Week 7: Coordination Chemistry 3 Overview Getting Started Techniques Procedure FAQ Full Lab Manual Introduction & Goals Chemistry & Background Key Questions Prelab Problems Safety Procedure In Your Write-up Experiments Index ChemLab Home		Deprotonation of a Coordinated Water Molecule Coordination of a water molecule to a metal center with a high positive charge greatly increases its acidity, compared to free water. For example, your cobalt complex acts as a weak acid in aqueous solution by donating a hydrogen ion to water: [Co(NH3)5(H2O)]3+ + H2O [Co(NH3)5(OH)]2+ + H3O+ (1) (2) Remember that the Ka for uncoordinated water is only 10-14, so the acidity of the coordinated water in the pentaamminecobalt complex is greatly increased. For comparison, the Ka for acetic acid is 1.8 x 10-5. The Ka value of 2.04 x 10-6 M quoted here is for an infinitely dilute solution, a condition that is not satisfied in this experiment. The presence of a significant concentration of highly charged ions in solution will result in a value which is somewhat smaller than 2.04 x 10-6 . In the following section, we show how a familiar acid-base titration can be used to analyze this acidity of the coordinated H2O. Addition of the base NaOH to a solution of the [Co(NH3)5(H2O)]3+ brings about the loss of a hydrogen ion from the coordinated water molecule. (3) aquapentaamminecobalt(III) hydroxopentaamminecobalt(III) The equilibrium constant for this process can be calculated by combining the acid dissociation constant for [Co(NH3)5(H2O)]3+ and the equilibrium constant for the autoionization of water. [Co(NH3)5(H2O)]3+ + H2O [Co(NH3)5(OH)]2+ + H3O+ Ka = 2.04 x 10-6 (4) H3O+ + OH - 2 H2O K5 = 1/Kw = 1014 (5) [Co(NH3)5(H2O)]3+ + OH - [Co(NH3)5(OH)]2+ + H2O K6 = Ka/Kw (6) If [Co(NH3)5(H2O)]3+ is titrated with NaOH, the large equilibrium constant K6 for (3) or (6) assures that the reaction will go essentially to completion at each stage of the titration. This fact is illustrated in the following sample calculation for a typical point in the titration before the end point. Example. A 4.0 mL volume of 0.100 M NaOH is added to 40 mL of 0.025 M [Co(NH3)5(H2O)]3+. What are the resulting concentrations of [Co(NH3)5(H2O)]3+, [Co(NH3)5(OH)]2+, and OH-? The initial number of mmol of aqua complex = (40 mL)(0.025 M) The number of mmol of aqua complex which have reacted with OH- = (4.0 mL)(0.10 M) The number of mmol of unreacted aqua complex = (40 mL)(0.025 M)-(4.0 mL)(0.10 M) The concentration of unreacted aqua complex = (40 mL)(0.025 M)-(4.0 mL)(0.10 M) 44 mL = 1.4 x 10-2 M The number of mmol of hydroxo complex formed = (4.0 mL)(0.10 M) The concentration of hydroxo complex formed = (4.0 mL)(0.10 M) 44 mL                            = 9.1 x 10-3 M The above considerations assume that the reaction goes to completion, i.e. there is no unreacted OH-. Although K6 is very large, there will be a small concentration of unreacted OH-. When this is taken into account, the following concentrations are obtained: Concentration of unreacted OH- = X Concentration of unreacted aqua complex, [Co(NH3)5(H2O)3+] = 1.4 x 10-2 M + X Concentration of complex formed, [Co(NH3)5(OH)2+] = 9.1 x 10-3 M - X These data are summarized below: [Co(NH3)5(H2O)]3+ +     OH-     [Co(NH3)5(OH)]2+ + H2O (1.4 x 10-2 M + X) X (9.1 x 10-3 M - X) If it is assumed that the concentration of unreacted OH-, (X), is negligibly small compared to both 1.4 x 10-2 M and 9.1 x 10-3 M, the following expression is obtained from K6: X = [OH-] = 3.3 x 10-9 M The results support the assumption of negligible [OH-] compared to [Co(NH3)5(H2O)]3+ or [Co(NH3)5(OH)]2+, indicating that the reaction has gone essentially to completion. If the assumption of complete reaction is invoked throughout the titration, a simple expression can be derived for the concentrations of both the aqua and hydroxo complexes as functions of the volume of base added. Let V1 = initial volume of the