1. 18-1 Chapter 18 Acid-Base Equilibria

2. 18-2 Acid-Base Equilibria 18.1 Acids and bases in water 18.2 Auto-ionization of water and the pH scale 18.3 Proton transfer and the Brønsted-Lowry acid-base definition 18.4 Solving problems involving weak acid equilibria 18.5 Weak bases and their relationships to weak acids 18.6 Molecular properties and acid strength 18.7 Acid-base properties of salt solutions 18.8 Generalizing the Brønsted-Lowry concept: The Leveling Effect 18.9 Electron-pair donations and the Lewis acid-base definition

3. 18-3 Figure 18.1 Etching with acids The inside surfaces of these light bulbs are etched with HF. Acids are used to wash away oxides of silicon and metals during the production of computer chips.

4. 18-4

5. 18-5 For reaction between a strong acid and strong base: HCl(aq) + NaOH(aq) NaCl(aq) + H 2 O(l) H + (aq) + OH - (aq) H 2 O(l) For acid dissociation: HA(g or l) + H 2 O(l) A - (aq) + H 3 O + (aq) In aqueous solution, the H + ions bind covalently to water to form solvated hydronium ions.

6. 18-6 Figure 18.2 The nature of the hydrated proton

7. 18-7 The Classical (Arrhenius) Definition of Acids and Bases An acid is a substance that has H in its formula and dissociates in water to yield H 3 O +. A base is a substance that has OH in its formula and dissociates in water to yield OH -. Arrhenius acids contain covalently bonded H atoms that ionize in water. Neutralization occurs when the H + ion from the acid and the OH - ion from the base combine to form water. H + (aq) + OH - (aq) H 2 O(l) ∆H o rxn = -55.9 kJ

8. 18-8 Defining the acid dissociation constant Weak acids dissociate very slightly into ions in water. Strong acids dissociate completely into ions in water. HA(g or l) + H 2 O(l) H 3 O + (aq) + A - (aq) HA(aq) + H 2 O(l) H 3 O + (aq) + A - (aq) K c >> 1 K c << 1 Kc =Kc = [H 3 O + ][A - ] [H 2 O][HA] K c [H 2 O] = K a = [H 3 O + ][A - ] [HA] weaker acid, lower [H 3 O + ], smaller K a stronger acid, higher [H 3 O + ], larger K a

9. 18-9 Figure 18.3 strong acid : HA( g or l ) + H 2 O( l ) H 3 O + ( aq ) + A - ( aq ) The extent of dissociation of strong acids Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display.

10. 18-10 Figure 18.3 The extent of dissociation of weak acids weak acid : HA( aq ) + H 2 O( l ) H 3 O + ( aq ) + A - ( aq ) Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display.

11. 18-11 Figure 18.4 Reaction of zinc with a strong and a weak acid 1 M HCl( aq )1 M CH 3 COOH( aq ) Zn(s) + 2H 3 O + (aq) Zn +2 (aq) + 2H 2 O(l) + H 2 (g)

12. 18-12 Acid strength decreases down the table (smaller K a values)

13. 18-13 SAMPLE PROBLEM 18.1 SOLUTION: Classifying acid and base strength from the chemical formula PROBLEM:Classify each of the following compounds as a strong acid, weak acid, strong base or weak base. (a) H 2 SeO 4 (b) (CH 3 ) 2 CHCOOH(c) KOH (d) (CH 3 ) 2 CHNH 2 PLAN:Pay attention to the text definitions of acids and bases. Look at O for acids and for the -COOH group; watch for amine groups and cations in bases. (a) strong acid - H 2 SeO 4 - the number of oxygen atoms exceeds the number of ionizable protons by a factor of 2. (b) weak acid - (CH 3 ) 2 CHCOOH is an organic acid having a -COOH group (a carboxylic acid). (c) strong base - KOH is a Group 1A hydroxide. (d) weak base - (CH 3 ) 2 CHNH 2 has a lone pair of electrons on the nitrogen and is an amine.

