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

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

Polyprotic Acid-Base Equilibria 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:

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

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

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] low pH high pH Carboxyl group Loses H+ ammonium group Loses H+

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 K1=4.70x10-3 K2=1.80x10-10 Assume H2L+ behaves as a monoprotic acid

Polyprotic Acid-Base Equilibria Diprotic Acids and Bases 3.) General Process to Determine pH 0.050 M leucine hydrochloride K1=4.70x10-3 + H+ H2L+ HL H+ 0.0500 - x x Determine [H+] from Ka: Determine pH from [H+]: Determine [H2L+]:

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

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

Polyprotic Acid-Base Equilibria Diprotic Acids and Bases 3.) General Process to Determine pH 0.050 M leucine salt (sodium leucinate) L- HL OH- 0.0500 - x x Determine [OH-] from Kb: Determine pH and [H+] from Kw: Determine [L-]:

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

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

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):

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+]

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:

Assume [HL]=F, minimal dissociation: 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)

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

Assume [HL]=F=0.0500M, minimal dissociation (K1 & K2 are small). 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: [HL]=0.0500M >> 9.36x10-6 [H2L+] & 1.02x10-5 [L-] Assumption Valid

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 Solution pH [H+] (M) [H2L+] (M) [HL] (M) [L-] (M) Acid form 0.0500 M H2A 1.88 1.32x10-2 3.68x10-2 1.80x10-10 Intermediate form 0.0500 M HA- 6.06 8.80x10-7 9.36x10-6 5.00x10-2 1.02x10-5 Basic form 0.0500 M HA2- 11.21 6.08x10-12 2.13x10-12 1.64x10-3 4.84x10-2

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:

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:

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

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

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?

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:

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

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” “End Forms” of Equilibria that Bracket Reactions are Treated as Monoprotic Use Kas that “bracket” or contain form, K2 & K3 Use Kas that “bracket” or contain form, K1 & K2

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 pH Major Species pH < pK1 H2A pK1 < pH < pK2 HA- pH > pK2 A2- Determine Principal Species by Comparing the pH of the Solution with the pKa Values

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:

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-:

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-:

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!

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!

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:

Polyprotic Acid-Base Equilibria Isoelectric and Isoionic pH 3.) Example: Determine isoelectric and isoionic pH for 0.10 M alanine (pK1 = 2.34, pK2=9.87). Solution: For isoionic point: For isoelectric point: Isoelectric and isoionic points for polyprotic acid are almost the same