TABLE 1: General Solubility Rules for Ionic Compounds at 25 C

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1 Experiment 16 version 1A-The Qualitative Process This lab explores how to use solubility reactions to identify specific ions. The lab is divided into three parts. In Part 1, you will perform characteristic tests for specific ions. These tests will help you determine the identity of unknown ions in Part 2. In Part 2, you will use your observations, combined with the tables provided (Tables 1 & 2) and general solubility rules to determine the identity of four ionic unknowns. In this lab, we are introduced to techniques used in qualitative analysis. Qualitative analysis is a branch of chemistry that is used to identify particular substances in a give sample of material. In Part 1 of this lab, you will observe characteristic chemical reactions of specific ions. In Part 2 of this lab, you will use the observations from Experiment 8, Part 1, and information from the pre-lab discussion for Part 2 to determine the identity of an unknown ionic solid. Although some aspects of qualitative analysis have been replaced by sophisticated instrumental methods in practical analysis, the process of using qualitative analysis to identify unknown compounds is useful because it provides an ideal context for the illustration and application of the principles of ionic solubility, it gives a framework for the discussion of the chemical properties of ions, and qualitative chemistry is fun. USEFUL TABLES TABLE 1: General Solubility Rules for Ionic Compounds at 25 C 1. A compound will be soluble if it contains at least one of the following ions: Li +, Na +, K +, NH 4 +, NO 3, or C 2H 3O A compound containing Cl, Br, or I is soluble unless the cation is Ag +, Pb 2+, or Hg Compounds that contain SO 4 2 are soluble except those sulfates that also contain Ba 2+, Ag +, Pb 2+, or Ca Most other ionic compounds are insoluble (in other words, the amount of solute that dissolves is extremely small and will be regarded as negligible). Compounds containing S 2, CO 3 2, PO 4 3, and OH are insoluble unless the cation is one of those listed in 1 above. NOTE: This is another statement of the general solubility rules listed in your textbook and the handout from lecture. You can use any one of these. Double displacement and qualitative analysis Page 1 of 10

2 Table 2: General Ion Solubilities Cl SO 4 2 C 2H 3O 2 CO 3 2, PO 4 3 OH,O 2 S 2 Na +, K +, NH 4 + S S S S S S Ba 2+ S I S A S S Ca 2+ S S S A S S Mg 2+ S S S A A S Fe 3+ (yellow) S S S A A A Fe 2+ (pale aqua) S S S A A A Cr 3+ (blue-violet) S S S A A A Al 3+ S S S A, B A, B A, B Ni 2+ (green) S S S A, N A, N A +, O + Co 2+ (pink) S S S A A A +, O + Zn 2+ S S S A, B, N A, B, N A Mn 2+ (light pink) S S S A A A Cu 2+ (blue) S S S A, N A, N O Cd 2+ S S S A, N A, N A +, O Ag + A +, N S, N S, N A, N A, N O Pb 2+ HW, B, A + B S A, B A, B O Note: Br and I have solubility properties similar to Cl, but they have significantly different oxidation potentials. Use chlorine water to test for bromide and iodide. Key: S soluble in water S slightly soluble in water HW soluble in hot water A A + O O + B soluble in acid (6 M HCl or other non-precipitating, non-oxidizing acid) soluble in 12 M HCl soluble in hot 6 M HNO 3 (an oxidizing acid) soluble in aqua regia (12 M HNO 3 and 12 M HCl mixed in a 1:3 ratio) soluble in 6 M NaOH N soluble in 6 M NH 3 I insoluble in any common reagent Double displacement and qualitative analysis Page 2 of 10

3 S TABLE 3: Interpreting the table of solubility properties The solubility in water is at least 0.1 M. For some compounds, shaking and stirring will be necessary. Compounds that are soluble in water are also soluble in acids, as long as the anion part of the acid does not precipitate with the cation part of the compound that you are trying to dissolve. S Some of the compound will dissolve in water, but its solubility is less than 0.1 M. HW A A + The compound will not dissolve in cold water, but its solubility in hot (near boiling) water is at least 0.1 M. The compound will not dissolve in pure water; in a 6 M solution of strong acid, its solubility is at least 0.1 M. An acid-base reaction is involved. Notice that most of the compounds