1. Chapter 15 Acids and Bases 2008, Prentice Hall Chemistry: A Molecular Approach, 1 st Ed. Nivaldo Tro

2. Tro, Chemistry: A Molecular Approach 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 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 Approach 3 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 Approach 4 Properties of Acids sour taste react with “active” metals – i.e., Al, Zn, Fe, but not Cu, Ag, or Au 2 Al + 6 HCl  AlCl 3 + 3 H 2 – corrosive react with carbonates, producing CO 2 – marble, baking soda, chalk, limestone CaCO 3 + 2 HCl  CaCl 2 + CO 2 + H 2 O change color of vegetable dyes – blue litmus turns red react with bases to form ionic salts

5. Tro, Chemistry: A Molecular Approach 5 Common Acids

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

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

8. Tro, Chemistry: A Molecular Approach 8 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 Approach 9 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 Approach 10 Common Bases

11. Tro, Chemistry: A Molecular Approach 11 Structure of Bases most ionic bases contain OH ions – NaOH, Ca(OH) 2 some contain CO 3 2- ions – CaCO 3 NaHCO 3 molecular bases contain structures that react with H + – mostly amine groups

12. Tro, Chemistry: A Molecular Approach 12 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 Approach 13 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 Approach 14 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 Approach 15 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 Approach 16 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 Approach 17 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 Approach 18 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 Approach 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 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 Approach 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 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 Approach 21 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 Approach 22 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 Approach 23 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 Approach 24 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 Approach 25 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 Approach 26 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 Approach 27 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 Approach 28 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

29. Tro, Chemistry: A Molecular Approach 29 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 -

30. Tro, Chemistry: A Molecular Approach 30 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 -

31. Tro, Chemistry: A Molecular Approach 31 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 

32. Tro, Chemistry: A Molecular Approach 32 Increasing Acidity Increasing Basicity

33. Tro, Chemistry: A Molecular Approach 33 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

34. Tro, Chemistry: A Molecular Approach 34 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

35. Tro, Chemistry: A Molecular Approach 35 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

36. 36

38. Tro, Chemistry: A Molecular Approach 38 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

39. Tro, Chemistry: A Molecular Approach 39 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

40. Tro, Chemistry: A Molecular Approach 40 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

41. 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  ]

42. Tro, Chemistry: A Molecular Approach 42 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

43. Tro, Chemistry: A Molecular Approach 43 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

44. Tro, Chemistry: A Molecular Approach 44 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

45. Tro, Chemistry: A Molecular Approach 45 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

46. 46 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)

47. 47 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

48. Tro, Chemistry: A Molecular Approach 48

49. 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

50. Tro, Chemistry: A Molecular Approach 50 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

51. Tro, Chemistry: A Molecular Approach 51 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

52. Tro, Chemistry: A Molecular Approach 52 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

53. Tro, Chemistry: A Molecular Approach 53 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

54. Tro, Chemistry: A Molecular Approach 54 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

55. Tro, Chemistry: A Molecular Approach 55 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

56. Tro, Chemistry: A Molecular Approach 56 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 +

57. Tro, Chemistry: A Molecular Approach 57 [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

58. Tro, Chemistry: A Molecular Approach 58 [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

59. Tro, Chemistry: A Molecular Approach 59 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

60. Tro, Chemistry: A Molecular Approach 60 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

61. Tro, Chemistry: A Molecular Approach 61 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

62. Tro, Chemistry: A Molecular Approach 62 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

63. Tro, Chemistry: A Molecular Approach 63 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

64. Tro, Chemistry: A Molecular Approach 64 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)

65. Tro, Chemistry: A Molecular Approach 65 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

66. Tro, Chemistry: A Molecular Approach 66 [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 +

67. Tro, Chemistry: A Molecular Approach 67 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 +

68. Tro, Chemistry: A Molecular Approach 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 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

69. Tro, Chemistry: A Molecular Approach 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 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

70. Tro, Chemistry: A Molecular Approach 70 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

71. Tro, Chemistry: A Molecular Approach 71 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

72. Tro, Chemistry: A Molecular Approach 72 [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 +

73. Tro, Chemistry: A Molecular Approach 73 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

74. Tro, Chemistry: A Molecular Approach 74 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

75. Tro, Chemistry: A Molecular Approach 75 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

76. Tro, Chemistry: A Molecular Approach 76 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

77. Tro, Chemistry: A Molecular Approach 77 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

78. Tro, Chemistry: A Molecular Approach 78 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

79. Tro, Chemistry: A Molecular Approach 79 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

80. Tro, Chemistry: A Molecular Approach 80 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

81. Tro, Chemistry: A Molecular Approach 81 [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

82. Tro, Chemistry: A Molecular Approach 82 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

83. Tro, Chemistry: A Molecular Approach 83 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

84. Tro, Chemistry: A Molecular Approach 84 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?

85. Tro, Chemistry: A Molecular Approach 85 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

86. Tro, Chemistry: A Molecular Approach 86 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

87. Tro, Chemistry: A Molecular Approach 87 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

88. 88 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

89. Tro, Chemistry: A Molecular Approach 89 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

90. Tro, Chemistry: A Molecular Approach 90 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

91. Tro, Chemistry: A Molecular Approach 91 [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

92. Tro, Chemistry: A Molecular Approach 92 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

93. Tro, Chemistry: A Molecular Approach 93 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

94. Tro, Chemistry: A Molecular Approach 94 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

95. Tro, Chemistry: A Molecular Approach 95 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

96. Tro, Chemistry: A Molecular Approach 96 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

97. Tro, Chemistry: A Molecular Approach 97 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)

98. 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

99. 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

100. Tro, Chemistry: A Molecular Approach 100 Practice - Calculate the pH of a 0.0010 M Ba(OH) 2 solution and determine if it is acidic, basic, or neutral

101. Tro, Chemistry: A Molecular Approach 101 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  ]

102. Tro, Chemistry: A Molecular Approach 102 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 -

103. Tro, Chemistry: A Molecular Approach 103