EXPERIMENT 13-14 Qualitative Analysis of Group II Cations THE GROUP II AND GROUP III PRECIPITATING AGENT Once the group I cations have been precipitated out of an unknown solution, and the precipitates removed, what remains is a dilute hydrochloric acid solution containing cations of groups II through V. When this aqueous solution is saturated with hydrogen sulfide, only the group II ions: copper (II), lead (II), bismuth (III), mercury (II), cadmium, arsenic (III), arsenic (V), antimony (III), tin (II), and tin (IV), precipitate as insoluble sulfides. The precipitating agent for the group II cations is hydrogen sulfide, but it is also the precipitating agent for the group III cations. 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+ (aq) + S 2- (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: K sp = [Cu 2+ ][S 2- ] = 8.5 10 34 M 2 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 8.5 10 36. Solving for the concentration of the sulfide ions to cause copper (II) sulfide to precipitate: [S 2- ] = K sp = 8.5 10 36 M 2 = 8.5 10 35 M [Cu 2+ ] 0.10 M The sulfide ion concentration must be at least this large 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+ (aq) + S 2- (aq) However, the solubility product constant for nickel (II) sulfide is K sp = [Ni 2+ ][S 2- ] = 2.0 10 21 M 2 To precipitate nickel (II) sulfide, the sulfide ion concentration must be a lot larger: [S 2- ] = K sp = 2.0 10 21 M 2 = 2.0 10 20 M [Ni 2+ ] 0.10 M 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 8.5 10 35 M but less than 2.0 10 20 M. 123
The source of the sulfide ions is hydrogen sulfide, produced by the hydrolysis (splitting of water) of thioacetamide. CH 3 CSNH 2 + H 2 O CH 3 CONH 2 + H 2 S This slow generation of hydrogen sulfide in the solution minimizes the concentration of this foul-smelling, highly toxic gas in the air. Hydrogen sulfide in water is a weak diprotic acid, H 2 S (aq) + 2H 2 O (l) 2H 3 O + (aq) + S 2- (aq) and an acid ionization constant expression can be written for it. The value of its acid ionization constant is given below: K a = [H 3 O + ] 2 [S 2- ] = 1.0 10 22 M 2 [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 sulfide concentration of about 1.0 10 22 M would work (larger than 8.5 10 35 M but less than 2.0 10 20 M). This can be accomplished by adjusting the ph (the hydronium ion concentration) of the solution. We can solve the acid ionization constant expression for the hydronium ion concentration that would give a sulfide ion concentration of 1.0 10 22 M, [H 3 O + ] = K a [H 2 S] ½ = (1.0 10 22 M 2 )(0.10 M) [S 2- ] M) (1.0 10 22 M) ½ = 0.32 M H 3 O + A hydronium ion concentration of 0.32 M corresponds to a ph of 0.50, which is very acidic. ph = log[h 3 O + ] = 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, and therefore, should precipitate out in a saturated hydrogen sulfide solution with a ph of 0.50. To verify this, the reaction quotient, Q, can be calculated for mercury (II) sulfide (knowing the solution is approximately 0.10 M in mercury (II) and 1.0 10 22 M in sulfide when the ph is adjusted to 0.50), and then comparing it to the K sp for mercury (II) sulfide: HgS (s) Hg 2+ (aq) + S 2- (aq) K sp = [Hg 2+ ][S 2- ] = 1.6 10 54 M 2 Q = [Hg 2+ ][S 2- ] = (0.10 M)(1.0 10 22 M) = 1.0 x 10-23 M 2 Because Q > K sp, the reverse reaction is spontaneous, and the mercury (II) sulfide will precipitate out from this solution. 124
