1. Chapter 14 Acids and Bases 2009, Prentice Hall

2. 14.2-14.4 Day One Tro's Introductory Chemistry, Chapter 14 2

3. 3 Types of Electrolytes Salts are water-soluble ionic compounds. All strong electrolytes. Acids form H +1 ions in water solution. Bases combine with H +1 ions in water solution. Increases the OH -1 concentration.  May either directly release OH -1 or pull H +1 off H 2 O.

4. Tro's Introductory Chemistry, Chapter 14 4 Properties of Acids Sour taste. React with “active” metals. I.e., Al, Zn, Fe, but not Cu, Ag or Au. 2 Al + 6 HCl  AlCl 3 + 3 H 2 Corrosive. React with carbonates, producing CO 2. Marble, baking soda, chalk, limestone. CaCO 3 + 2 HCl  CaCl 2 + CO 2 + H 2 O Change color of vegetable dyes. Blue litmus turns red. React with bases to form ionic salts.

5. Tro's Introductory Chemistry, Chapter 14 5 Common Acids

6. Tro's Introductory Chemistry, Chapter 14 6 Structures of Acids Binary acids have acid hydrogens attached to a nonmetal atom. HCl, HF Hydrofluoric acid

7. Tro's Introductory Chemistry, Chapter 14 7 Structure of Acids Oxyacids have acid hydrogens attached to an oxygen atom. H 2 SO 4, HNO 3

8. Tro's Introductory Chemistry, Chapter 14 8 Structure of Acids Carboxylic acids have COOH group. HC 2 H 3 O 2, H 3 C 6 H 5 O 3 Only the first H in the formula is acidic. The H is on the COOH.

9. Tro's Introductory Chemistry, Chapter 14 9 Properties of Bases Also known as alkalis. Taste bitter. Solutions feel slippery. Change color of vegetable dyes. Different color than acid. Red litmus turns blue. React with acids to form ionic salts. Neutralization.

10. Tro's Introductory Chemistry, Chapter 14 10 Common Bases

11. Tro's Introductory Chemistry, Chapter 14 11 Structure of Bases Most ionic bases contain OH ions. NaOH, Ca(OH) 2 Some contain CO 3 2- ions. CaCO 3 NaHCO 3 Molecular bases contain structures that react with H +. Mostly amine groups.

12. Tro's Introductory Chemistry, Chapter 14 12 Arrhenius Theory Bases dissociate in water to produce OH - ions and cations. Ionic substances dissociate in water. NaOH(aq) → Na + (aq) + OH – (aq) Acids ionize in water to produce H + ions and anions. Because molecular acids are not made of ions, they cannot dissociate. They must be pulled apart, or ionized, by the water. HCl(aq) → H + (aq) + Cl – (aq) In formula, ionizable H is written in front. HC 2 H 3 O 2 (aq) → H + (aq) + C 2 H 3 O 2 – (aq)

13. Tro's Introductory Chemistry, Chapter 14 13 Arrhenius Theory, Continued HCl ionizes in water, producing H + and Cl – ions. NaOH dissociates in water, producing Na + and OH – ions.

14. Tro's Introductory Chemistry, Chapter 14 14 Arrhenius Acid–Base Reactions The H + from the acid combines with the OH - from the base to make a molecule of H 2 O. The cation from the base combines with the anion from the acid to make a salt. acid + base → salt + water HCl(aq) + NaOH(aq) → NaCl(aq) + H 2 O(l)

15. Tro's Introductory Chemistry, Chapter 14 15 Problems with Arrhenius Theory Does not explain why molecular substances, like NH 3, dissolve in water to form basic solutions—even though they do not contain OH – ions. Does not explain acid–base reactions that do not take place in aqueous solution. H + ions do not exist in water. Acid solutions contain H 3 O + ions. H + = a proton! H 3 O + = hydronium ions.

16. Tro's Introductory Chemistry, Chapter 14 16 Arrow Conventions Chemists commonly use two kinds of arrows in reactions to indicate the degree of completion of the reactions. A single arrow indicates that all the reactant molecules are converted to product molecules at the end. A double arrow indicates that the reaction stops when only some of the reactant molecules have been converted into products.  in these notes.

