1. Chapter 15 Acids and Bases Roy Kennedy Massachusetts Bay Community College Wellesley Hills, MA Chemistry: A Molecular Approach, 2nd Ed. Nivaldo Tro Copyright  2011 Pearson Education, Inc.

2. Stomach Acid & Heartburn The stomach lining produces hydrochloric acid to break down food and kill bacteria HCl activates enzymes that break down food If stomach acid backs up into your esophagus, it irritates the esophagus, resulting in ordinary heartburn Chronic heartburn is a more serious problem GERD (gastroesophageal reflux disease) is chronic leaking of stomach acid into the esophagus 2 Curing Heartburn Heartburn is cured by neutralizing the acid saliva contains bicarbonate ion (NaHCO 3  Na + + HCO 3 - ) antacids contain hydroxide ions and/or carbonate ions

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 the color of vegetable dyes React with bases to form ionic salts 3

4. Common Acids 4

5. Structures of Acids: Binary Acids 5 Binary acids have an acidic hydrogen attached to a nonmetal atom (e.g. HCl, HF) Structure of Oxy Acids Oxy acids have acidic H’s attached to an oxygen atom H 2 SO 4, HNO 3

6. Structure of Acids: Oxy Acids 6 Carboxylic acids have COOH group HC 2 H 3 O 2 (acetic acid), H 3 C 6 H 5 O 7 (citric acid), H 2 C 4 H 3 O 5 (malic acid) Only the first H in the formula is acidic

7. Properties and Structure of Bases 7 Also known as alkalis, from alkaloid plants (often poisonous) