1. Chapter 15 Acids and Bases 2008, Prentice Hall Chemistry: A Molecular Approach, 1 st Ed. Nivaldo Tro Roy Kennedy Massachusetts Bay Community College Wellesley Hills, MA

2. Tro, Chemistry: A Molecular Approach2 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 if the stomach acid backs up into your esophagus, it irritates those tissues, resulting in heartburn acid reflux GERD = gastroesophageal reflux disease = chronic leaking of stomach acid into the esophagus

3. Tro, Chemistry: A Molecular Approach3 Curing 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

4. Tro, Chemistry: A Molecular Approach4 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, Chemistry: A Molecular Approach5 Common Acids

6. Tro, Chemistry: A Molecular Approach6 Structures of Acids binary acids have acid hydrogens attached to a nonmetal atom HCl, HF

7. Tro, Chemistry: A Molecular Approach7 Structure of Acids oxy acids have acid hydrogens attached to an oxygen atom H 2 SO 4, HNO 3

8. Tro, Chemistry: A Molecular Approach8 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

9. Tro, Chemistry: A Molecular Approach9 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

10. Tro, Chemistry: A Molecular Approach10 Common Bases

11. Tro, Chemistry: A Molecular Approach11 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, Chemistry: A Molecular Approach12 Indicators chemicals which change color depending on the acidity/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

13. Tro, Chemistry: A Molecular Approach13 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)

14. Tro, Chemistry: A Molecular Approach14 Arrhenius Theory HCl ionizes in water, producing H + and Cl – ions NaOH dissociates in water, producing Na + and OH – ions

15. Tro, Chemistry: A Molecular Approach15 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 a water molecule(s) 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 +

16. Tro, Chemistry: A Molecular Approach16 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)

17. Tro, Chemistry: A Molecular Approach17 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 how some ionic compounds, like Na 2 CO 3 or Na 2 O, dissolve in water to form basic solutions – even though they do not contain OH – ions does not explain why molecular substances, like CO 2, dissolve in water to form acidic solutions – even though they do not contain H + ions does not explain acid-base reactions that take place outside aqueous solution

18. Tro, Chemistry: A Molecular Approach18 Brønsted-Lowry Theory in a Brønsted-Lowry Acid-Base reaction, 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 base structure must contain an atom with an unshared pair of electrons in an acid-base reaction, the acid molecule gives an H + to the base molecule H–A + :B  :A – + H–B +

19. Tro, Chemistry: A Molecular Approach19 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 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) acidbase

20. Tro, Chemistry: A Molecular Approach20 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 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) base acid

21. Tro, Chemistry: A Molecular Approach21 Amphoteric Substances amphoteric substances can act as either an acid or a base have both transferable H and atom with lone pair 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)

22. Tro, Chemistry: A Molecular Approach22 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

23. Tro, Chemistry: A Molecular Approach23 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; and the original acid becomes the conjugate base

24. Tro, Chemistry: A Molecular Approach24 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

25. Tro, Chemistry: A Molecular Approach25 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

26. Tro, Chemistry: A Molecular Approach26 Ex 15.1a – 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 + acid baseconjugate conjugate base acid H 2 SO 4 + H 2 O  HSO 4 – +H 3 O + When the H 2 SO 4 becomes HSO 4 , it lost an H +  so H 2 SO 4 must be the acid and HSO 4  its conjugate base When the H 2 O becomes H 3 O +, it accepted an H +  so H 2 O must be the base and H 3 O + its conjugate acid

27. Tro, Chemistry: A Molecular Approach27 Ex 15.1b – Identify the Brønsted-Lowry Acids and Bases and Their Conjugates in the Reaction HCO 3 – + H 2 O  H 2 CO 3 +HO – base acidconjugate conjugate acid base HCO 3 – + H 2 O  H 2 CO 3 +HO – When the HCO 3  becomes H 2 CO 3, it accepted an H +  so HCO 3  must be the base and H 2 CO 3 its conjugate acid When the H 2 O becomes OH , it donated an H +  so H 2 O must be the acid and OH  its conjugate base

28. Tro, Chemistry: A Molecular Approach28 Practice – Write the formula for the conjugate acid of the following H 2 O NH 3 CO 3 2− H 2 PO 4 1−

29. Tro, Chemistry: A Molecular Approach29 Practice – Write the formula for the conjugate acid of the following H 2 OH 3 O + NH 3 NH 4 + CO 3 2− HCO 3 − H 2 PO 4 1− H 3 PO 4

30. Tro, Chemistry: A Molecular Approach30 Practice – Write the formula for the conjugate base of the following H 2 O NH 3 CO 3 2− H 2 PO 4 1−

31. Tro, Chemistry: A Molecular Approach31 Practice – Write the formula for the conjugate base of the following H 2 OHO − NH 3 NH 2 − CO 3 2− since CO 3 2− does not have an H, it cannot be an acid H 2 PO 4 1− HPO 4 2−

32. Tro, Chemistry: A Molecular Approach32 Arrow Conventions chemists commonly use two kinds of arrows in reactions to indicate the degree of completion of the reactions a single arrow indicates all the reactant molecules are converted to product molecules at the end a double arrow indicates the reaction stops when only some of the reactant molecules have been converted into products  in these notes

