Transcript Slide 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 Polyprotic Acid-Base Equilibria Introduction 1.) Polyprotic systems Amino acids (basic) (acidic) - Carboxyl group is stronger acid of ammonium group R is different group for each amino acid Overall charge is still neutral 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 Polyprotic Acid-Base Equilibria Diprotic Acids and Bases 2.) Multiple Equilibria Illustration with amino acid leucine (HL) high pH low pH Carboxyl group Loses H+ ammonium group Loses H+ Equilibrium reactions Diprotic acid: K a1 K1 K a2 K 2 Polyprotic Acid-Base Equilibria Diprotic Acids and Bases 2.) Multiple Equilibria Equilibrium reactions Diprotic base: K b1 K b2 Relationship between Ka and Kb K a1 K b2 Kw K a2 K b1 Kw 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 high pH low pH Carboxyl group Loses H+ Acid Form [H2L+] Basic Form [L-] Intermediate Form [HL] 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 K1=4.70x10-3 K2=1.80x10-10 H2L+ is a weak acid and HL is a very weak acid K1 K 2 Assume H2L+ behaves as a monoprotic acid K a K1 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 x Determine [H+] from Ka: K a 4.7 10 3 [HL][H ] x2 2 x 1 . 32 x 10 M [ HL ] [ H ] Fx [H2 L ] Determine pH from [H+]: pH log[ H ] log( 1.23 10 2 M ) 1.88 Determine [H2L+]: [ H2 L ] F x 3.68 10 2 M 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 K [HL] [H ][L ] K a2 [L ] a2 [HL] [H ] [L ] ( 1.80 10 -10 )( 1.32 10 - 2 ) ( 1.32 10 - 2 ) 1.80 10 -10 ( K a2 ) Approximation [H+] ≈ [HL], reduces Ka2 equation to [L-]=Ka2 [ L ] 1.80 10 10 1.32 10 2 [ HL ] Validates assumption Polyprotic Acid-Base Equilibria Diprotic Acids and Bases 3.) General Process to Determine pH For most diprotic acids, K1 >> K2 - Even if K1 is just 10x larger than K2 - Assumption that diprotic acid behaves as monoprotic is valid Ka ≈ Ka1 Error in pH is only 4% or 0.01 pH units Basic Form (L-) K b1 Kw / K a2 5.55 10 5 K b2 Kw / K a2 2.13 10 12 L- is a weak base and HL is an extremely weak base K b1 K b2 Assume L- behaves as a monoprotic base K b K b1 Polyprotic Acid-Base Equilibria Diprotic Acids and Bases 3.) General Process to Determine pH 0.050 M leucine salt (sodium leucinate) L- HL 0.0500 - x x OHx Determine [OH-] from Kb: K b 5.55 10 5 [HL][OH - ] [L- ] x2 x 1.64 x10 3 M [ HL ] [ OH ] Fx Determine pH and [H+] from Kw: [H ] Kw 1 10 [ OH ] 14 1.64 10 3 6.10 10 12 M pH 11.21 Determine [L-]: [ L ] F x 4.84 10 2 M 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 [H2 L ][OH ] [H2 L ]x K b2 [H2 L ] [HL] x [ H2 L ] 2.13 10 12 1.64 10 3 [ HL ] 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 K a K a2 1.80 10 10 K b K b2 2.13 10 12 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: K1 K b2 K2 K b1 Step 2: Charge Balance: [H ] [H2 L ] [L- ] [OH - ] Step 3: Mass Balance: F [ HL] [ H 2 L ] [ L- ] Step 4: Equilibrium constant expression (one for each reaction): [HL][H ] [H2 L ][OH - ] K [L- ][H ] K1 b2 K2 K b1 [HL] [HL] [H2 L ] [HL][OH - ] [L ] 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: - - [H2 L ] [L ] [H ] [OH ] 0 All Terms are related to [H+] [HL][H ] [L- ] [HL]K 2 [OH - ] Kw [H2 L ] [H ] [H ] K1 Kw [HL][H ] [HL]K 2 [ H ] 0 K1 [H ] [H ] Multiply by [H+] [HL][H ]2 [HL]K 2 [H ]2 Kw 0 K1 Polyprotic Acid-Base Equilibria Diprotic Acids and Bases 3.) General Process to Determine pH Intermediate Form (HL) Step 6: Solve: [HL][H ]2 [HL]K 2 [H ]2 Kw 0 K1 Factor out [H+]2: 2 [HL] [H ] 1 K 2 [HL] Kw K1 Rearrange: [H ] 2 K 2 [HL] Kw [HL] 1 K1 Polyprotic Acid-Base Equilibria Diprotic Acids and Bases 3.) General Process to Determine pH Intermediate Form (HL) Step 6: Solve: [H ] 2 K 2 [HL] Kw [HL] 1 K1 Multiply by K1 and take square-root: [H ] K 2 K1 [HL] K1Kw K1 [HL] Assume [HL]=F, minimal dissociation: (K1 & K2 are small) [H ] K 2 K1F K1Kw K1 F Polyprotic Acid-Base Equilibria Diprotic Acids and Bases 3.) General Process to Determine pH Intermediate Form (HL) Step 6: Solve: [H ] K 2 K1F K1Kw K1 F Calculate a pH: [H ] ( 4.70 x10 3 )( 1.80 x10 10 )( 0.0500 ) ( 4.70 x10 3 )( 1.0 x10 14 ) 4.70 x10 3 0.0500 8.80 x10 7 M pH 6.06 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][H ] ( 0.0500 )( 8.80 x10 7 ) 6 [H2 L ] 9 . 