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Chapter 9 Polyprotic Acid-Base Eqlilibria

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1 Chapter 9 Polyprotic Acid-Base Eqlilibria

2 9-1 Diprotic Acids and Bases
Amino acids Substituent (Side chain) R is different group for each amino acid Carboxyl group is stronger acid than ammonium group Amino group Carboxyl group Zwitterion At low pH, both ammonium and carboxy group are protonated At high pH, neither group is protonated Stabilized by interaction with solvent. Zwitterions are stabilized in solution by interactions of ammonium ion and carboxylic acid with water.

3 pKa of carboxy and ammonium group vary depending on substituents(R).

4 Multiple Equilibria for Diprotic Acid-Base
General Illustration Ka1 Ka2 H2L+ Kb2 HL Kb1 L- (H2A) (HA-) (A2-) low pH high pH ⇒ Conjugate acids/bases: H2L+/HL (or H2A/HA-) & HL/L- (or HA-/A2-) • Multiple equilibria for polyprotic acid-base - Stepwise dissociation of proton. ⇒ Number of proton=Number of equilibria(=Number of K values) - All equilibria must be satisfied simultaneously in any solution at equilibrium. - Only one concentration of H+ present in a solution.

5 Ex) Illustration with amino acid leucine(HL): diprotic acid
Ka1 Ka2 Kb2 Leucine Kb1 (H2L+) (HL) (L-) Carboxyl group loses H+ Ammonium group loses H+ low pH high pH Equilibrium reactions H2L+ = HL + H+ HL = L- + H+ (9-1) As Acids (9-2) L- + H2O = HL + OH- HL + H2O = H2L+ + OH- (9-3) As Bases (9-4) (9-5), (9-6) Calculation the pHs for diprotic acid - Three components to the process ① Acidic Form, H2L+ ② Basic Form, L- ③ Intermediate Form, HL

6 The Acid Form, H2L+ (H2L+) (HL) (L-)
K1=4.70×10-3 K2=1.80×10-10 Ka1 Ka2 Kb2 Leucine Kb1 (H2L+) (HL) (L-) ☞ H2L+ is a weak acid and HL is a very weak acid (∵K1≫K2) Assume H2L+ behaves as a monoprotic acid Ex. pH of M leucine hydrochloride (H2L+)?: [H+], [H2L+], [HL], [L-] ? H2L HL H+ 0.050-x x x ⇒ x=[H+]=1.32×10-2 M ⇒ pH=1.88 [H2L+]=F – [H+]=0.050 – 1.32×10-2=3.68×10-2 M ☞ What is the concentration of L- in the solution? : [L-] is very small (∵K1≫K2), but non-zero → Calculate from K2 (9-7) ∴ [L-] = 1.80×10-10 M Validates approximation: [H+]≒[HL], [HL]≫[L-]

7 The Basic Form, L- (H2L+) (L-) (HL) L- + H2O = HL + OH-
K1=4.70×10-3 K2=1.80×10-10 Ka1 Ka2 Kb2 Leucine Kb1 (H2L+) (HL) (L-) L- + H2O = HL + OH- HL + H2O = H2L+ + OH- ☞ L- is a weak base and HL is a very weak base (∵Kb1≫Kb2) Assume L- behaves as a monobasic base Ex. pH of M leucine salt(sodium leucinate, L-)?: [H+], [H2L+], [HL], [L-] ? L HL OH- 0.050-x x x ⇒ x=[OH-]=1.64×10-2 M=[HL] ∴[H+]=Kw/[OH-]=6.11×10-12 M ⇒ pH=11.21 [L-]=F – [OH-]=0.050 – 1.64×10-2=4.88×10-2 M ☞ What is the concentration of H2L+ in the solution? : [H2L+] is very small (∵K1≫K2), but non-zero → Calculate from Kb2 (or Ka1) ∴[H2L+] = 2.13×10-12 M Validates approximation: [OH-]≒[HL], [HL]≫[H2L+]

