Polyprotic Acids

1. Polyprotic Acids

4. Interesting Features of a plots • First, notice that the pH where two species concentrations are the same is around the pKa for that equilibrium. In fact, for polyprotic acids with pKa's that differ by over 3 to 4 units, the pH is equal to the pKa.

5. Take for example the point where [H3PO4]=[H2PO4-]. The equilibrium equation relating these two species is If we take the -log10, or "p", of this equation Since [H3PO4]=[H2PO4-], and log10(1) = 0, pH=pKa1 pKa1 pKa2 pKa3

6. Second, you might notice that the concentrations of the conjugate bases are maximum half-way between the pKa points. For example, the point where [H2PO4-] is a maximum lies half-way between between pKa1 and pKa2. Since H2PO4- is the major species present in solution, the major equilibrium is the disproportionation reaction. This equilibrium cannot be used to solve for pH because [H3O+] doesn't occur in the equilibrium equation. We solve the pH problem adding the first two equilibria equations +

7. Note that when we add chemical equilibria, we take the product of the equilibrium equations. Taking the -log10 of the last equation Since the disproportionation reaction predicts [H3PO4]=[HPO42-]

8. What about the “Z” word?? • Zwitterions – (German for “Double Ion”) – a molecule that both accepts and losses protons at the same time. • EXAMPLES??? • How about – AMINO ACIDS zwitterion Both groups protonated neutral amino-protonated carboxylic-deprotonated • why activity of proteins are pH dependent

10. Let’s look at the simplest of the amino acids, glycine K1 K2 Gly- glycinate H2Gly+ glycinium HGly In water the charge balance would be, Combining the autoprotolysis of water and the K1 and K2 expressions into the charge balance yields:

11. HGly Gly- H2Gly+

12. Diprotic Acids and Bases • 2.)Multiple Equilibria • Illustration with amino acid leucine (HL) • Equilibrium reactions high pH low pH Carboxyl group Loses H+ ammonium group Loses H+ Diprotic acid:

13. Diprotic Acids and Bases • 2.)Multiple Equilibriums • Equilibrium reactions Diprotic base: Relationship between Ka and Kb

14. pKa of carboxy and ammonium group vary depending on substituents Largest variations

15. Polyprotic Acid-Base Equilibria • Diprotic Acids and Bases • 3.)General Process to Determine pH • Three components to the process • Acid Form [H2L+] • Basic Form [L-] • Intermediate Form [HL] high pH low pH Carboxyl group Loses H+ ammonium group Loses H+

16. Polyprotic Acid-Base Equilibria • Diprotic Acids and Bases • 3.)General Process to Determine pH • Acid Form (H2L+) • Illustration with amino acid leucine • H2L+ is a weak acid and HL is a very weak acid K2=1.80x10-10 K1=4.70x10-3 Assume H2L+ behaves as a monoprotic acid

17. Polyprotic Acid-Base Equilibria • Diprotic Acids and Bases • 3.)General Process to Determine pH • 0.050 M leucine hydrochloride K1=4.70x10-3 + H+ Determine [H+] from Ka: Determine pH from [H+]: Determine [H2L+]:

18. Polyprotic Acid-Base Equilibria • Diprotic Acids and Bases • 3.)General Process to Determine pH • Acid Form (H2L+) What is the concentration of L- in the solution? [L-] is very small, but non-zero. Calculate from Ka2 Approximation [H+] ≈ [HL], reduces Ka2 equation to [L-]=Ka2 Validates assumption

19. Polyprotic Acid-Base Equilibria • Diprotic Acids and Bases • 3.)General Process to Determine pH • For most diprotic acids, K1 >> K2 • Assumption that diprotic acid behaves as monoprotic is valid • Ka ≈Ka1 • Even if K1 is just 10x larger than K2 • Error in pH is only 4% or 0.01 pH units • Basic Form (L-) • L- is a weak base and HL is an extremely weak base Assume L- behaves as a monoprotic base

20. Polyprotic Acid-Base Equilibria • Diprotic Acids and Bases • 3.)General Process to Determine pH • 0.050 M leucine salt (sodium leucinate) Determine [OH-] from Kb: Determine pH and [H+] from Kw: Determine [L-]:

21. Polyprotic Acid-Base Equilibria • Diprotic Acids and Bases • 3.)General Process to Determine pH • Basic Form (L-) What is the concentration of H2L+ in the solution? [H2L+] is very small, but non-zero. Calculate from Kb2 Validates assumption [OH-] ≈ [HL], Fully basic form of a diprotic acid can be treated as a monobasic, Kb=Kb1

