1. Slide 1 Chapter 14 Aqueous Equilibria: Acids and Bases

2. Slide 2 Acid–Base Concepts01 Arrhenius Acid: A substance which dissociates to form hydrogen ions (H + ) in solution. HA(aq)  H + (aq) + A – (aq) Arrhenius Base: A substance that dissociates in, or reacts with water to form hydroxide ions (OH – ). MOH(aq)  M + (aq) + OH – (aq)

3. Slide 3 Acid–Base Concepts02 Brønsted–Lowry Acid: Substance that can donate H + Brønsted–Lowry Base: Substance that can accept H + Chemical species whose formulas differ only by one proton are said to be conjugate acid–base pairs.

4. Slide 4 Strong vs. Weak acids03

5. Slide 5 Hydrated Protons and Hydronium Ions H 1+ (aq) + A 1- (aq)HA(aq) [H(H 2 O) n ] 1+ For our purposes, H 1+ is equivalent to H 3 O 1+. n = 4H 9 O 4 1+ n = 1H 3 O 1+ n = 2H 5 O 2 1+ n = 3H 7 O 3 1+ Due to high reactivity of the hydrogen ion, it is actually hydrated by one or more water molecules.

6. Slide 6 Acid–Base Concepts

7. Slide 7 Lewis Acid–Base Concepts

8. Slide 8 A Lewis Acid is an electron-pair acceptor. These are generally cations and neutral molecules with vacant valence orbitals, such as Al 3+, Cu 2+, H +, BF 3. A Lewis Base is an electron-pair donor. These are generally anions and neutral molecules with available pairs of electrons, such as H 2 O, NH 3, O 2–. The bond formed is called a coordinate bond. Acid–Base Concepts05

9. Slide 9 Acid–Base Concepts06 - +

10. Slide 10 Lewis Acids and Bases Lewis Base: An electron-pair donor. Lewis Acid: An electron-pair acceptor.

11. Slide 11 Lewis Acids and Bases Lewis Base: An electron-pair donor. Lewis Acid: An electron-pair acceptor.

12. Slide 12 Acid–Base Concepts07 Write balanced equations for the dissociation of each of the following Brønsted–Lowry acids. (a) H 2 SO 4 (b) HSO 4 – (c) H 3 O + Identify the Lewis acid and Lewis base in each of the following reactions: (a) SnCl 4 (s) + 2 Cl – (aq) æ SnCl 6 2– (aq) (b) Hg 2+ (aq) + 4 CN – (aq) æ Hg(CN) 4 2– (aq) (c) Co 3+ (aq) + 6 NH 3 (aq) æ Co(NH 3 ) 6 3+ (aq)

13. Slide 13 Dissociation of Water01 Water can act as an acid or as a base. H 2 O(l) æ H + (aq) + OH – (aq) K c = [H + ][OH – ] This is called the autoionization of water. H 2 O(l) + H 2 O(l) æ H 3 O + (aq) + OH – (aq)

14. Slide 14 Dissociation of Water02 This equilibrium gives us the ion product constant for water. K w = K c = [H + ][OH – ] = 1.0 x 10 –14 If we know either [H + ] or [OH – ] then we can determine the other quantity.

15. Slide 15 Dissociation of Water03 The concentration of OH – ions in a certain household ammonia cleaning solution is 0.0025 M. Calculate the concentration of H + ions. Calculate the concentration of OH – ions in a HCl solution whose hydrogen ion concentration is 1.3 M.

16. Slide 16 pH – A Measure of Acidity01 The pH of a solution is the negative logarithm of the hydrogen ion concentration (in mol/L). pH = –log [H + ], [H + ] = 10 -pH pH + pOH = 14 Acidic solutions:[H + ] > 1.0 x 10 –7 M, pH 7.00 Neutral solutions:[H + ] = 1.0 x 10 –7 M, pH = 7.00

17. Slide 17 pH – A Measure of Acidity02 Nitric acid (HNO 3 ) is used in the production of fertilizer, dyes, drugs, and explosives. Calculate the pH of a HNO 3 solution having a hydrogen ion concentration of 0.76 M. The pH of a certain orange juice is 3.33. Calculate the H + ion concentration. The OH – ion concentration of a blood sample is 2.5 x 10 –7 M. What is the pH of the blood?