8 The Intermediate Form, HL
K1=4.70×10-3 K2=1.80×10-10 Ka1 Ka2 Kb2 Leucine Kb1 (H2L+) (HL) (L-) As a base As an acid HL is both an acid and base ⇒ More complicated HL = L- + H+ (9-8) HL + H2O = H2L+ + OH- (9-9) - HL is amphiprotic – can both donate and accept a proton - Since Ka > Kb, expect solution to be acidic ⇒ But, can not ignore base equilibrium ⇒ Need to use complicate treatment of equilibrium

9 Ex. pH of 0.050 M leucine (HL)? : [H+], [H2L+], [HL], [L-] ?
K1=4.70×10-3 K2=1.80×10-10 Ka1 Ka2 Kb2 Leucine Kb1 (H2L+) (HL) (L-) Systematic treatment of equilibrium for HL Step 1: Pertinent reactions: H2L HL + H+ HL L- + H+ K1 L- + H2O HL + OH- HL + H2O H2L+ + OH- Kb1 K2 Kb2 Step 2: Charge Balance: Step 3: Mass Balance: Step 4: Equilibrium constant expression (one for each reaction):

10 Step 5: Solve: Substitute Acid Equilibrium Equations into Charge Balance : From CB: Rearrange: , : All Terms are related to [H+] Multiply by [H+]: Factor out [H+]2: Rearrange: Multiply by K1 to right term and take square-root: : [H+],[HL] ?? (9-10)

11 Assume [HL]=F (minimal dissociation of HL) (∵ Ka2 & Kb2 are small)
: [H+],[HL] 미지수 (9-10) Assume [HL]=F (minimal dissociation of HL) (∵ Ka2 & Kb2 are small) : [H+]외 모두 상수 (9-11) Ex) For M leucine(HL) - Calculate a pH: - Calculate [L-] & [H2L+] (from K1 & K2) ⇒Assumption valid [HL]=0.0500M ≫ [H2L+] & [L-] - [L-] ≈ [H2L+]  two equilibriums proceed equally even though K2>Kb2 - Nearly all leucine remained as HL Solution pH [H+] (M) [H2L+] (M) [HL] (M) [L-] (M) Acid form: M H2A 1.88 1.32×10-2 3.68×10-2 1.80×10-10 Intermediate form: M HA- 6.06 8.80×10-7 9.36×10-6 5.00×10-2 1.02×10-5 Basic form: M A2- 11.21 6.08×10-12 2.13×10-12 1.64×10-3 4.84×10-2

12 BOX 9-2 Successive Approximations
: to deal with difficult equations that do not have simple equations. Ex. 1.00×10-3M HM-(intermediate form of malic acid) ⇒[H+],[H2M],[M2-],[M2-]? H2M HM- M2- ☞ [HM-]≠FHM- (∵ K1 and K2 are nearly equal and F is small.) 1st approximation, assume [HM-]≒1.00×10-3 M 2nd approximation, revise estimate of [HM-] from mass balance 3rd approximation, revise estimate of [HM-] from mass balance [H+] [HM-] 1st app. 4.53×10-5 1.00×10-3 3rd app. 4.37×10-5 7.86×10-4

13 Monohydrogen phthalate HP– (Potassium hydrogen phthalate= K+HP–)
Simplified Calculation for the Intermediate Form (HL) (9-11) Assume K2F ≫ Kw: K1 H2L HL + H+ HL L- + H+ Assume K1≪ F: K2 Kb1 L- + H2O HL + OH- HL + H2O H2L+ + OH- Kb2 Cancel F: Take the -log: ☞pH of intermediate form(HL) of a diprotic acid is close to midway between pK1 and pK2 (9-12) ⇒ Independent for concentration of HL Example pH of the Intermediate Form of Diprotic Acid ⇒ Calculate the pH of 1)0.10M KHP and 2)0.010M KHP. Phthalic acid H2P Monohydrogen phthalate HP– (Potassium hydrogen phthalate= K+HP–) Phthalate P2– Solution 1) 0.10M KHP 2) 0.010M KHP