in the table that are labeled with an A contain the conjugate bases of weak acids. When choosing the strong acid, make sure that the anion of the acid does not precipitate with the cation of the compound that you are trying to dissolve. For example, Fe(OH) 3 is soluble in strong acid due to the acid-base reaction: Fe(OH) 3 (s) + 3 H 3O + (aq) Fe 3+ (aq) + 6 H 2O (l) Ag 2O is soluble in strong acid due to the acid-base reaction: Ag 2O (s) + 2 H 3O + (aq) 2 Ag + (aq) + 3 H 2O (l) Why would you need to use HNO 3 instead of HCl in the case of Ag 2O? When trying to react an oxide (rather than a hydroxide) with an acid, vigorous shaking and stirring and maybe even warming in a hot water bath will be necessary. The compound will not dissolve in pure water or in 6 M acids; in a 12 M HCl solution, its solubility is at least 0.1 M. A complex ion formation reaction is involved. Ag + and Pb 2+ form stable complex ions with chloride ions. Vigorous shaking and stirring will be necessary. Be careful! For example, PbCl 2 is soluble in excess 12 M HCl due to formation of a complex ion: PbCl 2(s) + 2 Cl (aq) PbCl 4 2- (aq) O The compound will not dissolve in pure water, nor will it dissolve in acids via ordinary acidbase reactions, but it will react with hot 6 M HNO 3 to form soluble ions. You will need to boil the solution. Use a boiling chip! An oxidation-reduction reaction is involved. Notice that most of the compounds in the table that are labeled with an O are sulfides. The sulfide ions (S 2- ) in these compounds will be oxidized to elemental sulfur (S). For example, while CuS is insoluble in water and even in acids, it can be converted to soluble Cu(NO 3) 2 by oxidation of its sulfide ion: O + B 3 CuS (s) + 2 NO 3 (aq) + 8 H + (aq) 3 Cu 2+ (aq) + 3 S (s) + 2 NO (g) + 4 H 2O (l) This is similar to O, except that the oxidizing reagents must be more concentrated for the oxidation-reduction reaction to go to completion. Hot 12 M HNO 3 is needed in combination with 12 M HCl, which serves as an additional source of hydronium ions. The compound will not dissolve in pure water; in a 6 M solution of NaOH, its solubility is at least 0.1 M. A complex ion formation reaction is involved. Al 3+, Zn 2+, and Pb 2+ form stable complex ions with hydroxide ions (OH ). Vigorous shaking and stirring will be necessary. For example, Zn(OH) 2 is soluble in NaOH due to formation of a complex ion: Double displacement and qualitative analysis Page 3 of 10

4 N I Zn(OH) 2(s) + 2 OH (aq) Zn(OH) 4Pb 2- (aq) The compound will not dissolve in pure water; in a 6 M solution of NH 3, its solubility is at least 0.1 M. A complex ion formation reaction is involved. Ni 2+, Zn 2+, Cu 2+, Cd 2+, and Ag + form stable complex ions with ammonia (NH 3). Vigorous shaking and stirring will be necessary. For example, Zn(OH) 2 is also soluble in NH 3 due to the formation of another complex ion: Zn(OH) 2(s) + 4 NH 3(aq) Zn(NH 3) 4 2+ (aq) + 2 OH (aq) The compound will not dissolve or react in any reagent available in the laboratory. Safety Precautions: Wear safety goggles. Silver nitrate solutions will stain skin and clothing. If you notice any silver nitrate splashing your skin or clothing, rinse it off immediately. Silver nitrate is a colorless solution that is lightsensitive. The stains don t show up immediately, but you will see them by the next day! Stains on your skin will eventually wear off as your skin wears off; stains on your clothing will be permanent. It is therefore best to wear old clothing for this lab. Waste Disposal: While you are doing the experiment, pour your waste into a beaker. When you are finished with the experiment, pour the contents of the waste beaker into the INORGANIC WASTE container (with a blue label) in the fume hood. Part 1: Spot tests for common ions: Carbonate, sulfate, phosphate, thiocyanate, chloride, acetate, ammonia, and halides. Now that you have observed double displacement reactions for common ions, you will perform specific tests on selected ions. A simple approach to the qualitative analysis of an unknown solution is to test for the presence of each possible ion by adding a reagent, which will cause the ion, if it is in the sample, to react in a characteristic way. This method involves a series of spot tests, one for each ion, carried out on separate samples of the unknown solution. When performing qualitative analysis tests, one must keep in mind that one ion may interfere with the analytical test for another. Although interferences are common, there are many ions that can be identified in mixtures by simple spot tests. The tests that you will perform, will involve simple acid-base, precipitation, complex ion formation or oxidation-reduction reactions. DON T WRITE NIE FOR THE TESTS IN THIS PART OF THE EXPERIMENT. 