Acid-Base Equilibria & Titrations

Chem 321 Lecture - Acid-Base Equilibria & Titrations Student Learning Objectives Continuing Problem: Titrate 25.0 mL of 0.10 M maleic acid (H2M) with 0.10 M NaOH and calculate the pH of the solution at various titrant volumes. 25.0 mL titrant (first equivalence point) At the first equivalence point, all the H2M has been converted to NaHM(aq) and there is a solution of the amphiprotic species HM-. Since the volume has doubled, the concentration of NaHM at this point is half the original [H2M], or 0.050 M. HM- can react with water in two ways. HM- + H2O º H3O+ + M2- Ka2 = 5.96 x 10-7 HM- + H2O º H2M + OH- Kb = Kw/Ka1 = 8.33 x 10-13 We expect the solution to be acidic because Ka2 >> Kb. If we ignore the base hydrolysis reaction and treat HM- solely as a weak acid, [H3O+]equil = 1.7 x 10-4 M and pH • 3.76. Check for Understanding 10.3 1. Solution Calculate the pH of a 0.050 M HM- solution assuming it acts solely as a weak acid. However, it turns out that the base hydrolysis reaction proceeds to nearly the same extent as the acid dissociation process because it is driven to the right by the H3O+ produced in the acid dissociation reaction. If one is doing careful calculations, one should consider both reactions and both equilibria. In these two equations there are 5 unknowns and in order to solve this problem exactly we need 3 additional equations. One equation comes from the water selfionization equilibrium. Another condition that must hold is charge balance. Since the solution is electrically neutral, the concentration of positive charges must equal the concentration of negative charges. In terms of the cations and anions in this titration solution this becomes Notice that the concentration of M2- is doubled because its charge is 2-. Since [Na+] can be determined at any point in the titration, no new unknowns have been introduced in this additional equation. The final relationship is based on mass conservation. Regardless of where you are in this titration, the maleic acid must exist as either H2M, HM- or M2-. This can be expressed as: Again, no new unknowns are involved. These 5 equations with 5 unknowns can be combined and solved exactly to get the [H3O+]. The rigorous solution gives Since the equilibrium constants for the acid dissociation and the base hydrolysis reactions are both very small, it is a reasonable assumption that neither reaction will alter [HM-] significantly, thus [HM-] ~ MNaHM. Consequently, the [H3O+] and pH can be evaluated. This rather complex expression for [H3O+] can be greatly simplified if certain conditions hold. Factoring out Ka1 in the numerator gives If Ka2 [HM-] >> Kw then Kw can be dropped in the numerator and the numerator becomes ([HM-] Ka1 Ka2). If [HM-] >> Ka1, then Ka1 can be dropped in the denominator and the overall expression becomes Taking the negative log of this equation gives: This rather simple expression for determining the pH of the amphiprotic substance (HM-) present at the first equivalence point is the basis for the method of estimating the equivalence point pH as an average of the pH plateaus before and after the equivalence point. Using these simple expressions, one calculates [H3O+] • 8.5 x 10-5 M and pH • 4.07 for the solution at the first equivalence point. However, since [HM-] • 0.050 M it is not >> Ka1 (= 0.012), so the denominator cannot be simplified and the more rigorous expression gives [H3O+] • 7.6 x 10-5 M and pH • 4.12. This value for [H3O+] differ by more than a factor of 2 from the result obtained by neglecting the base hydrolysis reaction. Check for Understanding 10.4 1. Solution Take into account both possible reactions of H2PO4- with water and calculate the pH of a 0.10 M KH2PO4 solution. Compare this to the previously calculated value assuming only weak acid behavior. 30.0 mL titrant This is 5.00 mL past the first equivalence point and 20% of the way to the second equivalence point. This means that 20% of the weak acid HM- has been converted to its conjugate base M2- and we have a HM-/M2- buffer solution. In order to assess the concentrations of HM- and M2- after the base has reacted, again prepare a mole table. initial moles of HM- = initial moles of H2M = (0.0250 L)(0.10 mol/L) = 0.00250 mol moles of NaOH added after 1st equiv point = (0.0050 L)(0.10 mol/L) = 0.00050 mol The mole table then looks like HM+ OH6 M2- + initial 0.00250 mol 0.00050 mol 0 after rxn 0.00200 mol 0 0.00050 mol Since the total solution volume is now 55.0 mL, [HM-]init = (0.00200 mol)/(0.0550 L) = 0.0364 M [M2-]init = (0.00050 mol)/(0.0550 L) = 0.00909 M The concentration table for the buffer then becomes: HM- + H2O initial 0.0364 M equil ~0.0364 M º H3O+ + M2~10-7 M 0.00909 M x ~0.00909 M where x is the equilibrium [H3O+]. Substituting gives Solving this gives x = 2.4 x 10-6 M = [H3O+] and pH • 5.62. H2O 50.0 mL titrant (second equivalence point) At the second equivalence point, all the HM- has been converted to Na2M(aq) and there is a solution of the weak base M2-. The total volume is now 75.0 mL, so the concentration of Na2M at this point is one third the original [H2M], or 0.0333 M. Base hydrolysis is the predominant reaction. M2- + H2O initial 0.0333 M equil 0.0333 - x º HM- + OH0 x Kb = Kw/Ka2 = 1.68 x 10-8 ~10-7 M ~x where x is the increase in the [HM-] and is it assumed that x >> 10-7 M. Substituting gives Solving this gives x = 2.4 x 10-5 M = [OH-] and pOH • 4.63. Thus, pH • 9.37. Note that the two assumptions regarding the value of x are valid. Maleic Acid Titration Curve The approaches discussed above can be used to make additional calculations of pH at various points for this maleic acid titration. A plot of pH versus titrant volume constitutes a titration curve, as shown below. Note the key points labeled on the titration curve. Determination of the pH halfway to the first equivalence point and halfway between the first and second equivalence points allows one to estimate pKa1 and pKa2, respectively, for the diprotic weak acid. mL NaOH pH 14 0.00 1.54 12 5.00 1.76 10 12.5 1.92 20.0 2.64 Maleic Acid Titration with NaOH pH ~ pKa2 8 pH 6 pH ~ pKa1 4 25.0 4.12 30.0 5.62 35.0 6.04 40.0 6.40 45.0 6.83 50.0 9.37 55.0 11.80 60.0 12.07 2nd equiv. pt. (soln of M2-) 1st equiv. pt. (soln of HM-) 2 0 0 Exercises for Acid-Base Equilibria & Titrations 10 20 30 40 mL NaOH 50 60 70