To precipitate out the cations of group III, the sulfide ion concentration would have to be higher than 2.0 10 19 M, so a concentration around 1.0 10 6 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: [H 3 O + ] = K a [H 2 S] ½ = (1.0 10 22 M 2 )(0.10 M) [S 2- ] M) (1.0 10 6 M) This corresponds to a ph of 8.50, which is basic. ph = log[h 3 O + ] = log(3.2 10 9 M) = 8.50 ½ = 3.2 10 9 M H 3 O + So once the group II cations are precipitated out, the group III cations can be precipitated out by making 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 reaction quotient, Q, can be calculated for manganese (II) sulfide at both ph s, and then comparing it to the K sp for manganese (II) sulfide: MnS (s) Mn 2+ (aq) + S 2- (aq) K sp = [Mn 2+ ][S 2- ] = 2.3 10 13 M 2 Q = [Hg 2+ ][S 2- ] = (0.10 M)(1.0 10 22 M) = 1.0 x 10-23 M 2 (at ph = 0.50) Q = [Hg 2+ ][S 2- ] = (0.10 M)(1.0 10 6 M) = 1.0 x 10-7 M 2 (at ph = 8.50) At ph = 0.50, because Q < K sp, the forward reaction is spontaneous, and the manganese (II) sulfide will completely dissolve in this solution. However, at ph = 8.50, because Q > K sp the reverse reaction is spontaneous, and the manganese (II) sulfide will precipitate out from this solution. Therefore, manganese (II) is a Group III cation. INTRODUCTION TO THE GROUP II CATIONS For this experiment we will deal with only four of the Group II cations: antimony (III), copper (II), tin (IV), and bismuth (III). In an acidic solution, hydrogen sulfide will precipitate antimony (III), copper (II), tin (IV), and bismuth (III). 2Sb 3+ (aq) + 3S 2- (aq) Sb 2 S 3 (s, orange) Cu 2+ (aq) + S 2- (aq) CuS (s, black) Sn 4+ (aq) + 2S 2- (aq) SnS 2 (s, yellow) 2Bi 3+ (aq) + 3S 2- (aq) Bi 2 S 3 (s, brown) 125
1. TIN (IV) ION. If the solution containing the precipitates of the Group II cations is made basic, the sulfide ion concentration will increase. The sulfides of antimony (III) and tin (IV) will then become soluble because antimony (III) and tin (IV) form stable complexes with sulfide, which are soluble in water, while the sulfides of copper (II) and bismuth (III) do not. This causes the antimony (III) sulfide and tin (IV) sulfide to dissolve, separating them from the copper (II) sulfide and the bismuth (III) sulfide. Sb 2 S 3 (s) + 3S 2- (aq) 2SbS 3 3- (aq) SnS 2 (s) + S 2- (aq) SnS 3 2- (aq) When the solution of the two complex ions is made acidic with hydrochloric acid, the insoluble sulfides reprecipitate. 2SbS 3 3- (aq) + 6H + (aq) 3H 2 S (g) + Sb 2 S 3 (s, orange) SnS 3 2- (aq) + 2H + (aq) H 2 S (g) + SnS 2 (s, yellow) With the further addition of hydrochloric acid, the precipitates redissolve, forming stable chlorocomplex ions. Sb 2 S 3 (s) + 6H + (aq) + 8Cl - (aq) 3H 2 S (g) + 2SbCl 4 - (aq) SnS 2 (s) + 4H + (aq) + 6Cl - (aq) 2H 2 S (g) + SnCl 6 2- (aq) To a sample of the acidic solution containing the chlorocomplexions, aluminum metal is added to reduce the tin (IV) to tin (II), and the excess aluminum is dissolved by the acid. 2Al (s) + 3SnCl 6 2- (aq) 3Sn 2+ (aq) + 2Al 3+ (aq) + 18Cl - (aq) Al (s) + 6H + (aq) 2Al 3+ (aq) + 3H 2 (g) Mercury (II) chloride is then added, and the tin (II) ions reduce the mercury (II) ions to mercury metal and mercury (I) ions. 2Sn 2+ (aq) + 3Hg 2+ (aq) + 14Cl - (aq) 2SnCl 6 2- (aq) + Hg (l, black) + Hg 2 Cl 2 (s, white) The finely divided mercury metal is black, and the mercury (I) chloride is white, combining to form a gray precipitate, which confirms the presence of tin (II) ions. 2. ANTIMONY (III) ION. To a second sample of the acidic solution containing the chlorocomplex ions, ammonia is added until the solution becomes basic to litmus, precipitating the antimony (III) and tin (IV). SbCl 4 - (aq) + 3NH 3 (aq) + 2H 2 O (l) SbOOH (s, white) + 3NH 4 + (aq) + 4Cl - (aq) SnCl 6 2- (aq) + 4NH 3 (aq) + 4H 2 O (l) Sn(OH) 4 (s, white) + 4NH 4 + (aq) + 6Cl - (aq) If the solution is made weakly acidic with acetic acid, then a crystal of sodium thiosulfate pentahydrate introduced and the solution heated, hydrogen sulfide is created which reacts with the antimony (III) oxyhydroxide. 2SbOOH (s, white) + 2H 2 S (aq) Sb 2 OS 2 (s, peach to orange-red) + 3H 2 O (l) The peach to orange-red precipitate of antimony (III) oxysulfide, which turns to orange-red upon standing, confirms the presence of antimony (III) ions. 126
3. COPPER (II) ION. Oxidation of the sulfide ion to sulfur by hot nitric acid dissolves the precipitated copper (II) sulfide and bismuth (III) sulfide. CuS (s) + 4H + (aq) + 2NO 3 - (aq) Cu 2+ (aq) + S (s) + 2NO 2 (g) + 2H 2 O (l) Bi 2 S 3 (s) + 12H + (aq) + 6NO 3 - (aq) 2Bi 3+ (aq) + 3S (s) + 6NO 2 (g) + 6H 2 O (l) The addition of aqueous ammonia to a solution of bismuth (III) and copper (II) ions precipitates bismuth as a white hydroxide, but complexes the copper (II) ions as the dark blue tetraamminecopper (II) complex. Bi 3+ (aq) + 3NH 3 (aq) + 3H 2 O (l) Bi(OH) 3 (s, white) + 3NH 4 + (aq) Cu 2+ (aq) + 4NH 3 (aq) Cu(NH 3 ) 4 2+ (aq, blue) The deep blue color of the copper (II) complex confirms the presence of copper (II) ions. 4. BISMUTH (III) ION. The presence of bismuth (III) is confirmed by the addition of sodium stannite (prepared by mixing solutions of tin (II) chloride and sodium hydroxide) to bismuth (III) hydroxide. This reduces the bismuth (III) ion to a finely divided black precipitate of elemental bismuth. Bi(OH) 3 (s) + 3Sn(OH) 3 - (aq) + 3OH - (aq) 2Bi (s, black) + 3Sn(OH) 6 2- (aq) PROCEDURE 0. Students will work individually for this experiment. Except for the laboratory handout, remove all books, purses, and such items from the laboratory bench top, and placed them in the storage area by the front door. For laboratory experiments you should be wearing closed-toe shoes. Tie back long hair, and do not wear long, dangling jewelry or clothes with loose and baggy sleeves. Open you lab locker. Put on your safety goggles, your lab coat, and gloves. 0. Your instructor will indicate how you will perform the qualitative analysis. You will either (1) use the first lab period (day one) to conduct the analysis on the known sample and then the second lab period (day two) to conduct the analysis on the unknown sample, or (2) use two full lab periods to anaylyze the known sample and the unknown sample simultaneously. Make sure to paste the unknown code label on your flow chart in the space provided. PART A SEPARATION OF GROUP II CATIONS 0. Prepare a gently-boiling water bath in your fume hood by half-filling a 250-mL beaker with water, placing it on your hot plate, and adjusting it to a setting that will allow the water to just boil. From the Qual Wall, observe individual solutions of Sn 4+, Sb 3+, Cu 2+, and Bi 3+, and record your observations in the flow chart. To analyze the sample solution on day one, take a centrifuge tube to the stock bottle of sample solution (containing Sn 4+, Sb 3+, Cu 2+, and Bi 3+ ), found on the balance table. Dispense 2 ml of the sample solution into the centrifuge tube using the pipet pump, and label it with your name and as sample. Observe the solution, and record your observations in the flow chart. On day two, or, if you are analyzing the known and unknown samples simulateously, you will place 2 ml of the unknown solution into a clean centrifuge tube, and label it with your name and as unknown. CAUTION: Wear gloves and use caution with the concentrated bases and acids as they are skin irritants. Do not let them contact your skin. 127
1. Add 6 M sodium hydroxide dropwise to the solution, mixing after each drop, until a precipitate remains after the mixing, or unitl you have added 20 drops, whichever comes first. Make the solution acidic to litmus paper by adding 6 M hydrochloric acid. Always mix the solution thoroughly before testing with litmus paper. Once the solution is acidic to litmus, add 2 more drops of 6 M hydrochloric acid. This should adjust the ph to about 0.5. Test a drop of the solution on a small piece of ph 0.0-3.0 paper, noting the first color you see, not a later color resulting from spreading, separating, or changing. Add drops of 6 M hydrochloric acid or 6 M sodium hydroxide until the color matches the yellow-green-color on the cover of the ph 0.0-3.0 paper container. Add 20 drops of 1 M thioacetamide to the centrifuge tube, and place it in the boiling water bath in your fume hood for at least five minutes. Do not let the contents of the test tube froth during the heating, this will result in the loss of solution by spillage. If all the Group II ions are present, a precipitate will form, and its color will