Chapter 10 Polyprotic AcidBase Equilibria Overview 10 1
Chapter 10 Polyprotic Acid-Base Equilibria
Overview 10 -1 Diprotic Acids and Bases 10 -2 Diprotic Buffers 10 -3 Polyprotic Acids and Bases 10 -4 Which Is the Principal Species? 10 -5 Fractional Composition Equations 10 -6 Isoelectric and Isoionic p. H
10 -1: Diprotic Acids and Bases • Polyprotic acids and bases are those that can donate or accept more than one proton. • Diprotic acids therefore can either donate or accept two protons. Types of polyprotic acids 1. H 2 B+, Amino acids, Example: Alanine – H 2 A+ Neutrally charged acids 2. H 2 A Example: H 2 SO 4, H 2 CO 3, H 2 C 2 O 4 3. H 3 A Example: H 3 PO 4
10 -1: Diprotic Acids and Bases • A common class of diprotic acids are amino acids, which are the building blocks of proteins. • They have an acidic carboxylic acid group, a basic amino group, and a variable substituent designated R. • The carboxyl group is a stronger acid than the ammonium group, so the nonionized form rearranges spontaneously to the zwitterion, which has both positive and negative sites.
Amino Acid Carboxyl group Amine (weak base) Substituent
Amino Acid p. Ka ~ 2. 5 Ammonium group p. Ka ~ 9. 5
10 -1: Diprotic Acids and Bases
10 -1: Diprotic Acids and Bases amphiprotic acid base
Diprotic Acid: Consider the Leucine System • HL is a Zwitterion. = HL
10 -2: Diprotic Buffers Calculate the p. H of the following 0. 050 M aqueous solutions: + H L 2 1. Leucine hydrochloride L 2. Sodium leucinate HL 3. Leucine p. Ka’s = 2. 328 and 9. 744 (1) Estimate answer (2) Calculate it!
10 -2: Diprotic Buffers •
10 -2: Diprotic Buffers •
10 -2: Diprotic Buffers • = L
10 -2: Diprotic Buffers •
10 -2: Diprotic Buffers •
10 -2: Diprotic Buffers A molecule that can both donate and accept a proton is said to be amphiprotic. HL + H 2 O ⇄ H 2 L+ + OH- HL + H 2 O ⇄ L- + H 3 O+
10 -2: Diprotic Buffers Using the systematic treatment of equilibrium, an equation for calculating the [H+] of amphiprotics can be derived (see text). p. H ~ ½ (p. Ka 1 + p. Ka 2)
10 -2: Diprotic Buffers What is the p. H of a 0. 05 M leucine solution? p. H =-Log(8. 79 × 10 -7 M) = 6. 06 p. H ~ ½ (p. Ka 1 + p. Ka 2) ~ ½(2. 328 + 9. 744) ~ 6. 04
Calculating p. H H 2 L+ or H 2 A monoprotic acid HL or HA- p. H ~ ½(p. Ka 1 + p. Ka 2) L- or A 2 - base hydrolysis
10 -2: Diprotic Buffers Calculate the p. H of the following 0. 10 M aqueous solutions: 1. Alanine chloride 2. Alanine H 2 A + 3. Sodium alanate p. Ka’s = 2. 34 and 9. 87 (1) Estimate answer (2) Calculate it! HA A-
Calculate the p. H of the following 0. 10 M aqueous solutions: Name Form Strategy Calculated p. H 1. 71 Alanine chloride Alanine H 2 A + Monoprotic acid HA Amphiprotic 6. 11 Sodium alanate A- Monoprotic base 11. 44
10 -2: Diprotic Buffers Describe how you would calculate the p. H of the following 0. 10 M aqueous solutions: • sodium monohydrogen phosphate • glycine hydrochloride • citric acid • trisodium citrate • potassium hydrogen phthalate, (KHP)
10 -2: Diprotic Buffers Describe how you would calculate the p. H of the following 0. 10 M aqueous solutions: Solution: Chemical Formula Acid or base form Strategy Na 2 HPO 4 H 2 GCl or HG. HCl H 3 C 6 H 5 O 7 Na 3 C 6 H 5 O 7 KHC 8 H 4 O 4 HPO 4 2 H 2 G H 3 C 6 H 5 O 73 HC 8 H 4 O 4 - amphiprotic, Ka 2, Ka 3 monoprotic acid, Ka 1 monoprotic base, Kb 1 amphiprotic, Ka 1, Ka 2
10 -2: Diprotic Buffers • A buffer made from a diprotic (or polyprotic) acid is treated in the same way as a buffer made from a monoprotic acid. • For the acid H 2 A, we can write two Henderson-Hasselbalch equations, both of which are always true. • If we know [H 2 A] and [HA-], then use the p. K 1 equation. If we know [HA-] and [A 2 -], use the p. K 2 equation. H 2 A ⇄ HA- + H+ p. K 1 HA- ⇄ A 2 - + H+ p. K 2
10 -2: Diprotic Buffers Example: A Diprotic Buffer System • Find the p. H of a solution prepared by dissolving 1. 00 g of potassium hydrogen phthalate and 1. 20 g of disodium phthalate in 50. 0 m. L of water.