18. Slide 18 pH – A Measure of Acidity04 Color of Tea: Polyphenols, Thearubigins Color of Red Cabbage: Anthocyanin

19. Slide 19 pH – A Measure of Acidity04

20. Slide 20 Strength of Acids and Bases03 HClO 4 HI HBr HCl H 2 SO 4 HNO 3 H 3 O + HSO 4 – HF HNO 2 HCOOH NH 4 + HCN H 2 O NH 3 ClO 4 – I – Br – Cl – HSO 4 – NO 3 – H 2 O SO 4 2– F – NO 2 – HCOO – NH 3 CN – OH – NH 2 – ACID CONJ. BASE Increasing Acid Strength

21. Slide 21 Strength of Acids and Bases04 Stronger acid + stronger base  weaker acid + weaker base Predict the direction of the following: HNO 2 (aq) + CN – (aq) æ HCN(aq) + NO 2 – (aq) HF(aq) + NH 3 (aq) æ F – (aq) + NH 4 + (aq)

22. Slide 22 Acid Ionization Constants01 Acid Ionization Constant: the equilibrium constant for the ionization of an acid. HA(aq) + H 2 O(l) æ H 3 O + (aq) + A – (aq) Or simply: HA(aq) æ H + (aq) + A – (aq)

23. Slide 23 Conjugate Base Ionization Const [HA] [OH − ] [A - ] K b = A - + H 2 O (l) HA( aq) + OH − (aq) K a  K b = K w [HA] [OH − ] [A - ] K b = K a  = K w

24. Slide 24 Acid Ionization Constants02 7.1 x 10 –4 4.5 x 10 –4 3.0 x 10 –4 1.7 x 10 –4 8.0 x 10 –5 6.5 x 10 –5 1.8 x 10 –5 4.9 x 10 –10 1.3 x 10 –10 HF HNO 2 C 9 H 8 O 4 (aspirin) HCO 2 H (formic) C 6 H 8 O 6 (ascorbic) C 6 H 5 CO 2 H (benzoic) CH 3 CO 2 H (acetic) HCN C 6 H 5 OH (phenol) F – NO 2 – C 9 H 7 O 4 – HCO 2 – C 6 H 7 O 6 – C 6 H 5 CO 2 – CH 3 CO 2 – CN – C 6 H 5 O – ACID K a CONJ. BASE K b 1.4 x 10 –11 2.2 x 10 –11 3.3 x 10 –11 5.9 x 10 –11 1.3 x 10 –10 1.5 x 10 –10 5.6 x 10 –10 2.0 x 10 –5 7.7 x 10 –5

25. Slide 25 Strength of Acids and Bases03 (a) Arrange the three acids in order of increasing value of K a. (b) Which acid, if any, is a strong acid? (c) Which solution has the highest pH, and which has the lowest? (4 2 /2) = 81 2 /5= 0.2 Very Large K =

26. Slide 26 HA æ H + +A (M):0.500.00 (M):–x+x+x Equilib (M): 0.50–xxx Acid Ionization Constants Determine the pH of 0.50 M HA solution at 25°C. K a = 7.1 x 10 –4 05 I nitial C hange E quilibrium T able :. Initial Change (aq) -

27. Slide 27 What is the pH of a 0.50 M Citric acid solution (at 25 0 C)? HA (aq) H + (aq) + A - (aq) K a = [H + ][A - ] [HA] = 7.1 x 10 -4 HA (aq) H + (aq) + A - (aq) Initial (M) Change (M) Equilibrium (M) 0.500.00 -x-x+x+x 0.50 - x 0.00 +x+x xx K a = x2x2 0.50 - x = 7.1 x 10 -4 Ka  Ka  x2x2 0.50 = 7.1 x 10 -4 0.50 – x  0.50 100K a < C o ? 100 x 7.1 x 10 -4 = 0.071 < 0.5 x 2 = 3.55 x 10 -4 x = 0.019 M [H + ] = [A - ] = 0.019 M pH = -log [H + ] = 1.72 [HA] = 0.50 – x = 0.48 M