be initially light, gradually darkening, and finally becoming black. Continue to heat the centrifuge tube for two minutes after the color has stopped changing. After the heating, remove the centrifuge tube from the water bath, allow it to cool for one minute, and centrifuge. Obtain two capillary pipets, one for your known and one for your unknown. With a capillary pipet, carefully decant out as much of the clear supernatant (absolutely no traces of precipitate) into a clean centrifuge tube. Save the precipitate for the group II analysis. If you are not going to test for Group III, IV and V cations today, discard the supernatant and continue on with step 2. If you will be testing for Group III, IV and V cations today, you must test for complete precipitation of the Group II cations as follows. To test for complete precipitation, test the ph of the supernatant to assure it is still 0.5, add 15 drops of thioacetamide, heat in the hot water bath for five minutes, cool for one minute, then centrifuge. If no more precipitate is formed, complete precipitation has been attained. If this produces more precipitate, the clear supernatant must be removed and the precipitate should be combined with its precipitate from the first thioacetamide addition. To combine the precipitates, add 10 drops of 1 M ammonium chloride and 10 drops of water to the new precipitate, mix, and pour the slurry into the capillary tube with the first portion of the Group II precipitates. To insure all of the Group II cations have been precipitated out, repeat (1) testing the ph of the supernatant, (2) adding 15 drops of thioacetamide, (3) heating in the hot water bath for five minutes, (4) cooling for one minute, and (5) centrifuging. When these steps produce no precipitate, complete precipitation has been attained. The precipitate contains the Group II insoluble sulfides, and it should be kept. The supernatant contains the cations of Groups III through V, and it should be labeled as Groups III, IV and V and saved. PART B TEST FOR TIN (IV) IONS 2. Wash the precipitate from step 1 with 2 ml of water, mix, centrifuge, and decant, discarding the wash liquid. Do this twice. To the precipitate, add 2 ml of 1 M sodium hydroxide and heat in the water bath with stirring for 2 minutes. Centrifuge and withdraw the supernatant into a clean centrifuge tube. Save the supernatant for step 3. Wash the precipitate with 2 ml of water, mix, centrifuge, and decant, discarding the wash liquid. Do this twice. Save the precipitate for step 8. 3. To the supernatant from step 2, add 10 drops of 6 M hydrochloric, or until acidic to litmus paper, mixing after each drop. Once acidic, a precipitate should be visible. Add 6 M hydrochloric acid until the ph is 0.5, testing with ph 0.0-3.0 paper. Mix well, centrifuge, and discard the supernatant. 128
4. To the precipitate from step 3, add 2 ml of 6 M hydrochloric acid. Mix, and transfer the mixture to a 50 ml beaker. Place the beaker on the hot plate and boil the liquid gently until the precipitate dissolves. 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. Once the precipitate has dissolved, remove the beaker from the hot plate, add 1 ml of 6 M hydrochloric and transfer into a separate, clean centrifuge tube. Centrifuge out any solid residue if present, and discard the solid. 5. Pour half of the solution from step 4 into a clean glass test tube, and add a small piece of aluminum 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. 