33. Tro, Chemistry: A Molecular Approach33 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, 

34. Tro, Chemistry: A Molecular Approach34 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 H’s 100% ionized in water strong electrolyte [H 3 O + ] = [strong acid] HCl  H + + Cl - HCl + H 2 O  H 3 O + + Cl -

35. Tro, Chemistry: A Molecular Approach35 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] HF  H + + F - HF + H 2 O  H 3 O + + F -

36. Tro, Chemistry: A Molecular Approach36 Polyprotic Acids often acid molecules 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 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 

37. Tro, Chemistry: A Molecular Approach37 Increasing Acidity Increasing Basicity

38. Tro, Chemistry: A Molecular Approach38 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 -1 + H 3 O +1 Base: + H 2 O  HBase +1 + OH -1 the farther the equilibrium position lies to 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

39. Tro, Chemistry: A Molecular Approach39 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 +1 > H 2 O > OH -1 ; NH 4 +1 > NH 3 > NH 2 -1 base trend opposite

40. Tro, Chemistry: A Molecular Approach40 Acid Ionization Constant, K a acid strength measured by the size of the equilibrium constant when react with H 2 O HAcid + H 2 O  Acid -1 + H 3 O +1 the equilibrium constant is called the acid ionization constant, K a larger K a = stronger acid

41. 41

43. Tro, Chemistry: A Molecular Approach43 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 – ] = 10 -7 M @ 25°C

44. Tro, Chemistry: A Molecular Approach44 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 – ] = K w = 1 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

45. Tro, Chemistry: A Molecular Approach45 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 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

46. Example 15.2b – Calculate the [OH  ] at 25°C when the [H 3 O + ] = 1.5 x 10 -9 M, and determine if the solution is acidic, basic, or neutral The units are correct. The fact that the [H 3 O + ] < [OH  ] means the solution is basic [H 3 O + ] = 1.5 x 10 -9 M [OH  ] Check: Solution: Concept Plan: Relationships: Given: Find: [H 3 O + ][OH  ]

47. Tro, Chemistry: A Molecular Approach47 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

48. Tro, Chemistry: A Molecular Approach48 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

49. Tro, Chemistry: A Molecular Approach49 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

50. Tro, Chemistry: A Molecular Approach50 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... since 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

51. 51 pH 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 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)

52. 52 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 (sat’d Mg(OH) 2 )10.5 household ammonia10.5 to 11.5 1.0 M NaOH14

53. Tro, Chemistry: A Molecular Approach53

54. Example 15.3b – Calculate the pH at 25°C when the [OH  ] = 1.3 x 10 -2 M, and determine if the solution is acidic, basic, or neutral pH is unitless. The fact that the pH > 7 means the solution is basic [OH  ] = 1.3 x 10 -2 M pH Check: Solution: Concept Plan: Relationships: Given: Find: [H 3 O + ][OH  ]pH

55. Tro, Chemistry: A Molecular Approach55 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

56. Tro, Chemistry: A Molecular Approach56 pH and pOH Complete the Table 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 pH pOH

57. Tro, Chemistry: A Molecular Approach57 pH and pOH Complete the Table 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 pH 0 1 3 5 7 9 11 13 14 pOH 14 13 11 9 7 5 3 1 0

58. Tro, Chemistry: A Molecular Approach58 Relationship between pH and pOH the sum of the pH and pOH of a solution = 14.00 at 25°C can use pOH to find pH of a solution

59. Tro, Chemistry: A Molecular Approach59 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

60. Tro, Chemistry: A Molecular Approach60 Finding the pH of a Strong Acid there are two sources of H 3 O + in an aqueous solution of a strong acid – the acid and the water for the strong acid, the contribution of the water to the total [H 3 O + ] is negligible shifts the K w equilibrium to the left so far that [H 3 O + ] water is too small to be significant  except in very dilute solutions, generally < 1 x 10 -4 M for a monoprotic strong acid [H 3 O + ] = [HAcid] for polyprotic acids, the other ionizations can generally be ignored 0.10 M HCl has [H 3 O + ] = 0.10 M and pH = 1.00

61. Tro, Chemistry: A Molecular Approach61 Finding the pH of a Weak Acid there are also two sources of H 3 O + in and 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 +

62. Tro, Chemistry: A Molecular Approach62 [HNO 2 ][NO 2 - ][H 3 O + ] initial change equilibrium Ex 15.6 Find the pH of 0.200 M HNO 2 (aq) solution @ 25°C 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 since no products initially, Q c = 0, and the reaction is proceeding forward HNO 2 + H 2 O  NO 2  + H 3 O + [HNO 2 ][NO 2 - ][H 3 O + ] initial 0.200 0 ≈ 0 change equilibrium

63. Tro, Chemistry: A Molecular Approach63 [HNO 2 ][NO 2 - ][H 3 O + ] initial 0.200 00 change equilibrium Ex 15.6 Find the pH of 0.200 M HNO 2 (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 +x+x +x+x xx 0.200  x xx

