Lecture 14 - Acid-Base Equilibria & Titrations

Chem 321 Lecture 14 - Acid-Base Equilibria & Titrations 10/15/13 Student Learning Objectives As a way to review most of the principles of acid-base chemistry, let us calculate the solution pH for various volumes of added strong base in the titration of the diprotic weak acid, maleic acid (cis-butenedioic acid). The focus will be on the set up of the appropriate calculation, so activities will not be used. Problem: Titrate 25.0 mL of 0.10 M maleic acid (H2M) with 0.10 M NaOH. Calculate the pH of the solution at various titrant volumes. (Note: For maleic acid, H2M, Ka1 = 1.20 x 10-2 and Ka2 = 5.96 x 10-7) At this point it is instructive to note the volume of base needed to reach the first equivalence point and the second equivalence point. Because the acid and base concentrations are equal, the first equivalence point volume of titrant is 25.0 mL, and 50.0 mL is needed to reach the second equivalence point. The titration reaction up to the first equivalence point is: H2M + OH- 6 HM- + H2O As is the custom, this reaction is not written as an equilibrium because it has such a large equilibrium constant and can be effectively considered to proceed only in the forward direction. The equilibrium constant (K1) for this reaction can be calculated from known quantities by noting that the reverse reaction is simply the base hydrolysis of HM-. Thus, The titration reaction from the first equivalence point to the second equivalence point is: HM- + OH- 6 M2- + H2O This reaction also has a very large equilibrium constant. Check for Understanding 10.1 1. Solution Calculate the equilibrium constant for the reaction occurring between the first and second equivalence points in the maleic acid titration. Acid-Base Equilibria & Titrations 10/15/13 page 2 At each point in the titration it is convenient (and justified) to assume that the added base will react completely with any available H2M or HM-. This neutralization reaction should be accounted for before deciding what chemical species are present in solution. 0.00 mL titrant At this point no base has been added so you have a solution of a diprotic weak acid. H2M + H2O º H3O+ + HM- Ka1 = 1.20 x 10-2 HM- + H2O º H3O+ + M2- Ka2 = 5.96 x 10-7 Since Ka2 << Ka1, we can neglect the effects of the second dissociation and treat this as a solution of a monoprotic weak acid. We will check this assumption later. The concentration table for the first acid dissociation looks like H2M + H2O initial 0.10 M equil 0.10 - x º H3O+ + HM~10-7 M ~x 0 x where x equals the increase in HM- concentration. Note that the equilibrium concentration of H3O+ is also approximated by x. This assumes that x >> 10-7 M. Substituting these equilibrium quantities into the reaction quotient and equating to Ka1 gives The approximation made in the denominator assumes that x << 0.10 M. Solving this gives x = 0.0346 M. This value for x is not at least a factor of 10 smaller than 0.10 M, so the simplification made in the denominator is not valid and the more complex quadratic equation must be solved. Recall that quadratic equations in Acid-Base Equilibria & Titrations 10/15/13 page 3 the form ax2 + bx + c = 0 have the two solutions given by: The exact solution using the quadratic formula is x = 0.0292 M (a difference of nearly 20%). However, one must be careful when using the quadratic formula because it is easy to make a mistake in substituting the various quantities. Alternatively, it is sometimes possible to use the method of successive approximations. Since the original value of x is close to satisfying the simplifying assumption that x << 0.10 M, one can use this value as a first estimate for x in the denominator term and solve the simpler quadratic equation for another estimate of x. This process is then repeated again and again until successive values of x are acceptably close. Although this calculation is repeated many times, it is much simpler to do and avoids the pitfalls of using the quadratic formula. Substituting x = 0.0346 into the denominator and solving for x gives Repeating this process gives Repeating this process gives At this point the successive values calculated for x are sufficiently close (~1% variation) that we can stop and use 0.0291 as our estimate for x. Note that one additional calculation will produce the exact solution of 0.0292. Thus, [H3O+] = 0.0292 M and pH • 1.54. Acid-Base Equilibria & Titrations 10/15/13 page 4 The calculated equilibrium concentrations for H3O+ and HM- can be used to check the assumption that the second dissociation can be neglected. The concentration table for the second dissociation is HM- + H2O initial 0.029 M equil 0.029 - x º H3O+ + M20.029 M 0.029 + x 0 x Substituting gives The approximations in the numerator and denominator were made because Ka2 is so small and very little dissociation is expected (x << 0.029). Solving this gives x = 5.96 x 10-7 M = [M2-]. This confirms our assumption that very little HM- dissociates. 5.00 mL titrant Since 25.0 mL is needed to reach the first equivalence point, the 5.00 mL mark is 20% of the way along. This means that 20% of the weak acid H2M has been converted to its conjugate base HM- and we have a H2M/HM- buffer solution. In order to assess the concentrations of H2M and HM- after the base has reacted, it is convenient to prepare a “mole table” for the titration reaction. The initial moles of weak acid and moles of added base are gotten by multiplying concentration times volume. initial moles of H2M = (0.0250 L)(0.10 mol/L) = 0.00250 mol moles of NaOH added = (0.0050 L)(0.10 mol/L) = 0.00050 mol The mole table then looks like H2M + OH6 HM- + initial 0.00250 mol 0.00050 mol 0 after rxn 0.00200 mol 0 0.00050 mol H2O Acid-Base Equilibria & Titrations 10/15/13 page 5 Since the total solution volume is now 30.0 mL, [H2M]init = (0.00200 mol)/(0.0300 L) = 0.0667 M [HM-]init = (0.00050 mol)/(0.0300 L) = 0.0167 M The concentration table for the buffer then becomes H2M + H2O initial 0.0667 M equil ~0.0667 M º H3O+ + HM~10-7 M 0.0167 M x ~0.0167 M The usual buffer assumption is to use the initial weak acid and conjugate base concentrations for their equilibrium concentrations, however, one of the “10-3 rules” does not hold for this buffer (Ka is not < 10-3). Consequently we must adjust these initial concentrations. How they are adjusted depends on which way this reaction must shift in order to reach equilibrium. Since this is an acidic buffer (pKa = 1.92), the [H3O+] must increase over the initial 10-7 M value so the reaction shifts to the right. Let x equal the increase in [HM-]. Then H2M + H2O initial 0.0667 M equil 0.0667 - x º H3O+ + HM~10-7 M 0.0167 M ~x 0.0167 + x This assumes that x >> 10-7 M. Substituting these equilibrium quantities into the reaction quotient and equating to Ka1 gives This quadratic equation cannot be simplified because, as noted above, we cannot use the initial weak acid and conjugate base concentrations for the equilibrium values, nor can we use the method of successive approximations here. Consequently, one must solve the quadratic equation using the quadratic formula. The exact solution is x = 1.74 x 10-2 M. Thus, [H3O+] = 0.0174 M and pH • 1.76. Acid-Base Equilibria & Titrations 10/15/13 page 6 12.5 mL titrant At this stage exactly half the base needed to reach the first equivalence point has been added and half the H2M has been converted to HM-. Thus, a buffer in which [H2M] • [HM-] is formed and pH • pKa1 = 1.92. Check for Understanding 10.2 1. Solution Calculate the pH at the 12.5 mL mark following the approach used for the 5.00 mL point, including use of the quadratic formula.