14. 18-14 The auto-ionization of water and the pH scale H 2 O( l ) H 3 O + ( aq )OH - ( aq ) + +

15. 18-15 Kc =Kc = [H 3 O + ][OH - ] [H 2 O] 2 K c [H 2 O] 2 = [H 3 O + ][OH - ] Defining the ion-product constant of water Kw =Kw = A change in [H 3 O + ] causes an inverse change in [OH - ]. = 1.0 x 10 -14 at 25 o C H 2 O(l) + H 2 O(l) H 3 O + (aq) + OH - (aq) In an acidic solution, [H 3 O + ] > [OH - ] In a basic solution, [H 3 O + ] < [OH - ] In a neutral solution, [H 3 O + ] = [OH - ]

16. 18-16 Figure 18.5 The relationship between [H 3 O + ] and [OH - ] and the relative acidity of solutions [H 3 O + ][OH - ] divide into K w acidic solution basic solution [H 3 O + ] > [OH - ] [H 3 O + ] = [OH - ][H 3 O + ] < [OH - ] neutral solution Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display.

17. 18-17 SAMPLE PROBLEM 18.2Calculating [H 3 O + ] and [OH - ] in an aqueous solution PROBLEM:A research chemist adds a measured amount of HCl gas to pure water at 25 o C and obtains a solution with [H 3 O + ] = 3.0 x 10 -4 M. Calculate [OH - ]. Is the solution neutral, acidic or basic? SOLUTION: PLAN:Use the K w at 25 o C and the [H 3 O + ] to find the corresponding [OH - ]. K w = 1.0 x 10 -14 = [H 3 O + ] [OH - ] [OH - ] = K w / [H 3 O + ] = 1.0 x 10 -14 /3.0 x 10 -4 = 3.3 x 10 -11 M [H 3 O + ] > [OH - ]; the solution is acidic.

18. 18-18 Figure 18.6 The pH values of some familiar aqueous solutions pH = -log [H 3 O + ] pOH = -log [OH - ] pK = -log K Related Expressions

19. 18-19 A low pK corresponds to a high K.

20. 18-20 Relationships between [H 3 O + ], pH, [OH - ] and pOH Figure 18.7

21. 18-21 SAMPLE PROBLEM 18.3Calculating [H 3 O + ], pH, [OH - ], and pOH PROBLEM: In a restoration project, a conservator prepares copper-plate etching solutions by diluting concentrated nitric acid, HNO 3, to 2.0 M, 0.30 M, and 0.0063 M HNO 3. Calculate [H 3 O + ], pH, [OH - ], and pOH of the three solutions at 25 o C. SOLUTION: PLAN:HNO 3 is a strong acid so [H 3 O + ] = [HNO 3 ]. Use K w to find the [OH - ] and then convert to pH and pOH. For 2.0 M HNO 3, [H 3 O + ] = 2.0 M and -log [H 3 O + ] = -0.30 = pH [OH - ] = K w / [H 3 O + ] = 1.0 x 10 -14 /2.0 = 5.0 x 10 -15 M; pOH = 14.30 [OH - ] = K w / [H 3 O + ] = 1.0 x 10 -14 /0.30 = 3.3 x 10 -14 M; pOH = 13.48 For 0.3 M HNO 3, [H 3 O + ] = 0.30 M and -log [H 3 O + ] = 0.52 = pH [OH - ] = K w / [H 3 O + ] = 1.0 x 10 -14 /6.3 x 10 -3 = 1.6 x 10 -12 M; pOH = 11.80 For 0.0063 M HNO 3, [H 3 O + ] = 0.0063 M and -log [H 3 O + ] = 2.20 = pH

22. 18-22 Figure 18.8 Methods for measuring the pH of an aqueous solution pH meter pH (indicator) paper Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display.

23. 18-23 The Brønsted-Lowry Definition of Acids and Bases An acid is a proton donor, that is, any species that donates an H + ion. All Arrhenius acids are Brønsted-Lowry acids. A base is a proton acceptor, that is, any species that accepts an H + ion. In the Brønsted-Lowry definition, an acid-base reaction is a proton transfer process.

24. 18-24 Figure 18.9 Proton transfer is the essential feature of a Brønsted- Lowry acid-base reaction (acid, H + donor)(base, H + acceptor) HClH2OH2O + Cl - H3O+H3O+ + lone pair binds H + (base, H + acceptor)(acid, H + donor) NH 3 H2OH2O + NH 4 + OH - + lone pair binds H + Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display.

