H 2 S(aq) + 2 H 2 O(l) S 2- (aq) + 2 H 3 O + (aq)

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1 Chem 1B Dr. White Saddleback College 1 Experiment 13/14: Separation and Identification of Group II Cations (The Acidic Sulfide Group: Cu 2+, Bi 3+, (Pb 2+ ), Sn 2+, and, Sb 3+ ) Objectives To understand the chemical reactions involved in the separation and identification of Cu 2+, Bi 3+, (Pb 2+ ), and Sn 2+, and Sb 3+ 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 K sp value. Table 1: K sp Values at 25 C for the sulfide solids of the Group II and Group III Cations Ionic Solid K sp CuS 8.5 x Bi 2 S x CdS 1.0 x Group II HgS 1.6 x PbS 1 x SnS 2 1 x Group III Sb 2 S x NiS 2.0 x CoS 1 x FeS 3.7 x MnS 2.3 x Cr 2 S 3 1 x Al 2 S x ZnS 2.5 x Referring to the K sp 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+ (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- ] = 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 Solving for the concentration of the sulfide ions to cause copper (II) sulfide to precipitate: [S 2 ] = K sp 8.5 x [Cu 2+ = = 8.5 x 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+ (aq) + S 2- (aq) However, the solubility product constant for nickel (II) sulfide is K sp = [Ni 2+ ][S 2- ] = To precipitate nickel (II) sulfide, the sulfide ion concentration must be a lot larger: [S 2 ] = K sp 2.0 x [Ni 2+ = = 2.0 x 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 M but less than 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, due to the small K a value for H 2 S (1.0 x ). This is seen in the equilibrium reaction below: H 2 S(aq) + 2 H 2 O(l) S 2- (aq) + 2 H 3 O + (aq) An acid ionization constant expression can be written for it. The value of its acid ionization constant is given below: K a = [H 3O + ][S 2 ] = 1.0 x [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 M would work (larger than M but less than M). This can be accomplished by adjusting the ph (the hydronium ion concentration) of the solution. We can solve the acid ionization constant expression

2 Chem 1B Dr. White Saddleback College 2 for the hydronium ion concentration that would give a sulfide ion concentration of M, [H 3 O + ] = K a [H 2 S] [S 2 ] 1/2 = (1.0 x )(0.10) 1.0 x /2 = 0.32M 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 (one we will not test for in our lab), and therefore, should precipitate out in a saturated hydrogen sulfide solution with a ph of To verify this, the ion product, Q sp, can be calculated for mercury (II) sulfide (knowing the solution is approximately 0.10 M in mercury (II) and 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- ] = Q sp = [Hg 2+ ][S 2- ] = (0.10)( ) = 1.0 x Because Q sp >K sp, 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 M, so a concentration around 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] [S 2 ] 1/2 = (1.0 x )(0.10) 1.0 x /2 = 3.2x10 9 M This corresponds to a ph of 8.50, which is basic. ph = log[h 3 O + ] = log( M) = 8.50 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 To verify this, the ion product, Q sp, 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- ] = Q sp = [Mn 2+ ][S 2- ] = (0.10)( ) = 1.0 x 10-7 At ph = 0.50, because Q sp <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 sp >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. After careful adjustment of the ph, the sulfide ion needed for precipitation within the solution is generated with the addition of thioacetamide (CH 3 CSNH 2 ) 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 will dissolve in hot 6 M HNO 3. Then, Cu 2+ is separated from Bi 3+ and Pb 2+ by the reaction with ammonia to form ammine complex ions of Cu 2+. 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, PbCl 2 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.

3 Chem 1B Dr. White Saddleback College 3 Reagents Available 0.3 M HCl 1M thioacetimide 6 M HCl 6 M NaOH 6 M NH3 1 M NH 4 Cl 1M NaOH 0.5 M NaC 2 H 3 O 2 Al (s) 0.1 M HgCl 2 12 M HCl 6 M HC 2 H 3 O 2 Na 2 S 2 O 3 (s) 6 M HNO M SnCl 2 Na 2 S 2 O 4 (s) Dilute, Known Solution that contains all the Group II cations (NOT Pb 2+ ) 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. 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 (excluding Pb 2+ ) 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 H 2 S (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 A ph of 0.5 gives a [S 2- ] of 1 x 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. (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 source of H 2 S is thioacetimide (CH 3 CSNH 2 ). The compound decomposes when heated to produce H 2 S: CH 3 CSNH 2 (aq) + 2H 2 O(l) + H + (aq) CH 3 COOH(aq) + NH 4 + (aq) + H 2 S(aq) The H 2 S then creates S 2- needed for precipitation of the Group II ions by the following equilibrium reaction: H 2 S (aq) + 2H 2 O (l) S 2- (aq) + 2H 3 O + (aq) Heating in step 1d increases the particle size of the