6. Immediately introduce a few drops of 0.1 M mercury (II) chloride to the test tube containing the solution from step 5. A white precipitate turning gray confirms the presence of tin. PART C TEST FOR ANTIMONY (III) IONS 7. To the other half of the solution from step 4, add 6 M ammonia unitl the solution become basic to litmus paper. Next add 6 M acetic acid until the solution become acidic to litmus paper, then add 10 more drops. Add a few crystals of solid sodium thiosulfate, and place 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 to orange-red color confirms the presence of antimony (III). PART D TEST FOR COPPER (II) IONS 8. Wash the precipitate from step 2 twice with 10 drops of 1 M ammonium chloride, stir, centrifuge, and discard the washings. Add 15 drops of 6 M nitric acid, mix, and transfer the mixture into a separate, clean glass test tube. With a test tube holder, heat the contents of the glass test tube over a very low burner flame. Allow the liquid to just barely boil, then remove it from the flame, allow it to boil, then remove it, and repeat this several times. Stop heating when the black precipitate has dissolved. A chunk of yellowish-black sulfur may appear in the test tube. If so, cool, and transfer the content of the glass test tube into a clean centrifuge tube. Centrifuge, and decant the liquid into a separate, clean centrifuge tube, discarding the yellowish-black sulfur. 9. Add drops of 15 M ammonia to the solution, mixing after each drop, until it is basic to litmus paper. CAUTION: Vapors from concentrated ammonia are very strong! The deep blue solution confirms the presence of copper. Centrifuge and decant out the copper (II) complex solution, and save the precipitate for step 10. PART E TEST FOR BISMUTH (III) IONS 10. In a separate glass test tube, place 10 drops of 0.1 M tin (II) chloride. Add 2 drops of 6 M sodium hydroxide, and a precipitate will form. Continue adding enough drops of 6 M sodium hydroxide, mixing after each drop, until the precipitate just dissolves. Mix thoroughly. Add several drops of this solution to the precipitate from step 9. The immediate formation of a black precipitate confirms the presence of bismuth. 11. Identify the cations present in your unknown and record them at the top of the Unkonwn Flow Chart. 12. All excess solutions should be disposed of in the Chem 1B Waste Container in the fume hood. 129
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EXPERIMENT 13-14 LAB REPORT Name: Student Lab Score: Date/Lab Start Time: Lab Station Number: KNOWN FLOW CHART 131
UNKNOWN IONS PRESENT: UNKNOWN FLOW CHART 132
QUESTIONS 1. Show that cadmium ion is a qualitative analysis Group II cation, and will precipitate out of a saturated hydrogen sulfide solution adjusted to ph 0.50. The K sp for cadmium sulfide is 1.0 x 10-28. 2. Determine if metal ion X 4+ is a qualitative analysis group II cation. The K sp for XS 2 is 9.0 x 10-37. 133
3. After silver mercury (I), and lead (II) ions have been removed, why can the tin (IV), copper (II), antimony (III) and bismuth (III) ions (the Group II cations) be separated from all other cations in solution by the addition of hydrochloric acid and thioacetimide? Indicate the reactions that occur. 4. Why can't tin (IV), copper (II),antimony (III) and bismuth (III) ions be separated from silver, mercury (I) and lead (II) ions in solution by the addition of hydrochloric acid and thioacetimide? 5. Why can tin (IV) sulfide and antimony (III) sulfide be separated from copper (II) sulfide and bismuth (III) sulfide by the addition of sodium hydroxide? Indicate the reaction that occurs. 134
6. Why can copper (II) ions be separated from bismuth (III) by the addition of aqueous ammonia? Indicate the reactions that occurs. 7. If no precipitate forms in step 1, what can you conclude? 8. If no precipitate forms in step 2, what can you conclude? 9. If no precipitate forms in step 3, what can you conclude? 10. What happens in step 9 if 6 M sodium hydroxide is substituted for 15 M ammonia? 11. Identify a reagent or reagents from the Group I or II qualitative labs that would do the following: (a) precipitate silver ions but not antimony (III) ions (b) precipitate tin (IV) ions but not iron (III) ions (c) precipitate bismuth (III) ions but not copper (II) ions 135
12. A solution contains only these ions: silver, mercury (I), copper (II), and bismuth (III). Prepare a flow chart to show how you would separate and identify the four ions in this solution. 136