25. 18-25 An acid reactant produces a base product and the two constitute an acid-base conjugate pair. The Conjugate Acid-Base Pair H 2 S + NH 3 HS - + NH 4 + H 2 S and HS - are a conjugate acid-base pair. HS - is the conjugate base of the acid H 2 S. NH 3 and NH 4 + are a conjugate acid-base pair. NH 4 + is the conjugate acid of the base NH 3. Every acid has a conjugate base, and every base has a conjugate acid.

26. 18-26 The conjugate base of the pair has one fewer H and one more negative charge than the acid. The conjugate acid of the pair has one more H and one less negative charge than the base. A Bronsted-Lowry acid-base reaction occurs when an acid and a base react to form their conjugate base and conjugate acid, respectively. acid 1 + base 2 base 1 + acid 2

27. 18-27 Table 18.4 Conjugate Pairs in Some Acid-Base Reactions base 1 acid 2 + acid 1 base 2 + conjugate pair reaction 4 H 2 PO 4 - OH - + reaction 5 H 2 SO 4 N2H5+N2H5+ + reaction 6 HPO 4 2 - SO 3 2 - + reaction 1HFH2OH2O+ F-F- H3O+H3O+ + reaction 3 NH 4 + CO 3 2 - + reaction 2HCOOH CN - + HCOO - HCN+ NH 3 HCO 3 - + HPO 4 2 - H2OH2O+ HSO 4 - N 2 H 6 2+ + PO 4 3 - HSO 3 - + Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display.

28. 18-28 SAMPLE PROBLEM 18.4 Identifying conjugate acid-base pairs PROBLEM:The following reactions are important environmental processes. Identify the conjugate acid-base pairs. (a) H 2 PO 4 - ( aq ) + CO 3 2 - ( aq ) HPO 4 2 - ( aq ) + HCO 3 - ( aq ) (b) H 2 O( l ) + SO 3 2 - ( aq ) OH - ( aq ) + HSO 3 - ( aq ) SOLUTION: PLAN:Identify proton donors (acids) and proton acceptors (bases). (a) H 2 PO 4 - ( aq ) + CO 3 2 - ( aq ) HPO 4 2 - ( aq ) + HCO 3 - ( aq ) proton donor proton acceptor proton donor conjugate pair 1 conjugate pair 2 (b) H 2 O( l ) + SO 3 2 - ( aq ) OH - ( aq ) + HSO 3 - ( aq ) conjugate pair 2 conjugate pair 1 proton donor proton acceptor proton donor

29. 18-29 Relative Acid-Base Strength and Reaction Direction General Rule: An acid-base reaction proceeds to the greater extent in the direction in which a stronger acid and stronger base form a weaker acid and a weaker base. H 2 S + NH 3 HS - + NH 4 + K c > 1 A competition for the proton between the two bases! HF + H 2 O F - + H 3 O + K c < 1 abba abba

30. 18-30 Figure 18.10 Strengths of conjugate acid- base pairs An acid-base reaction proceeds to the right if the acid reacts with a base that is lower on the list because this combination produces a weaker conjugate base and a weaker conjugate acid. A weaker acid has a stronger conjugate base.

31. 18-31 SAMPLE PROBLEM 18.5Predicting the net direction of an acid-base reaction PROBLEM:Predict the net direction and whether K a is greater or less than 1 for each of the following reactions (assume equal initial concentrations of all species): (b) H 2 O( l ) + HS - ( aq ) OH - ( aq ) + H 2 S( aq ) (a) H 2 PO 4 - ( aq ) + NH 3 ( aq ) HPO 4 2 - ( aq ) + NH 4 + ( aq ) SOLUTION: PLAN:Identify the conjugate acid-base pairs and then consult Figure 18.10 to determine the relative strength of each. The stronger the species, the more preponderant will be its conjugate. (a) H 2 PO 4 - ( aq ) + NH 3 ( aq ) HPO 4 2 - ( aq ) + NH 4 + ( aq ) stronger acidweaker acidstronger baseweaker base Net direction is to the right; K c > 1. (b) H 2 O( l ) + HS - ( aq ) OH - ( aq ) + H 2 S( aq ) stronger baseweaker basestronger acidweaker acid Net direction is to the left; K c < 1.

