1. Chapter 15 Acids and Bases

2. Stomach Acid & Heartburn The cells that line your stomach produce hydrochloric acid – to kill unwanted bacteria – to help break down food – to activate enzymes that break down food Acid Reflux = stomach acid backs up into the esophagus, irritating tissues, and causing heartburn Mild cases of heartburn can be cured by neutralizing the acid in the esophagus – swallowing saliva, which contains bicarbonate ion – taking antacids that contain hydroxide ions and/or carbonate ions Tro, Chemistry: A Molecular Approach, 2/e

3. 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 Tro, Chemistry: A Molecular Approach, 2/e

4. Common Acids Tro, Chemistry: A Molecular Approach, 2/e

5. Structures of Acids Binary acids have acid hydrogens attached to a nonmetal atom HCl, HF Tro, Chemistry: A Molecular Approach, 2/e

6. Structure of Acids Oxy-acids have hydrogen atoms attached to an oxygen atom H 2 SO 4, HNO 3 Tro, Chemistry: A Molecular Approach, 2/e

7. Structure of Acids Carboxylic acids have COOH group HC 2 H 3 O 2, H 3 C 6 H 5 O 7 Only the first H in the formula is acidic the H is on the COOH Tro, Chemistry: A Molecular Approach, 2/e

8. Properties of Bases Also known as alkalis Taste bitter alkaloids = plant product that is alkaline  often poisonous Solutions feel slippery Change color of vegetable dyes different color than acid red litmus turns blue React with acids to form ionic salts neutralization Tro, Chemistry: A Molecular Approach, 2/e

9. Common Bases Tro, Chemistry: A Molecular Approach, 2/e

10. 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 Tro, Chemistry: A Molecular Approach, 2/e

11. Indicators Chemicals that change color depending on the solution’s acidity or basicity Many vegetable dyes are indicators – anthocyanins Litmus – from Spanish moss – red in acid, blue in base Phenolphthalein – found in laxatives – red in base, colorless in acid Tro, Chemistry: A Molecular Approach, 2/e

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 written in front HC 2 H 3 O 2 (aq) → H + (aq) + C 2 H 3 O 2 − (aq) Tro, Chemistry: A Molecular Approach, 2/e

13. Arrhenius Theory HCl ionizes in water, producing H + and Cl – ions NaOH dissociates in water, producing Na + and OH – ions Tro, Chemistry: A Molecular Approach, 2/e

14. Hydronium Ion The H + ions produced by the acid are so reactive they cannot exist in water – H + ions are protons!! Instead, they react with water molecules to produce complex ions, mainly hydronium ion, H 3 O + H + + H 2 O  H 3 O + – there are also minor amounts of H + with multiple water molecules, H(H 2 O) n + Tro, Chemistry: A Molecular Approach, 2/e

15. Arrhenius Acid–Base Reactions The H + from the acid combines with the OH − from the base to make a molecule of H 2 O – it is often helpful to think of H 2 O as H-OH 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) Tro, Chemistry: A Molecular Approach, 2/e

16. Problems with Arrhenius Theory Why do molecular substances, such as NH 3, dissolve in water to form basic solutions – even though they do not contain OH – ions? How do some ionic compounds, such as Na 2 CO 3 or Na 2 O, dissolve in water to form basic solutions – even though they do not contain OH – ions? Why do molecular substances, such as CO 2, dissolve in water to form acidic solutions – even though they do not contain H + ions? How do acid–base reactions take place outside aqueous solutions? Tro, Chemistry: A Molecular Approach, 2/e

17. Another Definition: Brønsted-Lowry Acid-Base Theory Brønsted and Lowry redefined acids and bases based on what happens in a reaction. Any reaction that involves H + being transferred from one molecule to another is an acid–base reaction – regardless of whether it occurs in aqueous solution, or if there is OH − present The Brønsted-Lowry definition includes – all reactions that fit the Arrhenius definition. – Many others as well. Tro, Chemistry: A Molecular Approach, 2/e

18. Brønsted-Lowry Theory In a Brønsted-Lowry acid–base reaction, an H + is transferred The acid is an H donor The base is an H acceptor – base structure must contain an atom with an unshared pair of electrons In a Brønsted-Lowry acid-base reaction, the acid molecule gives an H + to the base molecule H–A + :B  :A – + H–B + Tro, Chemistry: A Molecular Approach, 2/e

19. Brønsted-Lowry Acids Brønsted-Lowry acids are H + donors – any material that has H can potentially be a Brønsted-Lowry acid – because of the molecular structure, often one H in the molecule is easier to transfer than others When HCl dissolves in water, the HCl is the acid 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) acidbase Tro, Chemistry: A Molecular Approach, 2/e

20. Brønsted-Lowry Bases Brønsted-Lowry bases are H + acceptors – any material that has atoms with lone pairs can potentially be a Brønsted-Lowry base – because of the molecular structure, often one atom in the molecule is more willing to accept H + transfer than others When NH 3 dissolves in water, the NH 3 (aq) is the base 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) base acid Tro, Chemistry: A Molecular Approach, 2/e

21. A Warning! Because chemists know common bonding patterns, we often do not draw lone pair electrons on our structures. You need to be able to recognize when an atom in a molecule has lone pair electrons and when it doesn’t! Tro, Chemistry: A Molecular Approach, 2/e

22. Practice – Draw structures of the following that include lone pairs of electrons HClO HCO 3 − Tro, Chemistry: A Molecular Approach, 2/e

23. Amphoteric Substances Amphoteric substances can act as either an acid or a base – because they have both a transferable H and an atom with lone pair electrons Water acts as base, accepting H + from HCl HCl(aq) + H 2 O(l) → Cl – (aq) + H 3 O + (aq) Water acts as acid, donating H + to NH 3 NH 3 (aq) + H 2 O(l)  NH 4 + (aq) + OH – (aq) Tro, Chemistry: A Molecular Approach, 2/e

24. Brønsted-Lowry Acid-Base Reactions One of the advantages of 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 will act as an acid in the reverse process And the original acid has a lone pair of electrons after the reaction – so it will act as a base in the reverse process :A – + H–B +  H–A + :B Tro, Chemistry: A Molecular Approach, 2/e

25. 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 its conjugate acid; and the original acid becomes its conjugate base Tro, Chemistry: A Molecular Approach, 2/e

26. 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 Tro, Chemistry: A Molecular Approach, 2/e

27. 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 Tro, Chemistry: A Molecular Approach, 2/e

28. Practice – Write the formula for the conjugate acid of the following H 2 O NH 3 CO 3 2− H 2 PO 4 1− H 2 OH 3 O + NH 3 NH 4 + CO 3 2− HCO 3 − H 2 PO 4 1− H 3 PO 4 Tro, Chemistry: A Molecular Approach, 2/e

29. Practice – Write the formula for the conjugate base of the following H 2 O NH 3 CO 3 2− H 2 PO 4 1− H 2 OHO − NH 3 NH 2 − CO 3 2− because CO 3 2− does not have an H, it cannot be an acid H 2 PO 4 1− HPO 4 2− Tro, Chemistry: A Molecular Approach, 2/e