14 Summary of Diprotic Acid Calculations
Ka1 Ka2 H2A HA A2- Kb2 Kb1 (H2L+) (HL) (L-) • Solution of H2A : Treat as a monoprotic weak acid (∵Ka1≫Ka2) Ka1 H2A HA- + H+ ∴x=[H+] F-x x x ☞ [H+], [H2A], [HA-] from Ka1 ; [A2-] from Ka2 • Solution of A2- : Treat as a monobasic weak base (∵Kb1≫Kb2) Kb1 A2-+ H2O HA- + OH- ∴x=[OH-] ⇒[H+]=Kw/[OH-] F-x x x ☞ [H+](=Kw/[OH-]), [A2-], [HA-] from Kb1 ; [H2A] from Ka1 • Solution of HA- : Systematic treatment Assume K2F≫Kw Assume K1≪F ☞ [HA-]≒F ; [H2A] from Ka1 ; [A2-] from Ka2

15 9-2 Diprotic Buffers ☞ Same approach as monoprotic buffer (HA+A-):
• For the acids of H2A - Write two Henderson-Hasselbalch equations - All equations are always true - Solution only has one pH Ka1 Ka2 H2A HA A2- Kb2 Kb1 Buffer solution 1: (H2A+HA-) Buffer solution 2: (HA-+A2-) - Choice of equation is based on buffering ranges (☞selecting buffer) i) Buffering pH range of pK1± ⇒ use pK1 equation ([H2A] and [HA-] mixture) ii) Buffering pH range of pK2±1 ⇒ use pK2 equation ([HA-] and [A2-] mixture) ([H2A] and [HA-] are given → use pK1 equation) ([HA-] and [A2-] are given → use pK2 equation)

16 Example A Diprotic Buffer System Dissolving 1.00g of KHP(FW , KHP) and 1.20g of disodium phthalate(FW , Na2P) in 50.0 mL water ⇒ pH= ? (pK1=2.950, pK2=5.408) Solution mole Na2P = 1.20/ = 5.71×10-3 mol mole KHP = 1.00/ = 4.90×10-3 mol

17 Preparing a Buffer in a Diprotic System
Example Preparing a Buffer in a Diprotic System (H2A) Ka1 Ka2 H2A HA A2- Kb2 Kb1 0.800M KOH x mL ? pH mL 3.38g H2A

18

19 9-3 Polyprotic Acids and Bases
Extend treatment of diprotic acids and bases to polyprotic systems ⇒ Equilibria for triprotic(H3A) system H3A H2A- + H+ Ka1 = K1 Acid H2A- HA2- + H+ Ka2 = K2 HA2- A3- + H+ Ka3 = K3 A3- + H2O HA2- + OH- Kb1 = Kw/Ka3 Base HA2- + H2O H2A- + OH- Kb2 = Kw/Ka2 H2A- + H2O H3A + OH- Kb3 = Kw/Ka1 Ka1 Ka2 Ka3 H3A H2A- HA2- A3- Kb3 Kb2 Kb1 ⇒ Ka1 × Kb3 = Ka2 × Kb2 = Ka3 × Kb1 = Kw

20 Summary of Triprotic Acid Calculations
Ka1 Ka2 Ka3 H3A H2A- HA2- A3- Kb3 Kb2 Kb1 ☞ Calculation of concentrations for [H+], [H3A], [H2A-], [HA2-], [A3-] 1. H3A is treated as a monoprotic acid, Ka = K1 2. H2A- is treated similarly as an intermediate form of a diprotic acid - Use K1 & K2 instead of K2 & K3 3. HA2- is also treated similarly as an intermediate form of a diprotic acid - Use K2 & K3, instead of K1 & K2 4. A3- is treated as monobasic, with Kb=Kb1=Kw/Ka3 (9-13) or pH=½(pK1+pK2) (9-14) or pH=½(pK2+pK3) For more complex system, just have additional intermediate forms between the two monoprotic acid and base forms at “ends”

