University:
Seattle Pacific University
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CHM 1211 | General Chemistry I
Academic year:

2020
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146

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23
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sonalhava9qv

Chapter 9 (Polyprotic Acids and Bases) Chapter 9 Diprotic Acids H2L+ + H2O ↔ HL + H3O+ Ka1 HL + H2O ↔ L- + H3O+ Ka2 L- + H2O ↔ HL + OH- Kb1 HL + H2O ↔ H2L+ + OH- Kb2 Kw = Ka1 Kb2 Kw = Ka2 Kb1 Zwitter ion (amino acids) Example 1 – (diprotic system, where neutral species is the most acidic form) Calculate the pH of 0.010 M malonic acid, H2M, solution. Ka1 = 1.42⋅10-3 Ka2 = 2.01⋅10-6 H2M + H2O ↔ HM- + H3O+ Ka1 = [HM-] [H3O+] / [H2M] ≈ x2/FH2M → x = 3.76⋅10-3 M, approx. 2 invalid!) Succ. Approx. → x = 0.006182 M, pH = 2.21 This assumes that the second equilibrium is unimportant!! Is it? Succ. Approx. → x = 0.006182 M, pH = 2.21 This assumes that the second equilibrium is unimportant!! Is it? Ka2 = [M2-][H3O+]/[HM-] [M2-] = Ka2 [HM-]/[H3O+] = (2.01⋅10-6)( 0.006182)/( 0.006182) = 2.01⋅10-6 M α2 = [M2-]/[HM-] = 2.01⋅10-6 / 0.006182 = 0.000325 or ≈ 0.03 % It is unimportant!!!! Example 2 (diprotic system, where neutral species is the most basic form) Calculate the pH of a 0.10 M piperazine (P) solution Ka1 = 4.65⋅10-6 Ka2 = 1.86⋅10-10 Kb1 = Kw / Ka2 = 5.43⋅10-5 Kb2 = Kw / Ka1 = 2.79⋅10-9 P + H2O ↔ HP+ + OHKb1 = [HP+] [OH-] / [P] ≈ x2/FP → x = [OH-] = 2.33⋅10-3 M approx. 2 is valid!) [H3O+] = 4.334⋅10-12 M, pH = 11.36 This assumes that the second equilibrium is unimportant!! Is it? [H3O+] = 4.334⋅10-12 M, pH = 11.36 This assumes that the second equilibrium is unimportant!! Is it? Kb2 = [H2P2+][OH-]/[HP+] [H2P2+] = Kb2 [HP+]/[OH-] = (2.79⋅10-9)( 2.33⋅103)/ (2.33⋅10-3) = 2.79⋅10-9 M α2 = [M2-]/[HM-] = 2.79⋅10-9/ 2.33⋅10-3 ≈ 1*10-6 The second equilibrium has a negligible effect on the pH!!!! This is almost always the case!!!!!!..... Unless Ka1 and Ka2 are very close, and even then the pH will not be off by much. For instance, Calculate the pH of a 0.010 M H2A, where Ka1 and Ka2 are 1.00⋅10-4 and 3.00⋅10-4, respectively. Ignoring the second equilibrium, → pH = 3.02 Solving strictly (we will learn how to do this when we cover chapter 9) → pH = 2.93 Example 3 (diprotic system, where neutral species is the intermediate form) Calculate the pH of a 0.050 M solution of Methionine (HM). Methionine is an amino acid. Its neutral form is the intermediate species, HM, of a diprotic system (Ka1 = 6.3⋅10-3 and Ka2 = 8.9⋅10-10). Two different equilibria are possible with HM in water. Ka2 = 8.9⋅10-10 HM + H2O ↔ M- + H3O+ HM + H2O ↔ H2M+ + OH- Kb2 = 1.6⋅10-12 Both will occur, and in fact based on LeChatlier’s principle they will drive one another, since H3O+ reacts rapidly with OH-. How can