32. 18-32 Weak Acid Equilibrium Problems Follow the general procedures described in Chapter 17 Two Key Assumptions [H 3 O + ] from the auto-ionization of water is negligible. [HA] eq = [HA] init - [HA] dissoc ≈ [HA] init (because a weak acid has a small K a ) [HA] dissoc is very small

33. 18-33 SAMPLE PROBLEM 18.6 Find the K a of a weak acid from the pH of its solution PROBLEM:Phenylacetic acid (C 6 H 5 CH 2 COOH, denoted as HPAc) builds up in the blood of people afflicted with phenylketonuria, an inherited genetic disorder that, if left untreated, causes mental retardation and death. A study of the acid shows that the pH of a 0.12 M solution of HPAc is 2.60. What is the K a of phenylacetic acid? PLAN:Write the dissociation equation. Use pH and solution concentration to find K a. K a = [H 3 O + ][PAc - ] [HPAc] Assumptions: At a pH of 2.60, [H 3 O + ] HPAc >> [H 3 O + ] water [PAc - ] eq ≈ [H 3 O + ] eq, and since HPAc is weak, [HPAc] eq ≈ [HPAc] initial = [HPAc] initial - [HPAc] dissociation SOLUTION:HPAc( aq ) + H 2 O( l ) H 3 O + ( aq ) + PAc - ( aq )

34. 18-34 SAMPLE PROBLEM 18.6(continued) concentration (M)HPAc( aq ) + H 2 O( l ) H 3 O + ( aq ) + PAc - ( aq ) initial0.12- 1 x 10 -7 0 change--x+x equilibrium-0.12 - xxx +(<1 x 10 -7 ) [H 3 O + ] = 10 -pH = 2.5 x 10 -3 M, which is >> 10 -7 ([H 3 O + ] from water) x ≈ 2.5 x 10 -3 M ≈ [H 3 O + ] ≈ [PAc - ][HPAc] eq = 0.12 - x ≈ 0.12 M K a = (2.5 x 10 -3 ) 0.12 = 5.2 x 10 -5 Testing the assumptions: = 4 x 10 -3 % x 100 [HPAc] dissn : 2.5 x 10 -3 M 0.12 M [H 3 O + ] water : 1 x 10 -7 M 2.5 x 10 -3 M x100 = 2.1 % (the 5% rule is not violated in either case)

35. 18-35 SAMPLE PROBLEM 18.7Determining concentrations from K a and initial [HA] PROBLEM:Propanoic acid (CH 3 CH 2 COOH, simplified as HPr) is an organic acid whose salts are used to retard mold growth in foods. What is the [H 3 O + ] of a 0.10 M aqueous solution of HPr (K a = 1.3 x 10 -5 )? SOLUTION: PLAN:Write the dissociation equation and K a expression; make assumptions about concentration that are valid; substitute. x = [HPr] diss = [H 3 O + ] from HPr = [Pr - ] Assumptions:For: HPr( aq ) + H 2 O( l ) H 3 O + ( aq ) + Pr - ( aq ) K a =[H 3 O + ][Pr - ] [HPr] HPr( aq ) + H 2 O( l ) H 3 O + ( aq ) + Pr - ( aq )concentration (M) initial0.10-00 change--x+x equilibrium-0.10 - xxx Since K a is small, we assume that x << 0.10

36. 18-36 SAMPLE PROBLEM 18.7(continued) (x) 2 0.10 1.3 x 10 -5 = [H 3 O + ][Pr - ] [HPr] = = 1.1 x 10 -3 M = [H 3 O + ] Checking assumptions: [HPr] diss : 1.1 x 10 -3 M / 0.10 M x 100 = 1.1% K a =

37. 18-37 percent HA dissociation = [HA] dissociated [HA] initial x 100 As the initial concentration of a weak acid decreases, the percent dissociation of the acid increases! In the prior problem: for 0.10 M HPr, 1.1% dissociation for 0.010 M HPr, 3.6% dissociation Rationale: larger solution volume accommodates more ions (analogous to pressure effects on gas equilibria when ∆n gas is non-zero)

