1. Prentice-Hall © 2007 General Chemistry: Chapter 16 Slide 1 of 52 Philip Dutton University of Windsor, Canada Prentice-Hall © 2007 CHEMISTRY Ninth Edition GENERAL Principles and Modern Applications Petrucci Harwood Herring Madura Chapter 16: Acids and Bases

2. Prentice-Hall © 2007 General Chemistry: Chapter 16 Slide 2 of 52 Contents 16-1The Arrhenius Theory: A Brief Review 16-2Brønsted-Lowry Theory of Acids and Bases 16-3The Self-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 Behaviour

3. Prentice-Hall © 2007 General Chemistry: Chapter 16 Slide 3 of 52 16-1 The Arrhenius Theory: A Brief Review 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 very well.

4. Prentice-Hall © 2007 General Chemistry: Chapter 16 Slide 4 of 52 16-2 Brønsted-Lowry Theory of Acids and Bases  An acid is a proton donor.  A base is a proton acceptor. NH 3 + H 2 O NH 4 + + OH - NH 4 + + OH - NH 3 + H 2 O baseacid baseacid conjugate acid conjugate base ??

5. Prentice-Hall © 2007 General Chemistry: Chapter 16 Slide 5 of 52 The Solvated Proton

6. Prentice-Hall © 2007 General Chemistry: Chapter 16 Slide 6 of 52 Base Ionization Constant NH 3 + H 2 O NH 4 + + OH - Kc=Kc= [NH 3 ][H 2 O] [NH 4 + ][OH - ] K b = K c [H 2 O] = [NH 3 ] [NH 4 + ][OH - ] = 1.8  10 -5 baseacid conjugate acid conjugate base

7. Prentice-Hall © 2007 General Chemistry: Chapter 16 Slide 7 of 52 Acid Ionization Constant CH 3 CO 2 H + H 2 O CH 3 CO 2 - + H 3 O + Kc=Kc= [CH 3 CO 2 H][H 2 O] [CH 3 CO 2 - ][H 3 O + ] K a = K c [H 2 O] = = 1.8  10 -5 [CH 3 CO 2 H] [CH 3 CO 2 - ][H 3 O + ] baseacid conjugate acid conjugate base

8. Prentice-Hall © 2007 General Chemistry: Chapter 16 Slide 8 of 52

9. Prentice-Hall © 2007 General Chemistry: Chapter 16 Slide 9 of 52 A Weak Base

10. Prentice-Hall © 2007 General Chemistry: Chapter 16 Slide 10 of 52 A Weak Acid

11. Prentice-Hall © 2007 General Chemistry: Chapter 16 Slide 11 of 52 A Strong Acid

12. Prentice-Hall © 2007 General Chemistry: Chapter 16 Slide 12 of 52 16-3 The Self-Ionization of Water and the pH Scale

13. Prentice-Hall © 2007 General Chemistry: Chapter 16 Slide 13 of 52 Ion Product of Water Kc=Kc= [H 2 O][H 2 O] [H 3 O + ][OH - ] H 2 O + H 2 O H 3 O + + OH - baseacid conjugate acid conjugate base K W = K c [H 2 O][H 2 O] = = 1.0  10 -14 [H 3 O + ][OH - ]

14. Prentice-Hall © 2007 General Chemistry: Chapter 16 Slide 14 of 52 pH and pOH  The potential of the hydrogen ion was defined in 1909 as the negative of the logarithm of [H + ]. pH = -log[H 3 O + ]pOH = -log[OH - ] -logK W = -log[H 3 O + ]-log[OH - ]= -log(1.0  10 -14 ) K W = [H 3 O + ][OH - ]= 1.0  10 -14 pK W = pH + pOH= -(-14) pK W = pH + pOH = 14

15. Prentice-Hall © 2007 General Chemistry: Chapter 16 Slide 15 of 52 pH and pOH Scales

16. Prentice-Hall © 2007 General Chemistry: Chapter 16 Slide 16 of 52 16-4 Strong Acids and Bases HCl CH 3 CO 2 H Thymol Blue Indicator pH < 1.2 < pH < 2.8 < pH

17. Prentice-Hall © 2007 General Chemistry: Chapter 16 Slide 17 of 52 16-5 Weak Acids and Bases Lactic AcidGlycine General Chemistry: Chapter 16Prentice-Hall © 2007 Acetic Acid

18. Prentice-Hall © 2007 General Chemistry: Chapter 16 Slide 18 of 52 Acetic Acid Weak Acids Ka=Ka= = 1.8  10 -5 [CH 3 CO 2 H] [CH 3 CO 2 - ][H 3 O + ] pK a = -log(1.8  10 -5 ) = 4.74 General Chemistry: Chapter 16Prentice-Hall © 2007