28. Slide 28 Acid Ionization Constants06 pH of a Weak Acid (Cont’d): 1. Substitute equilibrium concentrations into equilibrium expression. 2. If 100K a < C o then (C 0 – x) approximates to (C 0 ). 3. The equation can now be solved for x and pH. 4. If 100K a is not significantly smaller than C o the quadratic equation must be used to solve for x and pH.

29. Slide 29 Acid Ionization Constants07 The Quadratic Equation: The expression must first be rearranged to: The values are substituted into the quadratic and solved for a positive solution to x and pH.

30. Slide 30 Acid Ionization Constants09 Percent Dissociation: A measure of the strength of an acid. Stronger acids have higher percent dissociation. Percent dissociation of a weak acid decreases as its concentration increases. H 1+ (aq) + A 1- (aq)HA(aq)

31. Slide 31 Percent dissociation of a weak acid decreases as its concentration increases Concentration Dependence:

32. Slide 32 Weak Bases: Base Ionization Constants01 Base Ionization Constant: The equilibrium constant for the ionization of a base. The ionization of weak bases is treated in the same way as the ionization of weak acids. B(aq) + H 2 O(l) æ BH + (aq) + OH – (aq) Calculations follow the same procedure as used for a weak acid but [OH – ] is calculated, not [H + ].

33. Slide 33 Base Ionization Constants02 5.6 x 10 –4 4.4 x 10 –4 4.1 x 10 –4 1.8 x 10 –5 1.7 x 10 –9 3.8 x 10 –10 1.5 x 10 –14 C 2 H 5 NH 2 (ethylamine) CH 3 NH 2 (methylamine) C 8 H 10 N 4 O 2 (caffeine) NH 3 (ammonia) C 5 H 5 N (pyridine) C 6 H 5 NH 2 (aniline) NH 2 CONH 2 (urea) C 2 H 5 NH 3 + CH 3 NH 3 + C 8 H 11 N 4 O 2 + NH 4 + C 5 H 6 N + C 6 H 5 NH 3 + NH 2 CONH 3 + BASE K b CONJ. ACID K a 1.8 x 10 –11 2.3 x 10 –11 2.4 x 10 –11 5.6 x 10 –10 5.9 x 10 –6 2.6 x 10 –5 0.67 Note that the positive charge sits on the nitrogen. (caffeine)

34. Slide 34 Base Ionization Constants03 Product of K a and K b : multiplying out the expressions for K a and K b equals K w. K a  K b = K w

35. Slide 35 pH of Basic Solutions What is the pH of a 0.15 M solution of NH 3 ? [NH 4 + ] [OH − ] [NH 3 ] K b = = 1.8  10 −5 NH 3 (aq) + H 2 O (l) NH 4 + (aq) + OH − (aq) [NH 3 ], M[NH 4 + ], M[OH − ], M Initially0.1500 At Equilibrium0.15 - xxx

36. Slide 36 pH of Basic Solutions (1.8  10 −5 ) (0.15) = x 2 2.7  10 −6 = x 2 1.6  10 −3 = x 2 (x) 2 ( 0.15 - x ) 1.8  10 −5 = 100 x K b < C 0 ? 1.8  10 −3 < 0.15 0.15 –x = 0.15

37. Slide 37 pH of Basic Solutions Therefore, X = [OH − ] = 1.6  10 −3 M pOH = −log (1.6  10 −3 ) pOH = 2.80 pH = 14.00 − 2.80 pH = 11.20

38. Slide 38 Calculate the pH of a 0.10 M solution of Na 2 CO 3. Na + + H 2 O ---> neutral CO 3 2- +H 2 O HCO 3 - + OH - CO 3 2- +H 2 O æ HCO 3 - + OH - baseacid acidbase K b = 2.1 x 10 -4 K b = 2.1 x 10 -4 Acid-Base Properties of Salts

