacetamide. The difference is that this time we add the thioacetamide to a basic solution. The increased sulfide ion concentration results in the formation of a soluble SnS32- complex ion, while the other Group II ions remain as precipitated sulfide salts. The addition of concentrated HCl then converts the SnS32- ion into a soluble hexachloro complex of Sn4+, SnC6 2-:

SnS32- (aq) + 6HCl (aq) --> SnCl62- (aq) + 3H2S (aq)

The addition of aluminum to the solution containing the hexachloro tin complex ion reduces it to a soluble tetrachloro complex ion, SnCl42-. Now, this probably seems like a very roundabout way to approach a simple problem, but keep this in mind: We test for tin in our solution indirectly by observing the effect of the soluble tetrachloro complex ion on another reagent. When you perform the confirmatory test, what you are actually seeing is the product of the following reaction, in which SnCl42- plays an important role:

SnCl42- (aq) + 3HgCl2 (aq) --> HgCl2 (s) + Hg (l) + 2SnCl62- (aq)

Sn2+ is a strong reducing agent, in this case reducing Hg2+ to Hg- and triggering the precipitation of mercurous chloride.

B. Test for Tin Ion:
1.Wash the precipitate from part A.2(above) with 1 mL 0.1M HCl. Stir, centrifuge, and discard the supernatant. Repeat washing once.

2. To the washed precipitate, add 2 mL 6M KOH and 3 drops CH3CSNH2. Heat the solution in a hot water bath for 8 minutes. While the solution is still warm, centrifuge and decant the supernatant into another test tube. Save the precipitate for the next test in part C.

3.To the supernatant, add concentrated HCl drop by drop until the solution is just acid to blue litmus. Heat the solution in a hot water bath for 5 minutes. Centrifuge and discard the supernatant.

4.To the precipitate, add 2mL 6M HCl, stir, and heat in a hot water bath for 5 minutes. This serves to boil off all excess sulfide as H2S. You may notice a foul odor, but that is normal. After heating for 5 minutes, add 2mL deionized water and 2 drops 6M HCl to the solution. Place a piece of polished aluminum approximately .25" square into the solution. Continue heating until the aluminum dissolves completely. Add 1 drop 6M HCl and heat for 3 more minutes.

5.Add 5 drops 0.2M HgCl2 to the solution. A gray or black precipitate confirms the presence of tin.

Lead, Copper, and Bismuth: The sulfide salts of these ions remain as precipitates, and so must be dissolved in order to separate and identify them. This is accomplished with the addition of hot HN03, which oxidizes the sulfide ion to free (elemental) sulfur and forms NO gas as a reaction product. The resulting solution contains the remaining Group II ions.

The Lead Ion: As explained earlier, lead can be tricky to identify strictly with a Group I test. After dissolving the Group II precipitates, H2S04 is added to the solution and it is heated vigorously. This eliminates any excess HNO3 and forms a white PbSO4 precipitate. Then, with the addition of ammonium acetate and acetic acid, the PbSO4 is dissolved, and we can again add CrO42- to test for lead.

C. Test for Lead Ion:
1.Wash the precipitate from B.2 with 1 mL deionized water and 2 drops 1M NH4NO3. Centrifuge, and discard the supernatant. Now, add 15 drops 6M HNO3 to the precipitate and heat in a hot water bath for 5 minutes, stirring frequently. You may notice a yellow substance floating atop the solution. This is the free sulfur, which is normal. Centrifuge, and transfer the supernatant to an evaporating dish or crucible. Be careful not to get any free sulfur in with the supernatant.

2.Set up a bunsen burner in the fume hood. Add 5 drops concentrated H 2S04 to the solution and heat cautiously. As dense white S03 fumes appear remove the solution from the heat. There should be a moist residue remaining. If there is still too much solution, continue to heat cautiously until only a moist residue remains. Allow the residue to cool for a few minutes, then add 15 drops deionized water to it. Stir, and then carefully transfer the milky liquid to a clean, cetrifugable test tube. Centrifuge, and save the supernatant for the next test in part D.

3.Dissolve the precipitate with 10 drops 1M NH4C2 H302 and 2 drops 1M HC2H302. Heating may be required; if it is, use your hot water bath. Stir thoroughly, then add 2 drops 1M K2CrO4. A bright yellow precipitate confirms the presence of lead. Centrifuge if necessary.

The Copper Ion: Our test solution should now contain only copper and bismuth ions. By adding aqueous ammonia to the solution, we can precipitate bismuth as white Bi(OH)3. Conveniently, the ammonia also complexes the copper ion into soluble, deep-blue Cu(NH3)42+.

Cu2+ (aq) + 4 NH3(aq) --> Cu(NH3)42+(aq)

While this in and of itself confirms the presence of copper, a second confirmatory test is generally performed by adding potassium hexacyanoferrate. This results in the formation of the red-brown precipitate copper hexacyanoferrate.

2Cu(NH3)42+ (aq) + K4[Fe(CN)6](aq) --> Cu2[Fe(CN)6](s) + 8NH3 (aq) + 4K+ (aq)

D. Test for Copper Ion:
1. Add 5 drops concentrated NH3 to the solution from C.2 under the fume hood. A color change to deep blue in the solution confirms the presence of copper. Centrifuge, decant the supernatant into a clean test tube, and save the precipitate for the next test in part E.

2.Acidify the supernatant from D.1 to blue litmus with drops of 6M HC2H302. Add 3 drops O.2M K4[Fe(CN)6]. A red-brown precipitate reconfirms the presence of copper.

The Bismuth Ion: The Bi(OH)3 that remains is now dissolved and confirmed with the addition of freshly prepared sodium stannite, Na2Sn(OH)4 to a basic solution. The Bi3+ ion is reduced to black bismuth metal, confirming its presence in solution.
Bi3+ (aq) + 2NH3 (aq) +3H20 --> BI(OH)3 (s) + 3NH4+ (aq)

Bi(OH)3 (s) + 3/2 Sn(OH)42- (aq) --> Bi (s) + 3/2Sn(OH)62-

E. Test for Bismuth Ion:
1. Prepare a fresh Na2Sn(OH)4 solution by placing 2 drops 1M SnCl2 in a clean test tube, followed by drops of 6M NaOH. You will see the solution turn milky as you add the first drops of NaOH. Continue to add drops and agitate until the solution is clear again.

2. Add 5 drops of Na2Sn(OH)4 to the precipitate from D.1. The immediate formation of a black precipitate confirms the presence of bismuth in the sample.

Equipment & Materials Needed:

Materials from Previous Lab
Evaporating Dish
1M thioacetamide (CH3CSNH2)
concentrated HCl
0.1 M HCl 0.25" square polished Al pieces
0.2 M HgCl2
1 M NH4NO3
concentrated H2S04
1M NH4C2H302
1M HC2H302
1M K2CrO4
6M HC2H302
O.2M K4[Fe(CN)6
fresh Na2Sn(OH)4 solution (Prepare by placing 2 drops 1M SnCl2 in a test tube and adding drops of 6M NaOH until the Sn(OH)2 precipitate just dissolves. Agitate or stir the solution)

All contents copyrighted (c) 1998
Peter Jeschofnig, Ph.D., Professor of Science, Colorado Mountain College
All Rights reserved

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