22. Polyprotic Acid-Base Equilibria • Diprotic Acids and Bases • 3.)General Process to Determine pH • Intermediate Form (HL) • More complicated HL is both an acid and base • Amphiprotic – can both donate and accept a proton • Since Ka > Kb, expect solution to be acidic • Can not ignore base equilibrium • Need to use Systematic Treatment of Equilibrium

23. Polyprotic Acid-Base Equilibria • Diprotic Acids and Bases • 3.)General Process to Determine pH • Intermediate Form (HL) Step 1:Pertinent reactions: Step 2:Charge Balance: Step 3:Mass Balance: Step 4:Equilibrium constant expression (one for each reaction):

24. Polyprotic Acid-Base Equilibria • Diprotic Acids and Bases • 3.)General Process to Determine pH • Intermediate Form (HL) Step 6:Solve: Substitute Acid Equilibrium Equations into charge balance: All Terms are related to [H+] Multiply by [H+]

25. Polyprotic Acid-Base Equilibria • Diprotic Acids and Bases • 3.)General Process to Determine pH • Intermediate Form (HL) Step 6:Solve: Factor out [H+]2: Rearrange:

26. Polyprotic Acid-Base Equilibria • Diprotic Acids and Bases • 3.)General Process to Determine pH • Intermediate Form (HL) Step 6:Solve: Multiply by K1 and take square-root: Assume [HL]=F, minimal dissociation: (K1 & K2 are small)

27. Polyprotic Acid-Base Equilibria • Diprotic Acids and Bases • 3.)General Process to Determine pH • Intermediate Form (HL) Step 6:Solve: Calculate a pH:

28. Polyprotic Acid-Base Equilibria • Diprotic Acids and Bases • 3.)General Process to Determine pH • Intermediate Form (HL) Step 7:Validate Assumptions Assume [HL]=F=0.0500M, minimal dissociation (K1 & K2 are small). Calculate [L-] & [H2L+] from K1 & K2: Assumption Valid [HL]=0.0500M >> 9.36x10-6 [H2L+] & 1.02x10-5 [L-]

29. Polyprotic Acid-Base Equilibria • Diprotic Acids and Bases • 3.)General Process to Determine pH • Intermediate Form (HL) • Summary of results: • [L-] ≈ [H2L+]  two equilibriums proceed equally even though Ka>Kb • Nearly all leucine remained as HL • Range of pHs and concentrations for three different forms

30. Polyprotic Acid-Base Equilibria • Diprotic Acids and Bases • 3.)General Process to Determine pH • Simplified Calculation for the Intermediate Form (HL) Assume K2F >> Kw: Assume K1<< F:

31. Polyprotic Acid-Base Equilibria • Diprotic Acids and Bases • 3.)General Process to Determine pH • Simplified Calculation for the Intermediate Form (HL) Cancel F: Take the -log: pH of intermediate form of a diprotic acid is close to midway between pK1 and pK2 Independent of concentration:

32. Polyprotic Acid-Base Equilibria • Polyprotic Acids and Bases • 4.)Fractional Composition Equations • Diprotic Systems • Follows same process as monoprotic systems Fraction in the form H2A: Fraction in the form HA-: Fraction in the form A2-:

33. Polyprotic Acid-Base Equilibria • Isoelectric and Isoionic pH • 1.)Isoionic point – is the pH obtained when the pure, neutral polyprotic acid HA is dissolved in water • Neutral zwitterion • Only ions are H2A+, A-, H+ and OH- • Concentrations are not equal to each other pH obtained by simply dissolving alanine Isoionic point: Remember: Net Charge of Solution is Always Zero!

34. Polyprotic Acid-Base Equilibria • Isoelectric and Isoionic pH • 2.)Isoelectric point – is the pH at which the average charge of the polyprotic acid is 0 • pH at which [H2A+] = [A-] • Always some A- and H2A+ in equilibrium with HA • Most of molecule is in uncharged HA form • To go from isoionic point (all HA) to isoelectric point, add acid to decrease [A-] and increase [H2A+] until equal • pK1 < pK2 isoionic point is acidic excess [A-] Remember: Net Charge of Solution is Always Zero!

35. Polyprotic Acid-Base Equilibria • Isoelectric and Isoionic pH • 2.)Isoelectric point – is the pH at which the average charge of the polyprotic acid is 0 • isoelectric point: [A-] = [H2A+] Isoelectric point:

36. Polyprotic Acid-Base Equilibria • Isoelectric and Isoionic pH • 3.)Example: • Determine isoelectric and isoionic pH for 0.10 M alanine. Solution: For isoionic point:

37. Polyprotic Acid-Base Equilibria • Isoelectric and Isoionic pH • 3.)Example: • Determine isoelectric and isoionic pH for 0.10 M alanine. Solution: For isoelectric point: Isoelectric and isoionic points for polyprotic acid are almost the same