19. Prentice-Hall © 2007 General Chemistry: Chapter 16 Slide 19 of 52 Weak Bases Kb=Kb= = 4.3  10 -4 [CH 3 NH 2 ] [CH 3 NH 3 + ][HO - ] pK b = -log(4.2  10 -4 ) = 3.37

20. Prentice-Hall © 2007 General Chemistry: Chapter 16 Slide 20 of 52 Table 16.3 Ionization Constants of Weak Acids and Bases

21. Prentice-Hall © 2007 General Chemistry: Chapter 16 Slide 21 of 52 Determining a Value of K A from the pH of a Solution of a Weak Acid. Butyric acid, HC 4 H 7 O 2 (or CH 3 CH 2 CH 2 CO 2 H) is used to make compounds employed in artificial flavorings and syrups. A 0.250 M aqueous solution of HC 4 H 7 O 2 is found to have a pH of 2.72. Determine K A for butyric acid. HC 4 H 7 O 2 + H 2 O C 4 H 7 O 2 + H 3 O + K a = ? EXAMPLE 16-5

22. Prentice-Hall © 2007 General Chemistry: Chapter 16 Slide 22 of 52 HC 4 H 7 O 2 + H 2 O C 4 H 7 O 2 + H 3 O + Initial conc.0.250 M00 Changes-x M+x M+x M Equilibrium (0.250-x) M x Mx M Concentration EXAMPLE 16-5 Solution: For HC 4 H 7 O 2 K A is likely to be much larger than K W. Therefore assume self-ionization of water is unimportant.

23. Prentice-Hall © 2007 General Chemistry: Chapter 16 Slide 23 of 52 Log[H 3 O + ] = -pH = -2.72 HC 4 H 7 O 2 + H 2 O C 4 H 7 O 2 + H 3 O + [H 3 O + ] = 10 -2.72 = 1.9  10 -3 = x [H 3 O + ] [C 4 H 7 O 2 - ] [HC 4 H 7 O 2 ] Ka=Ka= 1.9  10 -3 · 1.9  10 -3 (0.250 – 1.9  10 -3 ) = K a = 1.5  10 -5 Check assumption: K a >> K W. EXAMPLE 16-5

24. Prentice-Hall © 2007 General Chemistry: Chapter 16 Slide 24 of 52 Percent Ionization 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%

25. Prentice-Hall © 2007 General Chemistry: Chapter 16 Slide 25 of 52 Percent Ionization K a = [H 3 O + ][A - ] [HA] K a = n H3O+H3O+ A-A- n HA n 1 V

26. Prentice-Hall © 2007 General Chemistry: Chapter 16 Slide 26 of 52 16-6 Polyprotic Acids H 3 PO 4 + H 2 O H 3 O + + H 2 PO 4 - H 2 PO 4 - + H 2 O H 3 O + + HPO 4 2- HPO 4 2- + H 2 O H 3 O + + PO 4 3- Phosphoric acid: A triprotic acid. K a = 7.1  10 -3 K a = 6.3  10 -8 K a = 4.2  10 -13

27. Prentice-Hall © 2007 General Chemistry: Chapter 16 Slide 27 of 52 Phosphoric Acid  K a1 >> K a2 ◦All H 3 O + is formed in the first ionization step.  H 2 PO 4 - essentially does not ionize further. ◦Assume [H 2 PO 4 - ] = [H 3 O + ].  [HPO 4 2- ]  K a2 regardless of solution molarity.

28. Prentice-Hall © 2007 General Chemistry: Chapter 16 Slide 28 of 52

29. Prentice-Hall © 2007 General Chemistry: Chapter 16 Slide 29 of 52 Calculating Ion Concentrations in a Polyprotic Acid Solution. For a 3.0 M H 3 PO 4 solution, calculate: (a) [H 3 O + ]; (b) [H 2 PO 4 - ]; (c) [HPO 4 2- ] (d) [PO 4 3- ] H 3 PO 4 + H 2 O H 2 PO 4 - + H 3 O + Initial conc.3.0 M00 Changes-x M+x M+x M Equilibrium(3.0-x) M x Mx M Concentration EXAMPLE 16-9

30. Prentice-Hall © 2007 General Chemistry: Chapter 16 Slide 30 of 52 H 3 PO 4 + H 2 O H 2 PO 4 - + H 3 O + [H 3 O + ] [H 2 PO 4 - ] [H 3 PO 4 ] Ka=Ka= x · x (3.0 – x) = Assume that x << 3.0 = 7.1  10 -3 x 2 = (3.0)(7.1  10 -3 ) x = 0.14 M [H 2 PO 4 - ] = [H 3 O + ] = 0.14 M EXAMPLE 16-9