17. Tro's Introductory Chemistry, Chapter 14 17 Brønsted–Lowry Theory A Brønsted-Lowry acid–base reaction is any reaction in which an H + is transferred. Does not have to take place in aqueous solution. Broader definition than Arrhenius. Acid is H + donor; base is H + acceptor. Since H + is a proton, acid is a proton donor and base is a proton acceptor. Base structure must contain an atom with an unshared pair of electrons to bond to H +. In the reaction, the acid molecule gives an H + to the base molecule. H–A + :B  :A – + H–B +

18. Tro's Introductory Chemistry, Chapter 14 18 Comparing Arrhenius Theory and Brønsted–Lowry Theory Arrhenius theory HCl(aq)  H + (aq) + Cl − (aq) HF(aq)  H + (aq) + F − (aq) NaOH(aq)  Na + (aq) + OH − (aq) NH 4 OH(aq)  NH 4 + (aq) + OH − (aq) Brønsted – Lowry theory HCl(aq) + H 2 O(l)  Cl − (aq) + H 3 O + (aq) HF(aq) + H 2 O(l)  F − (aq) + H 3 O + (aq) NaOH(aq) + H 2 O(l)  Na + (aq) + OH − (aq) + H 2 O(l) NH 3 (aq) + H 2 O(l)  NH 4 + (aq) + OH − (aq)

19. Tro's Introductory Chemistry, Chapter 14 19 Amphoteric Substances Amphoteric substances can act as either an acid or a base. They have both transferable H and an atom with a lone pair. HCl(aq) is acidic because HCl transfers an H + to H 2 O, forming H 3 O + ions. Water acts as base, accepting H +. HCl(aq) + H 2 O(l) → Cl – (aq) + H 3 O + (aq) NH 3 (aq) is basic because NH 3 accepts an H + from H 2 O, forming OH – (aq). Water acts as acid, donating H +. NH 3 (aq) + H 2 O(l)  NH 4 + (aq) + OH – (aq)

20. Tro's Introductory Chemistry, Chapter 14 20 Brønsted–Lowry Acid–Base Reactions One of the advantages of the Brønsted–Lowry theory is that it allows reactions to be reversible. H–A + :B → :A – + H–B + The original base has an extra H + after the reaction, so it can act as an acid in the reverse process. The original acid has a lone pair of electrons after the reaction, so it can act as a base in the reverse process. :A – + H–B + → H–A + :B A double arrow, , is usually used to indicate a process that is reversible.

21. Tro's Introductory Chemistry, Chapter 14 21 Conjugate Pairs In a Brønsted-Lowry acid-base reaction, the original base becomes an acid in the reverse reaction, and the original acid becomes a base in the reverse process. Each reactant and the product it becomes is called a conjugate pair. The original base becomes the conjugate acid; the original acid becomes the conjugate base.

22. Tro's Introductory Chemistry, Chapter 14 22 Brønsted–Lowry Acid–Base Reactions H–A + :B  :A – +H–B + AcidBaseConjugate Conjugate base acid HCHO 2 + H 2 O  CHO 2 – +H 3 O + Acid BaseConjugate Conjugate base acid H 2 O + NH 3  HO – +NH 4 + Acid BaseConjugate Conjugate base acid

23. Tro's Introductory Chemistry, Chapter 14 23 Conjugate Pairs In the reaction H 2 O + NH 3  HO – + NH 4 +: H 2 O and HO – constitute an acid/conjugate–base pair. NH 3 and NH 4 + constitute a base/conjugate – acid pair.

24. Tro's Introductory Chemistry, Chapter 14 24 Example—Identify the Brønsted – Lowry Acids and Bases and Their Conjugates in the Reaction. H 2 SO 4 + H 2 O  HSO 4 – +H 3 O +

25. Tro's Introductory Chemistry, Chapter 14 25 Practice—Write the Formula for the Conjugate Acid of the Following: H 2 O NH 3 CO 3 2− H 2 PO 4 1−

26. Tro's Introductory Chemistry, Chapter 14 26 Practice—Write the Formula for the Conjugate Base of the Following: H 2 O NH 3 CO 3 2− H 2 PO 4 1−

27. Tro's Introductory Chemistry, Chapter 14 27 Practice—Write the Equations for the Following Reacting with Water and Acting as a Monoprotic Acid. Label the Conjugate Acid and Base. HBr HSO 4 -1

28. Tro's Introductory Chemistry, Chapter 14 28 Practice—Write the Equations for the Following Reacting with Water and Acting as a Monoprotic- Accepting Base. Label the Conjugate Acid and Base. I − CO 3 2−

29. 14.5-14.7 Day Two Tro's Introductory Chemistry, Chapter 14 29

30. Tro's Introductory Chemistry, Chapter 14 30 Neutralization Reactions H + + OH -  H 2 O Acid + base  salt + water Double-displacement reactions. Salt = cation from base + anion from acid. Cation and anion charges stay constant. H 2 SO 4 + Ca(OH) 2 → CaSO 4 + 2 H 2 O Some neutralization reactions are gas evolving, where H 2 CO 3 decomposes into CO 2 and H 2 O. H 2 SO 4 + 2 NaHCO 3 → Na 2 SO 4 + 2 H 2 O + 2 CO 2

31. Tro's Introductory Chemistry, Chapter 14 31 Example 14.2—Write the Equation for the Reaction of Aqueous Nitric Acid with Aqueous Calcium Hydroxide.