aqua complex solution (mL) C1 = initial concentration of the aqua complex solution (M) V2 = volume of base added (mL) C2 = concentration of base (M) Before the end point: (7) where V1C1 = initial number of mmol of aqua complex, V2C2 = number of mmol of aqua complex that have reacted = number of mmol of hydroxo complex formed, V1+V2 = total volume of solution (in mL) and (8) The end point will be reached when V2C2 = V1C1, since the moles of base added equals the moles of acid. At, and after, the end point nearly all the cobalt will be present as the hydroxo complex, and its concentration will be given by (9): (9) Potentiometric End Point Detection Much ingenuity has gone into devising methods for detecting the end points of analytically useful titrations. Many of them are chemical techniques, generally involving indicators. In recent years, however, there has been increased reliance on the use of physical methods to detect end points. One reason for their popularity is that they often are able to circumvent restrictions which limit the applicability of chemical methods. Phenolphthalein would be unsuitable, for example, in the titration of the aqua complex with base since the indicator color change would be masked by the color of the complex. Another important reason for the increased use of physical methods is that they are intrinsically well-suited to automation. Most of the routine analyses of industrial and clinical importance are now done with some degree of automation. Automatic titrators are one example, and instruments which employ both pH and light absorption monitoring to locate the end point have been developed. In your titration, the pH of the solution will be monitored. The pH meter is a central fixture in research laboratories, whether they be concerned with chemistry, biology, geology, or medicine. Your pH meter will detect the end point by observing the sharp change in pH that accompanies it. Titration Curves By far the most general method of end point detection for acid-base titrations involves measuring the pH during the titration. One of the main advantages of the pH method is that it is capable of high precision since the pH changes very rapidly near the end point. The literature value for the acid dissociation constant of [Co(NH3)5(H2O)]3+ was provided in the previous section: The theoretical form of the titration curve is completely determined by the value of Ka and the concentrations of acid and base involved. Because of the importance of titration curves to acid-base equilibria, an example of a specific aquapentaamminecobalt(III) titration by sodium hydroxide will be worked through in detail. Figure 4 Titration curve for 100 mL of 0.01 M [Co(NH3)5(H2O)][NO3]3 with 0.10 M NaOH. Example. A 100 mL solution 0.010 M in [Co(NH3)5(H2O)][NO3]3 is titrated with 0.10 M NaOH. Four different cases occur during the titration which will be treated separately, since different approximations apply to each. I. Initial Solution. Since Ka is small, a negligible fraction of the weak acid [Co(NH3)5(H2O)]3+ is assumed to be dissociated. Equal concentrations of [Co(NH3)5(OH)]2+ and H3O+ are produced by this dissociation: [Co(NH3)5(H2O)]3+ +  H2O   [Co(NH3)5(OH)]2+ +   H3O+ (0.01 M - X) X X X is assumed to be much smaller than 0.01 M. Thus, Solving for X yields X = [H3O+] = 1.4 x 10-4 M. Thus pH = -log10 [H3O+] = 3.85. Note that the value X = 1.4 x 10-4 M is consistent with the original assumption that X<<0.01 M. Only (1.4 x 10-4 M/10-2 M) x 100 = 1.4% of the aqua complex is dissociated. Also, any concern you might have that the calculated [H3O+] is in error because of a failure to consider the dissociation of water is easily satisfied. Since Kw = [H3O+][OH-] = 10-14 and since water dissociation by itself produces equal concentrations of H3O+ and OH-, it only contributes significantly to the hydrogen ion concentration of acidic solutions when they approach neutrality. In