1. GenChem Ch 162013/03/03TMHsiung 1/60 Chapter 16 Acids and Bases

2. GenChem Ch 162013/03/03TMHsiung 2/60 Contents in Chapter 16 16-1Arrhenius Theory: A Brief Review 16-2Brønsted–Lowry Theory of Acids and Bases 16-3Self-Ionization of Water and the pH Scale 16-4Strong Acids and Strong Bases 16-5Weak Acids and Weak Bases 16-6Polyprotic Acids 16-7Ions as Acids and Bases 16-8Molecular Structure and Acid–Base Behavior 16-9Lewis Acids and Bases

3. GenChem Ch 162013/03/03TMHsiung 3/60 HCl(g) → H + (aq) + Cl - (aq) NaOH(s) → Na + (aq) + OH - (aq) H2OH2O H2OH2O Na + (aq) + OH - (aq) + H + (aq) + Cl - (aq) → H 2 O(l) + Na + (aq) + Cl - (aq) H + (aq) + OH - (aq) → H 2 O(l) Arrhenius theory did not handle non OH – bases such as ammonia (NH 3 ). Neutralization reaction: Combination of hydrogen ions (H + ) and hydroxide ions (OH – ) to form water. 16-1Arrhenius Theory: A Brief Review Acid: A substance that provides H + in aqueous solution, e.g., HCl Base (alkalis): A substance that provides OH – in aqueous solution, e.g., NaOH

4. GenChem Ch 162013/03/03TMHsiung 4/60 16-2Brønsted–Lowry Theory of Acids and Bases***  Definition: Acid: the substance act as H + donor Base: the substance act as H + acceptor Conjugate base of NH 4 + Conjugate Acid of NH 3 Conjugate base of H 2 O Conjugate acid of OH −

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6. GenChem Ch 162013/03/03TMHsiung 6/60 CH 3 COOH + H 2 O  CH 3 COO – + H 3 O + Conjugate acid Conjugate base NH 3 + H 2 O  NH 4 + + OH – Conjugate base Conjugate acid  Ionization constants*** Acid ionization constant Base ionization constant *H 2 O is an amphiprotic (amphoteric) substance, act as either an acid or a base.

7. GenChem Ch 162013/03/03TMHsiung 7/60  Strengths of conjugate acid-base pairs The stronger an acid, the weaker its conjugate base. (The stronger a base, the weaker its conjugate acid.) An acid-base reaction is favored in the direction from the stronger member to the weaker member of each conjugate acid-base pair.

8. GenChem Ch 162013/03/03TMHsiung 8/60 Brønsted–Lowry acid–base reaction: weak acid CH 3 COOH is only slightly ionized. Reverse reaction proceeds to a greater extent than does the forward reaction. H 3 O + is a stronger acid than CH 3 COOH and CH 3 COO − is a stronger base than H 2 O.

9. GenChem Ch 162013/03/03TMHsiung 9/60 Brønsted–Lowry acid–base reaction: strong acid HCl is essentially completely ionized. The forward reaction proceeds almost to completion. H 3 O + is a weaker acid than HCl and Cl − is a much weaker base than H 2 O.

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11. GenChem Ch 162013/03/03TMHsiung 11/60  More about strengths of acid and bases K a (K b ) values are used to compare the strengths of weak acids (bases). Leveling (solvent) effect: The solvent's ability to level the effect of a strong acid (or strong base) dissolved in it. e.g., HI and HBr are leveled to the same acidic strength in H 2 O. Differentiating (solvent) effect: The solvent's ability to differentiate the acidic (or basic) strength. Example:

12. GenChem Ch 162013/03/03TMHsiung 12/60 16-3Ionization of Water and the pH Scale  Self ionization of water Ion-product of water, K W : K W = [H 3 O + ][OH – ] At 25 o C, K W = 1x10 –14 K W applies to all aqueous solutions—acids, bases, salts, and nonelectrolytes—not just to pure water.

13. GenChem Ch 162013/03/03TMHsiung 13/60  p function: –log pH = –log [H 3 O + ][H 3 O + ] = 10 –pH pOH = –log [OH – ][OH – ] = 10 –pOH Acidic solution:[H 3 O + ] > [OH – ]pH < pOH Basic solution:[H 3 O + ] pOH Aqueous solution at 25 o C: pK W = pH + pOH = 14.00 Aqueous solution at 25 o C: acidic solution:[H 3 O + ] > 1.0×10 –7 pH < 7.00 basic solution:[H 3 O + ] 7.00 neutral solution:[H 3 O + ] = 1.0×10 –7 pH = 7.00

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15. GenChem Ch 162013/03/03TMHsiung 15/60  The pH scale and pH values of some common materials pH = −1 ([H 3 O + ] ≈ 10 M) and pH = 15 ([OH − ] ≈ 10 M) are possible. The pH scale is useful only in the range 2 < pH < 12, because the molarities of H 3 O + and OH − in concentrated acids and bases may differ from their true activities.