32. Tro's Introductory Chemistry, Chapter 14 32 Practice – Complete Each Reaction. Ca(OH) 2 (s) + H 2 SO 3 (aq)  HClO 3 (aq) + Pb(OH) 4 (s)  CaCO 3 (s) + HNO 3 (aq)  Mg(HCO 3 ) 2 (aq) + HC 2 H 3 O 2 (aq) 

33. Tro's Introductory Chemistry, Chapter 14 33 Acid Reactions: Acids React with Metals Acids react with many metals. But not all!! When acids react with metals, they produce a salt and hydrogen gas. 3 H 2 SO 4 (aq) + 2 Al(s) → Al 2 (SO 4 ) 3 (aq) + 3 H 2 (g)

34. Tro's Introductory Chemistry, Chapter 14 34 Titration Titration is a technique that uses reaction stoichiometry to determine the concentration of an unknown solution. Titrant (unknown solution) is added from a buret. Indicators are chemicals that are added to help determine when a reaction is complete. The endpoint of the titration occurs when the reaction is complete.

35. Tro's Introductory Chemistry, Chapter 14 35 Titration, Continued

36. Tro's Introductory Chemistry, Chapter 14 36 Acid–Base Titration The base solution is the titrant in the buret. As the base is added to the acid, the H + reacts with the OH – to form water. But there is still excess acid present, so the color does not change. At the titration’s endpoint, just enough base has been added to neutralize all the acid. At this point, the indicator changes color.

37. Tro's Introductory Chemistry, Chapter 14 37 Example 14.4—What Is the Molarity of an HCl Solution if 10.00 mL Is required to Titrate 12.54 mL of 0.100 M NaOH? NaOH(aq) + HCl(aq)  NaCl(aq) + H 2 O(l)

38. Tro's Introductory Chemistry, Chapter 14 38 Practice—What Is the Molarity of a Ba(OH) 2 Solution if 37.6 mL Is Required to Titrate 43.8 mL of 0.107 M HCl? Ba(OH) 2 (aq) + 2 HCl(aq)  BaCl 2 (aq) + 2 H 2 O(l)

39. Tro's Introductory Chemistry, Chapter 14 39 Strong or Weak A strong acid is a strong electrolyte. Practically all the acid molecules ionize, →. A strong base is a strong electrolyte. Practically all the base molecules form OH – ions, either through dissociation or reaction with water, →. A weak acid is a weak electrolyte. Only a small percentage of the molecules ionize, . A weak base is a weak electrolyte. Only a small percentage of the base molecules form OH – ions, either through dissociation or reaction with water, .

40. Tro's Introductory Chemistry, Chapter 14 40 Strong Acids The stronger the acid, the more willing it is to donate H. Use water as the standard base. Strong acids donate practically all their Hs. 100% ionized in water. Strong electrolyte. [H 3 O + ] = [strong acid]. [ ] = molarity. HCl  H + + Cl - HCl + H 2 O  H 3 O + + Cl -

41. Tro's Introductory Chemistry, Chapter 14 41 Strong Acids, Continued Hydrochloric acidHCl Hydrobromic acidHBr Hydroiodic acidHI Nitric acidHNO 3 Perchloric acidHClO 4 Sulfuric acidH 2 SO 4

42. Tro's Introductory Chemistry, Chapter 14 42 Strong Acids, Continued Pure waterHCl solution

43. Tro's Introductory Chemistry, Chapter 14 43 Weak Acids Weak acids donate a small fraction of their Hs. Most of the weak acid molecules do not donate H to water. Much less than 1% ionized in water. [H 3 O + ] << [weak acid]. HF  H + + F - HF + H 2 O  H 3 O + + F -

44. Tro's Introductory Chemistry, Chapter 14 44 Weak Acids, Continued Hydrofluoric acidHF Acetic acidHC 2 H 3 O 2 Formic acidHCHO 2 Sulfurous acidH 2 SO 3 Carbonic acidH 2 CO 3 Phosphoric acidH 3 PO 4

