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How Can Eleven Colorless Solutions Be Chemically Identified?

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How Can Eleven Colorless Solutions Be Chemically Identified? EXPERIMENT 1 How Can Eleven Colorless Solutions be Chemically Identified? Developed by Department of Chemistry, Brown University INTRODUCTION In this experiment, the following eleven colorless solutions will be supplied and labeled only with numbers. The numbered solutions will be identified by mixing samples of the solutions with each other, analyzing pH, and conducting flame tests. No other chemicals need to be used. 1.0 M sodium thiosulfate (Na2S2O3) 1.0 M ammonium chloride (NH4Cl) 1.0 M hydrochloric acid (HCl) 0.1 M zinc chloride (ZnCl2) 1.0 M sodium carbonate (Na2CO3) 0.1 M silver nitrate (AgNO3) 0.1 M sodium oxalate (Na2C2O4) 0.1 M aluminum nitrate (Al(NO3)3) 0.1 M barium nitrate (Ba(NO3)2) 3% hydrogen peroxide (H2O2) 0.1 M manganese (II) nitrate (Mn(NO3)2) To prepare for this experiment, draw on the following information to develop a systematic and logical flow chart that will be used to identify the solutions. The most efficient way to start the analysis is by using a single unknown solution. When some unknowns have been identified, they can be used as reagents to identify others. A similar problem using a different set of unknowns is shown at the end of the Introduction. Once all solutions have been tentatively identified, you are required to perform addional tests for verifying the solutions. Flame Tests When a drop of some of the unknown solutions is placed on a wire loop and vaporized in a flame, the flame becomes brightly colored. The intense heat of the flame releases free (not bonded) atoms from molecules. The electrons of these free atoms are raised to high-energy states by the hot flame. When the electron returns to the ground state, light with a discrete frequency (color) characteristic of the element is emitted. Among the ions used in this lab, there are only two strong light emitters. Metal Flame color Sodium yellow/orange Barium green The sodium flame is so intense that only traces of sodium (e.g., from finger prints on the wire) give somewhat yellow flames. Therefore it is important to inspect the flame from the wire before adding each chemical. Remember how yellow the flame is just from the wire. When there is sodium present in the sample the change in intensity of the orange flame color will be significant. PH Tests pH Paper Color Interpretation Solution (See label on vial) strong acid HCl weak acid Al(NO3)3 strong base Na2CO3 + + - While Al(NO3)3 and Na2CO3 solutions do not obviously contain excess H (H3O in water) or OH , they do 1 react with water to change the pH: 3+ 3+ Al (aq) + 6 H2O(l) → [Al(H2O)6] (aq) (complex ion formation) (Eq. 1) 3+ 2+ + [Al(H2O)6] (aq) + H2O(l) [Al(OH)(H2O)5] (aq) + H3O (aq) (Eq. 2) 2- - - CO3 (aq) + H2O(l) HCO3 (aq) + OH (aq) (Eq. 3) All the other unknown solutions are nearly neutral in pH and produce only ambiguous results with pH paper. Precipitation Reactions When two of the solutions are mixed, a compound may form from the cation in one solution and the anion in the other (double displacement reaction). If the compound is present at a concentration higher than its 2+ 2- solubility, it will precipitate. For example, the Ba from Ba(NO3)2 and the SO4 from Na2SO4 react to form insoluble BaSO4. Ba(NO3)2(aq) + Na2SO4(aq) → BaSO4(s) + 2 NaNO3(aq) (Eq. 4) Most of the anions and cations in the list form insoluble compounds with at least some of the other ions. + + - Those that form only soluble salts are Na , NH4 , and NO3 . Since some precipitates form slowly, a general rule is to wait for about 2 to 5 minutes before discarding the test mixture. Some precipitates take even longer to form and are noted in the accompanying table. Since the dissolution and precipitation of salts is an equilibrium reaction, the concentrations of the ions will affect the equilibrium position. Adding one more drop of a solution can produce a precipitate where there was none before. + A few precipitates formed from Ag ions will darken and finally turn black after several minutes due to the formation of free metallic silver. Examples are Ag2CO3 and Ag2S2O3. This blackening, which requires - light, is essentially the same process that occurs in photographic film. Also note that the OH ion precipitates silver oxide. + - 2 Ag (aq) + 2 OH (aq) Ag2O(s) + H2O(l) (Eq. 5) The precipitate is dark brown or black when formed, and may become blacker after a few minutes. This reaction might be expected when Na2CO3 is added to AgNO3, since