39. Slide 39 Calculate the pH of a 0.10 M solution of Na 2 CO 3. Na + + H 2 O ---> neutral CO 3 2- +H 2 O HCO 3 - + OH - CO 3 2- +H 2 O æ HCO 3 - + OH - baseacid acidbase K b = 2.1 x 10 -4 K b = 2.1 x 10 -4 Step 1.Set up ICE table [CO 3 2- ] [HCO 3 - ][OH - ] initial 0.1 0 0 [CO 3 2- ] æ [HCO 3 - ][OH - ] initial 0.1 0 0 change-x+x+x equilib0.10 - xxx equilib0.10 - xxx Acid-Base Properties of Salts

40. Slide 40 Calculate the pH of a 0.10 M solution of Na 2 CO 3. Na + + H 2 O ---> neutral CO 3 2- +H 2 O e HCO 3 - + OH - baseacid acidbase K b = 2.1 x 10 -4 K b = 2.1 x 10 -4 Acid-Base Properties of Salts Assume 0.10 - x ≈ 0.10, because 100K b < C o x = [HCO 3 - ] = [OH - ] = 0.0046 M Step 2.Solve the equilibrium expression

41. Slide 41 Calculate the pH of a 0.10 M solution of Na 2 CO 3. Na + + H 2 O ---> neutral CO 3 2- + H 2 O e HCO 3 - + OH - baseacid acidbase K b = 2.1 x 10 -4 K b = 2.1 x 10 -4 Acid-Base Properties of Salts Step 3.Calculate the pH [OH - ] = 0.0046 M pOH = - log [OH - ] = 2.34 pH + pOH = 14, so pH = 11.66, and the solution is Basic.

42. Slide 42 Diprotic & Polyprotic Acids01 Diprotic and polyprotic acids yield more than one hydrogen ion per molecule. One proton is lost at a time. Conjugate base of first step is acid of second step. Ionization constants decrease as protons are removed. H 2 SO 4 H 3 PO 4

43. Slide 43 Diprotic & Polyprotic Acids02 Very Large 1.3 x 10 –2 6.5 x 10 –2 6.1 x 10 –5 1.3 x 10 –2 6.3 x 10 –8 4.2 x 10 –7 4.8 x 10 –11 9.5 x 10 –8 1 x 10 –19 7.5 x 10 –3 6.2 x 10 –8 4.8 x 10 –13 H 2 SO 4 HSO 4 – C 2 H 2 O 4 C 2 HO 4 – H 2 SO 3 HSO 3 – H 2 CO 3 HCO 3 – H 2 S HS – H 3 PO 4 H 2 PO 4 – HPO 4 2– ACID K a CONJ. BASE K b HSO 4 – SO 4 2– C 2 HO 4 – C 2 O 4 2– HSO 3 – SO 3 2– HCO 3 – CO 3 2– HS – S 2– H 2 PO 4 – HPO 4 2– PO 4 3– Very Small 7.7 x 10 –13 1.5 x 10 –13 1.6 x 10 –10 7.7 x 10 –13 1.6 x 10 –7 2.4 x 10 –8 2.1 x 10 –4 1.1 x 10 –7 1 x 10 –5 1.3 x 10 –12 1.6 x 10 –7 2.1 x 10 –2

44. Slide 44 Molecular Structure and Acid Strength01 The strength of an acid depends on its tendency to ionize. For general acids of the type H–X: 1. The stronger the bond, the weaker the acid. 2. The more polar the bond, the stronger the acid. For the hydrohalic acids, bond strength plays the key role giving: HF < HCl < HBr < HI 299 kJ/mol for HI 567 kJ/mol for HF

45. Slide 45 Molecular Structure and Acid Strength02 The electrostatic potential maps show all the hydrohalic acids are polar. The variation in polarity is less significant than the bond strength which decreases from 567 kJ/mol for HF to 299 kJ/mol for HI.

