4.  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 [H 3 PO 4 ]=[H 2 PO 4 - ]. The equilibrium equation relating these two species is If we take the -log10, or "p", of this equation Since [H 3 PO 4 ]=[H 2 PO 4 - ], and log10(1) = 0, pH=pK a1 pKa 1 pKa 2 pKa 3

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 [H 2 PO 4 - ] is a maximum lies half-way between between pK a1 and pK a2. Since H 2 PO 4 - is the major species present in solution, the major equilibrium is the disproportionation reaction. This equilibrium cannot be used to solve for pH because [H 3 O + ] doesn't occur in the equilibrium equation. We solve the pH problem adding the first two equilibria equations +

7. Since the disproportionation reaction predicts [H 3 PO 4 ]=[HPO 4 2- ] Note that when we add chemical equilibria, we take the product of the equilibrium equations. Taking the -log10 of the last equation

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

10. Let’s look at the simplest of the amino acids, glycine H 2 Gly + glycinium HGly Gly - glycinate K1K1 K2K2 In water the charge balance would be, Combining the autoprotolysis of water and the K 1 and K 2 expressions into the charge balance yields:

11. HGly Gly - H 2 Gly +

12. Diprotic Acids and Bases 2.)Multiple Equilibria  Illustration with amino acid leucine (HL)  Equilibrium reactions low pH high 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 K a and K b

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 [H 2 L + ]  Basic Form [L - ]  Intermediate Form [HL] low pH high 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 (H 2 L + )  Illustration with amino acid leucine  H 2 L + is a weak acid and HL is a very weak acid K 1 =4.70x10 -3 K 2 =1.80x10 -10 Assume H 2 L + 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 + H + H2L+H2L+ HLH+H+ 0.0500 - xxx K 1 =4.70x10 -3 Determine [H + ] from K a : Determine pH from [H + ]: Determine [H 2 L + ]:

18. Polyprotic Acid-Base Equilibria Diprotic Acids and Bases 3.)General Process to Determine pH  Acid Form (H 2 L + ) What is the concentration of L - in the solution? [L - ] is very small, but non-zero. Calculate from K a2 Approximation [H + ] ≈ [HL], reduces K a2 equation to [L - ]=K a2 Validates assumption

19. Polyprotic Acid-Base Equilibria Diprotic Acids and Bases 3.)General Process to Determine pH  For most diprotic acids, K 1 >> K 2 - Assumption that diprotic acid behaves as monoprotic is valid - K a ≈ K a1  Even if K 1 is just 10x larger than K 2 - 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) L - HLOH - 0.0500 - xx x Determine [OH - ] from K b : Determine pH and [H + ] from K w : 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 H 2 L + in the solution? [H 2 L + ] is very small, but non-zero. Calculate from K b2 Validates assumption [OH - ] ≈ [HL], Fully basic form of a diprotic acid can be treated as a monobasic, K b =K b1

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 K a > K b, expect solution to be acidic - Can not ignore base equilibrium  Need to use Systematic Treatment of Equilibrium

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

24. Step 6: Solve: Polyprotic Acid-Base Equilibria Diprotic Acids and Bases 3.)General Process to Determine pH  Intermediate Form (HL) 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 K 1 and take square-root: Assume [HL]=F, minimal dissociation: (K 1 & K 2 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 (K 1 & K 2 are small). Calculate [L - ] & [H 2 L + ] from K 1 & K 2 : [HL]=0.0500M >> 9.36x10 -6 [H 2 L + ] & 1.02x10 -5 [L - ] Assumption Valid

29. Polyprotic Acid-Base Equilibria Diprotic Acids and Bases 3.)General Process to Determine pH  Intermediate Form (HL) Summary of results:  [L - ] ≈ [H 2 L + ]  two equilibriums proceed equally even though K a >K b  Nearly all leucine remained as HL  Range of pHs and concentrations for three different forms SolutionpH[H + ] (M)[H 2 L + ] (M)[HL] (M)[L - ] (M) Acid form0.0500 M H 2 A1.881.32x10 -2 3.68x10 -2 1.32x10 -2 1.80x10 -10 Intermediate form0.0500 M HA - 6.068.80x10 -7 9.36x10 -6 5.00x10 -2 1.02x10 -5 Basic form0.0500 M HA 2- 11.216.08x10 -12 2.13x10 -12 1.64x10 -3 4.84x10 -2

30. Polyprotic Acid-Base Equilibria Diprotic Acids and Bases 3.)General Process to Determine pH  Simplified Calculation for the Intermediate Form (HL) Assume K 2 F >> K w : Assume K 1 << 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 pK 1 and pK 2 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 H 2 A: Fraction in the form HA - : Fraction in the form A 2- :

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 H 2 A +, A -, H + and OH - - Concentrations are not equal to each other Isoionic point: pH obtained by simply dissolving alanine 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 [H 2 A + ] = [A - ] - Always some A- and H 2 A+ 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 [H 2 A + ] until equal - pK 1 < pK 2  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 - ] = [H 2 A + ] 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