1. Acids and Bases Chapter 14 Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display.

2. Acids Have a sour taste. Vinegar owes its taste to acetic acid. Citrus fruits contain citric acid. React with certain metals to produce hydrogen gas. React with carbonates and bicarbonates to produce carbon dioxide gas Have a bitter taste. Feel slippery. Many soaps contain bases. Bases 4.3

3. Arrhenius acid is a substance that produces H + (H 3 O + ) in water A Brønsted acid is a proton donor A Lewis acid is a substance that can accept a pair of electrons A Lewis base is a substance that can donate a pair of electrons Definition of An Acid H+H+ H O H + OH - acidbase N H H H H+H+ + acidbase 15.12 N H H H H +

4. Arrhenius acid is a substance that produces H + (H 3 O + ) in water Arrhenius base is a substance that produces OH - in water 4.3

5. A Brønsted acid is a proton donor A Brønsted base is a proton acceptor acidbaseacidbase 15.1 acid conjugate base base conjugate acid

6. Lewis Acids and Bases N H H H acidbase F B F F + F F N H H H No protons donated or accepted! 15.12

7. Water as an Acid and a Base Water is amphoteric (it can behave either as an acid or a base). H 2 O + H 2 O  H 3 O + + OH  conj conj acid 1 base 2 acid 2 base 1 K w = 1 × 10  14 at 25°C Copyright©2000 by Houghton Mifflin Company. All rights reserved. 7

8. O H H+ O H H O H HH O H - + [] + Acid-Base Properties of Water H 2 O (l) H + (aq) + OH - (aq) H 2 O + H 2 O H 3 O + + OH - acid conjugate base base conjugate acid 15.2 autoionization of water

9. H 2 O (l) H + (aq) + OH - (aq) The Ion Product of Water K c = [H + ][OH - ] [H 2 O] [H 2 O] = constant K c [H 2 O] = K w = [H + ][OH - ] The ion-product constant (K w ) is the product of the molar concentrations of H + and OH - ions at a particular temperature. At 25 0 C K w = [H + ][OH - ] = 1.0 x 10 - 14 [H + ] = [OH - ] [H + ] > [OH - ] [H + ] < [OH - ] Solution Is neutral acidic basic 15.2

10. What is the concentration of OH - ions in a HCl solution whose hydrogen ion concentration is 1.3 M? K w = [H + ][OH - ] = 1.0 x 10 - 14 [H + ] = 1.3 M [OH - ] = KwKw [H + ] 1 x 10 -14 1.3 = = 7.7 x 10 -15 M 15.2

11. pH – A Measure of Acidity pH = - log [H + ] [H + ] = [OH - ] [H + ] > [OH - ] [H + ] < [OH - ] Solution Is neutral acidic basic [H + ] = 1 x 10 -7 [H + ] > 1 x 10 -7 [H + ] < 1 x 10 -7 pH = 7 pH < 7 pH > 7 At 25 0 C pH[H + ] 15.3

12. pOH = -log [OH - ] [H + ][OH - ] = K w = 1.0 x 10 -14 -log [H + ] – log [OH - ] = 14.00 pH + pOH = 14.00

13. The pH of rainwater collected in a certain region of the northeastern United States on a particular day was 4.82. What is the H + ion concentration of the rainwater? pH = - log [H + ] [H + ] = 10 -pH = 10 -4.82 = 1.5 x 10 -5 M The OH - ion concentration of a blood sample is 2.5 x 10 -7 M. What is the pH of the blood? pH + pOH = 14.00 pOH = -log [OH - ]= -log (2.5 x 10 -7 )= 6.60 pH = 14.00 – pOH = 14.00 – 6.60 = 7.40 15.3

14. pH calculations – Solving for H+ If the pH of Coke is 3.12, [H + ] = ??? Because pH = - log [H + ] then - pH = log [H + ] - pH = log [H + ] Take antilog (10 x ) of both sides and get 10 -pH = [H + ] [H + ] = 10 -3.12 = 7.6 x 10 -4 M *** to find antilog on your calculator, look for “Shift” or “2 nd function” and then the log button *** to find antilog on your calculator, look for “Shift” or “2 nd function” and then the log button

15. Strong Electrolyte – 100% dissociation NaCl (s) Na + (aq) + Cl - (aq) H2OH2O Weak Electrolyte – not completely dissociated CH 3 COOH CH 3 COO - (aq) + H + (aq) Strong Acids are strong electrolytes HCl (aq) + H 2 O (l) H 3 O + (aq) + Cl - (aq) HNO 3 (aq) + H 2 O (l) H 3 O + (aq) + NO 3 - (aq) HClO 4 (aq) + H 2 O (l) H 3 O + (aq) + ClO 4 - (aq) H 2 SO 4 (aq) + H 2 O (l) H 3 O + (aq) + HSO 4 - (aq) 15.4