38. 18-38 Polyprotic acids Acids that contain more than one ionizable proton H 3 PO 4 ( aq ) + H 2 O( l ) H 2 PO 4 - ( aq ) + H 3 O + ( aq ) H 2 PO 4 - ( aq ) + H 2 O( l ) HPO 4 2 - ( aq ) + H 3 O + ( aq ) HPO 4 2 - ( aq ) + H 2 O( l ) PO 4 3 - ( aq ) + H 3 O + ( aq ) K a1 = [H 3 O + ] [H 2 PO 4 - ] [H 3 PO 4 ] K a2 = [H 3 O + ] [HPO 4 2- ] [H 2 PO 4 - ] K a3 = [H 3 O + ] [PO 4 3 - ] [HPO 4 2 - ] K a1 > K a2 > K a3 = 7.2 x 10 -3 = 6.3 x 10 -8 = 4.2 x 10 -13 phosphoric acid, H 3 PO 4

39. 18-39 Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display.

40. 18-40 Since successive acid dissociation constants typically differ by several orders of magnitude, pH calculations for aqueous solutions of the free acids can be simplified by neglecting H 3 O + generated by subsequent dissociations. Why are the K a values successively smaller as more H + ions dissociate from a polyprotic acid like phosphoric acid?

41. 18-41 SAMPLE PROBLEM 18.8Calculating equilibrium concentrations for a polyprotic acid PROBLEM:Ascorbic acid (H 2 C 6 H 6 O 6 ; abbreviated H 2 Asc), known as vitamin C, is a diprotic acid (K a1 = 1.0 x 10 -5 and K a2 = 5 x 10 -12 ) found in citrus fruit. Calculate [H 2 Asc], [HAsc - ], [Asc 2 - ], and the pH of a 0.050 M aqueous solution of H 2 Asc. SOLUTION: PLAN:Write out expressions for both dissociations and make assumptions. K a1 >> K a2 so the first dissociation produces virtually all of the H 3 O +. K a1 is small, thus [H 2 Asc] initial ≈ [H 2 Asc] eq After finding concentrations of the various species for the first dissociation, use them as initial concentrations for the second dissociation. H 2 Asc( aq ) + H 2 O (l ) HAsc - ( aq ) + H 3 O + ( aq ) K a1 = [HAsc - ][H 3 O + ] [H 2 Asc] = 1.0 x 10 -5 HAsc - ( aq ) + H 2 O (l ) Asc 2 - ( aq ) + H 3 O + ( aq ) K a2 = [Asc 2 - ][H 3 O + ] [HAsc - ] = 5 x 10 -12

42. 18-42 SAMPLE PROBLEM 18.8(continued) H 2 Asc( aq ) + H 2 O (l ) HAsc - ( aq ) + H 3 O + ( aq )concentration (M) initial0.050-00 change- x-+ x equilibrium0.050 - x-xx K a1 = [HAsc - ][H 3 O + ]/[H 2 Asc] = 1.0 x 10 -5 = (x)(x)/0.050 M pH = -log(7.1 x 10 -4 ) = 3.15 HAsc - ( aq ) + H 2 O (l ) Asc 2 - ( aq ) + H 3 O + ( aq ) 7.1 x 10 -4 - 00 change- x-+ x equilibrium7.1 x 10 -4 - x-xx initial x x = 7.1 x 10 -4 M = [HAsc - ] concentration (M) x = [(7.1 x 10 -4 )(5 x 10 -12 )] 0.5 = 6 x 10 -8 M = [H 3 O + ]

43. 18-43 We can ignore the hydronium ion generated by second ionization. Also: 7.1 x 10 -4 /0.050 x 100 = 1.4% This value is < 5%, thus the assumption made in the analysis of the first dissociation reaction is justified. [Asc - 2 ] = (K a2 x HAsc - )/[H 3 O + ] = 5 x 10 -12 M

44. 18-44 Weak Bases: Their Relationship to Weak Acids B(aq) + H 2 O(l) BH + (aq) + OH - (aq) [BH + ][OH - ] [B][H 2 O] Base-dissociation constant, K b [BH + ][OH - ] [B] pK b decreases with increasing K b (i.e., increasing base strength) K c = K b =

45. 18-45 Main Classes of Weak Bases: nitrogen-containing molecules (ammonia and amines) and anions of weak acids. Ammonia: NH 3 (aq) + H 2 O(l) NH 4 + (aq) + OH - (aq) K b = 1.76 x 10 -5 (25 o C) Amines: RNH 2, R 2 NH and R 3 N: all have a lone pair of electrons that can bind a proton donated by an acid