we deal with this? Charge balance: [OH-] + [M-] = [H3O+] + [H2M+] make substitutions from Ka2 and Kb2 expressions!!! Ka2 = [M-] [H3O+]/[HM] → Kb2 = [H2M+] [OH-]/[HM] → [M-] = Ka2 [HM]/ [H3O+] [H2M+] = Kb2 [HM]/ [OH-] [OH-] = Kw / [H3O+] [OH-] + Ka2 [HM]/ [H3O+] = [H3O+] + Kb2 [HM]/ [OH-] Kw/[H3O+] + Ka2[HM]/[H3O+] = [H3O+] + Kb2 [HM] [H3O+]/kw Kw + Ka2 [HM] = [H3O+]2 + Kb2 [HM] [H3O+]2/ kw (Kw + Ka2 [HM]) = [H3O+]2 (1 + Kb2 [HM]/ kw) (Kw + Ka2 [HM]) / (1 + Kb2 [HM]/ Kw) = [H3O+]2 [H3O+] = [(Kw + Ka2 [HM]) / (1 + Kb2 [HM]/ Kw)]1/2 [H3O+] = [(Kw + Ka2 [HM]) / (1 + Kb2 [HM]/ Kw)]1/2 simplify [H3O+] = [(Kw + Ka2 [HM]) / (1 + [HM]/ Ka1)]1/2 [H3O+] = [Ka1(Kw + Ka2 [HM]) / (Ka1 + [HM])]1/2 [H3O+] = [Ka1Kw + Ka1Ka2 [HM]) / (Ka1 + [HM])]1/2 Now, Ka2 and Kb2 are small, so you can make the following approximation. FHM ≈ [HM], and therefore, [H3O+] = [Ka1Kw + Ka1Ka2 FHM/ (Ka1 + FHM)]1/2 So, [H3O+] = [(6.3⋅10-3)(1.01⋅10-14) + (6.3⋅10-3) (8.9⋅10-10)(.050)/ (6.3⋅10-3 + .050)]1/2 2.23⋅10-6 M pH = 5.65 Is the approximation a good one? Kb2 = [H2M+] [OH-]/[HM] → [H2M+] = Kb2 [HM]/[OH-] = (1.6⋅10-12)(0.050)/(4.53⋅10-9) = 1.8⋅10-5 M Ka2 = [M-] [H3O+]/[HM] → [M-] = Ka2 [HM]/ [H3O+] = (8.9⋅10-10)(0.050)/(2.23⋅10-6) = 2.0⋅10-5 M Approximation was valid!! Note that [H2M+] ≈ [M-]. This is the case even though Ka2 is greater that Kb2 by almost three orders of magnitude!!! Polyprotic systems H3PO4 NaH2PO4 Na2HPO4 K3PO4 weak acid equilibrium; using Ka1 Ka1 = x2/F, where x = [H3O]+ intermediate; using Ka1 and Ka2 pH ≈ (pKa1 + pKa2)/2 intermediate; using Ka2 and Ka3 pH ≈ (pKa2 + pKa3)/2 weak base equilibrium; using Kb1 Kb1 = x2/F, where x = [OH]- Remember Kb1 = Kw/Ka3 Polyprotic buffers A pH 7 phosphate buffer contains approximately equal amounts of H3PO4 NaH2PO4 Na2HPO4 K3PO4 Ka1 = 7.11⋅10-3, pKa1 = 2.148 Ka2 = 6.34⋅10-8, pKa1 = 7.198 Ka3 = 4.22⋅10-13, pKa1 = 12.375 Problem 9-13 How many grams of Na2CO3 (FM 105.99) should be mixed with 5.00 g NaHCO3 (FM 84.01) to produce 100 mL of a buffer with a pH of 10.00. Carbonic acid system: H2CO3 pKa1 = 6.351 NaHCO3 pKa1 = 10.329 Na2CO3 10.00 = 10.329 + log {(mass Na2CO3/105.99)/(5.00/84.01) Solve for mass Na2CO3 mass Na2CO3 = 2.96 g Which is the Principle Species? Histidine pKa1 pKa2 pKa3 (H3His2+, H2His+, HHis, His-) = 1.7 = 6.02 = 9.08 Principle species pH 1 2 3 4 5 6 7 8 9 10 11 12 13 14 answer Principle species pH 1 2 3 4 5 6 7 8 9 10 11 12 13 14 H3His2+ H2His+ H2His+ H2His+ H2His+ H2His+, HHis HHis HHis HHis, HisHisHis His His His Alpha Fractions Monoprotic systems (HA or B) Monoprotic acid αHA = [HA]/FHA, where FHA = [HA] + [A-], or [HA] / ([HA] + [A-]) αA- = [A-]/FHA, where FHA = [HA] + [A-], or [A-] / ([HA] + [A-]) Monoprotic base αB = [B]/FB, where FHA = [B] + [HB+], or [B] / ([B] + [HB+]) αHB+ = [HB+]/FB, where FB = [B] + [BH+], or [HB+] / ([B] + [HB+]) Monoprotic