64. Tro, Chemistry: A Molecular Approach64 Ex 15.6 Find the pH of 0.200 M HNO 2 (aq) solution @ 25°C determine the value of K a from Table 15.5 since K a is very small, approximate the [HNO 2 ] eq = [HNO 2 ] init and solve for x K a for HNO 2 = 4.6 x 10 -4 [HNO 2 ][NO 2 - ][H 3 O + ] initial 0.200 0 ≈ 0 change -x-x+x+x+x+x equilibrium 0.200xx 0.200  x

65. Tro, Chemistry: A Molecular Approach65 Ex 15.6 Find the pH of 0.200 M HNO 2 (aq) solution @ 25°C K a for HNO 2 = 4.6 x 10 -4 [HNO 2 ][NO 2 - ][H 3 O + ] initial 0.200 0 ≈ 0 change -x-x+x+x+x+x equilibrium 0.200 xx check if the approximation is valid by seeing if x < 5% of [HNO 2 ] init the approximation is valid x = 9.6 x 10 -3

66. Tro, Chemistry: A Molecular Approach66 Ex 15.6 Find the pH of 0.200 M HNO 2 (aq) solution @ 25°C K a for HNO 2 = 4.6 x 10 -4 [HNO 2 ][NO 2 - ][H 3 O + ] initial 0.200 0 ≈ 0 change -x-x+x+x+x+x equilibrium 0.200-x xx x = 9.6 x 10 -3 substitute x into the equilibrium concentration definitions and solve [HNO 2 ][NO 2 - ][H 3 O + ] initial 0.200 0 ≈ 0 change -x-x+x+x+x+x equilibrium 0.1900.0096

67. Tro, Chemistry: A Molecular Approach67 Ex 15.6 Find the pH of 0.200 M HNO 2 (aq) solution @ 25°C K a for HNO 2 = 4.6 x 10 -4 substitute [H 3 O + ] into the formula for pH and solve [HNO 2 ][NO 2 - ][H 3 O + ] initial 0.200 0 ≈ 0 change -x-x+x+x+x+x equilibrium 0.1900.0096

68. Tro, Chemistry: A Molecular Approach68 Ex 15.6 Find the pH of 0.200 M HNO 2 (aq) solution @ 25°C K a for HNO 2 = 4.6 x 10 -4 [HNO 2 ][NO 2 - ][H 3 O + ] initial 0.200 0 ≈ 0 change -x-x+x+x+x+x equilibrium 0.1900.0096 check by substituting the equilibrium concentrations back into the equilibrium constant expression and comparing the calculated K a to the given K a though not exact, the answer is reasonably close

69. Tro, Chemistry: A Molecular Approach69 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)

70. Tro, Chemistry: A Molecular Approach70 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

71. Tro, Chemistry: A Molecular Approach71 [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 +

72. Tro, Chemistry: A Molecular Approach72 determine the value of K a since K a is very small, approximate the [HA] eq = [HA] init and solve for x [HA][A 2 - ][H 3 O + ] initial 0.012 0 ≈ 0 change -x-x+x+x+x+x equilibrium 0.012xx 0.012  x 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 HC 6 H 4 NO 2 + H 2 O  C 6 H 4 NO 2  + H 3 O +

73. Tro, Chemistry: A Molecular Approach73 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 K a for HC 6 H 4 NO 2 = 1.4 x 10 -5 check if the approximation is valid by seeing if x < 5% of [HC 6 H 4 NO 2 ] init the approximation is valid x = 4.1 x 10 -4 [HA][A 2 - ][H 3 O + ] initial 0.012 0 ≈ 0 change -x-x+x+x+x+x equilibrium 0.012 xx

74. Tro, Chemistry: A Molecular Approach74 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 x = 4.1 x 10 -4 substitute x into the equilibrium concentration definitions and solve [HA][A 2 - ][H 3 O + ] initial 0.012 0 ≈ 0 change -x-x+x+x+x+x equilibrium 0.012-x xx

75. Tro, Chemistry: A Molecular Approach75 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 2 - ][H 3 O + ] initial 0.012 0 ≈ 0 change -x-x+x+x+x+x equilibrium 0.0120.00041

76. Tro, Chemistry: A Molecular Approach76 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 2 - ][H 3 O + ] initial 0.012 0 ≈ 0 change -x-x+x+x+x+x equilibrium 0.0120.00041

77. Tro, Chemistry: A Molecular Approach77 [HClO 2 ][ClO 2 - ][H 3 O + ] initial 0.100 0≈ 0 change equilibrium Ex 15.7 Find the pH of 0.100 M HClO 2 (aq) solution @ 25°C 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 HClO 2 + H 2 O  ClO 2  + H 3 O +

78. Tro, Chemistry: A Molecular Approach78 Ex 15.7 Find the pH of 0.100 M HClO 2 (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 [HClO 2 ][ClO 2 - ][H 3 O + ] initial 0.100 0≈ 0 change -x-x+x+x+x+x equilibrium 0.100-xxx

79. Tro, Chemistry: A Molecular Approach79 Ex 15.7 Find the pH of 0.100 M HClO 2 (aq) solution @ 25°C determine the value of K a from Table 15.5 since K a is very small, approximate the [HClO 2 ] eq = [HClO 2 ] init and solve for x K a for HClO 2 = 1.1 x 10 -2 [HClO 2 ][ClO 2 - ][H 3 O + ] initial 0.100 0≈ 0 change -x-x+x+x+x+x equilibrium 0.100-xxx