30. Practice – Identify the Brønsted-Lowry acid, base, conjugate acid, and conjugate base in the following reaction HSO 4 − (aq) + HCO 3 − (aq)  SO 4 2− (aq) + H 2 CO 3(aq) Base Conjugate Acid Conjugate Base Tro, Chemistry: A Molecular Approach, 2/e

31. Practice—Write the equations for the following reacting with water and acting as a monoprotic acid & label the conjugate acid and base HBr + H 2 O  Br − + H 3 O + Acid Base Conj. Conj. base acid HSO 4 − + H 2 O  SO 4 2− + H 3 O + Acid Base Conj. Conj. base acid HBr HSO 4 − Tro, Chemistry: A Molecular Approach, 2/e

32. Practice—Write the equations for the following reacting with water and acting as a monoprotic-accepting base and label the conjugate acid and base I − + H 2 O  HI + OH − Base Acid Conj. Conj. acid base I − CO 3 2− CO 3 2− + H 2 O  HCO 3 − + OH − Base Acid Conj. Conj. acid base Tro, Chemistry: A Molecular Approach, 2/e

33. Comparing Arrhenius Theory and Brønsted-Lowry Theory 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)  Na + (aq) + OH − (aq) NH 3 (aq) + H 2 O(l)  NH 4 + (aq) + OH − (aq) 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) Tro, Chemistry: A Molecular Approach, 2/e

34. 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,  Tro, Chemistry: A Molecular Approach, 2/e

35. Strong Acids HCl  H + + Cl − HCl + H 2 O  H 3 O + + Cl − 0.10 M HCl = 0.10 M H 3 O + The stronger the acid, the more willing it is to donate H we use water as the standard base to donate H to Strong acids donate practically all their H’s 100% ionized in water strong electrolyte [H 3 O + ] = [strong acid] [X] means the molarity of X Tro, Chemistry: A Molecular Approach, 2/e

36. HF  H + + F − HF + H 2 O  H 3 O + + F − 0.10 M HF ≠ 0.10 M H 3 O + Weak Acids Weak acids donate a small fraction of their H’s most of the weak acid molecules do not donate H to water much less than 1% ionized in water [H 3 O + ] << [weak acid] Tro, Chemistry: A Molecular Approach, 2/e

37. Strong & Weak Acids Tro, Chemistry: A Molecular Approach, 2/e

39. Strengths of Acids & Bases Commonly, acid or base strength is measured by determining the equilibrium constant of a substance’s reaction with water HAcid + H 2 O  Acid − + H 3 O + Base: + H 2 O  HBase + + OH − The farther the equilibrium position lies toward the products, the stronger the acid or base The position of equilibrium depends on the strength of attraction between the base form and the H + – stronger attraction means stronger base or weaker acid Tro, Chemistry: A Molecular Approach, 2/e

40. General Trends in Acidity The stronger an acid is at donating H, the weaker the conjugate base is at accepting H Higher oxidation number = stronger oxyacid – H 2 SO 4 > H 2 SO 3 ; HNO 3 > HNO 2 Cation stronger acid than neutral molecule; neutral stronger acid than anion – H 3 O + > H 2 O > OH − ; NH 4 + > NH 3 > NH 2 − – trend in base strength opposite Tro, Chemistry: A Molecular Approach, 2/e

41. Acid Ionization Constant, K a Acid strength measured by the size of the equilibrium constant when reacts with H 2 O HAcid + H 2 O  Acid − + H 3 O + The equilibrium constant for this reaction is called the acid ionization constant, K a – larger K a = stronger acid Tro, Chemistry: A Molecular Approach, 2/e

43. Autoionization of Water Water is actually an extremely weak electrolyte – therefore there must be a few ions present About 2 out of every 1 billion 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 – ] = 10 −7 M @ 25 °C Tro, Chemistry: A Molecular Approach, 2/e

44. 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 – aka the Dissociation Constant of Water [H 3 O + ] x [OH – ] = K w = 1.00 x 10 −14 @ 25 °C – if you measure one of the concentrations, you can calculate the other As [H 3 O + ] increases the [OH – ] must decrease so the product stays constant – inversely proportional Tro, Chemistry: A Molecular Approach, 2/e

45. Acidic and Basic Solutions All aqueous solutions contain both H 3 O + and OH – ions Neutral solutions have equal [H 3 O + ] and [OH – ] – [H 3 O + ] = [OH – ] = 1.00 x 10 −7 Acidic solutions have a larger [H 3 O + ] than [OH – ] – [H 3 O + ] > 1.00 x 10 −7 ; [OH – ] < 1.00 x 10 −7 Basic solutions have a larger [OH – ] than [H 3 O + ] – [H 3 O + ] 1.00 x 10 −7 Tro, Chemistry: A Molecular Approach, 2/e

46. Practice – Complete the table [H + ] vs. [OH − ] OH − H+H+ H+H+ H+H+ H+H+ H+H+ [OH − ] [H + ] 10 0 10 −1 10 − 3 10 −5 10 −7 10 −9 10 −11 10 −13 10 −14 Tro, Chemistry: A Molecular Approach, 2/e

47. Practice – Complete the table [H + ] vs. [OH − ] OH − H+H+ H+H+ H+H+ H+H+ H+H+ [OH − ]10 −14 10 −13 10 −11 10 −9 10 −7 10 −5 10 −3 10 −1 10 0 [H + ] 10 0 10 −1 10 −3 10 −5 10 −7 10 −9 10 −11 10 −13 10 −14 Even though it may look like it, neither H + nor OH − will ever be 0 The sizes of the H + and OH − are not to scale because the divisions are powers of 10 rather than units Acid Base Tro, Chemistry: A Molecular Approach, 2/e

48. Practice – Determine the [H 3 O + ] when the [OH − ] = 2.5 x 10 −9 M the units are correct; the fact that the [H 3 O + ] > [OH  ] means the solution is acidic [OH  ] = 2.5 x 10 −9 M [H 3 O + ] Check: Solution: Conceptual Plan: Relationships: Given: Find: [OH  ][H 3 O + ] Tro, Chemistry: A Molecular Approach, 2/e

49. Measuring Acidity: pH The acidity or basicity of a solution is often expressed as pH pH = −log[H 3 O + ] exponent on 10 with a positive sign pH water = −log[10 −7 ] = 7 need to know the [H 3 O + ] concentration to find pH pH 7 is basic, pH = 7 is neutral [H 3 O + ] = 10 −pH Tro, Chemistry: A Molecular Approach, 2/e

50. Sig. Figs. & Logs When you take the log of a number written in scientific notation, the digit(s) before the decimal point come from the exponent on 10, and the digits after the decimal point come from the decimal part of the number log(2.0 x 10 6 ) = log(10 6 ) + log(2.0) = 6 + 0.30303… = 6.30303... Because the part of the scientific notation number that determines the significant figures is the decimal part, the sig figs are the digits after the decimal point in the log log(2.0 x 10 6 ) = 6.30 Tro, Chemistry: A Molecular Approach, 2/e

51. What Does the pH Number Imply? 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 of pH is 0 to 14 pH 0 is [H 3 O + ] = 1 M, pH 14 is [OH – ] = 1 M pH can be negative (very acidic) or larger than 14 (very alkaline) Tro, Chemistry: A Molecular Approach, 2/e