1. Carry out the test for each of the ions as described below. 2. In some of the tests, a boiling water bath will be needed, so set that up before proceeding. When performing a test, if no reaction is immediately apparent, stir the mixture with your stirring rod to mix the reagents. 3. These tests can easily be used to detect the anions at concentrations of 0.02 M or greater, but in dilute solutions, careful observation may be required. Double displacement and qualitative analysis Page 4 of 10

5 Carbonate Ion. Carbonate ion reacts with a strong acid to produce bubbles of carbon dioxide gas. Carbon dioxide is colorless and odorless. Procedure: Cautiously add 1 ml of 6 M HCl to 1 ml of 1 M Na 2CO 3 in a small test tube. With concentrated solutions, bubbles of carbon dioxide gas are immediately evolved. Sulfate Ion. Sulfate ions react with barium ions to form insoluble barium sulfate, which is a white, finely divided precipitate. While two of the other anions in this experiment also form precipitates with barium, addition of HCl makes these compounds soluble. Barium sulfate is the only precipitate of barium that is not soluble in strong acid. Procedure: Add 1 ml of 6 M HCl to 1 ml of 0.5 M Na 2SO 4. Add a few drops of 1 M BaCl 2. Phosphate Ion. Phosphate ions react with ammonium molybdate to produce a characteristic yellow precipitate, ammonium phosphomolybdate ((NH 4) 3PO 4 12 MoO 4). Procedure: Add 1 ml 6 M HNO 3 to 1 ml of 0.5 M Na 2HPO 4. Then add 1 ml of 0.5 M (NH 4) 2MoO 4 and stir thoroughly. The ammonium phosphomolybdate precipitate may form slowly, particularly in dilute solutions; if a solid precipitate does not appear promptly, put the test tube in the boiling water bath for a few minutes. Often the formation of a transparent yellow solution indicates a positive test. Thiocyanate Ion. Thiocyanate ions react with ferric ions to form the complex ion FeSCN 2+, which has a characteristic deep red color. Two other anions in this experiment form precipitates with iron(iii) ions, which would prevent the formation of the complex ion. However, addition of nitric acid makes these compounds soluble, so the FeSCN 2+ complex ion can be observed. Procedure: Add 1 ml 6 M HNO 3 to 1 ml of 0.5 M KSCN and stir. Add one or two drops of 0.1 M Fe(NO 3) 3. Chloride Ion. Chloride ions react with silver ions to form an insoluble, white compound, silver chloride. Unfortunately, many inferences are possible in tests involving silver ions. Thiocyanate ions, and two of the other anions in this experiment also form precipitates with dilute silver solutions. While addition of nitric acid makes two of these compounds soluble, both silver chloride and silver thiocyanate are insoluble in strong acid. You are going to test for chloride ion using silver nitrate. Procedure: Add 1 ml of 6 M HNO 3 to 1 ml of 0.5 M NaCl. Add 2-3 drops of 0.1 M AgNO 3. Acetate Ion. Acetate ion reacts with strong acids to form acetic acid, which has the characteristic odor of vinegar. Procedure: Add 1 ml 3M H 2SO 4 to 1 ml of 1 M NaC 2H 3O 2 and stir. Cautiously smell the mixture. The intensity of the vinegar odor is enhanced if the tube is warmed for 30 seconds in the boiling water bath. Double displacement and qualitative analysis Page 5 of 10