21 9-4 Which is the Principal Species?
Principal species depends on the pH and the pKa values of monoprotic acid(HA). Similar for polyprotic(H2A, H3A…), but several pKa values. 1) Monoprotic acid HA (HA = H+ + A-) More acidic pH More ® basic Predominant form HA A- pKa Ex) Major species of Benzoic acid at different pHs. pH<pKa ⇒ [A-]<[HA] (Ex. pH=3.20 ⇒ [A-]/[HA]=0.1 ) pH=pKa ⇒ [A-]=[HA] (Ex. pH=4.20=pKa ⇒ [A-]/[HA]=1) pH>pKa ⇒ [A-]>[HA] (Ex. pH=5.20 ⇒ [A-]/[HA]=10 ) Example Principal Species-Which One and How Much? Predominant form of NH4+ (pKa=9.24) at pH 7.0? Approximate fraction? Solution NH4+ = NH3 + H+ ∴ Predominant form = NH4+ (☞~99% NH4+, ~1% NH3)

22 2) Diprotic acid (H2A) H2A HA- A2- pH H2A HA- A2- pK1 pK2
Kb1 Kb2 Ka2 Ka1 More acidic pH More ® basic pH Major Species pH < pK1 H2A pK1 < pH < pK2 HA- pH > pK2 A2- Predominant form H2A HA- A2- pK1 pK2 3) Triprotic acid (H3A) H3A H2A- HA2- Kb2 Kb3 Ka2 Ka1 A3- Kb1 Ka3 Determine principal species by comparing the pH of the solution with the pKa values

23 9-5 Fractional Composition Equations
• Fraction(α) of each species at a given pH (⇒ pH vs. α) • Useful for acid-base titrations, EDTA titrations, electrochemical equilibria… • Combine Mass Balance and Equilibrium Constant Monoprotic System Mass balance: F=[HA]+[A-] Fraction in the form HA(αHA) : (9-16) Fraction in the form A-(αA-) : Fractional Composition Equation of HA (9-15) (9-17) (9-18) (αHA+αA-=1)

24 Fractional composition (α) diagram of HA
(pKa=5.00) (αHA+αA-=1) More acidic pH More ® basic Predominant form HA A- pKa

25 Diprotic System … … (9-19) (9-20) (9-21) (αH2A+αHA-+αA2-=1)
Mass balance: F=[H2A]+[HA-]+[A2-] : F=Total concentration of H2A Factor out [H2A]: (9-19) Same procedure as αH2A (9-20) (9-21)

26 Fractional composition (α) diagram of H2A
(αH2A+αHA-+αA2-=1) α 일반식, HnA - 분모(D)는 모든 α에 대해 같음: D=[H+]n+K1[H+]n-1+K1K2[H+]n-2…+K1K2…Kn - 분자: α(HnA) = [H+]n α(Hn-1A) = K1[H+]n-1 α(Hn-2A) = K1K2[H+]n-2 More acidic More ® basic Predominant form H2A HA- A2- α(Hn-jA)=K1K2…Kj[H+]n-j pK1 pK2 pH Major Species pH < pK1 H2A pK1 < pH < pK2 HA- pH > pK2 A2-

27 Fractional composition (α) diagram of H3A
(Ex. H3PO4) F = [H3PO4]+[H2PO4 -] ]+[HPO4 2-]+[PO4 3-] αH3PO4 αH2PO4- αHPO42- αPO43-

28 9-6 Isoelectric and Isoionic pH
Isoionic point (or isoionic pH): the pH obtained when the pure, neutral polyprotic acid HA (Neutral zwitterion) is dissolved in water. - The only ions are H2A+, A-, H+ and OH- - [H2A+]≠ [A-] Ex) Alanine Alanine cation (H2A+) Natural zwitterion (HA) Alanine anion (A-) For HA(intermediate form of H2A+) Isoionic point: (9-22) Ex) Isoionic pH for 0.10 M alanine? ∴ pH=6.11 [H2A+]=1.68×10-5 M (from K1) [A-]=1.76×10-5 M (from K2)

29 - pH at which [H2A+] = [A-]
Isoelectric point (or isoelectic pH, pI): the pH at which the average charge of the polyprotic acid is 0 - pH at which [H2A+] = [A-] Alanine cation (H2A+) Natural zwitterion (HA) Alanine anion (A-) Isoelectric point : [H2A+] = [A-] Isoelectric point: (9-23) Ex. Isoelectric pH of 0.10 M alanine (pK1=2.34, pK2=9.87) ? ☞ Isoelectric and isoionic points for polyprotic acid are almost the same


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