52. Practice – Determine the pH @ 25 ºC of a solution that has [OH − ] = 2.5 x 10 −9 M pH is unitless; the fact that the pH < 7 means the solution is acidic [OH  ] = 2.5 x 10 −9 M pH Check: Solution: Conceptual Plan: Relationships: Given: Find: [H 3 O + ][OH  ]pH Tro, Chemistry: A Molecular Approach, 2/e

53. Practice – Determine the [OH − ] of a solution with a pH of 5.40 because the pH < 7, [OH − ] should be less than 1 x 10 −7 ; and it is pH = 5.40 [OH − ], M Check: Solution: Conceptual Plan: Relationships: Given: Find: [H 3 O + ]pH[OH  ] Tro, Chemistry: A Molecular Approach, 2/e

54. pOH Another way of expressing the acidity/basicity of a solution is pOH pOH = −log[OH  ], [OH  ] = 10 −pOH pOH water = −log[10 −7 ] = 7 need to know the [OH  ] concentration to find pOH pOH 7 is acidic, pOH = 7 is neutral pH + pOH = 14.0 Tro, Chemistry: A Molecular Approach, 2/e

55. pH and pOH Practice – Complete the table pH pOH OH − H+H+ H+H+ H+H+ H+H+ H+H+ [OH − ]10 −14 10 −13 10 −11 10 −9 10 −7 10 −5 10 −3 10 −1 10 0 [H + ] 10 0 10 −1 10 −3 10 −5 10 −7 10 −9 10 −11 10 −13 10 −14 Tro, Chemistry: A Molecular Approach, 2/e

56. pH and pOH Practice – Complete the table pH 0 1 3 5 7 9 11 13 14 pOH 14 13 11 9 7 5 3 1 0 OH − H+H+ H+H+ H+H+ H+H+ H+H+ [OH − ]10 −14 10 −13 10 −11 10 −9 10 −7 10 −5 10 −3 10 −1 10 0 [H + ] 10 0 10 −1 10 −3 10 −5 10 −7 10 −9 10 −11 10 −13 10 −14 Acid Base Tro, Chemistry: A Molecular Approach, 2/e

57. Relationship between pH and pOH pH + pOH = 14.00 – at 25 °C – you can use pOH to find pH of a solution Tro, Chemistry: A Molecular Approach, 2/e

58. Practice – Determine the pOH @ 25 ºC of a solution that has [H 3 O + ] = 2.5 x 10 −9 M pH is unitless; the fact that the pH < 7 means the solution is acidic [H 3 O + ] = 2.5 x 10 −9 M pOH Check: Solution: Conceptual Plan: Relationships: Given: Find: pH[H 3 O + ]pOH Tro, Chemistry: A Molecular Approach, 2/e

59. pKpK A way of expressing the strength of an acid or base is pK pK a = −log(K a ), K a = 10 −pKa pK b = −log(K b ), K b = 10 −pKb The stronger the acid, the smaller the pK a – larger K a = smaller pK a because it is the –log The stronger the base, the smaller the pK b – larger K b = smaller pK b Tro, Chemistry: A Molecular Approach, 2/e

60. [H 3 O + ] and [OH − ] in a Strong Acid or Strong Base Solution There are two sources of H 3 O + in an aqueous solution of a strong acid – the acid and the water There are two sources of OH − in an aqueous solution of a strong acid – the base and the water For a strong acid or base, the contribution of the water to the total [H 3 O + ] or [OH − ] is negligible – the [H 3 O + ] acid shifts the K w equilibrium so far that [H 3 O + ] water is too small to be significant except in very dilute solutions, generally < 1 x 10 −4 M Tro, Chemistry: A Molecular Approach, 2/e

61. Finding pH of a Strong Acid or Strong Base Solution For a monoprotic strong acid [H 3 O + ] = [HAcid] – for polyprotic acids, the other ionizations can generally be ignored for H 2 SO 4, the second ionization cannot be ignored – 0.10 M HCl has [H 3 O + ] = 0.10 M and pH = 1.00 For a strong ionic base, [OH − ] = (number OH − )x[Base] – for molecular bases with multiple lone pairs available, only one lone pair accepts an H, the other reactions can generally be ignored – 0.10 M Ca(OH) 2 has [OH − ] = 0.20 M and pH = 13.30 Tro, Chemistry: A Molecular Approach, 2/e

62. Finding the pH of a Weak Acid There are also two sources of H 3 O + in an aqueous solution of a weak acid – the acid and the water However, finding the [H 3 O + ] is complicated by the fact that the acid only undergoes partial ionization Calculating the [H 3 O + ] requires solving an equilibrium problem for the reaction that defines the acidity of the acid HAcid + H 2 O  Acid  + H 3 O + Tro, Chemistry: A Molecular Approach, 2/e

63. Practice – What is the pH of a 0.012 M solution of nicotinic acid, HC 6 H 4 NO 2 ? write the reaction for the acid with water construct an ICE table for the reaction enter the initial concentrations – assuming the [H 3 O + ] from water is ≈ 0 HC 6 H 4 NO 2 + H 2 O  C 6 H 4 NO 2  + H 3 O + [HA][A − ][H 3 O + ] initial 0.012 0 ≈ 0 change equilibrium Tro, Chemistry: A Molecular Approach, 2/e

64. [HA][A − ][H 3 O + ] initial 0.012 00 change equilibrium Practice – What is the pH of a 0.012 M solution of nicotinic acid, HC 6 H 4 NO 2 ? represent the change in the concentrations in terms of x sum the columns to find the equilibrium concentrations in terms of x substitute into the equilibrium constant expression +x+x +x+x xx 0.012  x xx HC 6 H 4 NO 2 + H 2 O  C 6 H 4 NO 2  + H 3 O + Tro, Chemistry: A Molecular Approach, 2/e

65. Practice – What is the pH of a 0.012 M solution of nicotinic acid, HC 6 H 4 NO 2 ? K a = 1.4 x 10 −5 @ 25 °C determine the value of K a because K a is very small, approximate the [HA] eq = [HA] init and solve for x [HA][A − ][H 3 O + ] initial 0.012 0 ≈ 0 change −x−x+x+x+x+x equilibrium 0.012xx 0.012  x HC 6 H 4 NO 2 + H 2 O  C 6 H 4 NO 2  + H 3 O + Tro, Chemistry: A Molecular Approach, 2/e

66. Practice – What is the pH of a 0.012 M solution of nicotinic acid, HC 6 H 4 NO 2 ? K a = 1.4 x 10 −5 @ 25 °C check if the approximation is valid by seeing if x < 5% of [HC 6 H 4 NO 2 ] init K a for HC 6 H 4 NO 2 = 1.4 x 10 −5 the approximation is valid x = 4.1 x 10 −4 [HA][A − ][H 3 O + ] initial 0.012 0 ≈ 0 change −x−x+x+x+x+x equilibrium 0.012 xx Tro, Chemistry: A Molecular Approach, 2/e