10 -3: Polyprotic Acids and Bases Example: A Triprotic System • Find the p. H of 0. 10 M H 3 His 2+, 0. 10 M H 2 His+, 0. 10 M HHis, and 0. 10 M His+, where His stands for the amino acid histidine.
10 -3: Polyprotic Acids and Bases Example: A Triprotic System • Find the p. H of 0. 10 M H 3 His 2+, 0. 10 M H 2 His+, 0. 10 M HHis, and 0. 10 M His+, where His stands for the amino acid histidine.
10 -4: Which Is the Principal Species? What is the principal form of benzoic acid (p. Ka 4. 20) at p. H 8? From the Henderson-Hasselbalch equation: p. H = 4. 20, [HA] = [A-] ⇠acidic p. H basic⇢ p. H < 4. 20, [HA] < [A-] p. H > 4. 20, [HA] < [A-] At p. H 8. 0, base form A- predominates. HA Ap. Ka 4. 20
10 -4: Which Is the Principal Species?
What is the principal form of alanine (p. Ka 2. 34 and 9. 87) at p. H 8? p. H 8 H 2 A+ HA (amphiprotic) 2. 34 A- 9. 87 The principal form of alanine p. H 8 is the amphiprotic form, HA.
10 -4: Which Is the Principal Species?
10 -5: Fractional Composition Equations We can derive equations that give the fraction of each species of acid or base at a given p. H. Monoprotic Systems: Example: What fraction of benzoic acid exists as benzoate at p. H 8. 0? At p. H 8. 0, almost all of the benzoic acid exists in the basic form!
10 -5: Fractional Composition Equations • Fractional composition diagram of a monoprotic system with p. Ka = 5. 00. Below p. H 5, HA is the dominant form, whereas, above p. H 5, A- dominates
10 -5: Fractional Composition Equations Diprotic Systems:
10 -5: Fractional Composition Equations
10 -6: Isoelectric and Isoionic p. H
10 -6: Isoelectric and Isoionic p. H • The isoionic point (or isoionic p. H): – p. H obtained when the pure, neutral polyprotic acid HA (the neutral zwitterion) is dissolved in water. – The only ions are H 2 A+, A-, H+, and OH-. – Most alanine is in the form HA, and the concentrations of H 2 A+ and A - are not equal to each other. – For neutral alanine, HA, dissolved in water, the p. H would be somewhere between 2. 234 and 9. 87. The [A-] would be slightly larger than the [H 2 A+]. This is the isoionic p. H.
10 -6: Isoelectric and Isoionic p. H • The isoelectric point (or isoelectric p. H): – p. H at which the average charge of the polyprotic acid is 0. – Most of the molecules are in the uncharged form HA, and the concentrations of H 2 A+ and A- are equal to each other. [H 2 A+ ] = [A-] • If a pure sample of neutral alanine (p. K 1 2. 34, p. K 2 9. 87) is dissolved in water, the [A-] would be slightly larger than the [H 2 A+ ]. • By adding a small amount of acid, some A- would be converted to H 2 A+ until the concentrations are equal. This is the isoelectric p. H = ½ (p. Ka 1 + p. Ka 2) • Can be used to separate proteins from one another.
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