31. Prentice-Hall © 2007 General Chemistry: Chapter 16 Slide 31 of 52 H 2 PO 4 - + H 2 O HPO 4 2- + H 3 O + [H 3 O + ] [HPO 4 2- ] [H 2 PO 4 - ] Ka=Ka= y · (0.14 + y) (0.14 - y) = = 6.3  10 -8 Initial conc.0.14 M00.14 M Changes-y M+y M+y M Equilibrium (0.14 - y) M y M(0.14 +y) M Concentration y << 0.14 M y = [HPO 4 2- ] = 6.3  10 -8 EXAMPLE 16-9

32. Prentice-Hall © 2007 General Chemistry: Chapter 16 Slide 32 of 52 HPO 4 - + H 2 O PO 4 3- + H 3 O + [H 3 O + ] [HPO 4 2- ] [H 2 PO 4 - ] Ka=Ka= (0.14)[PO 4 3- ] 6.3  10 -8 = = 4.2  10 -13 M [PO 4 3- ] = 1.9  10 -19 M EXAMPLE 16-9

33. Prentice-Hall © 2007 General Chemistry: Chapter 16 Slide 33 of 52 Sulfuric Acid Sulfuric acid: A diprotic acid. H 2 SO 4 + H 2 O H 3 O + + HSO 4 - HSO 4 - + H 2 O H 3 O + + SO 4 2- K a = very large K a = 1.96

34. Prentice-Hall © 2007 General Chemistry: Chapter 16 Slide 34 of 52 General Approach to Solution Equilibrium Calculations  Identify species present in any significant amounts in solution (excluding H 2 O).  Write equations that include these species.  Number of equations = number of unknowns. ◦Equilibrium constant expressions. ◦Material balance equations. ◦Electroneutrality condition.  Solve the system of equations for the unknowns.

35. Prentice-Hall © 2007 General Chemistry: Chapter 16 Slide 35 of 52 16-7 Ions as Acids and Bases NH 4 + + H 2 O NH 3 + H 3 O + base acid CH 3 CO 2 - + H 2 O CH 3 CO 2 H + OH - base acid [NH 3 ] [H 3 O + ] [OH - ] Ka=Ka= [NH 4 + ] [OH - ] [NH 3 ] [H 3 O + ] Ka=Ka= [NH 4 + ] = ? = KWKW KbKb = 1.0  10 -14 1.8  10 -5 = 5.6  10 -10 K a K b = K w

36. Prentice-Hall © 2007 General Chemistry: Chapter 16 Slide 36 of 52 Hydrolysis  Water (hydro) causing cleavage (lysis) of a bond. Na + + H 2 O → Na + + H 2 O NH 4 + + H 2 O → NH 3 + H 3 O + Cl - + H 2 O → Cl - + H 2 O No reaction Hydrolysis

37. Prentice-Hall © 2007 General Chemistry: Chapter 16 Slide 37 of 52 16-8 Molecular Structure and Acid-Base Behavior  Why is HCl a strong acid, but HF is a weak one?  Why is CH 3 CO 2 H a stronger acid than CH 3 CH 2 OH?  There is a relationship between molecular structure and acid strength.  Bond dissociation energies are measured in the gas phase and not in solution.

38. Prentice-Hall © 2007 General Chemistry: Chapter 16 Slide 38 of 52 Strengths of Binary Acids HIHBrHClHF 160.9>141.4>127.4>91.7 pm 297<368<431<569 kJ/mol Bond length Bond energy 10 9 >10 8 >1.3  10 6 >> 6.6  10 -4 Acid strength HF + H 2 O → [F - ·····H 3 O + ] F - + H 3 O + ion pair H-bonding free ions

39. Prentice-Hall © 2007 General Chemistry: Chapter 16 Slide 39 of 52 Strengths of Oxoacids  Factors promoting electron withdrawal from the OH bond to the oxygen atom:  High electronegativity (EN) of the central atom.  A large number of terminal O atoms in the molecule. H-O-ClH-O-Br EN Cl = 3.0EN Br = 2.8 K a = 2.9  10 -8 K a = 2.1  10 -9

40. Prentice-Hall © 2007 General Chemistry: Chapter 16 Slide 40 of 52 Strengths of Oxoacids S O O O O HH ·· - 2+ ·· - S O O O HH - + S O O O O HH S O O O HH K a  10 3 K a =1.3  10 -2

41. Prentice-Hall © 2007 General Chemistry: Chapter 16 Slide 41 of 52 Strengths of Organic Acids C O C O HH ·· H H O C HH H H C H H K a = 1.8  10 -5 K a =1.3  10 -16 acetic acidethanol