16. GenChem Ch 162013/03/03TMHsiung 16/60 16-4Strong Acids and Strong Bases The contribution due to the self-ionization of water can generally be ignored (unless the solution is extremely dilute), i.e., for strong acids and bases, dissociated completely. Therefore, [H 3 O + ]  C HCl C HCl : initial concentration of HCl [OH – ]  C NaOH C NaOH : initial concentration of NaOH.

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18. GenChem Ch 162013/03/03TMHsiung 18/60

19. GenChem Ch 162013/03/03TMHsiung 19/60 For extremely dilute solution of a strong acid and strong base, 1.0 x 10 –8 M HCl for example:

20. GenChem Ch 162013/03/03TMHsiung 20/60 16-5Weak Acids and Weak Bases  Identifying Weak Acids and Bases

21. GenChem Ch 162013/03/03TMHsiung 21/60  Percent Ionization (A weak acid HA for example) HA + H 2 O H 3 O + + A - Degree of ionization = [H 3 O + ] from HA [HA] originally Percent ionization = [H 3 O + ] from HA [HA] originally  100%

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23. GenChem Ch 162013/03/03TMHsiung 23/60  Equilibrium of monoprotic acid, HA HA + H 2 O  H 3 O + + A – C HA – x x x Therefore Assume C HA – x  C HA *C HA – x  C HA, using: 5% rule:

24. GenChem Ch 162013/03/03TMHsiung 24/60  Equilibrium of monobasic base, B Assume C B – x  C B *C B – x  C B, using: 5% rule: B + H 2 O  HB + + OH – C B – x x x Therefore

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28. GenChem Ch 162013/03/03TMHsiung 28/60 16-6Polyprotic Acids

29. GenChem Ch 162013/03/03TMHsiung 29/60  Diprotic acid H 2 A + H 2 O  HA – + H 3 O + HA – + H 2 O  A 2– + H 3 O + For conjugate base: A 2– + H 2 O  HA – + OH – HA – + H 2 O  H 2 A + OH – K a1 x K b2 = K w K a2 x K b1 = K w

30. GenChem Ch 162013/03/03TMHsiung 30/60  Triprotic acid K a1 x K b3 = K w K a2 x K b2 = K w K a3 x K b1 = K w Phosphoric acid for example: H 3 PO 4 + H 2 O  H 2 PO 4 – + H 3 O + H 2 PO 4 – + H 2 O  HPO 4 2– + H 3 O + HPO 4 2– + H 2 O  PO 4 3– + H 3 O + Ionization constants for polyprotic acid progressively decrease: K a1 > K a2 > K a3 > ….. Except in very dilute solutions, essentially all of the H 3 O + ions come from the first ionization step alone.

31. GenChem Ch 162013/03/03TMHsiung 31/60 For a 3.0 M H 3 PO 4, calculate (a) [H 3 O + ]; (b) [H 2 PO 4 – ]; (c) [HPO 4 2– ]; (d) [PO 4 3– ]. Solve (a) assume that all H 3 O + the forms in the first ionization step H 3 PO 4 + H 2 O  H 2 PO 4 – + H 3 O + Initial conc. 3.0 - - Change –x ＋ x ＋ x Equilibrium (3.0 – x) x x x 2 = 0.021 x = [H 3 O + ] = 0.14 M Check: (x/C H 3 PO 4 ) x 100% = 4.7% < 5%, OK!! EXAMPLE 16-9 Calculating Ion Concentrations in a Polyprotic Acid Solution

32. GenChem Ch 162013/03/03TMHsiung 32/60 (b) H 3 PO 4 + H 2 O  H 2 PO 4 – + H 3 O + Equilibrium (3.0 – x) x x [H 2 PO 4 – ]  [H 3 O + ] = 0.14 M [HPO 4 2– ] = 6.3 x 10 –8 M (c) H 2 PO 4 – + H 2 O  HPO 4 2– + H 3 O + Since [H 2 PO 4 – ]  [H 3 O + ], therefore

33. GenChem Ch 162013/03/03TMHsiung 33/60 Since [H 3 O + ] = 0.14 M and [HPO 4 2– ] = 6.3 x 10 –8 M (d) HPO 4 2– + H 2 O  PO 4 3– + H 3 O + [PO 4 3– ] = 1.9x 10 –19 M