80. Tro, Chemistry: A Molecular Approach80 Ex 15.7 Find the pH of 0.100 M HClO 2 (aq) solution @ 25°C K a for HClO 2 = 1.1 x 10 -2 check if the approximation is valid by seeing if x < 5% of [HNO 2 ] init the approximation is invalid x = 3.3 x 10 -2 [HClO 2 ][ClO 2 - ][H 3 O + ] initial 0.100 0≈ 0 change -x-x+x+x+x+x equilibrium 0.100-xxx

81. Tro, Chemistry: A Molecular Approach81 Ex 15.7 Find the pH of 0.100 M HClO 2 (aq) solution @ 25°C K a for HClO 2 = 1.1 x 10 -2 if the approximation is invalid, solve for x using the quadratic formula

82. Tro, Chemistry: A Molecular Approach82 Ex 15.7 Find the pH of 0.100 M HClO 2 (aq) solution @ 25°C K a for HClO 2 = 1.1 x 10 -2 x = 0.028 substitute x into the equilibrium concentration definitions and solve [HClO 2 ][ClO 2 - ][H 3 O + ] initial 0.100 0≈ 0 change -x-x+x+x+x+x equilibrium 0.100-xxx [HClO 2 ][ClO 2 - ][H 3 O + ] initial 0.100 0≈ 0 change -x-x+x+x+x+x equilibrium 0.0720.028

83. Tro, Chemistry: A Molecular Approach83 Ex 15.7 Find the pH of 0.100 M HClO 2 (aq) solution @ 25°C K a for HClO 2 = 1.1 x 10 -2 substitute [H 3 O + ] into the formula for pH and solve [HClO 2 ][ClO 2 - ][H 3 O + ] initial 0.100 0≈ 0 change -x-x+x+x+x+x equilibrium 0.0720.028

84. Tro, Chemistry: A Molecular Approach84 Ex 15.7 Find the pH of 0.100 M HClO 2 (aq) solution @ 25°C K a for HClO 2 = 1.1 x 10 -2 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 [HClO 2 ][ClO 2 - ][H 3 O + ] initial 0.100 0≈ 0 change -x-x+x+x+x+x equilibrium 0.0720.028

85. Tro, Chemistry: A Molecular Approach85 Ex 15.8 - What is the K a of a weak acid if a 0.100 M solution has a pH of 4.25? 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 0.100 0 ≈ 0 change equilibrium 5.6E-05

86. Tro, Chemistry: A Molecular Approach86 [HA][A - ][H 3 O + ] initial 0.100 00 change equilibrium Ex 15.8 - What is the K a of a weak acid if a 0.100 M solution has a pH of 4.25? 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 +5.6E-05 −5.6E-05 0.100  5.6E-05 HA + H 2 O  A  + H 3 O + 0.100

87. Tro, Chemistry: A Molecular Approach87 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 since [ionized acid] equil = [H 3 O + ] equil

88. Tro, Chemistry: A Molecular Approach88 Ex 15.9 - What is the percent ionization of a 2.5 M HNO 2 solution? Write the reaction for the acid with water Construct an ICE table for the reaction Enter the Initial Concentrations Define the Change in Concentration in terms of x Sum the columns to define the Equilibrium Concentrations HNO 2 + H 2 O  NO 2  + H 3 O + [HNO 2 ][NO 2 - ][H 3 O + ] initial change equilibrium [HNO 2 ][NO 2 - ][H 3 O + ] initial 2.5 0 ≈ 0 change equilibrium +x+x +x+x xx 2.5  x xx

89. Tro, Chemistry: A Molecular Approach89 determine the value of K a from Table 15.5 since K a is very small, approximate the [HNO 2 ] eq = [HNO 2 ] init and solve for x K a for HNO 2 = 4.6 x 10 -4 [HNO 2 ][NO 2 - ][H 3 O + ] initial 2.5 0≈ 0 change -x-x+x+x+x+x equilibrium 2.5-x ≈2.5xx Ex 15.9 - What is the percent ionization of a 2.5 M HNO 2 solution?

90. Tro, Chemistry: A Molecular Approach90 Ex 15.9 - What is the percent ionization of a 2.5 M HNO 2 solution? HNO 2 + H 2 O  NO 2  + H 3 O + [HNO 2 ][NO 2 - ][H 3 O + ] initial 2.5 0 ≈ 0 change -x-x+x+x+x+x equilibrium 2.50.034 2.5  x xx substitute x into the Equilibrium Concentration definitions and solve x = 3.4 x 10 -2

91. Tro, Chemistry: A Molecular Approach91 Ex 15.9 - What is the percent ionization of a 2.5 M HNO 2 solution? HNO 2 + H 2 O  NO 2  + H 3 O + [HNO 2 ][NO 2 - ][H 3 O + ] initial 2.5 0 ≈ 0 change -x-x+x+x+x+x equilibrium 2.50.034 Apply the Definition and Compute the Percent Ionization since the percent ionization is < 5%, the “x is small” approximation is valid

92. Tro, Chemistry: A Molecular Approach92 Relationship Between [H 3 O + ] equilibrium & [HA] initial increasing the initial concentration of acid results in increased H 3 O + concentration 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

