1Qualitative Analysis BACKGROUND INFORMATION One task often faced by chemists is the characterization of a sample. If the sample is a mixture, this involves the identification of its components. This type of analysis, which is concerned with “what” is in a mixture rather than “how much” is called qualitative analysis. In this experiment, you will be required to identify the cations present in an aqueous solution of three simple salts. A possible unknown solution is a solution containing Fe(NO3)3, Al(NO3)3 and Ba(NO3)2. A correct analysis of this solution would identify the presence of the ions: Fe3+, Al3+, and Ba2+. The procedure is to add to the solution certain reagents which react in a specific, known way with each of the possible cations in the solution. To illustrate the method consider an unknown solution which contains only one of the cations: Fe3+, Al3+ or Ba2+. How do we determine which one is present? One approach is to add a solution of NaOH, a strong base. Both Fe3+, and Al3+ form insoluble hydroxides, and the following precipitation reactions occur: Al(OH)3(s) →← Al3+ + 3 OH− Ksp = 1.9 x 10-33 (1) Fe(OH)3(s) →← Fe3+ + 3 OH− Ksp = 2.6 x 10-39 (2) Barium hydroxide is much more soluble than the other two and does not precipitate. Hence, the formation of a precipitate indicates the presence of either Fe3+ or Al3+. How do we distinguish between these two ions? In the cation is Al;3+, further addition of NaOH will cause the precipitate to dissolve due to complex formation: Al(OH)3(s) + OH− →← Al(OH)4−(aq) Kf = 2 x 1033 (3) Therefore, if upon addition of NaOH, a precipitate forms which dissolves in excess NaOH the presence of Al3+ is indicated. If the precipitate does not dissolve, then Fe3+ is present. If no precipitate forms, then Ba2+ is present. The presence of Ba2+ can be confirmed by the addition of NaSO4(aq) which results in the precipitation of highly insoluble barium sulfate: BaSO4(s) →← Ba2+(aq) + SO42− Ksp = 1.0 x 10-10 (4) You will encounter several different kinds of chemical reactions in the experiment: precipitation reactions, complex ion formation reactions (reaction (3)), acid-base reactions and oxidation-reduction reactions. Precipitation reactions Examples of these are reactions (1), (2) and (4), above. A salt will precipitate when its ion product equals or exceeds its Ksp. For example, barium sulfate has Ksp = 1.0 x 10-10. Barium sulfate will not precipitate unless [Ba2+][SO42-] ≥ Ksp. The smaller Ksp the smaller the solubility. Some Ksp values are given below.2 Solubility Product Constants at 25oC Compound Ksp Al(OH)3(s) 1.9 x 10-33 Ba(OH)2(s) 5.0 x 10-3 Ca(OH)2(s) 4.7 x 10-6 Co(OH)2(s) 1.1 x 10-15 Cu(OH)2(s) 1.6 x 10-19 Fe(OH)3(s) 2.6 x 10-39 Mg(OH)2(s) 5.6 x 10-12 Mn(OH)2(s) 2.1 x 10-13 BaCO3(s) 2.6 x10-9 CaCO3(s) 5.0 x 10-9 CuCO3(s) 2.5 x 10-10 MgCO3(s) 6.8 x 10-6 MnCO3(s) 2.2 x 10-11 MgNH4PO4(s) 2.5 x 10-13 Acid-base reactions The net ionic equation for the reaction of a strong acid with a strong base is shown below. The large value of K indicates that this reaction goes to completion. H3O+ + OH- → 2 H2O K = 1/Kw = 1014 (5) The reaction of a strong base with a weak acid, HA, equation (6), goes to completion since the equilibrium constant is equal to the reciprocal of Kb, a large number, since A- is a weak base, and Kb is small. OH- + HA → H2O + A- K = 1/Kb = Ka/Kw >>1 The same is true of the reaction of a strong acid with a weak base (here, the weak base, A-, is the anion of the weak acid, HA, and the equilibrium constant is equal to the reciprocal of Ka, again a large number since HA is weak and Ka is small): H3O+ + A- → H2O + HA K = 1/Ka = Kb/Kw >>13Dissolving salts of weak acids: A salt whose anion is the conjugate base of a weak acid tends to be more soluble in acid than in pure water. Metal hydroxides dissolve in acid because the hydroxide ion is the conjugate base of H2O. For example, iron (III) hydroxide is very insoluble in pure water, but dissolves readily in acid due to the sequence of reactions: Fe(OH)3(s) →← Fe3+ + 3 OH− Ksp = 2.6 x 10-39 (7) 3 H3O+ + 3 OH- → 6 H2O______________ K = (1/Kw)3 = 1042 (8) Net reaction: Fe(OH)3(s) + 3 H3O+ →← Fe3+ + 6 H2O Knet = 2.6 x 103 The large value of Knet indicates that for the net reaction the position of equilibrium lies largely on the product side. The addition of strong acid results in the removal of hydroxide ion by the acid-base reaction (8). According to Le Chatelier’s Principle, the removal of OH- drives reaction (7) forward, and the Fe(OH)3(s) dissolves. Oxidation-reduction reactions (electron transfer reactions) One example, in our analytical scheme, is the reaction of iodide ion with Cu2+. The copper (II) ion is reduced to copper (I) iodide, and the iodide is oxidized to I2. The products of the reaction are a distinctive tan colored precipitate of CuI and a brown solution of I2. The occurrence of this reaction confirms the presence of Cu2+. 2 Cu2+ + 4 I- →← 2 CuI(s) + I2(aq) The half-reactions are: 2 I- + 2 Cu2+ + 2 e- → 2 CuI(s) reduction 2 I- → I2 + 2 e- oxidation Complex ion formation reactions A complex ion consists of a central metal ion to which are bonded two, four or six neutral or ionic species called ligands. An example is the aluminum hydroxide complex ion, Al(OH)4+ The stability of a complex ion is indicated by its formation constant, Kf. For Al(OH)4+, the expression for Kf is the equilibrium constant for the reaction Al(OH)3(s) + OH− →←Al(OH)4−(aq) -4f3+ - 4[Al(OH) ]K=[Al ][OH ] Thus, the bigger Kf, the more stable the complex ion. Some formation constants important to this experiment are shown in the table below.4 Formation Constants for Complex Ions at 25 oC Complex Ion Kf Al(OH)4− 2 x 1033 Co(NH3)62+ 1.7 x 1035 Cu(NH3)42+ 5.6 x 1011 Ni(NH3)62+ 2.0 x 108 LABORATORY PROCEDURES FOR QUALITATIVE ANALYSIS In this experiment, you will be analyzing small volumes (1mL) of solutions, and will be adding correspondingly small amounts of reagents to them. You will be working on what is called the “semimicro” scale. Most tests will be carried out in test tubes. Keys to success in