6 Ammonium Ion. Ammonium ion reacts with strong bases to form ammonia, which has a characteristic odor. Procedure: Add 1 ml of 6 M NaOH to 1 ml of 0.5 M NH 4Cl. Cautiously smell the mixture to determine if ammonia is present. For a more sensitive test, pour the 1 ml of sample into a small beaker, moisten a piece of red litmus paper and put it on the bottom of a watch glass; cover the beaker with the watch glass and gently heat the liquid to the boiling point. Do not boil it and be careful that no liquid comes in contact with the litmus paper. The litmus paper will gradually turn blue if it is exposed to the vapors of evolving NH 3. Remove the watch glass and try to smell the ammonia. Halide ions: Halides undergoes several distinct reactions. You will not be testing fluoride. Bromide and iodide are oxidized to their elemental form by an aqueous solution of chlorine. All halides react with silver nitrate to give a precipitate. The color of the precipitate indicates the identity of the ion. Procedure 1: Clean 3 test tubes. Add 1 ml of silver nitrate to each test tube. Add a small amount of NaCl (aq) to test tube #1; Add a small amount of NaBr (aq) to test tube #2; Add a small amount of NaI (aq) to test tube #3. Note the color changes against a white background. [Hint: they are NOT the same color] All are soluble in 6M NH 3, silver chloride being the most soluble, and silver iodide being the least. Procedure 2: Clean 3 test tubes. Add 1 ml of chlorine water to each test tube. Add a small amount of NaCl (aq) to test tube #1; Add a small amount of NaBr (aq) to test tube #2; Add a small amount of NaI (aq) to test tube #3. Note the color changes against a white background. Part 2 Pre-lab Discussion Before we begin Part 2: Identification of an Unknown Solid, we will break into small groups to discuss how to use Table 2: Solubility properties of ions and solids and your results from Part 1 to devise strategies for determining the identity of an unknown. The samples that we will discuss are listed below. You need to come to me so that I can give each person in your group an unknown to evaluate. You cannot ask for particular unknowns. After receiving your unknown example, think about the properties of the ionic compounds in the sample. You should be able to answer the following questions: 1. Can I use solubility as a method of identifying my unknown? If both of the ionic compounds are soluble, then solubility is not an option. 2. Do they have the same cation? Look at the anion. Think about the tests that you did in Part 1. These tests will help you determine the identity of the anion, and, therefore, the identity of the ionic compound. 3. Are the cations and the anions different? You might have to do a selective test for a cation. Look at Table 2. Is there a difference in the solubility profile of your two cations? Can you precipitate out a cation, or use a strong acid or base to dissolve the Double displacement and qualitative analysis Page 6 of 10

7 unknown. Generally, one of the pair in the unknown will be reactive and the other not so much. Pre-lab questions for Part 2 1. A student wants to determine the difference between cadmium hydroxide and silver hydroxide. When she looks at Table 2, She notices that they are both insoluble in distilled water, but the addition of 6M HCl or 6M NH 3 will dissolve the solid. Unfortunately, she can t use the results of the test to distinguish between the two cations. Feeling frustrated and stymied, she pouts and considers breaking her test tubes in a fit of pique, when Table 2 again captures her eye. She sees the solution to her problem. She takes her test tube that contains the unknown dissolved in 6M NH 3 and adds enough ----What does she add to see a positive test for either silver ions or cadmium ions? 2. Compounds containing carbonate and phosphate are insoluble in distilled water, but the identity of the unknown can be determined by solubility. Looking at TABLE 2, what is the common feature of all but one of the carbonates? If you had to distinguish between aluminum carbonate and nickel carbonate, what would you add to the test tube? 3. Now complete the following using your knowledge of solubility, Table 2, and the tests you ran in Part 1. There are some examples listed below that will help you with formulating an answer. READ THEM FIRST BEFORE YOU ATTEMPT TO DESCRIBE HOW YOU WOULD DETERMINE THE IDENTITY OF THE FOLLOWING PAIRS BASED ON TESTS. a) Na 2CO 3 or NaCl b) CaSO 4 or Na 2SO 4 c) MgCO 3 or BaSO 4 d) NiCl 2 or BaCl 2 e) KCl or NH 4Cl f) ZnCO 3 or CdCO 3 g) NaOH or Mg(OH) 2 h) AgCl or Ag 2O i) KC 2H 3O 2 or Na 2SO 4 j) ZnS or CuS k) NaOH or Na 2S l) NaCl or NaBr m) FeCl 3 or FeCl 2 n) NiCO 3 or MnCO 3 