16. HF (aq) + H 2 O (l) H 3 O + (aq) + F - (aq) Weak Acids are weak electrolytes HNO 2 (aq) + H 2 O (l) H 3 O + (aq) + NO 2 - (aq) HSO 4 - (aq) + H 2 O (l) H 3 O + (aq) + SO 4 2- (aq) H 2 O (l) + H 2 O (l) H 3 O + (aq) + OH - (aq) Strong Bases are strong electrolytes NaOH (s) Na + (aq) + OH - (aq) H2OH2O KOH (s) K + (aq) + OH - (aq) H2OH2O Ba(OH) 2 (s) Ba 2+ (aq) + 2OH - (aq) H2OH2O 15.4

17. F - (aq) + H 2 O (l) OH - (aq) + HF (aq) Weak Bases are weak electrolytes NO 2 - (aq) + H 2 O (l) OH - (aq) + HNO 2 (aq) Conjugate acid-base pairs: The conjugate base of a strong acid has no measurable strength. H 3 O + is the strongest acid that can exist in aqueous solution. The OH - ion is the strongest base that can exist in aqeous solution. 15.4

19. Strong AcidWeak Acid 15.4

20. HA (aq) + H 2 O (l) H 3 O + (aq) + A - (aq) Weak Acids (HA) and Acid Ionization Constants HA (aq) H + (aq) + A - (aq) K a = [H + ][A - ] [HA] K a is the acid ionization constant KaKa weak acid strength 15.5

22. What is the pH of a 0.5 M HF solution (at 25 0 C)? HF (aq) H + (aq) + F - (aq) K a = [H + ][F - ] [HF] = 7.1 x 10 -4 HF (aq) H + (aq) + F - (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 K a ≈ x2x2 0.50 = 7.1 x 10 -4 0.50 – x ≈ 0.50K a << 1 x 2 = 3.55 x 10 -4 x = 0.019 M [H + ] = [F - ] = 0.019 M pH = -log [H + ] = 1.72 [HF] = 0.50 – x = 0.48 M 15.5

23. When can I use the approximation? 0.50 – x ≈ 0.50K a << 1 When x is less than 5% of the value from which it is subtracted. x = 0.019 0.019 M 0.50 M x 100% = 3.8% Less than 5% Approximation ok. What is the pH of a 0.05 M HF solution (at 25 0 C)? K a ≈ x2x2 0.05 = 7.1 x 10 -4 x = 0.006 M 0.006 M 0.05 M x 100% = 12% More than 5% Approximation not ok. Must solve for x exactly using quadratic equation or method of successive approximation. 15.5

24. Solving weak acid ionization problems: 1.Identify the major species that can affect the pH. In most cases, you can ignore the autoionization of water. Ignore [OH - ] because it is determined by [H + ]. 2.Use ICE to express the equilibrium concentrations in terms of single unknown x. 3.Write K a in terms of equilibrium concentrations. Solve for x by the approximation method. If approximation is not valid, solve for x exactly. 4.Calculate concentrations of all species and/or pH of the solution. 15.5

25. What is the pH of a 0.122 M monoprotic acid whose K a is 5.7 x 10 -4 ? HA (aq) H + (aq) + A - (aq) Initial (M) Change (M) Equilibrium (M) 0.1220.00 -x-x+x+x 0.122 - x 0.00 +x+x xx K a = x2x2 0.122 - x = 5.7 x 10 -4 Ka ≈ Ka ≈ x2x2 0.122 = 5.7 x 10 -4 0.122 – x ≈ 0.122K a << 1 x 2 = 6.95 x 10 -5 x = 0.0083 M 0.0083 M 0.122 M x 100% = 6.8% More than 5% Approximation not ok. 15.5

26. K a = x2x2 0.122 - x = 5.7 x 10 -4 x 2 + 0.00057x – 6.95 x 10 -5 = 0 ax 2 + bx + c =0 -b ± b 2 – 4ac √ 2a2a x = x = 0.0081x = - 0.0081 HA (aq) H + (aq) + A - (aq) Initial (M) Change (M) Equilibrium (M) 0.1220.00 -x-x+x+x 0.122 - x 0.00 +x+x xx [H + ] = x = 0.0081 M pH = -log[H + ] = 2.09 15.5