45. Tro's Introductory Chemistry, Chapter 14 45 Weak Acids, Continued Pure waterHF solution

46. Tro's Introductory Chemistry, Chapter 14 46 Degree of Ionization The extent to which an acid ionizes in water depends in part on the strength of the bond between the acid H + and anion compared to the strength of the bond between the acid H + and the O of water. HA(aq) + H 2 O(l)  A − (aq) + H 3 O + (aq)

47. Tro's Introductory Chemistry, Chapter 14 47 Example 14.5—Determine the [H 3 O + ] in the Following Solutions: 1.5 M HCl Since HCl is a strong acid, [H 3 O + ] = [HCl] = 1.5 M. 3.0 M HC 2 H 3 O 2 Since HC 2 H 3 O 2 is a weak acid, [H 3 O + ] << [HC 2 H 3 O 2 ]. Therefore, [H 3 O + ] << 3.0 M.

48. Practice—Determine the [H 3 O + ] in the Following Solutions: 0.5 M HI 0.1 M HCHO 2 Hydrochloric acid HCl Hydrobromic acid HBr Hydroiodic acid HI Nitric acid HNO 3 Perchloric acid HClO 4 Sulfuric acidH 2 SO 4 Hydrofluoric acid HF Acetic acid HC 2 H 3 O 2 Formic acid HCHO 2 Sulfurous acid H 2 SO 3 Carbonic acid H 2 CO 3 Phosphoric acid H 3 PO 4 Strong Weak

49. Tro's Introductory Chemistry, Chapter 14 49 Strong Bases The stronger the base, the more willing it is to accept H. Use water as the standard acid. Strong bases, practically all molecules are dissociated into OH – or accept Hs. Strong electrolyte. Multi-OH bases completely dissociated. [HO – ] = [strong base] x (# OH). NaOH  Na + + OH -

50. Tro's Introductory Chemistry, Chapter 14 50 Strong Bases, Continued Lithium hydroxideLiOH Sodium hydroxideNaOH Potassium hydroxideKOH Calcium hydroxideCa(OH) 2 Strontium hydroxideSr(OH) 2 Barium hydroxideCa(OH) 2

51. Tro's Introductory Chemistry, Chapter 14 51 Weak Bases In weak bases, only a small fraction of molecules accept Hs. Weak electrolyte. Most of the weak base molecules do not take H from water. Much less than 1% ionization in water. [HO – ] << [strong base]. NH 3 + H 2 O  NH 4 + + OH -

52. Tro's Introductory Chemistry, Chapter 14 52 Weak Bases, Continued Ammonia NH 3 (aq) + H 2 O(l)  NH 4 + (aq) + OH − (aq) Pyridine C 5 H 5 N(aq) + H 2 O(l)  C 5 H 5 NH + (aq) + OH − (aq) Methyl amine CH 3 NH 2 (aq) + H 2 O(l)  CH 3 NH 3 + (aq) + OH − (aq) Ethyl amine C 2 H 5 NH 2 (aq) + H 2 O(l)  C 2 H 5 NH 3 + (aq) + OH − (aq) Bicarbonate HCO 3 − (aq) + H 2 O(l)  H 2 CO 3 (aq) + OH − (aq)

53. Tro's Introductory Chemistry, Chapter 14 53 Example 14.6—Determine the [OH − ] in the Following Solutions: 2.25 M KOH Since KOH is a strong base, [OH − ] = [KOH] x 1 = 2.25 M. 0.35 M CH 3 NH 2 Since CH 3 NH 2 is a weak base, [OH − ] << [CH 3 NH 2 ]. Therefore, [OH − ] << 0.35 M. 0.025 M Sr(OH) 2 Since Sr(OH) 2 is a strong base, [OH − ] = [Sr(OH) 2 ] x 2 = 0.050 M.

54. Practice—Determine the [OH − ] in the Following Solutions: 0.05 M Ba(OH) 2 0.01 M C 5 H 5 N Lithium hydroxide LiOH Sodium hydroxide NaOH Potassium hydroxide KOH Calcium hydroxide Ca(OH) 2 Strontium hydroxide Sr(OH) 2 Barium hydroxide Ba(OH) 2 AmmoniaNH 3 PyridineC5H5NC5H5N Methyl amine CH 3 NH 2 Ethyl amineC 2 H 5 NH 2 BicarbonateH 2 CO 3 Strong Weak

55. 14.8-14.9 Day Three Tro's Introductory Chemistry, Chapter 14 55

56. Tro's Introductory Chemistry, Chapter 14 56 Autoionization of Water Water is actually an extremely weak electrolyte. Therefore, there must be a few ions present. About 1 out of every 10 million water molecules form ions through a process called autoionization. H 2 O  H + + OH – H 2 O + H 2 O  H 3 O + + OH – All aqueous solutions contain both H 3 O + and OH –. The concentration of H 3 O + and OH – are equal in water. [H 3 O + ] = [OH – ] = 1 x 10 -7 M at 25 °C in pure water.