67. Practice – What is the pH of a 0.012 M solution of nicotinic acid, HC 6 H 4 NO 2 ? K a = 1.4 x 10 −5 @ 25 °C substitute x into the equilibrium concentration definitions and solve x = 4.1 x 10 −4 [HA][A − ][H 3 O + ] initial 0.012 0 ≈ 0 change −x−x+x+x+x+x equilibrium 0.012−x xx Tro, Chemistry: A Molecular Approach, 2/e

68. Practice – What is the pH of a 0.012 M solution of nicotinic acid, HC 6 H 4 NO 2 ? K a = 1.4 x 10 −5 @ 25 °C substitute [H 3 O + ] into the formula for pH and solve [HA][A − ][H 3 O + ] initial 0.012 0 ≈ 0 change −x−x+x+x+x+x equilibrium 0.0120.00041 Tro, Chemistry: A Molecular Approach, 2/e

69. Practice – What is the pH of a 0.012 M solution of nicotinic acid, HC 6 H 4 NO 2 ? K a = 1.4 x 10 −5 @ 25 °C check by substituting the equilibrium concentrations back into the equilibrium constant expression and comparing the calculated K a to the given K a the values match [HA][A − ][H 3 O + ] initial 0.012 0 ≈ 0 change −x−x+x+x+x+x equilibrium 0.0120.00041 Tro, Chemistry: A Molecular Approach, 2/e

70. Practice – What is the K a of nicotinic acid, HC 6 H 4 NO 2, if a 0.012 M solution of nicotinic acid has a pH of 3.40? use the pH to find the equilibrium [H 3 O + ] write the reaction for the acid with water construct an ICE table for the reaction enter the initial concentrations and [H 3 O + ] equil HA + H 2 O  A  + H 3 O + [HA][A − ][H 3 O + ] initial change equilibrium [HA][A − ][H 3 O + ] initial 0.012 0 ≈ 0 change equilibrium 4.0E-04 Tro, Chemistry: A Molecular Approach, 2/e

71. [HA][A − ][H 3 O + ] initial 0.012 00 change equilibrium 0.012 Practice – What is the K a of nicotinic acid, HC 6 H 4 NO 2, if a 0.012 M solution of nicotinic acid has a pH of 3.40? fill in the rest of the table using the [H 3 O + ] as a guide if the difference is insignificant, [HA] equil = [HA] initial substitute into the K a expression and compute K a +4.0E-04 −4.0E-04 0.012  4.0E-04 HA + H 2 O  A  + H 3 O + Tro, Chemistry: A Molecular Approach, 2/e

72. Polyprotic Acids Acid molecules often have more than one ionizable H – these are called polyprotic acids – the ionizable H’s may have different acid strengths or be equal – 1 H = monoprotic, 2 H = diprotic, 3 H = triprotic – HCl = monoprotic, H 2 SO 4 = diprotic, H 3 PO 4 = triprotic Polyprotic acids ionize in steps – each ionizable H is removed sequentially Removing of the first H automatically makes removal of the second H harder – H 2 SO 4 is a stronger acid than HSO 4  Tro, Chemistry: A Molecular Approach, 2/e

73. Percent Ionization Another way to measure the strength of an acid is to determine the percentage of acid molecules that ionize when dissolved in water – this is called the percent ionization – the higher the percent ionization, the stronger the acid Because [ionized acid] equil = [H 3 O + ] equil Tro, Chemistry: A Molecular Approach, 2/e

74. Practice – What is the percent ionization of a 0.012 M solution of nicotinic acid, HC 6 H 4 NO 2 ? write the reaction for the acid with water construct an ICE table for the reaction enter the initial concentrations – assuming the [H 3 O + ] from water is ≈ 0 HC 6 H 4 NO 2 + H 2 O  C 6 H 4 NO 2  + H 3 O + [HA][A − ][H 3 O + ] initial 0.012 0 ≈ 0 change equilibrium Tro, Chemistry: A Molecular Approach, 2/e

75. [HA][A − ][H 3 O + ] initial 0.012 00 change equilibrium Practice – What is the percent ionization of a 0.012 M solution of nicotinic acid, HC 6 H 4 NO 2 ? represent the change in the concentrations in terms of x sum the columns to find the equilibrium concentrations in terms of x substitute into the equilibrium constant expression +x+x +x+x xx 0.012  x xx HC 6 H 4 NO 2 + H 2 O  C 6 H 4 NO 2  + H 3 O + Tro, Chemistry: A Molecular Approach, 2/e

76. Practice – What is the percent ionization of a 0.012 M solution of nicotinic acid, HC 6 H 4 NO 2 ? K a = 1.4 x 10 −5 determine the value of K a because K a is very small, approximate the [HA] eq = [HA] init and solve for x [HA][A − ][H 3 O + ] initial 0.012 0 ≈ 0 change −x−x+x+x+x+x equilibrium 0.012xx 0.012  x HC 6 H 4 NO 2 + H 2 O  C 6 H 4 NO 2  + H 3 O + Tro, Chemistry: A Molecular Approach, 2/e

77. Practice – What is the percent ionization of a 0.012 M solution of nicotinic acid, HC 6 H 4 NO 2 ? substitute x into the equilibrium concentration definitions and solve x = 4.1 x 10 −4 [HA][A − ][H 3 O + ] initial 0.012 0 ≈ 0 change −x−x+x+x+x+x equilibrium 0.012−x xx Tro, Chemistry: A Molecular Approach, 2/e

78. Practice – What is the percent ionization of a 0.012 M solution of nicotinic acid, HC 6 H 4 NO 2 ? apply the definition and compute the percent ionization because the percent ionization is < 5%, the “x is small” approximation is valid [HA][A − ][H 3 O + ] initial 0.012 0 ≈ 0 change −x−x+x+x+x+x equilibrium 0.0124.1E-04 Tro, Chemistry: A Molecular Approach, 2/e

79. Relationship Between [H 3 O + ] equilibrium & [HA] initial Increasing the initial concentration of acid results in increased [H 3 O + ] at equilibrium Increasing the initial concentration of acid results in decreased percent ionization This means that the increase in [H 3 O + ] concentration is slower than the increase in acid concentration Tro, Chemistry: A Molecular Approach, 2/e

80. Why doesn’t the increase in H 3 O + keep up with the increase in HA? The reaction for ionization of a weak acid is HA (aq) + H 2 O (l)  A − (aq) + H 3 O + (aq) According to Le Châtelier’s Principle, if we reduce the concentrations of all the (aq) components, the equilibrium should shift to the right to increase the total number of dissolved particles – we can reduce the (aq) concentrations by using a more dilute initial acid concentration The result will be a larger [H 3 O + ] in the dilute solution compared to the initial acid concentration This will result in a larger percent ionization Tro, Chemistry: A Molecular Approach, 2/e

81. Finding the pH of Mixtures of Acids Generally, you can ignore the contribution of the weaker acid to the [H 3 O + ] equil For a mixture of a strong acid with a weak acid, the complete ionization of the strong acid provides more than enough [H 3 O + ] to shift the weak acid equilibrium to the left so far that the weak acid’s added [H 3 O + ] is negligible For mixtures of weak acids, generally only need to consider the stronger for the same reasons – as long as one is significantly stronger than the other, and their concentrations are similar Tro, Chemistry: A Molecular Approach, 2/e