p) Ni(OH) 2 or Al(OH) 3 Here are some examples for how to think through your unknown and answer pre-lab question 3: Many of the solids can be identified on the basis of their anions. Use the spot tests in Part 2 to find the correct formula. In other cases the compounds may both be insoluble or may both have the same anion, so that other approaches will be necessary. One very useful procedure is to consider the solubilities of the two compounds in different reagents or the solubilities of products that can be made from those substances. Probably the most important solvent to consider is water. One of the compounds may be soluble in water and the other may not be. TABLE 2: Solubility Properties of Ions and Solids and TABLE 3: Interpreting the Table of Solubility Properties tabulate the solubility properties of many ionic solids in water and other common laboratory reagents. This information will allow you to make many decisions regarding solubilities of ionic substances under different conditions. EXAMPLE 1: The vial contains either MgSO 4 or BaSO 4. Looking at the tables, you would find that MgSO 4 is soluble in water. In the table, you would also find that BaSO 4 is insoluble in all common reagents. Thus to distinguish MgSO 4 from BaSO 4, we need merely to add water. Double displacement and qualitative analysis Page 7 of 10

8 Differences in solubility in solvents other than water can often be used to good advantage. Although most carbonates, phosphates, hydroxides, and oxides are insoluble in water, they are soluble in 6 M solutions of strong acids. from CuO. EXAMPLE 2 & 3: Solubility in acid will allow you to distinguish BaSO 4 from BaCO 3 or CuS EXAMPLE 4: Another useful distinguishing solvent is 6 M NH 3, which dissolves many insoluble compounds by forming complex ions with the cation in the compound. Given a sample that might be Mg(OH) 2 or Zn(OH) 2, we notice that both water insoluble, and both soluble in strong acids, one could dissolve Zn(OH) 2 in 6 M NH 3. Mg 2+, however, does not form an ammonia complex ion, so Mg(OH) 2 would not dissolve in excess 6 M NH 3. If both compounds are soluble in water or another common solvent and contain the same anion, then a precipitating reagent that reacts with one of the cations can be very helpful. EXAMPLE 5 & 6: You needed to distinguish between MgCl 2 and CdCl 2 (Example 5), both of which are water-soluble. From the solubility tables, you could determine that MgS is water-soluble while CdS is not. Addition of 1 M (NH 4) 2S to the water solution of the unknown (in the fume hood!) should allow you to identify the substance present. In a more complicated case, with both compounds insoluble in water but soluble in a common solvent, it might be possible to find a selective precipitating reagent. If the unknown might be CuS or PbS considerable examination of solubility tables would reveal that, although both substances are very insoluble, they both dissolve in hot 6 M HNO 3. Following treatment of the solid, which would put Cu 2+ or Pb 2+ into the solution, one could identify the ion by its color, since Cu 2+ is blue, or by precipitation on addition of 3 M H 2SO 4, which would precipitate PbSO 4 but not CuSO 4. EXAMPLE 7: Sometimes acidic or basic properties can be applied very easily to distinguish compounds. Anions like CO 3 2 or OH will always make their solutions basic. HSO 4 ion is highly acidic. Litmus paper becomes a useful tool to distinguish acids and bases. Color can sometimes be employed, as with the 2+ in Example 6. EXAMPLE 8: Although oxidation-reduction reactions are not often needed for distinguishing between compounds, they are sometimes very useful, as is the case with CuS and PbS (Example 6). Halide ions in solution can be treated with chlorine water, which will oxidize colorless Br to pale yellow-orange Br 2 and colorless I to dark yellow I 2. Nitric acid is a strong enough oxidizing agent to oxidize I, but not Br. Although it is not feasible to present a procedure that will work for every possible pair of compounds, we have presented in the chart below a general outline that should be effective for dealing with most pairs. It essentially summarizes the material presented earlier in this discussion. Table 4. Example Procedures for Distinguishing Between Two Possible Solids (These are the same reactions as stated above in table form.) Example 1: MgSO 4 (S according to appendix) or BaSO 4 (I according to appendix) Procedure Expected result for MgSO 4 Expected result for BaSO 4 Mix a small amount of solid with water. Would dissolve to give a Would not dissolve. Double displacement and qualitative analysis Page 8 of 10