34. GenChem Ch 162013/03/03TMHsiung 34/60 (1)H 2 SO 4 + H 2 O  HSO 4 – + H 3 O + K a1  10 3 C H 2 SO 4 C H2SO4 C H2SO4 (2)HSO 4 – + H 2 O  SO 4 2– + H 3 O + C H2SO4 － x x x Concentrated solutions (>0.5 M H 2 SO 4 ): H 3 O + is predominated by first ionization step. e.g., 1.00 M H 2 SO 4, [H 3 O + ]  1.00 M. Very dilute solutions (< 0.001 M H 2 SO 4 ): both ionization steps are nearly completely dissociated, e.g., 0.001 M H 2 SO 4, [H 3 O + ]  0.002 M, [SO 4 2– ]  0.001 M. Intermediate concentrations (0.001 M< C H2SO4 <0.5 M), first ionization step is completely dissociated, the second ionization step is partially dissociated.  A Somewhat Different Case: H 2 SO 4

35. GenChem Ch 162013/03/03TMHsiung 35/60 Check: (x/C H 2 SO 4 ) x 100% = 2.2% < 5%, OK!!

36. GenChem Ch 162013/03/03TMHsiung 36/60 16-7Ions as Acids and Bases  Hydrolysis: The reaction between an ion and water. (1)NaCl (aq)  Na + (aq) + Cl – (aq) Neutral Na + (aq) + H 2 O  Cl – (aq) + H 2 O  (2)NH 4 Cl (aq)  NH 4 + (aq) + Cl – (aq) Acidic NH 4 + (aq) + H 2 O  NH 3(aq) + H 3 O + (aq) Cl – (aq) + H 2 O  (3)CH 3 COONa (aq)  Na + (aq) + CH 3 COO – (aq) Basic Na + (aq) + H 2 O  CH 3 COO – (aq) + H 2 O  CH 3 COOH (aq) + OH – (aq) (4)CH 3 COONH 4(aq)  NH 4 + (aq) + CH 3 COO – (aq) ????? NH 4 + (aq) + H 2 O  NH 3(aq) + H 3 O + (aq) CH 3 COO – (aq) + H 2 O  CH 3 COOH (aq) + OH – (aq) X X X X

37. GenChem Ch 162013/03/03TMHsiung 37/60 Type of SaltExample Ion That Undergo Hydrolysis pH of Solution Cation from strong base/ Anion from strong acid NaCl, KI, KNO 3, RbBr, BaCl 2 None 77 Cation from strong base/ Anion from weak acid CH 3 COONa, KNO 2, NaOCl Anion>7 Cation from weak base/ Anion from strong acid NH 4 Cl, NH 4 NO 3 Cation<7 Cation from weak base/ Anion from weak acid NH 4 NO 2, NH 4 CN, CH 3 COONH 4 Anion and Cation <7 if K b <K a  7 if K b  K a >7 if K b >K a Cation is small, highly charged/ Anion from strong acid AlCl 3, Fe(NO 3 ) 3 Hydrated Cation <7  The pH of Salt Solutions

38. GenChem Ch 162013/03/03TMHsiung 38/60 (a)NaOCl (aq)  Na + (aq) + OCl – (aq) Basic Na + (aq) + H 2 O  OCl – (aq) + H 2 O  HOCl (aq) + OH – (aq) X (b)KCl (aq)  Na + (aq) + Cl – (aq) Neutral K + (aq) + H 2 O  Cl – (aq) + H 2 O  X X (c)NH 4 NO 3(aq)  NH 4 + (aq) + NO 3 – (aq) Acidic NH 4 + (aq) + H 2 O  NH 3(aq) + H 3 O + (aq) NO 3 – (aq) + H 2 O  X

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41. GenChem Ch 162013/03/03TMHsiung 41/60 16-8 Molecular Structure and Acid–Base Behavior  Strengths of Binary Acids Homolytic dissociation vs. Heterolytic dissociation Homolytic dissociation HX  H + XD(H–X) Heterolytic dissociation HX  H + + X – D(H + X – ) Bond dissociation energy for the gas phase ionization reaction D(H + X – ) = D(H–X) + IE(H) + ΔH ea

42. GenChem Ch 162013/03/03TMHsiung 42/60 Bond dissociation energies (kJ/mol ) and Ka values for some binary acids

43. GenChem Ch 162013/03/03TMHsiung 43/60 Comparing binary acids of X in the same row* MoleculeCH 4 NH 3 H 2 OHF ΔEN0.40.91.41.9 Acidity:CH 4 <NH 3 <H 2 O <HF The higher polarity of the bond (the larger ΔEN (electronegativity difference)), the stronger acid. Small ΔEN has more covalent character Large ΔEN has more ionic character