93. 93 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

94. Tro, Chemistry: A Molecular Approach94 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

95. Tro, Chemistry: A Molecular Approach95 Ex 15.10 Find the pH of a mixture of 0.150 M HF(aq) solution and 0.100 M HClO 2 (aq) Write the reactions for the acids with water and determine their K a s If the K a s are sufficiently different, use the strongest acid to construct an ICE table for the reaction Enter the initial concentrations – assuming the [H 3 O + ] from water is ≈ 0 HF + H 2 O  F  + H 3 O + K a = 3.5 x 10 -4 [HF][F - ][H 3 O + ] initial 0.150 0 ≈ 0 change equilibrium HClO + H 2 O  ClO  + H 3 O + K a = 2.9 x 10 -8 H 2 O + H 2 O  OH  + H 3 O + K w = 1.0 x 10 -14

96. Tro, Chemistry: A Molecular Approach96 [HF][F - ][H 3 O + ] initial 0.150 00 change equilibrium Ex 15.10 Find the pH of a mixture of 0.150 M HF(aq) solution and 0.100 M HClO 2 (aq) 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 0.150  x xx

97. Tro, Chemistry: A Molecular Approach97 Ex 15.10 Find the pH of a mixture of 0.150 M HF(aq) solution and 0.100 M HClO 2 (aq) determine the value of K a for HF since K a is very small, approximate the [HF] eq = [HF] init and solve for x K a for HF = 3.5 x 10 -4 [HF][F - ][H 3 O + ] initial 0.150 0 ≈ 0 change -x-x+x+x+x+x equilibrium 0.150xx 0.150  x

98. Tro, Chemistry: A Molecular Approach98 Ex 15.10 Find the pH of a mixture of 0.150 M HF(aq) solution and 0.100 M HClO 2 (aq) K a for HF = 3.5 x 10 -4 [HF][F - ][H 3 O + ] initial 0.150 0 ≈ 0 change -x-x+x+x+x+x equilibrium 0.150 xx check if the approximation is valid by seeing if x < 5% of [HF] init the approximation is valid x = 7.2 x 10 -3

99. Tro, Chemistry: A Molecular Approach99 Ex 15.10 Find the pH of a mixture of 0.150 M HF(aq) solution and 0.100 M HClO 2 (aq) K a for HF = 3.5 x 10 -4 [HF][F - ][H 3 O + ] initial 0.150 0 ≈ 0 change -x-x+x+x+x+x equilibrium 0.150-x xx x = 7.2 x 10 -3 substitute x into the equilibrium concentration definitions and solve [HF][F - ][H 3 O + ] initial 0.150 0 ≈ 0 change -x-x+x+x+x+x equilibrium 0.1430.0072

100. Tro, Chemistry: A Molecular Approach100 Ex 15.10 Find the pH of a mixture of 0.150 M HF(aq) solution and 0.100 M HClO 2 (aq) K a for HF = 3.5 x 10 -4 substitute [H 3 O + ] into the formula for pH and solve [HF][F - ][H 3 O + ] initial 0.150 0 ≈ 0 change -x-x+x+x+x+x equilibrium 0.1430.0072

101. Tro, Chemistry: A Molecular Approach101 Ex 15.10 Find the pH of a mixture of 0.150 M HF(aq) solution and 0.100 M HClO 2 (aq) K a for HF = 3.5 x 10 -4 check by substituting the equilibrium concentrations back into the equilibrium constant expression and comparing the calculated K a to the given K a though not exact, the answer is reasonably close [HF][F - ][H 3 O + ] initial 0.150 0 ≈ 0 change -x-x+x+x+x+x equilibrium 0.1430.0072

102. Tro, Chemistry: A Molecular Approach102 NaOH  Na + + OH - Strong Bases the stronger the base, the more willing it is to accept H use water as the standard acid for strong bases, practically all molecules are dissociated into OH – or accept H’s strong electrolyte multi-OH strong bases completely dissociated [HO – ] = [strong base] x (# OH)

103. Example 15.11b – Calculate the pH at 25°C of a 0.0015 M Sr(OH) 2 solution and determine if the solution is acidic, basic, or neutral pH is unitless. The fact that the pH > 7 means the solution is basic [Sr(OH) 2 ] = 1.5 x 10 -3 M pH Check: Solution: Concept Plan: Relationships: Given: Find: [H 3 O + ][OH  ]pH[Sr(OH) 2 ] [OH  ]=2[Sr(OH) 2 ] [OH  ] = 2(0.0015) = 0.0030 M

104. Example 15.11b (alternative) – Calculate the pH at 25°C of a 0.0015 M Sr(OH) 2 solution and determine if the solution is acidic, basic, or neutral pH is unitless. The fact that the pH > 7 means the solution is basic [Sr(OH) 2 ] = 1.5 x 10 -3 M pH Check: Solution: Concept Plan: Relationships: Given: Find: pOH[OH  ]pH[Sr(OH) 2 ] [OH  ]=2[Sr(OH) 2 ] [OH  ] = 2(0.0015) = 0.0030 M

105. Tro, Chemistry: A Molecular Approach105 Practice - Calculate the pH of a 0.0010 M Ba(OH) 2 solution and determine if it is acidic, basic, or neutral