9 Example 2: BaSO 4 (I according to appendix) or BaCO 3 (A according to appendix) Procedure Expected result for BaSO 4 Expected result for BaCO 3 Mix a small amount of solid with 6 M HCl. Would not dissolve. Example 3: CuO (A according to appendix) or CuS (O according to appendix) Would dissolve with formation of bubbles to give a BaCO 3(s) + 2 H 3O + (aq) Ba 2+ (aq) + H 2O (l) + CO 2 (g) Procedure Expected result for CuO Expected result for CuS Add a small amount of solid to 6 M HCl. Stir well, while heating in a water bath. Would dissolve to give a blue solution. CuO (s) + 2 H 3O + (aq) Cu 2+ (aq) + 3 H 2O (l) Would not dissolve. Example 4: Mg(OH) 2 (A according to appendix) or Zn(OH) 2 (A, B, N according to appendix) Procedure Add a small amount of solid to 6 M NH3. Stir very well. Expected result for Mg(OH) 2 Would not dissolve. Expected result for Zn(OH) 2 Would dissolve to give a Zn(OH) 2(s) + 4 NH 3(aq) Zn(NH 3) 4 2+ (aq) + 2 OH (aq) Example 5: MgCl 2 (S according to appendix) or CdCl 2 (S according to appendix) Procedure Expected result for MgCl 2 Expected result for CdCl 2 (a) Add a small amount of solid to water. (b) Add a few drops of (NH 4) 2S in fume hood. (b)no change. (b)a precipitate would form. Cd 2+ (aq) + S 2 (aq) CdS (s) Example 6: CuS (O according to appendix) or PbS (O according to appendix) Procedure Expected result for CuS Expected result for PbS (a) Add a small amount of solid to 6 M HNO3. Boil the solution in the fume hood. (b)cool, filter out the sulfur, and add a few drops of Na 2SO 4. blue solution and finely divided solid sulfur. 3 CuS (s) + 2 NO 3 (aq) + 8 H + (aq) 3 Cu 2+ (aq) + 3 S (s) + 2 NO (g) + 4 H 2O (l) (b)no change. colorless solution and finely divided solid sulfur. 3 PbS (s) + 2 NO 3 (aq) + 8 H + (aq) 3 Pb 2+ (aq) + 3 S (s) + 2 NO (g) + 4 H 2O (l) (b)a precipitate would form. Pb 2+ (aq) + SO 4 2 (aq) PbSO 4(s) Example 7: NaOH (S according to appendix) or NaHSO 4 Procedure Expected result for NaOH Expected result for NaHSO 4 (a) Add a small amount of solid to water. (b)test solution with litmus paper. (b)would turn red litmus blue; OH is a strong base. (b)would turn blue litmus red. HSO 4 (aq)+ H 2O (l) SO 4 2 (aq) + H 3O + (aq) Double displacement and qualitative analysis Page 9 of 10

10 Example 8: NaCl or NaBr Procedure Expected result for NaCl Expected result for NaBr (a) Add a small amount of solid to water. (b)add a few ml of chlorine water (Cl 2). (b)no change. (b)would turn pale yelloworange. 2 Br (aq) + Cl 2(aq) Br 2(aq) + 2 Cl (aq) Part 2: Identification of an unknown solid. In this part of the experiment you will design and perform tests to identify ionic solids. You will make use of your knowledge of general solubility rules, acid-base chemistry, and solubility rules. A variety of pure ionic solids will be available in numbered bottles. Each bottle will have two chemical formulas written on it. Your charge will be to determine which of the two chemical formulas on the bottle is correct. Analysis Give the unknown number and chemical formula of each compound you identified. Explain your reasoning in a TYPED paragraph form for each compound identified. This means 4, four, IV, evaluations. Post lab questions 1. What was your unknown sample in the pre-lab discussion? What strategy did you use to determine the identity of the ionic compound? 2. Using the solubility charts, which of the other anions would also form precipitates with barium ions? 3. Using the solubility charts, which of the other anions would form precipitates with iron(iii) ions? 4. Which of the other anions (other than SCN ) would form precipitates with silver ions? 5. Write the net ionic equation for the reaction that occurs when the H 2SO 4 solution is added to the sodium acetate (CH 3COONa or NaC 2H 3O 2) solution. Double displacement and qualitative analysis Page 10 of 10