82. Practice – Determine the pH @ 25 ºC of a solution that is a mixture of 0.045 M HCl and 0.15 M HF pH is unitless; the fact that the pH < 7 means the solution is acidic [HCl] = 4.5 x 10 −2 M, [HF] = 0.15 M pH Check: Solution: Conceptual Plan: Relationships: Given: Find: [H 3 O + ][HCl]pH Because HCl is a strong acid and HF is a weak acid, [H 3 O + ] = [HCl] Tro, Chemistry: A Molecular Approach, 2/e

83. Strong Bases NaOH  Na + + OH − The stronger the base, the more willing it is to accept H use water as the standard acid For ionic bases, practically all units are dissociated into OH – or accept H’s strong electrolyte multi-OH strong bases completely dissociated [HO – ] = [strong base] x (# OH) Tro, Chemistry: A Molecular Approach, 2/e

84. Practice – Calculate the pH of the following strong acid or base solutions 0.0020 M HCl 0.0015 M Ca(OH) 2 [H 3 O + ] = [HCl] = 2.0 x 10 −3 M pH = −log(2.0 x 10 −3 ) = 2.70 [OH − ] = 2 x [Ca(OH) 2 ] = 3.0 x 10 −3 M pOH = −log(3.0 x 10 −3 ) = 2.52 pH = 14.00 − pOH = 14.00 − 2.52 pH = 11.48

85. Weak Bases NH 3 + H 2 O  NH 4 + + OH − In weak bases, only a small fraction of molecules accept H’s weak electrolyte most of the weak base molecules do not take H from water much less than 1% ionization in water [HO – ] << [weak base] Finding the pH of a weak base solution is similar to finding the pH of a weak acid Tro, Chemistry: A Molecular Approach, 2/e

86. Base Ionization Constant, K b Base strength measured by the size of the equilibrium constant when react with H 2 O :Base + H 2 O  OH − + H:Base + The equilibrium constant is called the base ionization constant, K b – larger K b = stronger base Tro, Chemistry: A Molecular Approach, 2/e

88. Structure of Amines Tro, Chemistry: A Molecular Approach, 2/e

89. Practice – Find the pH of a 0.0015 M morphine solution write the reaction for the base with water construct an ICE table for the reaction enter the initial concentrations – assuming the [OH  ] from water is ≈ 0 because no products initially, Q c = 0, and the reaction is proceeding forward B + H 2 O  BH + + OH  [B][BH + ][OH  ] initial 0.0015 0 ≈ 0 change equilibrium Tro, Chemistry: A Molecular Approach, 2/e

90. [B][BH + ][OH  ] initial 0.0015 00 change equilibrium Practice – Find the pH of a 0.0015 M morphine solution represent the change in the concentrations in terms of x sum the columns to find the equilibrium concentrations in terms of x substitute into the equilibrium constant expression +x+x +x+x xx 0.0015  x xx B + H 2 O  BH + + OH  Tro, Chemistry: A Molecular Approach, 2/e

91. Practice – Find the pH of a 0.0015 M morphine solution determine the value of K b because K b is very small, approximate the [B] eq = [B] init and solve for x K b for morphine = 1.6 x 10 −6 [B][BH + ][OH  ] initial 0.0015 0 ≈ 0 change −x−x+x+x+x+x equilibrium 0.0015xx 0.0015  x Tro, Chemistry: A Molecular Approach, 2/e

92. Practice – Find the pH of a 0.0015 M morphine solution check if the approximation is valid by seeing if x < 5% of [B] init the approximation is valid x = 4.9 x 10 −5 K b for morphine = 1.6 x 10 −6 [B][BH + ][OH  ] initial 0.0015 0 ≈ 0 change −x−x+x+x+x+x equilibrium 0.0015 xx Tro, Chemistry: A Molecular Approach, 2/e

93. Practice – Find the pH of a 0.0015 M morphine solution substitute x into the equilibrium concentration definitions and solve x = 4.9 x 10 −5 K b for morphine = 1.6 x 10 −6 [B][BH + ][OH  ] initial 0.0015 0 ≈ 0 change −x−x+x+x+x+x equilibrium 0.0015  x xx [B][BH + ][OH  ] initial 0.0015 0 ≈ 0 change −x−x+x+x+x+x equilibrium 0.00154.9E−5 Tro, Chemistry: A Molecular Approach, 2/e

94. Practice – Find the pH of a 0.0015 M morphine solution use the [OH − ] to find the [H 3 O + ] using K w substitute [H 3 O + ] into the formula for pH and solve K b for morphine = 1.6 x 10 −6 [B][BH + ][OH  ] initial 0.0015 0 ≈ 0 change −x−x+x+x+x+x equilibrium 0.00154.9E−5 Tro, Chemistry: A Molecular Approach, 2/e

95. Practice – Find the pH of a 0.0015 M morphine solution use the [OH − ] to find the pOH use pOH to find pH K b for morphine = 1.6 x 10 −6 [B][BH + ][OH  ] initial 0.0015 0 ≈ 0 change −x−x+x+x+x+x equilibrium 0.00154.9E−5 Tro, Chemistry: A Molecular Approach, 2/e

96. Practice – Find the pH of a 0.0015 M morphine solution check by substituting the equilibrium concentrations back into the equilibrium constant expression and comparing the calculated K b to the given K b the answer matches the given K b K b for morphine = 1.6 x 10 −6 [B][BH + ][OH  ] initial 0.0015 0 ≈ 0 change −x−x+x+x+x+x equilibrium 0.00154.9E−5 Tro, Chemistry: A Molecular Approach, 2/e

97. Acid–Base Properties of Salts Salts are water-soluble ionic compounds Salts that contain the cation of a strong base and an anion that is the conjugate base of a weak acid are basic – NaHCO 3 solutions are basic Na + is the cation of the strong base NaOH HCO 3 − is the conjugate base of the weak acid H 2 CO 3 Salts that contain cations that are the conjugate acid of a weak base and an anion of a strong acid are acidic – NH 4 Cl solutions are acidic NH 4 + is the conjugate acid of the weak base NH 3 Cl − is the anion of the strong acid HCl Tro, Chemistry: A Molecular Approach, 2/e

98. Anions as Weak Bases Every anion can be thought of as the conjugate base of an acid Therefore, every anion can potentially be a base – A − (aq) + H 2 O(l)  HA(aq) + OH − (aq) The stronger the acid is, the weaker the conjugate base is An anion that is the conjugate base of a strong acid is pH neutral Cl − (aq) + H 2 O(l)  HCl(aq) + OH − (aq) An anion that is the conjugate base of a weak acid is basic F − (aq) + H 2 O(l)  HF(aq) + OH − (aq) Tro, Chemistry: A Molecular Approach, 2/e

99. Example 15.13: Use the table to determine if the given anion is basic or neutral a)NO 3 − the conjugate base of a strong acid, therefore neutral b)NO 2 − the conjugate base of a weak acid, therefore basic Tro, Chemistry: A Molecular Approach, 2/e

100. Relationship between K a of an Acid and K b of Its Conjugate Base Many reference books only give tables of K a values because K b values can be found from them when you add equations, you multiply the K’s Tro, Chemistry: A Molecular Approach, 2/e