46. 18-46 + Figure 18.11 Abstraction of a proton from water by methylamine + lone pair binds H + Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display. CH 3 NH 2 methylamine H2OH2O CH 3 NH 3 + methylammonium ion OH -

47. 18-47 Base strength decreases down the table (smaller K b values)

48. 18-48 SAMPLE PROBLEM 18.9Determining pH from K b and initial [B] PROBLEM:Dimethylamine, (CH 3 ) 2 NH, a key intermediate in detergent manufacture, has a K b = 5.9 x 10 -4. What is the pH of a 1.5 M aqueous solution of (CH 3 ) 2 NH? SOLUTION: PLAN:Perform this calculation as done for acids. Keep in mind that you are working with K b and a base. (CH 3 ) 2 NH( aq ) + H 2 O( l ) (CH 3 ) 2 NH 2 + ( aq ) + OH - ( aq ) Assumptions: [(CH 3 ) 2 NH 2 + ] = [OH - ] = x [(CH 3 ) 2 NH] eq ≈ [(CH 3 ) 2 NH] initial K b >> K w so [OH - ] water is negligible initial1.5000- change- x-+ x equilibrium1.50 - x-xx (CH 3 ) 2 NH( aq ) + H 2 O( l ) (CH 3 ) 2 NH 2 + ( aq ) + OH - ( aq )concentration

49. 18-49 SAMPLE PROBLEM 18.9(continued) K b = 5.9 x 10 -4 = [(CH 3 ) 2 NH 2 + ][OH - ] [(CH 3 ) 2 NH] 5.9 x 10 -4 = (x) 1.5 M x = 3.0 x 10 -2 M = [OH - ] Check assumption:[3.0 x 10 -2 M / 1.5 M] x 100 = 2% (error is < 5%; thus, assumption is justified) [H 3 O + ] = K w /[OH - ] = 1.0 x 10 -14 /3.0 x 10 -2 = 3.3 x 10 -13 M pH = -log (3.3 x 10 -13 ) = 12.48

50. 18-50 Anions of Weak Acids as Weak Bases A - (aq) + H 2 O(l) HA(aq) + OH - (aq) F - (aq) + H 2 O(l) HF(aq) + OH - (aq) K b = ([HF][OH - ]) / [F - ] Rationale for the basicity of A - (aq): e.g., 1 M NaF Acidity is determined by [OH - ] generated from the F - reaction and [H 3 O + ] generated from the auto-ionization of water; since [OH - ] >> [H 3 O + ], the solution is basic. K a x K b = K w For a conjugate acid-base pair:

51. 18-51 SAMPLE PROBLEM 18.10 Determining the pH of a solution of A - PROBLEM:Sodium acetate (CH 3 COONa, abbreviated NaAc) has applications in photographic development and textile dyeing. What is the pH of a 0.25 M aqueous solution of NaAc? K a of acetic acid (HAc) is 1.8 x 10 - 5. SOLUTION: PLAN:Sodium salts are soluble in water so [Ac - ] = 0.25 M. Use K a to find K b. initial0.25-00 change-x+x+x+x- equilibrium-0.25 - xxx Ac - ( aq ) + H 2 O( l ) HAc( aq ) + OH - ( aq )concentration K b = [HAc][OH - ] [Ac - ] = KwKw KaKa = 5.6 x 10 - 10 K b = 1.0 x 10 - 14 1.8 x 10 - 5

52. 18-52 SAMPLE PROBLEM 18.10(continued) K b = [HAc][OH - ] [Ac - ] [Ac - ] = 0.25 M - x ≈ 0.25 M (since K b is small) 5.6 x 10 - 10 ≈ x 2 /0.25 M x ≈ 1.2 x 10 - 5 M = [OH - ] Check assumption:[1.2 x 10 - 5 M / 0.25 M] x 100 = 4.8 x 10 - 3 % [H 3 O + ] = K w /[OH - ] = 1.0 x 10 - 14 /1.2 x 10 - 5 = 8.3 x 10 - 10 M pH = -log (8.3 x 10 - 10 M) = 9.08

53. 18-53 Molecular Properties and Acid Strength Non-metal hydrides: two factors determine acid strength, namely, the electronegativity of the central non-metal atom E and the strength of the E-H bond Non-metal hydride acid strength increases across a period (E electronegativity effect) Non-metal hydride acid strength increases down a group (E-H bond strength effect)

54. 18-54 Figure 18.12 The effect of atomic and molecular properties on non-metal hydride acidity 6A(16) H2OH2O H2SH2S H 2 Se H 2 Te 7A(17) HF HCl HBr HI electronegativity increases, acidity increases bond strength decreases, acidity increases Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display.