4 Chem 1B Dr. White Saddleback College 4 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 NaC 2 H 3 O 2 drop by drop with stirring until the ph is again 0.5. A brown or yellow precipitate may form which indicates that not all the Group II ions precipitated the 1 st time. precipitate and this makes it easier to separate from the solution. CuS, and PbS are black precipitates; Bi 2 S 3 is brown; SnS 2 is yellow; and Sb 2 S 5 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 NH 4 Cl 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 NH 4 Cl 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 NH 4 Cl 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 4+ and Sb 3+ by the formation of complex ions: Washing the precipitate with NH 4 Cl 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 with NaOH. Sn(OH) 6 2- (aq) and Sb(OH) 4 - (aq) form. *Do this step carefully because insufficient extraction will cause the Sn and Sb tests to fail. 3. Re-precipitation of Sn 4+ and Sb 3+ : The Sn(OH) 6 2- (aq) and Sb(OH) 4 - (aq) are precipitated by adding an acid. The H 3 O + reacts with the OH - in the complex and the sulfide solid is formed again. The balanced net ionic chemical equation, for the reprecipitation of SnS 2 and Sb 2 S 3 follow: Sn(OH) 6 2- (aq) + 6H 3O + (aq) + 2S 2- (aq) SnS 2 (s) + 12 H 2O (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 H 2 S and dissolve the sulfides. (Don t boil to dryness). 2Sb(OH) 2-4 (aq) + 8H 3O + (aq) + 3S 2- (aq) Sb 2S 3 (s) + 16 H 2O(l) 4. Re-dissolving the SnS 2 (s) and Sb 2 S 3 (s): The SnS 2 (s) and Sb 2 S 3 (s) readily dissolve in excess HCl at 100 C with the formation of the chloro complexes SnCl 2-6 and SbCl 3-6. (c) Add 1 ml 6 M HCl and then pour the solution

5 Chem 1B Dr. White Saddleback College 5 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 24- guage 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. 5. Confirmation of the presence of tin. The aluminum wire causes Sn 4+ to be reduced to Sn 2+ : 3Sn 4+ (aq) + 2Al(s) 3Sn 2+ (aq) + 2Al 3+ (aq) The confirmation test for tin involves the following redox reaction: 2Hg 2+ (aq) + Sn 2+ (aq) + 2Cl - (aq) Hg 2 Cl 2 (s) + Sn 4+ (aq) The Hg 2 Cl 2 (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 HgCl 2. 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 NH 3 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 (NH 4 OH) causes antimony to precipitate: SbCl 6 3- (aq) + 4NH 4 OH (aq) SbOOH (s) + 3NH 4 + (aq) + 6Cl - (aq) + H 2 O (l) (b) Now add 6 M HC 2 H 3 O 2 until the mixture becomes acidic to litmus. Then add 0.5 ml more. (c) Add a little (spatula tip full) of solid Na 2 S 2 O 3 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 HNO 3. 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 the supernatant which may contain Cu 2+, Bi 3+ [and Pb 2+ ] into a centrifuge tube. LABEL IT. You may discard any precipitate 8.(a) To the supernatant from step (7b), add 6M NH 3 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). The thiosulfate ion, S 2 O 3 2- reacts sparingly to yield H 2 S: S 2 O 3 2- (aq) + H 2 O (l) SO 4 2- (aq) + H 2 S (aq) The H 2 S reacts with the SbOOH (s) to create a peach precipitate of Sb 2 OS 2 : 2SbOOH (s) + 2H 2 S (aq) Sb 2 OS 2 (s) + 3H 2 O 7. Dissolving CuS (s), Bi 2 S 3 (s) [and PbS (s)] CuS, Bi 2 S 3, and PbS readily dissolve in hot 6 M HNO 3 in a redox reaction. An example reaction is: Bi 2 S 3 (s) + 8H + (aq) + 2NO - 3 (aq) 2Bi 3+ (aq) + 3S(s) + 2NO(g) + 4 H 2 O (l) Elemental sulfur (yellow or tan) will be produced as a side product, but will be removed. 8. Separating the Bi 3+ [and the Pb 2+ ] from the Cu 2+ Upon addition of NH 3, the complex ions Cu(NH 3 ) 4 2+ (aq, blue) and Cd(NH 3 ) 4 2+ (aq, colorless) form. The NH 3 also raises the ph and the [OH - ] in solution so that Bi(OH) 3 (s) and Pb(OH) 2 (s) form.

