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Polyprotic Acid-Base Equilibria

Polyprotic Acid-Base Equilibria. Introduction 1.) Polyprotic systems Acid or bases that can donate or accept more than one proton Proteins are a common example of a polyprotic system why activity of proteins are pH dependent Polymer of amino acids

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Polyprotic Acid-Base Equilibria

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  1. Polyprotic Acid-Base Equilibria • Introduction • 1.)Polyprotic systems • Acid or bases that can donate or accept more than one proton • Proteins are a common example of a polyprotic system • why activity of proteins are pH dependent • Polymer of amino acids • Some amino acids have acidic or basic substituents

  2. Polyprotic Acid-Base Equilibria • Introduction • 1.)Polyprotic systems • Amino acids • Carboxyl group is stronger acid of ammonium group • R is different group for each amino acid • Amino acids are zwitterion – molecule with both positive and negative charge • At low pH, both ammonium and carboxy group are protonated • At high pH, neither group is protonated • Stabilized by interaction with solvent (basic) (acidic) Overall charge is still neutral

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

  4. Polyprotic Acid-Base Equilibria • Diprotic Acids and Bases • 2.)Multiple Equilibriums • Equilibrium reactions Diprotic base: Relationship between Ka and Kb

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

  6. 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+

  7. 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

  8. 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+]:

  9. 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

  10. 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

  11. 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-]:

  12. 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

  13. 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

  14. 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):

  15. 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+]

  16. 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:

  17. 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)

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

  19. 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-]

  20. 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

  21. 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:

  22. 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:

  23. Polyprotic Acid-Base Equilibria • Diprotic Buffers • 1.)Same Approach as Monoprotic Buffer • Write two Henderson-Hasselbalch equations • Both Equations are always true • Solution only has one pH • Choice of equation is based on what is known • [H2A] and [HA-] known use pK1 equation • [HA-] and [A2-] known use pK2 equation

  24. Polyprotic Acid-Base Equilibria • Diprotic Buffers • 1.)Example 1: • How many grams of Na2CO3 (FM 105.99) should be mixed with 5.00 g of NaHCO3 (FM 84.01) to produce 100 mL of buffer with pH 10.00? FM = 62.03 FM = 84.01 FM = 105.99

  25. Polyprotic Acid-Base Equilibria • Diprotic Buffers • 1.)Example 2: • How many milliliters of 0.202 M NaOH should be added to 25.0 mL of 0.0233 M of salicylic acid (2-hydroxybenzoic acid) to adjust the pH to 3.50?

  26. Polyprotic Acid-Base Equilibria • Polyprotic Acids and Bases • 1.)Extend Treatment of Diprotic Acids and Bases to Polyprotic Systems • Equilibria for triprotic system • For a polyprotic system, would simply contain n such equilibria Acid equilibria: Base equilibria:

  27. Polyprotic Acid-Base Equilibria • Polyprotic Acids and Bases • 1.)Extend Treatment of Diprotic Acids and Bases to Polyprotic Systems • Rules for triprotic system • H3A is treated as a monoprotic acid, Ka = K1 • H2A- is treated similarly as an intermediate form of a diprotic acid • HA2- is also treated similarly as an intermediate form of a diprotic acid • Surrounded by H2A- and A3- • Use K2 & K3, instead of K1 & K2 • A3- is treated as monobasic, with Kb=Kb1=Kw/Ka3

  28. Polyprotic Acid-Base Equilibria • Polyprotic Acids and Bases • 1.)Extend Treatment of Diprotic Acids and Bases to Polyprotic Systems • Rules for triprotic system Treat as Monoprotic acid: Treat as Intermediate Forms Treat as Intermediate Forms Treat as Monoprotic base: For more complex system, just have additional intermediate forms in-between the two monoprotic acid and base forms at “ends” Use Kas that “bracket” or contain form, K1 & K2 Use Kas that “bracket” or contain form, K2 & K3 “End Forms” of Equilibria that Bracket Reactions are Treated as Monoprotic

  29. Polyprotic Acid-Base Equilibria • Polyprotic Acids and Bases • 2.)Example • How many milliters of 1.00 M KOH should be added to 100 mL of solution containing 10.0 g of histidine hydrochloride (His.HCl FM 191.62) to get a pH of 9.30?.

  30. Polyprotic Acid-Base Equilibria • Polyprotic Acids and Bases • 3.)Which is the Principal Species? • Depends on the pH of the Sample and the pKa values • At pKa, 1:1 mixture of HA and A- • For monoprotic, A- is predominant when pH > pKa • For monoprotic, HA is predominant when pH < pKa • Similar for polyprotic, but several pKa values Triprotic acid Diprotic acid Determine Principal Species by Comparing the pH of the Solution with the pKa Values

  31. Polyprotic Acid-Base Equilibria • Polyprotic Acids and Bases • 4.)Fractional Composition Equations • Fraction of Each Species at a Given pH • Useful for: • Acid-base titrations • EDTA titrations • Electrochemical equilibria • Combine Mass Balance and Equilibrium Constant Rearrange:

  32. Polyprotic Acid-Base Equilibria • Polyprotic Acids and Bases • 4.)Fractional Composition Equations • Combine Mass Balance and Equilibrium Constant Recall: fraction of molecule in the form HA is: Divide by F: Fraction in the form HA: Fraction in the form A-:

  33. 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-:

  34. 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!

  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 • 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!

  36. 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:

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

  38. 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

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