106. Tro, Chemistry: A Molecular Approach106 Practice - Calculate the pH of a 0.0010 M Ba(OH) 2 solution and determine if it is acidic, basic, or neutral [H 3 O + ] = 1.00 x 10 -14 2.0 x 10 -3 = 5.0 x 10 -12 M pH > 7 therefore basic Ba(OH) 2 = Ba 2+ + 2 OH - therefore [OH - ] = 2 x 0.0010 = 0.0020 = 2.0 x 10 -3 M pH = -log [H 3 O + ] = -log (5.0 x 10 -12 ) pH = 11.30 K w = [H 3 O + ][OH  ]

107. Tro, Chemistry: A Molecular Approach107 Weak Bases 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 NH 3 + H 2 O  NH 4 + + OH -

108. Tro, Chemistry: A Molecular Approach108

109. Tro, Chemistry: A Molecular Approach109 Structure of Amines

110. Tro, Chemistry: A Molecular Approach110 [NH 3 ][NH 4 + ][OH  ] initial change equilibrium Ex 15.12 Find the pH of 0.100 M NH 3 (aq) 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 since no products initially, Q c = 0, and the reaction is proceeding forward NH 3 + H 2 O  NH 4 + + OH  [NH 3 ][NH 4 + ][OH  ] initial 0.100 0 ≈ 0 change equilibrium

111. Tro, Chemistry: A Molecular Approach111 [NH 3 ][NH 4 + ][OH  ] initial 0.100 00 change equilibrium Ex 15.12 Find the pH of 0.100 M NH 3 (aq) 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 0.100  x xx

112. Tro, Chemistry: A Molecular Approach112 Ex 15.12 Find the pH of 0.100 M NH 3 (aq) solution determine the value of K b from Table 15.8 since K b is very small, approximate the [NH 3 ] eq = [NH 3 ] init and solve for x K b for NH 3 = 1.76 x 10 -5 [NH 3 ][NH 4 + ][OH  ] initial 0.100 0 ≈ 0 change -x-x+x+x+x+x equilibrium 0.100xx 0.100  x

113. Tro, Chemistry: A Molecular Approach113 Ex 15.12 Find the pH of 0.100 M NH 3 (aq) solution K b for NH 3 = 1.76 x 10 -5 [NH 3 ][NH 4 + ][OH  ] initial 0.100 0 ≈ 0 change -x-x+x+x+x+x equilibrium 0.100 xx check if the approximation is valid by seeing if x < 5% of [NH 3 ] init the approximation is valid x = 1.33 x 10 -3

114. Tro, Chemistry: A Molecular Approach114 Ex 15.12 Find the pH of 0.100 M NH 3 (aq) solution substitute x into the equilibrium concentration definitions and solve K b for NH 3 = 1.76 x 10 -5 [NH 3 ][NH 4 + ][OH  ] initial 0.100 0 ≈ 0 change -x-x+x+x+x+x equilibrium 0.100  x xx x = 1.33 x 10 -3 [NH 3 ][NH 4 + ][OH  ] initial 0.100 0 ≈ 0 change -x-x+x+x+x+x equilibrium 0.0991.33E-3

115. Tro, Chemistry: A Molecular Approach115 Ex 15.12 Find the pH of 0.100 M NH 3 (aq) 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 NH 3 = 1.76 x 10 -5 [NH 3 ][NH 4 + ][OH  ] initial 0.100 0 ≈ 0 change -x-x+x+x+x+x equilibrium 0.0991.33E-3

116. Tro, Chemistry: A Molecular Approach116 Ex 15.12 Find the pH of 0.100 M NH 3 (aq) solution use the [OH - ] to find the pOH use pOH to find pH K b for NH 3 = 1.76 x 10 -5 [NH 3 ][NH 4 + ][OH  ] initial 0.100 0 ≈ 0 change -x-x+x+x+x+x equilibrium 0.0991.33E-3

117. Tro, Chemistry: A Molecular Approach117 Ex 15.12 Find the pH of 0.100 M NH 3 (aq) solution K b for NH 3 = 1.76 x 10 -5 [NH 3 ][NH 4 + ][OH  ] initial 0.100 0 ≈ 0 change -x-x+x+x+x+x equilibrium 0.0991.33E-3 check by substituting the equilibrium concentrations back into the equilibrium constant expression and comparing the calculated K b to the given K b though not exact, the answer is reasonably close

118. Tro, Chemistry: A Molecular Approach118 Practice – Find the pH of a 0.0015 M morphine solution, K b = 1.6 x 10 -6

119. Tro, Chemistry: A Molecular Approach119 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 since 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

120. Tro, Chemistry: A Molecular Approach120 [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

121. Tro, Chemistry: A Molecular Approach121 Practice – Find the pH of a 0.0015 M morphine solution determine the value of K b since 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