101. If a 0.0015 M NaA solution has a pOH of 5.45, what is the K a of HA? Na + is the cation of a strong base – pOH neutral. Because pOH is < 7, the solutinon is basic. A − is basic. write the reaction for the anion with water construct an ICE table for the reaction enter the initial concentrations – assuming the [OH  ] from water is ≈ 0 A − + H 2 O  HA + OH  [A − ][HA][OH  ] initial 0.1000≈ 0 change equilibrium

102. Practice – If a 0.15 M NaA solution has a pOH of 5.45, what is the K a of HA? use the pOH to find the [OH − ] use [OH − ] to fill in other items [A − ][HA][OH  ] initial 0.150≈ 0 change −3.6E−6 + 3.6E−6 equilibrium 0.153.6E-6 [A − ][HA][OH  ] initial 0.150≈ 0 change equilibrium Tro, Chemistry: A Molecular Approach, 2/e

103. calculate the value of K b of A − [A − ][HA][OH  ] initial 0.150≈ 0 change − 3.6E−6 + 3.6E−6 equilibrium 0.153.6E-6 Practice – If a 0.15 M NaA solution has a pOH of 5.45, what is the K a of HA? Tro, Chemistry: A Molecular Approach, 2/e

104. use K b of A − to find K a of HA Practice – If a 0.15 M NaA solution has a pOH of 5.45, what is the K a of HA? [A − ][HA][OH  ] initial 0.150≈ 0 change − 3.6E−6 + 3.6E−6 equilibrium 0.153.6E-6 Tro, Chemistry: A Molecular Approach, 2/e

105. Polyatomic Cations as Weak Acids Some cations can be thought of as the conjugate acid of a weak base – others are the counter-ions of a strong base Therefore, some cations can potentially be acidic – MH + (aq) + H 2 O(l)  MOH(aq) + H 3 O + (aq) The stronger the base is, the weaker the conjugate acid is – a cation that is the counter-ion of a strong base is pH neutral – a cation that is the conjugate acid of a weak base is acidic NH 4 + (aq) + H 2 O(l)  NH 3 (aq) + H 3 O + (aq) because NH 3 is a weak base, the position of this equilibrium favors the right Tro, Chemistry: A Molecular Approach, 2/e

106. Metal Cations as Weak Acids Cations of small, highly charged metals are weakly acidic – alkali metal cations and alkali earth metal cations are pH neutral – cations are hydrated Al(H 2 O) 6 3+ (aq) + H 2 O(l)  Al(H 2 O) 5 (OH) 2+ (aq) + H 3 O + (aq) Tro, Chemistry: A Molecular Approach, 2/e

107. Example 15.15: Determine if the given cation Is acidic or neutral a)C 5 N 5 NH 2 + the conjugate acid of the weak base pyridine, therefore acidic b)Ca 2+ the counter-ion of the strong base Ca(OH) 2, therefore neutral c)Cr 3+ a highly charged metal ion, therefore acidic Tro, Chemistry: A Molecular Approach, 2/e

108. Classifying Salt Solutions as Acidic, Basic, or Neutral If the salt cation is the counter-ion of a strong base and the anion is the conjugate base of a strong acid, it will form a neutral solution – NaCl Ca(NO 3 ) 2 KBr If the salt cation is the counter-ion of a strong base and the anion is the conjugate base of a weak acid, it will form a basic solution – NaF Ca(C 2 H 3 O 2 ) 2 KNO 2 Tro, Chemistry: A Molecular Approach, 2/e

109. Classifying Salt Solutions as Acidic, Basic, or Neutral If the salt cation is the conjugate acid of a weak base and the anion is the conjugate base of a strong acid, it will form an acidic solution – NH 4 Cl If the salt cation is a highly charged metal ion and the anion is the conjugate base of a strong acid, it will form an acidic solution – Al(NO 3 ) 3 Tro, Chemistry: A Molecular Approach, 2/e

110. Classifying Salt Solutions as Acidic, Basic, or Neutral If the salt cation is the conjugate acid of a weak base and the anion is the conjugate base of a weak acid, the pH of the solution depends on the relative strengths of the acid and base – NH 4 F because HF is a stronger acid than NH 4 +, K a of NH 4 + is larger than K b of the F − ; therefore the solution will be acidic Tro, Chemistry: A Molecular Approach, 2/e

111. the solution is pH neutral Co 3+ is a highly charged cation, pH acidic Practice – Determine whether a solution of the following salts is acidic, basic, or neutral KNO 3 CoCl 3 Ba(HCO 3 ) 2 CH 3 NH 3 NO 3 K + is the counter-ion of a strong base, pH neutral NO 3 − is the counter-ion of a strong acid, pH neutral Cl − is the counter-ion of a strong acid, pH neutral the solution is pH acidic Ba 2+ is the counter-ion of a strong base, pH neutral HCO 3 − is the conjugate of a weak acid, pH basic the solution is pH basic CH 3 NH 3 + is the conjugate of a weak base, pH acidic NO 3 − is the counter-ion of a strong acid, pH neutral the solution is pH acidic Tro, Chemistry: A Molecular Approach, 2/e

112. Ionization in Polyprotic Acids Because polyprotic acids ionize in steps, each H has a separate K a K a1 > K a2 > K a3 Generally, the difference in K a values is great enough so that the second ionization does not happen to a large enough extent to affect the pH – most pH problems just do first ionization – except H 2 SO 4  use [H 2 SO 4 ] as the [H 3 O + ] for the second ionization [A 2− ] = K a2 as long as the second ionization is negligible Tro, Chemistry: A Molecular Approach, 2/e

114. Practice – What is the pH of a 0.12 M solution of carbonic acid, H 2 CO 3 ? write the reactions for the acid with water one H at a time construct an ICE table for the reaction enter the initial concentrations – assuming the second ionization is negligible H 2 CO 3 + H 2 O  HCO 3  + H 3 O + [HA][A − ][H 3 O + ] initial 0.12 0 ≈ 0 change equilibrium HCO 3 − + H 2 O  CO 3 2− + H 3 O + Tro, Chemistry: A Molecular Approach, 2/e

115. [HA][A − ][H 3 O + ] initial 0.12 00 change equilibrium Practice – What is the pH of a 0.12 M solution of carbonic acid, H 2 CO 3 ? represent the change in the concentrations in terms of x sum the columns to find the equilibrium concentrations in terms of x substitute into the equilibrium constant expression +x+x +x+x xx 0.12  x xx H 2 CO 3 + H 2 O  HCO 3  + H 3 O + Tro, Chemistry: A Molecular Approach, 2/e

116. Practice – What is the pH of a 0.12 M solution of carbonic acid, H 2 CO 3 ? determine the value of K a1 because K a1 is very small, approximate the [HA] eq = [HA] init and solve for x [HA][A − ][H 3 O + ] initial 0.12 0 ≈ 0 change −x−x+x+x+x+x equilibrium 0.012xx 0.12  x K a1 = 4.3 x 10 −7, K a2 = 5.6 x 10 −11 Tro, Chemistry: A Molecular Approach, 2/e