36 x 10 K1 4.70 x10 3 - [L ] [HL]K 2 [H ] ( 0.0500 )( 1.80 x10 10 ) 8.80 x10 7 [HL]=0.0500M >> 9.36x10-6 [H2L+] & 1.02x10-5 [L-] 1.02 x10 5 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 Solution pH [H+] (M) [H2L+] (M) [HL] (M) [L-] (M) Acid form 0.0500 M H2A 1.88 1.32x10-2 3.68x10-2 1.32x10-2 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 Range of pHs and concentrations for three different forms Polyprotic Acid-Base Equilibria Diprotic Acids and Bases 3.) General Process to Determine pH Simplified Calculation for the Intermediate Form (HL) [H ] K 2 K1F K1Kw K1 F Assume K2F >> Kw: [H ] K 2 K1F K1 F Assume K1<< F: [H ] K 2 K1F F Polyprotic Acid-Base Equilibria Diprotic Acids and Bases 3.) General Process to Determine pH Simplified Calculation for the Intermediate Form (HL) [H ] K 2 K1F F Cancel F: [H ] K2 K1 Take the -log: - log[H ] 2 ( log K1 log K2 ) 1 Independent of concentration: pH 2 ( pK1 pK2 ) 1 pH of intermediate form of a diprotic acid is close to midway between pK1 and pK2 Polyprotic Acid-Base Equilibria Diprotic Buffers 1.) Same Approach as Monoprotic Buffer Write two Henderson-Hasselbalch equations [HA ] pH pK1 log [H2 A] [ A2 ] pH pK 2 log [HA ] 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 pK a1 6.351 FM = 105.99 pK a 2 10.329 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? pK a1 2.972 pK a 2 13.7 Polyprotic Acid-Base Equilibria Polyprotic Acids and Bases 1.) Extend Treatment of Diprotic Acids and Bases to Polyprotic Systems Equilibria for triprotic system Acid equilibria: Base equilibria: For a polyprotic system, would simply contain n such equilibria Polyprotic Acid-Base Equilibria Polyprotic Acids and Bases 1.) Extend Treatment of Diprotic Acids and Bases to Polyprotic Systems Rules for triprotic system 1. 2. H3A is treated as a monoprotic acid, Ka = K1 H2A- is treated similarly as an intermediate form of a diprotic acid [H ] 3. HA2- is also treated similarly as an intermediate form of a diprotic acid a. b. Surrounded by H2A- and A3Use K2 & K3, instead of K1 & K2 [H ] 4. K 2 K1F K1Kw K1 F K 2 K 3 F K 2 Kw K2 F 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 Treat asacid: Intermediate Forms Treat Treatas as Intermediate Monoprotic Forms base: For more complex system, just have additional intermediate forms K s that “bracket” or contain form, “End Forms” Use of Equilibria that Bracket Reactions are KTreated 1 & K3 2 as Monoprotic 2 in-between the atwo monoprotic acid and base forms at “ends” 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 AFor 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 - [H ][ A ] Ka [HA] - F [HA ] [ A ] [ A ] F [HA ] [H ](F-[HA]) Ka [HA] Rearrange: [HA] [H ]F [H ] K a 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: [HA] HA [H ]F [H ] K a Divide by F: [HA] [H ] Fraction in the form HA: HA F [H ] K a Fraction in the form A-: K [ A ] A a F [H ] K a [HA] [HA] F [HA] [ 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: [H2 A] [H ]2 H2 A 2 F [H ] [H ]K1 K1K 2 Fraction in the form HA-: K1 [H ] [HA ] HA 2 F [H ] [H ]K1 K1K 2 Fraction in the form A2-: A2 K1K 2 [ A2 ] 2 F [H ] [H ]K1 K1K 2 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 OHConcentrations are not equal to each other Isoionic point: [H ] K1K 2 F K1Kw K1 F pH obtained by simply dissolving alanine 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+] [HA][H ] [H2 A ] K1 [ A- ] K 2 [HA] [H ] [HA][H ] K 2 [HA] [ H ] K1K 2 K1 [H ] Isoelectric point: pH 2 ( pK1 pK2 ) 1 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: [H ] K1K 2 F K1Kw ( 10 2.34 )( 10 9.87 )( 0.10 ) ( 10 2.34 )( 1 10 14 ) K1 F 10 2.34 0.10 7.7 10 7 M pH 6.11 For isoelectric point: pH 2 ( pK1 pK2 ) 2 ( 2.34 9.87 ) 6.10 1 1 Isoelectric and isoionic points for polyprotic acid are almost the same