122. Tro, Chemistry: A Molecular Approach122 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

123. Tro, Chemistry: A Molecular Approach123 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

124. Tro, Chemistry: A Molecular Approach124 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

125. Tro, Chemistry: A Molecular Approach125 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

126. Tro, Chemistry: A Molecular Approach126 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

127. Tro, Chemistry: A Molecular Approach127 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

128. Tro, Chemistry: A Molecular Approach128 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)  since HCl is a strong acid, this equilibrium lies practically completely to the left an anion that is the conjugate base of a weak acid is basic F − (aq) + H 2 O(l)  HF(aq) + OH − (aq)  since HF is a weak acid, the position of this equilibrium favors the right

129. Tro, Chemistry: A Molecular Approach129 Ex 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

130. 130 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

131. Tro, Chemistry: A Molecular Approach131 Ex 15.14 Find the pH of 0.100 M NaCHO 2 (aq) solution Na + is the cation of a strong base – pH neutral. The CHO 2 − is the anion of a weak acid – pH 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 CHO 2 − + H 2 O  HCHO 2 + OH  [CHO 2 − ][HCHO 2 ][OH  ] initial 0.1000≈ 0 change equilibrium

132. 132 0.100  x Ex 15.14 Find the pH of 0.100 M NaCHO 2 (aq) solution represent the change in the concentrations in terms of x sum the columns to find the equilibrium concentrations in terms of x Calculate the value of K b from the value of K a of the weak acid from Table 15.5 substitute into the equilibrium constant expression +x+x+x+x xx xx [CHO 2 − ][HCHO 2 ][OH  ] initial 0.1000≈ 0 change equilibrium

133. Tro, Chemistry: A Molecular Approach133 Ex 15.14 Find the pH of 0.100 M NaCHO 2 (aq) solution since K b is very small, approximate the [CHO 2 − ] eq = [CHO 2 − ] init and solve for x K b for CHO 2 − = 5.6 x 10 -11 [CHO 2 − ][HCHO 2 ][OH  ] initial 0.1000≈ 0 change -x-x+x+x+x+x equilibrium 0.100xx 0.100  x

134. Tro, Chemistry: A Molecular Approach134 Ex 15.14 Find the pH of 0.100 M NaCHO 2 (aq) solution check if the approximation is valid by seeing if x < 5% of [CHO 2 − ] init the approximation is valid x = 2.4 x 10 -6 K b for CHO 2 − = 5.6 x 10 -11 [CHO 2 − ][HCHO 2 ][OH  ] initial 0.1000≈ 0 change -x-x+x+x+x+x equilibrium 0.100 xx

135. Tro, Chemistry: A Molecular Approach135 Ex 15.14 Find the pH of 0.100 M NaCHO 2 (aq) solution substitute x into the equilibrium concentration definitions and solve x = 2.4 x 10 -6 K b for CHO 2 − = 5.6 x 10 -11 [CHO 2 − ][HCHO 2 ][OH  ] initial 0.1000≈ 0 change -x-x+x+x+x+x equilibrium 0.100 −x xx [CHO 2 − ][HCHO 2 ][OH  ] initial 0.1000≈ 0 change -x-x+x+x+x+x equilibrium 0.1002.4E-6

136. Tro, Chemistry: A Molecular Approach136 Ex 15.14 Find the pH of 0.100 M NaCHO 2 (aq) 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 CHO 2 − = 5.6 x 10 -11 [CHO 2 − ][HCHO 2 ][OH  ] initial 0.1000≈ 0 change -x-x+x+x+x+x equilibrium 0.1002.4E-6

137. Tro, Chemistry: A Molecular Approach137 Ex 15.14 Find the pH of 0.100 M NaCHO 2 (aq) solution use the [OH - ] to find the pOH use pOH to find pH K b for CHO 2 − = 5.6 x 10 -11 [CHO 2 − ][HCHO 2 ][OH  ] initial 0.1000≈ 0 change -x-x+x+x+x+x equilibrium 0.1002.4E-6

138. Tro, Chemistry: A Molecular Approach138 Ex 15.14 Find the pH of 0.100 M NaCHO 2 (aq) solution check by substituting the equilibrium concentrations back into the equilibrium constant expression and comparing the calculated K b to the given K b though not exact, the answer is reasonably close K b for CHO 2 − = 5.6 x 10 -11 [CHO 2 − ][HCHO 2 ][OH  ] initial 0.1000≈ 0 change -x-x+x+x+x+x equilibrium 0.1002.4E-6

139. Tro, Chemistry: A Molecular Approach139 Polyatomic Cations as Weak Acids some cations can be thought of as the conjugate acid of a base others are the counterions of a strong base therefore, some cation can potentially be an acid 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 counterion 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)  since NH 3 is a weak base, the position of this equilibrium favors the right

140. Tro, Chemistry: A Molecular Approach140 Metal Cations as Weak Acids cations of small, highly charged metals are weakly acidic alkali metal cations and alkali earth metal cations 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)

141. Tro, Chemistry: A Molecular Approach141 Ex 15.15 - Determine if the Given Cation Is Acidic or Neutral a) C 5 N 5 NH 2 + the conjugate acid of a weak base, therefore acidic b) Ca 2+ the counterion of a strong base, therefore neutral c) Cr 3+ a highly charged metal ion, therefore acidic

142. Tro, Chemistry: A Molecular Approach142 Classifying Salt Solutions as Acidic, Basic, or Neutral if the salt cation is the counterion 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 counterion 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