6 Chem 1B Dr. White Saddleback College 6 (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 NH 3. 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 contain lead. (If there is no Pb 2+ 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 SnCl 2 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 Na 2 S 2 O 4 (not Na 2 S 2 O 3 ). If the blue color is not gone, add more Na 2 S 2 O 4. 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 Na 2 S 2 O 4 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 HC 2 H 3 O 2. Heat gently if necessary to dissolve any lead containing salts. (b) Add 1 ml DI water and 1 ml 6 M H 2 SO 4. A white precipitate confirms the presence of lead. Cu 2+ (aq) + 4NH 3 (aq) Cu(NH 3 ) 4 + (aq) The net ionic equation for the reaction that occurs when the bismuth hydroxide precipitate forms: Bi 3+ (aq) + 3H 2 O(l) + 3NH 3 (aq) Bi(OH) 3 (s) + 3NH + 4 (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): Bi(OH) 3 (s) + 3H + (aq) Bi 3+ (aq) + 3H 2 O (l) The HCl reacts with Pb(OH) 2 to form PbCl 2 (s, white). 10. Confirmation of Bi 3+ There are 2 confirmation tests for bismuth. (a) The reaction of H 2 O with Bi 3+ in HCl gives BiOCl a white solid. Cl - (aq) + H 2 O (l) + Bi 3+ (aq) BiOCl (s) + 2H + (aq) (b) In another test for the presence of Bi 3+, the solution is made basic by adding NaOH. This precipitates Bi 3+ as Bi(OH) 3. SnCl 2 is then added. This causes Bi(OH) 3 to be reduce to elemental Bi 0, a black solid. The unbalanced net ionic chemical equation for this reaction is: Bi(OH) 3 (s) + Sn(OH) 3 - (aq) + OH - (aq) Bi 0 (s) + Sn(OH) 6 2- (aq) 11. Confirmation of Cu 2+ To test for the presence of Cu 2+, Na 2 S 2 O 4 is added and a redox reaction occurs. The Cu 2+ is reduced to metallic copper (Cu 0 ) in a basic solution and the reaction is shown below. 5Cu 2+ (aq) + S 2 O 4 2- (aq) + 10OH - (aq) 5Cu (s) + 2SO 4 2- (aq) + 5H 2 O (l) 11. Confirmation of Pb 2+ Lead is dissolved in acetic acid: PbCl 2 (s) + 2HC 2 H 3 O 2 (aq) Pb 2+ (aq) + 2C 2 H 3 O 2 - (aq) + 2Cl - (aq) + 2H + (aq) Then it is confirmed by adding sufate: Pb 2+ (aq) + SO 4 2- (aq) PbSO 4 (s) If it is present, a white precipitate will form.

7 Chem 1B Dr. White Saddleback College 7 Outline of Procedure for Group II Cations (Group II Flow Chart)

8 Chem 1B Dr. White Saddleback College 8 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 K sp values given in Table 1 of the Introduction to this lab, calculate the molar solubility of a) CuS b) Al 2 S 3 2. If the [Cu 2+ ] in solution is 0.10 M, what must the [S 2- ] be greater than in order for CuS to precipitate? 3. If the [Al 3+ ] in solution is 0.10 M, what must the [S 2- ] be greater than in order for Al 2 S 3 to precipitate?

9 Chem 1B Dr. White Saddleback College 9 Name: 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.

10 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. (HINT: think Le Chatlier s Principle for this reaction: H 2 S(aq) + 2 H 2 O(l) S 2- (aq) + 2 H 3 O + (aq) 2. In step 2, the tin (IV) and antimony (III) ions form soluble complex ions with NaOH. Sn(OH) 6 2- (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 SnS 2 (s): b. Write the balanced net ionic chemical equation, including phase labels, for the reaction that occurs when NaOH(aq) is added to Sb 2 S 3 (s): 3. In step 4, the SnS 2 (s) and Sb 2 S 3 (s) dissolve in excess HCl at 100 C with the formation of the chloro- complexes SnCl 6 2- and SbCl a. Write the balanced net ionic chemical equation, including phase labels, for the reaction that occurs when HCl(aq) is added to SnS 2 (s) to form SnCl 6 2- : b. Write the balanced net ionic chemical equation, including phase labels, for the reaction that occurs when HCl(aq) is added to Sb 2 S 3 (s) to form SbCl 6 3- :

11 Chem 1B Dr. White Saddleback College In step 7, CuS, Bi 2 S 3, and PbS are dissolved in hot nitric acid. Balance the following redox reaction that occurs in acid below: CuS (s) + NO 3 - (aq) Cu 2+ (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. Bi(OH) 3 (s) + Sn(OH) 3 - (aq) Bi 0 (s) + Sn(OH) 6 2- (aq) 6. For a Cation Group II unknown solution, determine from the following observations whether each of the ions in our Group II analysis (Cu 2+, Bi 3+, Sn 2+, and Sb 3+ ) 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 HNO 3. Addition of an excess of 6 M NH 3 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:

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