the basic Na2CO3 solution contains - - OH ions. However, white Ag2CO3 is obtained instead of Ag2O because [OH ] in the Na2CO3 solution is too 2- low to produce Ag2O in the presence of CO3 . After awhile this white Ag2CO3 precipitate may turn slightly tan/gray. Gas-Producing Reactions Reaction Odor Bubbles + - NH4 (aq) + OH (aq) → NH3(g) + H2O(l) (Eq. 6) ammonia no 2- + 8 S2O3 (aq) + 16 H (aq) → S8(s) + 8 SO2(g) + 8 H2O(l) (Eq. 7) SO2 no 2- + CO3 (aq) + 2 H (aq) → CO2(g) + H2O(l) (Eq. 8) none yes Oxidation-Reduction Reactions Reaction Observation 2+ - Mn (aq) + H2O2(aq) + 2 OH (aq) → MnO2(s) + 2 H2O(l) (Eq. 9) dark brown ppt + - 2 Ag (aq) + H2O2(aq) + 2 OH (aq) → 2 AgO(s) + 2 H2O(l) (Eq. 10) black ppt 2 These precipitates (above) catalyze the breakdown of additional H2O2, producing O2 bubbles that may or may not be visible. 2- The reaction of S2O3 in acid, discussed in the preceding section, is also an oxidation-reduction reaction. Complex Ion Formation Reaction Observation 2- 3- - AgCl(s) + 2 S2O3 (aq) → Ag(S2O3)2 (aq) + Cl (aq) (Eq. 11) ppt. dissolves Solubility of ionic compounds in water at 20 OC Chloride ion Nitrate ion Carbonate Oxalate ion Thiosulfate - - 2- 2- 2- Cl NO3 - ion CO3 - C2O4 ion S2O3 Sodium NaCl Soluble NaNO3 Na2CO3 Na2C2O4 Na2S2O3 + ion Na Soluble Soluble Soluble Soluble Ammoniu NH4Cl NH4NO3 (NH4)2CO3 (NH4)2C2O4 (NH4)2S2O3 m ion Soluble Soluble Soluble* Soluble Soluble 4+ NH Silver ion AgCl AgNO3 Ag2CO3 Ag2C2O4 Ag2S2O3 + Ag Insoluble Soluble Insoluble Insoluble Insoluble** Barium BaCl2 Ba(NO3)2 BaCO3 BaC2O4 BaS2O3 2+ ion Ba Soluble Soluble Insoluble Insoluble Slightly Soluble*** Manganes MnCl2 Mn(NO3)2 MnCO3 MnC2O4 MnS2O3 2+ e ion Mn Soluble Soluble Insoluble**** Insoluble***** Soluble Zinc ion ZnCl2 Zn(NO3)2 ZnCO3 ZnC2O4 ZnS2O3 2+ Zn Soluble Soluble Insoluble Insoluble*** Soluble Aluminum AlCl3 Al(NO3)3 Al2(CO3)3 Al2(C2O4)3 Al2(S2O3)3 3+ ion Al Soluble Soluble Insoluble***# Soluble# Insoluble*** * (NH4)2CO3 is unstable, and releases NH3 gas. + 2- ** An equal number of drops of Ag and S2O3 solutions give a white precipitate that vanishes rapidly. In that case, 2- + add one drop of S2O3 to 10 drops of Ag . This should produce a yellow precipitate that turns brown and then black in one to two minutes. *** Precipitate takes several minutes to form, and often the solution becomes only slightly turbid. **** Precipitate slowly turns tan as it oxidizes in air to dark brown MnO2. ***** Precipitation may not be observed. + # These tests are sometimes inconclusive. Al3 can be identified more clearly by another test. A SOLVED ELEVEN SOLUTION PROBLEM This problem is given as an example to show a logical approach to identify unknown solutions. These are not the solutions that will be used in the actual experiment. Because the solutions are not the same, there are a few more reactions involved. 2- + SO3 (aq) + 2 H (aq) → SO2(g) + H2O(l) - - I is oxidized to dark brown I2 in acid solution by H2O2 or IO3 - 2- 2- In acid solution, I2 is reduced to I by SO3 or S2O3 (thiosulfate) 3 Solutions of NaOH, H2SO4, Na2CO3, HCl, BaCl2, H2O, Zn(NO3) 2, NaI, NaIO3, Na2SO3 and Zn(NO3)2 can be distinguished by the following tests: 1. pH tests. Two strongly basic solutions (NaOH and Na2CO3) and two strongly acidic solutions (H2SO4 and HCl) and seven nearly neutral solutions (BaCl2, H2O, H2O2, Zn(NO3) 2, NaI, NaIO3 and Na2SO3). 2. Flame test. Find BaCl2 from the green flame and the group of five sodium compounds from the bright yellow flame. 3. Mix a drop of BaCl2 with drops of each of the acidic solutions. A precipitate of BaSO4 identifies H2SO4; the other acid must be HCl. 4. Mix a drop of BaCl2 with drops of each of the basic solutions. A precipitate of BaCO3 identifies Na2CO3; the other base must be NaOH. 5. Mix a drop of NaOH with a drop of each of the remaining neutral solutions. A precipitate of Zn(OH) 2 picks out Zn(NO3) 2. 6. Mix a drop of Zn(NO3) 2.with a drop of each of the rest of the remaining neutral solutions and precipitates of Zn(NO3) 2 and ZnSO3 form. 7. One of the neutral solutions (H2O, H2O2, NaI) that did not react with Zn(NO3) 2 is identified as NaI because it gives a positive Na flame test. 8. Mix a drop of NaI with a drop of acid and add a drop of each of the two solutions that reacted with Zn(NO3) 2. NaIO3 gives a brown solution of I2. Since two out of the three sodium containing solutions have now been identified by precipitation-reactions, the remaining solution with no reaction must be Na2SO3. 9. Mix a drop of NaI with a drop of acid, and add a drop of the solutions left over from step 7— H2O or H2O2.
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