55. 18-55 HOIHOBrHOCl>> HO O O O << Figure 18.13 The relative strengths of oxoacids  HOCl   Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display. Electronegativity increases, acidity increases number of O atoms increases, acidity increases

56. 18-56 Acidity of Hydrated Metal Ions Aqueous solutions of certain metal ions are acidic because the hydrated metal ion transfers an H + ion to water. Generalized Reactions M(NO 3 ) n (s) + xH 2 O(l) M(H 2 O) x n + (aq) + nNO 3 - (aq) M(H 2 O) x n + (aq) + H 2 O(l) M(H 2 O) x - 1 OH (n - 1) + (aq) + H 3 O + (aq)

57. 18-57 Al(H 2 O) 5 OH 2 + Al(H 2 O) 6 3 + Figure 18.14 The acidic behavior of the hydrated Al 3+ ion H2OH2O H3O+H3O+ electron density drawn toward Al 3 + solvent H 2 O acts as base Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display.

58. 18-58 high charge and small size enhance acidity

59. 18-59 Acid-Base Properties of Salt Solutions A. Salts That Yield Neutral Solutions: the anion of a strong acid and the cation of a strong base (the ions do not react with water) The anion of a strong acid is a much weaker base than water (HNO 3 ). The cation of a strong base only becomes hydrated (NaOH). HNO 3 (l) + H 2 O(l) NO 3 - (aq) + H 3 O + (aq) NaOH(s) Na + (aq) + OH - (aq)

60. 18-60 B. Salts That Yield Acidic Solutions: (1) the anion of a strong acid and the cation of a weak base (the cation acts as a weak acid); (2) small, highly charged metal ions; (3) cations of strong bases and anions of polyprotic acids with another ionizable proton. NH 4 Cl(s) NH 4 + (aq) + Cl - (aq) NH 4 + (aq) + H 2 O(l) NH 3 (aq) + H 3 O + (aq) Case 1 Fe(NO 3 ) 3 (s) + 6H 2 O(l) Fe(H 2 O) 6 3 + (aq) + 3NO 3 - (aq) Fe(H 2 O) 6 3 + (aq) + H 2 O(l) Fe(H 2 O) 5 OH 2 + (aq) + H 3 O + (aq) Case 2 NaH 2 PO 4 (s) Na + (aq) + H 2 PO 4 - (aq) H 2 PO 4 - (aq) + H 2 O(l) HPO 4 2 - (aq) + H 3 O + (aq) Case 3

61. 18-61 C. Salts That Yield Basic Solutions: the anion of a weak acid and the cation of a strong base (the anion acts as a weak base) CH 3 COONa(s) Na + (aq) + CH 3 COO - (aq) CH 3 COO - (aq) + H 2 O(l) CH 3 COOH(aq) + OH - (aq)

62. 18-62

63. 18-63 SAMPLE PROBLEM 18.11 Predicting the relative acidity of salt solutions PROBLEM:Predict whether aqueous solutions of the following compounds are acidic, basic, or neutral (write an equation for the reaction of the appropriate ion with water to explain pH effect). (a) potassium perchlorate, KClO 4 (b) sodium benzoate, C 6 H 5 COONa (c) chromium trichloride, CrCl 3 (d) sodium hydrogen sulfate, NaHSO 4 SOLUTION: PLAN:Consider the acid-base nature of the anions and cations. Strong acid-strong base combinations produce a neutral solution; strong acid-weak base, acidic; weak acid-strong base, basic. (a) The ions are K + and ClO 4 -, which come from a strong base (KOH) and a strong acid (HClO 4 ). The salt solution will be neutral. (b) Na + comes from the strong base NaOH while C 6 H 5 COO - is the anion of a weak organic acid. The salt solution will be basic. (c) Cr 3 + is a small cation with a large + charge, so its hydrated form will react with water to produce H 3 O +. Cl - comes from the strong acid HCl. The salt solution will be acidic. (d) Na + comes from a strong base. HSO 4 - can react with water to form H 3 O +. The salt solution will be acidic.