46. Slide 46 (pm)

47. Slide 47 Molecular Structure and Acid Strength03 For binary acids in the same group, H–A bond strength decreases with increasing size of A, so acidity increases. For binary acids in the same row, H–A polarity increases with increasing electronegativity of A, so acidity increases.

48. Slide 48 Molecular Structure and Acid Strength04 For oxoacids bond polarity is more important. If we consider the main element (Y): Y–O–H If Y is an electronegative element, the Y–O bond will pull more electrons, the O–H bond will be more polar and the acid will be stronger.

49. Slide 49 Molecular Structure and Acid Strength05 For oxoacids with different central atoms that are from the same group of the periodic table and that have the same oxidation number, acid strength increases with increasing electronegativity.

50. Slide 50 Polar Covalent Bonds02 Pauling Electronegativities Detailed List of Electronegativity; http://environmentalchemistry.com/yogi/periodic/electronegativity.html Electronegativity

51. Slide 51 Molecular Structure and Acid Strength07 Oxoacids of Chlorine:

52. Slide 52 Molecular Structure and Acid Strength08 Predict the relative strengths of the following groups of oxoacids: a) HClO, HBrO, and HIO. b) HNO 3 and HNO 2. c) H 3 PO 3 and H 3 PO 4.

53. Slide 53 Acid-Base Properties of Salts

54. Slide 54 Strong bases Strong bases: The following metals make strong hydroxy base Alkali metal cations of group 1A Alkaline earth metal cations of group 2A except for Be

55. Slide 55 Acid–Base Properties of Salts01 Salts that produce neutral solutions are those formed from strong acids and strong bases. Salts that produce basic solutions are those formed from weak acids and strong bases. Salts that produce acidic solutions are those formed from strong acids and weak bases.

56. Slide 56 The pH of an ammonium carbonate solution, (NH 4 ) 2 CO 3, depends on the relative acid strength of the cation and the relative base strength of the anion. Is it acidic or basic? Salts That Contain Cation from a Weak Base and anion from a Weak Base

57. Slide 57 Acid-Base Properties of Salts Salts That Contain Acidic Cations and Basic Anions HCO 3 1- (aq) + OH 1- (aq)CO 3 2- (aq) + H 2 O(l) KbKb H 3 O 1+ (aq) + NH 3 (aq)NH 4 1+ (aq) + H 2 O(l) KaKa (NH 4 ) 2 CO 3 : Three possibilities: K a > K b : The solution will contain an excess of H 3 O 1+ ions, Acidic solution, (pH < 7). K a 7). K a ≈ K b : The solution will contain approximately equal concentrations of H 3 O 1+ and OH 1- ions (pH ≈ 7).

58. Slide 58 Salts That Contain Cation from a Weak Acid and anion from a Weak Base HCO 3 1- (aq) + OH 1- (aq)CO 3 2- (aq) + H 2 O(l) KbKb H 3 O 1+ (aq) + NH 3 (aq)NH 4 1+ (aq) + H 2 O(l) KaKa (NH 4 ) 2 CO 3 : = 1.8 x 10 -4 5.6 x 10 -11 1.0 x 10 -14 K b for CO 3 2- = K a for HCO 3 1- KwKw = = 5.6 x 10 -10 1.8 x 10 -5 1.0 x 10 -14 K a for NH 4 1+ = K b for NH 3 KwKw = Basic, K a < K b

59. Slide 59 Acid-Base Properties of Salts

60. Slide 60 Hydrated Cation of Al 3+

61. Slide 61 Acid–Base Properties of SaltsAcid–Base Properties of Salts03 Acid–Base Properties of SaltsAcid–Base Properties of Salts03 Metal Ion Hydrolysis:

62. Slide 62 Acid–Base Properties of Salts04 Calculate the pH of a 0.020 M Al(NO 3 ) 3 solution K a = 1.4 x 10 -5. Predict whether the following solutions will be acidic, basic, or nearly neutral: (a) NH 4 I (b) CaCl 2 (c) KCN (d) Fe(NO 3 ) 3