117. Practice – What is the pH of a 0.12 M solution of carbonic acid, H 2 CO 3 ? check if the approximation is valid by seeing if x < 5% of [H 2 CO 3 ] init K a1 for H 2 CO 3 = 4.3 x 10 −7 the approximation is valid x = 2.27 x 10 −4 [HA][A − ][H 3 O + ] initial 0.12 0 ≈ 0 change −x−x+x+x+x+x equilibrium 0.12 xx Tro, Chemistry: A Molecular Approach, 2/e

118. Practice – What is the pH of a 0.12 M solution of carbonic acid, H 2 CO 3 ? substitute x into the equilibrium concentration definitions and solve x = 2.3 x 10 −4 [HA][A − ][H 3 O + ] initial 0.12 0 ≈ 0 change −x−x+x+x+x+x equilibrium 0.12−x xx Tro, Chemistry: A Molecular Approach, 2/e

119. Practice – What is the pH of a 0.12 M solution of carbonic acid, H 2 CO 3 ? substitute [H 3 O + ] into the formula for pH and solve [HA][A − ][H 3 O + ] initial 0.12 0 ≈ 0 change −x−x+x+x+x+x equilibrium 0.120.00023 Tro, Chemistry: A Molecular Approach, 2/e

120. Practice – What is the pH of a 0.12 M solution of carbonic acid, H 2 CO 3 ? check by substituting the equilibrium concentrations back into the equilibrium constant expression and comparing the calculated K a to the given K a the values match within sig figs [HA][A − ][H 3 O + ] initial 0.12 0 ≈ 0 change −x−x+x+x+x+x equilibrium 0.120.00023 K a1 for H 2 CO 3 = 4.3 x 10 −7 Tro, Chemistry: A Molecular Approach, 2/e

121. Practice – What is the [CO 3 2− ] in a 0.12 M solution of carbonic acid, H 2 CO 3 ? write the reactions for the acid with water one H at a time construct an ICE table for the reaction enter the initial concentrations for the second ionization using the equilibrium concentrations from first ionization H 2 CO 3 + H 2 O  HCO 3  + H 3 O + [HCO 3 − ][CO 3 2− ][H 3 O + ] initial 0.00023 0 change equilibrium HCO 3 − + H 2 O  CO 3 2− + H 3 O + Tro, Chemistry: A Molecular Approach, 2/e

122. [HCO 3 − ][CO 3 2− ][H 3 O + ] initial 0.00023 0 change equilibrium Practice – What is the [CO 3 2− ] in a 0.12 M solution of carbonic acid, H 2 CO 3 ? represent the change in the concentrations in terms of x sum the columns to find the equilibrium concentrations in terms of x substitute into the equilibrium constant expression +x+x +x+x xx 2.3E−4  x x HCO 3 − + H 2 O  CO 3 2− + H 3 O + 2.3E−4 +x Tro, Chemistry: A Molecular Approach, 2/e

123. Practice – What is the [CO 3 2− ] in a 0.12 M solution of carbonic acid, H 2 CO 3 ? determine the value of K a2 because K a2 is very small, approximate the [HA] eq = [HA] init, [H 3 O + ] eq = [H 3 O + ] init, and solve for x using this approximation, it is seen that x = K a2. Therefore [CO 3 2− ] = K a2 K a1 = 4.3 x 10 −7, K a2 = 5.6 x 10 −11 [HCO 3 − ][CO 3 2− ][H 3 O + ] initial 0.00023 0 change −x−x+x+x+x+x equilibrium 2.3E−4 − x x 2.3E−4 + x [HCO 3 − ][CO 3 2− ][H 3 O + ] initial 0.00023 0 change −x−x+x+x+x+x equilibrium 2.3E−4x Tro, Chemistry: A Molecular Approach, 2/e

124. Ionization in H 2 SO 4 The ionization constants for H 2 SO 4 are H 2 SO 4 + H 2 O  HSO 4  + H 3 O + strong HSO 4  + H 2 O  SO 4 2  + H 3 O + K a2 = 1.2 x 10 −2 For most sulfuric acid solutions, the second ionization is significant and must be accounted for Because the first ionization is complete, use the given [H 2 SO 4 ] = [HSO 4 − ] initial = [H 3 O + ] initial Tro, Chemistry: A Molecular Approach, 2/e

125. Example 15.18: Find the pH of 0.0100 M H 2 SO 4 (aq) solution @ 25 °C write the reactions for the acid with water construct an ICE table for the second ionization reaction enter the initial concentrations – assuming the [HSO 4 − ] and [H 3 O + ] is ≈ [H 2 SO 4 ] [HSO 4  ][SO 4 2  ][H 3 O + ] initial 0.01000 change equilibrium HSO 4  + H 2 O  SO 4 2  + H 3 O + H 2 SO 4 + H 2 O  HSO 4  + H 3 O + Tro, Chemistry: A Molecular Approach, 2/e

126. Example 15.18: Find the pH of 0.0100 M H 2 SO 4 (aq) solution @ 25 °C represent the change in the concentrations in terms of x sum the columns to find the equilibrium concentrations in terms of x substitute into the equilibrium constant expression [HSO 4  ][SO 4 2  ][H 3 O + ] initial 0.01000 change −x−x+x+x+x+x equilibrium 0.0100 −x x Tro, Chemistry: A Molecular Approach, 2/e126

127. Example 15.18: Find the pH of 0.0100 M H 2 SO 4 (aq) solution @ 25 °C expand and solve for x using the quadratic formula K a for HSO 4 − = 0.012 Tro, Chemistry: A Molecular Approach, 2/e127

128. Example 15.18: Find the pH of 0.0100 M H 2 SO 4 (aq) solution @ 25 °C substitute x into the equilibrium concentration definitions and solve x = 0.0045 [HSO 4  ][SO 4 2  ][H 3 O + ] initial 0.01000 change −x−x+x+x+x+x equilibrium 0.0100 −xx K a for HSO 4 − = 0.012 [HSO 4  ][SO 4 2  ][H 3 O + ] initial 0.01000 change −x−x+x+x+x+x equilibrium 0.00550.00450.0145 Tro, Chemistry: A Molecular Approach, 2/e128

129. Example 15.18: Find the pH of 0.0100 M H 2 SO 4 (aq) solution @ 25 °C substitute [H 3 O + ] into the formula for pH and solve K a for HSO 4 − = 0.012 [HSO 4  ][SO 4 2  ][H 3 O + ] initial 0.01000 change −x−x+x+x+x+x equilibrium 0.00550.00450.0145 Tro, Chemistry: A Molecular Approach, 2/e129

130. Example 15.18: Find the pH of 0.0100 M H 2 SO 4 (aq) solution @ 25 °C check by substituting the equilibrium concentrations back into the equilibrium constant expression and comparing the calculated K a to the given K a the answer matches K a for HSO 4 − = 0.012 [HSO 4  ][SO 4 2  ][H 3 O + ] initial 0.01000 change −x−x+x+x+x+x equilibrium 0.00550.00450.0145 Tro, Chemistry: A Molecular Approach, 2/e

131. Strengths of Binary Acids The more  + H─X  − polarized the bond, the more acidic the bond The stronger the H─X bond, the weaker the acid Binary acid strength increases to the right across a period – acidity: H─C < H─N < H─O < H─F Binary acid strength increases down the column – acidity: H─F < H─Cl < H─Br < H─I Tro, Chemistry: A Molecular Approach, 2/e