64. 18-64 Salts of Weakly Acidic Cations and Weakly Basic Anions Both components react with water! Acidity is determined by the relative acid strength and base strength of the separated ions. example: NH 4 HS NH 4 + (aq) + H 2 O(l) NH 3 (aq) + H 3 O + (aq) HS - (aq) + H 2 O(l) H 2 S(aq) + OH - (aq) K a (NH 4 + ) = 5.7 x 10 - 10 K b (HS - ) = 1 x 10 - 7 Since K b > K a, the solution is basic.

65. 18-65 SAMPLE PROBLEM 18.12Predicting the relative acidity of salt solutions from K a and K b of the ions PROBLEM:Determine whether an aqueous solution of zinc formate, Zn(HCOO) 2, is acidic, basic, or neutral. SOLUTION: PLAN:Both Zn 2 + and HCOO - come from weak conjugates. In order to find the relatively acidity, write the dissociation reactions and use the information in Tables 18.2 and 18.7. K a Zn(H 2 O) 6 2 + = 1 x 10 - 9 K a HCOOH = 1.8 x 10 - 4 ; K b = K w /K a = 1.0 x 10 -14 /1.8 x 10 - 4 = 5.6 x 10 - 11 K a for Zn(H 2 O) 6 2 + > K b HCOO - ; the solution is acidic. Zn(H 2 O) 6 2 + ( aq ) + H 2 O( l ) Zn(H 2 O) 5 OH + ( aq ) + H 3 O + ( aq ) HCOO - ( aq ) + H 2 O( l ) HCOOH( aq ) + OH - ( aq )

66. 18-66 The Leveling Effect In water, the strongest acid possible is H 3 O + and the strongest base possible is OH -. Any acid stronger than H 3 O + donates its proton to H 2 O, and any base stronger than OH - accepts a proton from H 2 O; thus, water exerts a leveling effect (levels the strengths of all strong acids and bases). To rank strong acids: must dissolve in a solvent that is a weaker base than water (i.e., one that accepts their protons less readily). HCl(g) + CH 3 COOH(l) Cl - (acet) + CH 3 COOH 2 + (acet) HBr(g) + CH 3 COOH(l) Br - (acet) + CH 3 COOH 2 + (acet) HI(g) + CH 3 COOH(l) I - (acet) + CH 3 COOH 2 + (acet) K HI > K HBr > K HCl

67. 18-67 Three Definitions of Acids and Bases The Arrhenius Definition The Brønsted-Lowry Definition The Lewis Definition

68. 18-68 acidbaseadduct An acid is an electron-pair acceptor.A base is an electron-pair donor. M2+M2+ H 2 O(l) M(H 2 O) 4 2 + (aq) adduct The Lewis Acid-Base Definition The adduct contains a new covalent bond.

69. 18-69 Lewis Acids with Electron-Deficient Atoms Lewis Acids with Polar Multiple Bonds Metal Cations as Lewis Acids ROR’ + AlCl 3 R-O-R’ AlCl 3 baseacid adduct SO 2 + H 2 O H 2 SO 3 M 2 + + 4H 2 O M(H 2 O) 4 2 + baseacidadduct baseacid Metal ions act as Lewis acids when dissolved in water. Lewis acids contain (or can generate) a vacant orbital.

70. 18-70 Figure 18.15 Mg 2 + ion as a Lewis acid in the chlorophyll molecule Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display. The ion accepts electron pairs from four nitrogen atoms comprising part of the chlorophyll molecule

71. 18-71 SAMPLE PROBLEM 18.13 Identifying Lewis acids and bases PROBLEM:Identify the Lewis acids and Lewis bases in the following reactions: (a) H + + OH - H 2 O (b) Cl - + BCl 3 BCl 4 - (c) K + + 6H 2 O K(H 2 O) 6 + SOLUTION: PLAN:Look for electron pair acceptors (acids) and donors (bases). (a) H + + OH - H 2 O acceptor donor (b) Cl - + BCl 3 BCl 4 - donor acceptor (c) K + + 6H 2 O K(H 2 O) 6 + acceptor donor