27. percent ionization = Ionized acid concentration at equilibrium Initial concentration of acid x 100% For a monoprotic acid HA Percent ionization = [H + ] [HA] 0 x 100% [HA] 0 = initial concentration 15.5

28. What is the pH of a 2 x 10 -3 M HNO 3 solution? HNO 3 is a strong acid – 100% dissociation. HNO 3 (aq) + H 2 O (l) H 3 O + (aq) + NO 3 - (aq) pH = -log [H + ] = -log [H 3 O + ] = -log(0.002) = 2.7 Start End 0.002 M 0.0 M What is the pH of a 1.8 x 10 -2 M Ba(OH) 2 solution? Ba(OH) 2 is a strong base – 100% dissociation. Ba(OH) 2 (s) Ba 2+ (aq) + 2OH - (aq) Start End 0.018 M 0.036 M0.0 M pH = 14.00 – pOH = 14.00 + log(0.036) = 12.6 15.4

29. Weak Bases

30. NH 3 (aq) + H 2 O (l) NH 4 + (aq) + OH - (aq) Weak Bases and Base Ionization Constants K b = [NH 4 + ][OH - ] [NH 3 ] K b is the base ionization constant KbKb weak base strength 15.6 Solve weak base problems like weak acids except solve for [OH-] instead of [H + ].

31. Equilibrium Constants for Weak Bases Weak base has K b < 1 Leads to small [OH - ] and a pH of 12 - 7

32. Equilibria Involving A Weak Base You have 0.010 M NH 3. Calc. the pH. NH 3 + H 2 O ↔ NH 4 + + OH - NH 3 + H 2 O ↔ NH 4 + + OH - K b = 1.8 x 10 -5 Step 1. Define equilibrium concs. in ICE table [NH 3 ][NH 4 + ][OH - ] [NH 3 ][NH 4 + ][OH - ]initialchangeequilib 0.01000 -x+x+x 0.010 - xx x

33. Equilibria Involving A Weak Base You have 0.010 M NH 3. Calc. the pH. NH 3 + H 2 O  NH 4 + + OH - NH 3 + H 2 O  NH 4 + + OH - K b = 1.8 x 10 -5 Step 2. Solve the equilibrium expression Assume x is small, so x = [OH - ] = [NH 4 + ] = 4.2 x 10 -4 M x = [OH - ] = [NH 4 + ] = 4.2 x 10 -4 M and [NH 3 ] = 0.010 - 4.2 x 10 -4 ≈ 0.010 M The approximation is valid!

34. Equilibria Involving A Weak Base You have 0.010 M NH 3. Calc. the pH. NH 3 + H 2 O  NH 4 + + OH - NH 3 + H 2 O  NH 4 + + OH - K b = 1.8 x 10 -5 Step 3. Calculate pH [OH - ] = 4.2 x 10 -4 M so pOH = - log [OH - ] = 3.37 Because pH + pOH = 14, pH = 10.63

35. 15.6

36. 15.7 Ionization Constants of Conjugate Acid-Base Pairs HA (aq) H + (aq) + A - (aq) A - (aq) + H 2 O (l) OH - (aq) + HA (aq) KaKa KbKb H 2 O (l) H + (aq) + OH - (aq) KwKw K a K b = K w Weak Acid and Its Conjugate Base Ka =Ka = KwKw KbKb Kb =Kb = KwKw KaKa

37. 15.8

38. Molecular Structure and Acid Strength H X H + + X - The stronger the bond The weaker the acid HF << HCl < HBr < HI 15.9

39. Molecular Structure and Acid Strength Z O HZ O-O- + H + -- ++ The O-H bond will be more polar and easier to break if: Z is very electronegative or Z is in a high oxidation state 15.9

40. Molecular Structure and Acid Strength 1. Oxoacids having different central atoms (Z) that are from the same group and that have the same oxidation number. Acid strength increases with increasing electronegativity of Z H O Cl O O H O Br O O Cl is more electronegative than Br HClO 3 > HBrO 3 15.9

41. Molecular Structure and Acid Strength 2. Oxoacids having the same central atom (Z) but different numbers of attached groups. Acid strength increases as the oxidation number of Z increases. HClO 4 > HClO 3 > HClO 2 > HClO 15.9

42. Acid-Base Properties of Salts EXAMINE CATION OF SALT: If CATION came from a strong BASE it would make a neutral soln weak BASE it would make an Acidic soln EXAMINE ANION OF SALT: If ANION came from a strong ACID it would make a neutral soln weak ACID it would make an Basic soln