132. Relationship between Bond Strength and Acidity Acid Bond Energy kJ/mol Type of Acid HF565weak HCl431strong HBr364strong Tro, Chemistry: A Molecular Approach, 2/e

133. Strengths of Oxyacids, H–O–Y The more electronegative the Y atom, the stronger the oxyacid – HClO > HIO – acidity of oxyacids decreases down a group same trend as binary acids – helps weakens the H–O bond The larger the oxidation number of the central atom, the stronger the oxyacid – H 2 CO 3 > H 3 BO 3 – acidity of oxyacids increases to the right across a period opposite trend of binary acids The more oxygens attached to Y, the stronger the oxyacid – further weakens and polarizes the H–O bond – HClO 3 > HClO 2 Tro, Chemistry: A Molecular Approach, 2/e

134. Relationship Between Electronegativity and Acidity Acid H─O─Y Electronegativity of Y KaKa H─O─Cl3.02.9 x 10 −8 H─O─Br2.82.0 x 10 −9 H─O─IH─O─I2.52.3 x 10 −11 Tro, Chemistry: A Molecular Approach, 2/e

135. Relationship Between Number of Oxygens on the Central Atom and Acidity Tro, Chemistry: A Molecular Approach, 2/e

136. Practice – Order the Following By Acidity (Least to Most) H 3 PO 4 HNO 3 H 3 PO 3 H 3 AsO 3 By Acidity (Least to Most) HClHBrH 2 SHS − By Basicity (Least to Most) CO 3 2− NO 3 − HCO 3 − BO 3 3− Tro, Chemistry: A Molecular Approach, 2/e136

137. Practice – Order the Following By Acidity (Least to Most) H 3 PO 4 HNO 3 H 3 PO 3 H 3 AsO 3 H 3 AsO 3 < H 3 PO 3 < H 3 PO 4 < HNO 3 By Acidity (Least to Most) HClHBrH 2 SHS − HS − < H 2 S < HCl < HBr By Basicity (Least to Most) CO 3 2− NO 3 − HCO 3 − BO 3 3− NO 3 − < HCO 3 − < CO 3 2− < BO 3 3− Tro, Chemistry: A Molecular Approach, 2/e137

138. Lewis Acid–Base Theory Lewis Acid–Base theory focuses on transferring an electron pair – lone pair  bond – bond  lone pair Does NOT require H atoms The electron donor is called the Lewis Base – electron rich, therefore nucleophile The electron acceptor is called the Lewis Acid – electron deficient, therefore electrophile Tro, Chemistry: A Molecular Approach, 2/e

139. Lewis Bases Lewis Base has electrons it is willing to give away to or share with another atom Lewis Base must have lone pair of electrons on it that it can donate Anions are better Lewis Bases than neutral atoms or molecules – N: < N: − Generally, the more electronegative an atom, the less willing it is to be a Lewis Base – O: < S: Tro, Chemistry: A Molecular Approach, 2/e

140. Lewis Acids Electron deficient, either from – being attached to electronegative atom(s) – not having an octet Must have empty orbital willing to accept the electron pair H + has empty 1s orbital B in BF 3 has empty 2p orbital and an incomplete octet Many small, highly charged metal cations have empty orbitals they can use to accept electrons Atoms that are attached to highly electronegative atoms and have multiple bonds can be Lewis Acids Tro, Chemistry: A Molecular Approach, 2/e

141. Lewis Acid–Base Reactions The base donates a pair of electrons to the acid Generally results in a covalent bond forming H 3 N: + BF 3  H 3 N─BF 3 The product that forms is called an adduct Arrhenius and Brønsted-Lowry acid–base reactions are also Lewis Tro, Chemistry: A Molecular Approach, 2/e

142. Examples of Lewis Acid–Base Reactions Tro, Chemistry: A Molecular Approach, 2/e

143. Examples of Lewis Acid–Base Reactions Ag + (aq) + 2 :NH 3(aq)  Ag(NH 3 ) 2 + (aq) Lewis Acid Lewis Base Adduct Tro, Chemistry: A Molecular Approach, 2/e

144. Practice – Identify the Lewis Acid and Lewis Base in Each Reaction Lewis Base Lewis Base Lewis Acid Lewis Acid Tro, Chemistry: A Molecular Approach, 2/e

145. U.S. Fuel Consumption Over 85% of the energy use in the United States comes from the combustion of fossil fuels – oil, natural gas, coal Combustion of fossil fuels produces CO 2 CH 4 (g) + 2 O 2 (g) → CO 2 (g) + 2 H 2 O(g) Natural fossil fuels also contain small amounts of S that burn to produce SO 2 (g) S(s) + O 2 (g) → SO 2 (g) The high temperatures of combustion allow N 2 (g) in the air to combine with O 2 (g) to form oxides of nitrogen N 2 (g) + 2 O 2 (g) → 2 NO 2 (g) Tro, Chemistry: A Molecular Approach, 2/e

146. What Is Acid Rain? Natural rain water has a pH of 5.6 – naturally slightly acidic due mainly to CO 2 Rain water with a pH lower than 5.6 is called acid rain Acid rain is linked to damage in ecosystems and structures Tro, Chemistry: A Molecular Approach, 2/e

147. What Causes Acid Rain? Many natural and pollutant gases dissolved in the air are nonmetal oxides – CO 2, SO 2, NO 2 Nonmetal oxides are acidic CO 2 (g) + H 2 O(l)  H 2 CO 3 (aq) 2 SO 2 (g) + O 2 (g) + 2 H 2 O(l)  2 H 2 SO 4 (aq) 4 NO 2 (g) + O 2 (g) + 2 H 2 O(l)  4 HNO 3 (aq) Processes that produce nonmetal oxide gases as waste increase the acidity of the rain – natural – volcanoes and some bacterial action – man-made – combustion of fuel Tro, Chemistry: A Molecular Approach, 2/e

148. pH of Rain in Different Regions Tro, Chemistry: A Molecular Approach, 2/e148

149. Weather Patterns The prevailing winds in the United States travel west to east Weather patterns may cause rain to be acidic in regions other than where the nonmetal oxide is produced Much of the northeast United States has rain of very low pH, even though it has very low sulfur emissions, due in part to the general weather patterns Tro, Chemistry: A Molecular Approach, 2/e

150. Sources of SO 2 from Utilities Tro, Chemistry: A Molecular Approach, 2/e 150

151. Damage from Acid Rain Acids react with metals, and materials that contain carbonates Acid rain damages bridges, cars, and other metallic structures Acid rain damages buildings and other structures made of limestone or cement Acidifying lakes affects aquatic life Soil acidity causes more dissolving of minerals and leaching more minerals from soil – making it difficult for trees Tro, Chemistry: A Molecular Approach, 2/e

152. Damage from Acid Rain Tro, Chemistry: A Molecular Approach, 2/e

153. Acid Rain Legislation 1990 Clean Air Act attacks acid rain – forces utilities to reduce SO 2 Result is acid rain in the Northeast stabilized and beginning to be reduced Tro, Chemistry: A Molecular Approach, 2/e