57. Tro's Introductory Chemistry, Chapter 14 57 Ion Product of Water The product of the H 3 O + and OH – concentrations is always the same number. The number is called the ion product of water and has the symbol K w. [H 3 O + ] x [OH – ] = 1 x 10 -14 = K w. As [H 3 O + ] increases, the [OH – ] must decrease so the product stays constant. Inversely proportional.

58. Tro's Introductory Chemistry, Chapter 14 58 Acidic and Basic Solutions Neutral solutions have equal [H 3 O + ] and [OH – ]. [H 3 O + ] = [OH – ] = 1 x 10 -7 Acidic solutions have a larger [H 3 O + ] than [OH – ]. [H 3 O + ] > 1 x 10 -7 ; [OH – ] < 1 x 10 -7 Basic solutions have a larger [OH – ] than [H 3 O + ]. [H 3 O + ] 1 x 10 -7

59. Tro's Introductory Chemistry, Chapter 14 59 Example—Determine the [H 3 O + ] for a 0.00020 M Ba(OH) 2 and Determine Whether the Solution Is Acidic, Basic, or Neutral.

60. Tro's Introductory Chemistry, Chapter 14 60 Practice—Determine the [H 3 O + ] Concentration and Whether the Solution Is Acidic, Basic, or Neutral for the Following: [OH – ] = 0.000250 M [OH – ] = 3.50 x 10 -8 M Ca(OH) 2 = 0.20 M

61. Tro's Introductory Chemistry, Chapter 14 61 pH The acidity/basicity of a solution is often expressed as pH. pH = ─log[H 3 O + ], [H 3 O + ] = 10 −pH Exponent on 10 with a positive sign. pH water = −log[10 -7 ] = 7. Need to know the [H + ] concentration to find pH. pH 7 is basic; pH = 7 is neutral.

62. 62 pH, Continued The lower the pH, the more acidic the solution; the higher the pH, the more basic the solution. 1 pH unit corresponds to a factor of 10 difference in acidity. Normal range is 0 to 14. pH 0 is [H + ] = 1 M, pH 14 is [OH – ] = 1 M. pH can be negative (very acidic) or larger than 14 (very alkaline).

63. 63 pH of Common Substances SubstancepH 1.0 M HCl0.0 0.1 M HCl1.0 Stomach acid1.0 to 3.0 Lemons2.2 to 2.4 Soft drinks2.0 to 4.0 Plums2.8 to 3.0 Apples2.9 to 3.3 Cherries3.2 to 4.0 Unpolluted rainwater5.6 Human blood7.3 to 7.4 Egg whites7.6 to 8.0 Milk of magnesia (saturated Mg(OH) 2 ) 10.5 Household ammonia10.5 to 11.5 1.0 M NaOH14

64. Tro's Introductory Chemistry, Chapter 14 64 Example—Calculate the pH of a 0.0010 M Ba(OH) 2 Solution and Determine if It Is Acidic, Basic, or Neutral.

65. Tro's Introductory Chemistry, Chapter 14 65 Practice—Calculate the pH of the Following Strong Acid or Base Solutions. 0.0020 M HCl 0.0050 M Ca(OH) 2 0.25 M HNO 3

66. Tro's Introductory Chemistry, Chapter 14 66 Calculate the Concentration of [H 3 O + ] for a Solution with pH 3.7.

67. Tro's Introductory Chemistry, Chapter 14 67 pOH The acidity/basicity of a solution may also be expressed as pOH. pH = ─log[OH − ], [OH − ] = 10 −pOH Exponent on 10 with a positive sign. pOH water = −log[10 −7 ] = 7. Need to know the [OH − ] concentration to find pOH. pOH 7 is basic, pOH = 7 is neutral.

68. Tro's Introductory Chemistry, Chapter 14 68 pOH, Continued The lower the pOH, the more basic the solution; the higher the pOH, the more acidic the solution. 1 pOH unit corresponds to a factor of 10 difference in basicity. Normal range is 0 to 14. pOH 0 is [OH − ] = 1 M; pOH 14 is [H 3 O + ] = 1 M. pOH can be negative (very basic) or larger than 14 (very acidic). pH + pOH = 14.00.

69. Tro's Introductory Chemistry, Chapter 14 69 Practice—Calculate the pOH and pH of the Following Strong Acid or Base Solutions. 0.0020 M KOH 0.0050 M Ca(OH) 2 0.25 M HNO 3