acid Ka = [H3O+][A-]/[HA] = [H3O+](FHA-[HA])/[HA] A little algebra [HA] Ka + [H3O+][HA] = [H3O+]FHA [HA] (Ka + [H3O+]) = [H3O+]FHA [HA] / FHA = [H3O+] / (Ka + [H3O+]) αHA = [HA]/FHA = [H3O+] / (Ka + [H3O+]) similarly, you can derive αA- = [A-]/FHA = Ka / (Ka + [H3O+]) or for a monoprotic base, αB = [B]/FB = Ka / (Ka + [H3O+]) αHB+ = [HB+]/FB = [H3O+] / (Ka + [H3O+]) alpha fractions for a Diprotic System H2A ↔ HA- + H+ HA- ↔ A2- + H+ (K1) (K2) K1 = [H+][HA-]/[H2A] → [HA-] =[H2A]{K1/[H+]} (1) K2 = [H+][A2-]/[HA-] → [A2-] =[HA-]{K2/[H+]} (2) Mass balance equation F = [H2A] + [HA-] + [A2-] Mass balance equation F = [H2A] + [HA-] + [A2-] First solve for the αH2A. To do these we need to write expression that puts the three other species in terms of H2A. We already have eq 1 [HA-] = [H2A]K1/[H+] Now substitute this into eq 2 to get [A2-] = [H2A]K1K2/[H+]2 Substitute these three new equations into the mass balance equation F = [H2A] + [H2A]K1/[H+] + [H2A]K1K2/[H+]2 F = [H2A] (1+ K1/[H+] + K1K2/[H+]2) F/[H3A] = 1+ K1/[H+] + K1K2/[H+]2 Get a common denominator F/[H3A] = ([H+]2+ [H+]K1 + K1K2 ) / [H+]2 Flip the numerator and denominator on both sides αH2A = [H2A]/F = [H+]2/([H+]2+ [H+]K1 + K1K2) Next I will solve for αHA-. To do these we need to write expression that put the three other species in terms of HA-. rearranging eq 1we get [H2A] = [HA-][H+]/K1 eq 2 gives us the expression for A2[A2-] = [HA-]K2/[H+] Substitute these into the mass balance equation F = [HA-][H+]/K1 + [HA-] + [HA-]K2/[H+] F = [HA-] ([H+]/K1 + 1 + K2/[H+]) F/[HA-] = [H+]/K1 + 1 + K2/[H+] Get a common denominator F/[HA-] = ([H+]2+ [H+]K1 + K1K2) / K1[H+] Flip the numerator and denominator on both sides αHA- = [HA-]/F = (K1[H+]) / ([H+]2+ [H+]K1 + K1K2 + K1K2) Next I will solve for αA2-. To do these we need to write expression that put the three other species in terms of A2-. Eq 2 gives us an expression for A-2 [HA-] = [A2-][H+]/K2 Rearrange eq 1to get [H2A] = [HA-][H+]/K1 Rearrange eq 1 and substitute [H2A] = [A2-][H+]2/K1K2 F = [A2-][H+]2/K1K2 + [A2-][H+]/K2 + [A-2] F = [A-2] ([H+]2/K1K2 + [H+]/K2 + 1) F/[A2-] = [H+]2/K1K2 + [H+]/K2 + 1 Get a common denominator F/[A2-] = ([H+]2+ [H+]K1 + K1K2) / K1K2 Flip the numerator and denominator on both sides αA2- = [A2-]/F = (K1K2) / ([H+]2+ [H+]K1 + K1K2) Isoelectric point and electrophoresis Isoelectric point = pH = (pKa1 + pKa2) isoionic pt; [H3O+] = [Ka1Kw + Ka1Ka2 FHM/ (Ka1 + FHM)]1/2 Separation of peptides Ions move in response to an electric field. If we place our peptide mixture in a capillary column and then create a pH gradient across the capillary, the peptides will migrate until they reach a pH where the average charge on the peptide is zero (the isoelectric point)

Chapter 9 Polyprotic Acids and Bases

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