more acidic media, some other source of H3O+ dominates. II. Buffer Region. Nearly the entire titration curve before the equivalence point can be treated by the simple assumption that the reaction between the added base and [Co(NH3)5(H2O)]3+ goes stoichiometrically to completion. Let V2 = the volume of base added in milliliters for the net titration reaction [Co(NH3)5(H2O)]3+ + OH- [Co(NH3)5(OH)]2+ + H2O. Assuming complete reaction, and Thus, An important special case occurs at V2 = 5 mL, the half-equivalence point. When this volume of base has been added, the numerator and the denominator in the equation for [H3O+] are equal, and hence [H3O+] = Ka. The concentrations of [Co(NH3)5(H2O)]3+ and [Co(NH3)5(OH)]2+ and the pH at this stage of the titration are calculated as follows: [H3O+] = Ka           pH = -log10 (2.04 x 10-6) = 5.69 = pKa Experimental values of Ka for weak acids are typically calculated from the pH at the half-equivalence point. You will use this technique to determine the experimental Ka of your cobalt complex. III. The Equivalence Point. Here nearly all the cobalt is present as the conjugate base[Co(NH3)5(OH)]2+. A small concentration of [Co(NH3)5(H2O)]3+ arises from the following equilibrium (known as hydrolysis): With [OH-] = [Co(NH3)5(H2O)] 3+= X and, with V2 = 10 mL, since X is going to be small. The equilibrium constant for this reaction (call it Kb because [Co(NH3)5(OH)]2+ acts like a base in the above reaction) is given by which, if multiplied in the numerator and denominator by [H3O+], becomes Thus or, solving for X, X = [OH-] = 6.7 x 10-6 M,     [H3O+] = 10-14/[OH-] = 1.5 x 10-9 M and pH = -log10 (1.5 x 10-9) = 8.82, which is decidedly basic. IV. Past Equivalence. All but a negligible fraction of cobalt exists as [Co(NH3)5(OH)]2+. Excess hydroxide is present and the concentration of this excess base determines the pH. The numerator in the equation for [OH-] is the total mmol of NaOH added minus the mmol NaOH required for equivalence. The denominator is the volume of the solution in milliliters. For example, at V2 = 12 mL, [OH-] = (1.2 - 1.0 mmol)/112 mL = 1.8 x 10-3 M and, therefore, [H3O+] = 10-14/[OH-] = 5.6 x 10-12 M and pH = -log10 (5.6 x 10-12) = 11.3. How nearly do the results in I through IV produce a full pH titration curve? The answer depends in general on concentrations and the Ka value involved. For the present rather typical case, the results are excellent, as is shown in Figure 4. The solid curve in Figure 4 represents an exact computer-generated curve for the whole titration. The discrete points have been calculated at 0.5 mL intervals using cases I, II, III, or IV as appropriate. The rapid rise in pH at the equivalence point allows precise determination of the titration end point. Again it should be stressed that an understanding of the calculations of cases I, II, III and IV is of great importance to building a useful familiarity with the concepts of acid-base equilibrium. The experimental method is conveniently summarized by rearranging equation (2) above: [H3O+] is measured with a pH meter. The initial concentration of the aquo complex (numerator) is known since a measured mass is dissolved in a measured volume (see below). The concentrations of the two Co complexes present during titration can be calculated easily since the titration reaction goes to completion. Preliminary Kinetics Measurements In addition to the acid-base titration of your complex, you and your partner will perform preliminary kinetics measurements this week. The mechanism of the ligand exchange reaction between the water ligand of your complex and the nitrite ion, NO2-, will be examined by following the reaction colorimetrically. Details of the reaction kinetics will be further investigated next week. Introductory material and procedural information can be found in the pages for Week 7 Part 2.
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