44. GenChem Ch 162013/03/03TMHsiung 44/60 Comparing binary acids of X the same group The larger bond length (the larger X radius, the weaker H—X bond) the stronger acid. MoleculeHFHClHBrHI BE (kJ/mol)565431364297 Anion radius (pm)136181195216 K a 6.6x10 –4 ~10 6 ~10 8 ~10 9 Acidity:HF < HCl <HBr <HI Other example: Acidity:H 2 O <H 2 S <H 2 Se <H 2 Te *** HF (aq) is a weak acid

45. GenChem Ch 162013/03/03TMHsiung 45/60  Strengths of Oxoacids (H–O–EO n ) MoleculeH–OIH–OBrH–OCl EN2.52.83.0 K a 2.3x10 –11 2.5x10 –9 2.9x10 –8 Acid strengthHOI <HOBr <HOCl More examples: Acid strength:H 2 SeO 3 < H 2 SO 3 HBrO 4 < HClO 4 Comparing the EN of E The larger EN (electronegativity) of E, the weaker H– O bond, the stronger acid.

46. GenChem Ch 162013/03/03TMHsiung 46/60 Comparing the n of O n The more number (n) of terminal O, the weaker protonated H–O bond, the stronger acid. *O is the element has second higher electronegativity. More examples: K a1 (H 2 SO 3 ) < K a1 (H 2 SO 4 )

47. GenChem Ch 162013/03/03TMHsiung 47/60  Strengths of R–COOH vs. R–OH Ethoxide ion is a much stronger base than is acetate ion. The stronger the conjugate base, the weaker the corresponding acid.

48. GenChem Ch 162013/03/03TMHsiung 48/60  Strengths of carboxylic acids (R–COOH) Comparing electron-donating ability of R Electron-donating ability: –C 2 H 5 > –CH 3 > –H The higher electron-donating ability, the weaker acid strength. Example Acidity:CH 3 CH 2 –COOH < CH 3 –COOH < H–COOH K a :CH 3 CH 2 –COOH < CH 3 –COOH < H–COOH pK a :CH 3 CH 2 –COOH > CH 3 –COOH > H–COOH

49. GenChem Ch 162013/03/03TMHsiung 49/60 Comparing electron-withdrawing ability of R Electron withdrawing ability: Cl > Br > I The higher electron withdrawing ability, the stronger acid strength Example

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51. GenChem Ch 162013/03/03TMHsiung 51/60  Strengths of amines (–NH 2 ) as bases Effect of electron-withdrawing (halogen, –OH, and – NO 2 ) group Electronegative group withdraws electron density from the N atom, i.e., the lone-pair electrons cannot bind a proton as strongly, and the base is weaker. Example Base strength:weaker

52. GenChem Ch 162013/03/03TMHsiung 52/60 Effect of electron-donating group Electron-donating ability: –C 2 H 5 > –CH 3 > –H The higher electron-donating, the stronger base strength Example Base strength: Stronger

53. GenChem Ch 162013/03/03TMHsiung 53/60 Effect of aromatic group π electrons in the benzene ring of an aromatic molecule are delocalized and can involve the N’s lone-pair electrons in the resonance hybrid. Aromatic amines are much weaker bases than aliphatic amines. Example 1 Example 2 The Weaker base

54. GenChem Ch 162013/03/03TMHsiung 54/60 16-9Lewis Acids and Bases  Glossary and Definition Lewis acid: The species act as electron-pair acceptor. Lewis base: The species act as electron-pair donor. *In organic chemistry: Lewis acids called electrophiles (electron-loving) Lewis bases called nucleophiles (nucleus-loving) Example: Lewis base Lewis acid Complex (adduct)

55. GenChem Ch 162013/03/03TMHsiung 55/60 Example: CaO (s) reacts with SO 2(g) to form CaSO 3(s) Ans: O 2– act as Lewis base, SO 2 act as Lewis acid Ca 2+ O2–O2– +S O O O S O O 2–2–

56. GenChem Ch 162013/03/03TMHsiung 56/60  Summary of acid-base definition AcidBase ArrheniusH + donorOH – donor Bronsted-LowryH + donorH + acceptor Lewise – acceptore – donor

57. GenChem Ch 162013/03/03TMHsiung 57/60  Hydrated metal ion Al 3+ + 6H 2 O  Al(H 2 O) 6 3+

58. GenChem Ch 162013/03/03TMHsiung 58/60  Hydrolysis of hydrated metal ion Lewis base Lewis acid

59. GenChem Ch 162013/03/03TMHsiung 59/60 ionic charge ρ = charge density = ionic volume The higher charge density, the higher acid strength.

60. GenChem Ch 162013/03/03TMHsiung 60/60 End of Chapter 16