43. Acid-Base Properties of Salts Copyright©2000 by Houghton Mifflin Company. All rights reserved. 43 Salt comes from SA & SB makes Neutral Salt Salt comes from SA & WB makes Acidic Salt Salt comes from WA & SB makes Basic Salt Salt comes from WA & WB makes ? Salt need to check Ka vs Kb

44. CATIONANIONACIDIC OR BASIC EXAMPLE NEUTRAL NaCl NEUTRALCONJUGATE BASE OF A WEAK ACID BASICNaF CONJUGATE ACID OF A WEAK BASE NEUTRALACIDICNH 4 Cl CONJUGATE ACID OF A WEAK BASE CONJUGATE BASE OF A WEAK ACID DEPENDS ON Ka & Kb VALUES Al2(SO4)3

45. Acid-Base Properties of Salts Neutral Solutions: Salts containing an alkali metal or alkaline earth metal ion (except Be 2+ ) and the conjugate base of a strong acid (e.g. Cl -, Br -, and NO 3 - ). NaCl (s) Na + (aq) + Cl - (aq) H2OH2O Basic Solutions: Salts derived from a strong base and a weak acid. NaCH 3 COO (s) Na + (aq) + CH 3 COO - (aq) H2OH2O CH 3 COO - (aq) + H 2 O (l) CH 3 COOH (aq) + OH - (aq) 15.10

46. Acid-Base Properties of Salts Acid Solutions: Salts derived from a strong acid and a weak base. NH 4 Cl (s) NH 4 + (aq) + Cl - (aq) H2OH2O NH 4 + (aq) NH 3 (aq) + H + (aq) Salts with small, highly charged metal cations (e.g. Al 3+, Cr 3+, and Be 2+ ) and the conjugate base of a strong acid. Al(H 2 O) 6 (aq) Al(OH)(H 2 O) 5 (aq) + H + (aq) 3+2+ 15.10

47. Acid-Base Properties of Salts Solutions in which both the cation and the anion hydrolyze: K b for the anion > K a for the cation, solution will be basic K b for the anion < K a for the cation, solution will be acidic K b for the anion ≈ K a for the cation, solution will be neutral 15.10

48. Polyprotic Acids... can furnish more than one proton (H + ) to the solution. Copyright©2000 by Houghton Mifflin Company. All rights reserved. 48

49. Calculating Concentrations of Species in a Solution of a Diprotic Acid Ascorbic acid (vitamin C) is a diprotic acid, H 2 C 6 H 6 O 6. What is the pH of a 0.10 M solution? What is the concentration of ascorbate ion, C 6 H 6 O 6 2- ? The acid ionization constants are Ka 1 = 7.9 x 10 -5 and Ka 2 = 1.6 x 10 -12.

50. Consider first ionization H 2 Asc(aq) H+(aq) + Hasc-(aq) I 0.10 0 0 C -x +x +x E 0.10 - x x x [H3O+] [HAsc-] = Ka1 [H 2 Asc] x 2 = 7.9 x 10-5 0.10 - x x = 2.8 x 10 -3 = 0.0028 = [H+] = [HAsc-] pH = -log [H+] = -log(0.0028) = 2.55

51. Hasc- (aq) H+ (aq) + Asc- (aq) I 0.0028 0.0028 0 C -x +x +x E 0.0028- x 0.0028+ x x [H3O+][Asc2-] = Ka 2 [HAsc-] (0.0028) x = 1.6 x 10 -12 0.0028 x = 1.6 x 10 -12 = [H+] The 2 nd [H+] always equals Ka 2

52. From 1 st Ionization we get [H+] = 2.8 x 10 -3 M From 2 nd Ionization we get [H+] = 1.6 x 10 -12 M If you add together: [H+] = 2.8000000016 x 10 -3 M Same pH, 1 st Ionization always contributes most [H+]

53. OXIDES Acidic Oxides (Acid Anhydrides):  O  X bond is strong and covalent. SO 2, NO 2, CrO 3 Basic Oxides (Basic Anhydrides):  O  X bond is ionic. K 2 O, CaO

54. Oxides of the Representative Elements In Their Highest Oxidation States 15.11 CO 2 (g) + H 2 O (l) H 2 CO 3 (aq) N 2 O 5 (g) + H 2 O (l) 2HNO 3 (aq)

55. Chemistry In Action: Antacids and the Stomach pH Balance NaHCO 3 (aq) + HCl (aq) NaCl (aq) + H 2 O (l) + CO 2 (g) Mg(OH) 2 (s) + 2HCl (aq) MgCl 2 (aq) + 2H 2 O (l)