143. Tro, Chemistry: A Molecular Approach143 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

144. Tro, Chemistry: A Molecular Approach144 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 since 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

145. Tro, Chemistry: A Molecular Approach145 Ex 15.16 - Determine whether a solution of the following salts is acidic, basic, or neutral a) SrCl 2 Sr 2+ is the counterion of a strong base, pH neutral Cl − is the conjugate base of a strong acid, pH neutral solution will be pH neutral b) AlBr 3 Al 3+ is a small, highly charged metal ion, weak acid Cl − is the conjugate base of a strong acid, pH neutral solution will be acidic c) CH 3 NH 3 NO 3 CH 3 NH 3 + is the conjugate acid of a weak base, acidic NO 3 − is the conjugate base of a strong acid, pH neutral solution will be acidic

146. Tro, Chemistry: A Molecular Approach146 Ex 15.16 - Determine whether a solution of the following salts is acidic, basic, or neutral d) NaCHO 2 Na + is the counterion of a strong base, pH neutral CHO 2 − is the conjugate base of a weak acid, basic solution will be basic e) NH 4 F NH 4 + is the conjugate acid of a weak base, acidic F − is the conjugate base of a weak acid, basic K a (NH 4 + ) > K b (F − ); solution will be acidic

147. Tro, Chemistry: A Molecular Approach147 Polyprotic Acids since 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

148. 148

149. Tro, Chemistry: A Molecular Approach149 [HSO 4  ][SO 4 2  ][H 3 O + ] initial 0.0100 0 change equilibrium Ex 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 reaction Enter the initial concentrations – assuming the [HSO 4 − ] and [H 3 O + ] is ≈ [H 2 SO 4 ] HSO 4  + H 2 O  SO 4 2  + H 3 O + H 2 SO 4 + H 2 O  HSO 4  + H 3 O +

150. Tro, Chemistry: A Molecular Approach150 Ex 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 −xx

151. Tro, Chemistry: A Molecular Approach151 Ex 15.18 Find the pH of 0.0100 M H 2 SO 4 (aq) solution @ 25°C K a for HSO 4 − = 0.012 expand and solve for x using the quadratic formula

152. Tro, Chemistry: A Molecular Approach152 Ex 15.18 Find the pH of 0.0100 M H 2 SO 4 (aq) solution @ 25°C x = 0.0045 substitute x into the equilibrium concentration definitions and solve [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

153. Tro, Chemistry: A Molecular Approach153 Ex 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

154. Tro, Chemistry: A Molecular Approach154 Ex 15.7 Find the pH of 0.100 M HClO 2 (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

155. Tro, Chemistry: A Molecular Approach155 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 H-C < H-N < H-O < H-F binary acid strength increases down the column H-F < H-Cl < H-Br < H-I

156. Tro, Chemistry: A Molecular Approach156 Strengths of Oxyacids, H-O-Y the more electronegative the Y atom, the stronger the acid helps weakens the H-O bond the more oxygens attached to Y, the stronger the acid further weakens and polarizes the H-O bond

157. Tro, Chemistry: A Molecular Approach157 Lewis Acid - Base Theory electron sharing electron donor = Lewis Base = nucleophile must have a lone pair of electrons electron acceptor = Lewis Acid = electrophile electron deficient when Lewis Base gives electrons from lone pair to Lewis Acid, a covalent bond forms between the molecules Nucleophile : + Electrophile  Nucleophile : Electrophile product called an adduct other acid-base reactions also Lewis

158. Tro, Chemistry: A Molecular Approach158 Example - Complete the Following Lewis Acid-Base Reactions Label the Nucleophile and Electrophile OH HCH  + OH -1  OH HCH OH HCH  + OH -1  Electrophile Nucleophile

159. Tro, Chemistry: A Molecular Approach159 Practice - Complete the Following Lewis Acid-Base Reactions Label the Nucleophile and Electrophile BF 3 + HF  CaO + SO 3  KI + I 2 

160. Tro, Chemistry: A Molecular Approach160 Practice - Complete the Following Lewis Acid-Base Reactions Label the Nucleophile and Electrophile BF 3 + HF  H +1 BF 4 -1 CaO + SO 3  Ca +2 SO 4 -2 KI + I 2  KI 3 F B F F H F + F B F F F H +1 Nuc Ele c O S O O Ca +2 O -2 + O S O O O -2 Ca +2 ElecNuc I K +1 I -1 + ElecNuc K +1 I I I -1

161. Tro, Chemistry: A Molecular Approach161 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

162. Tro, Chemistry: A Molecular Approach162 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 + H 2 O  H 2 CO 3 2 SO 2 + O 2 + 2 H 2 O  2 H 2 SO 4 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 weather patterns may cause rain to be acidic in regions other than where the nonmetal oxide is produced

163. pH of Rain in Different Regions

164. Sources of SO 2 from Utilities

165. Tro, Chemistry: A Molecular Approach165 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 affecting aquatic life dissolving and leaching more minerals from soil making it difficult for trees

166. Tro, Chemistry: A Molecular Approach166 Acid Rain Legislation 1990 Clean Air Act attacks acid rain force utilities to reduce SO 2 result is acid rain in northeast stabilized and beginning to be reduced

167. 167 Damage from Acid Rain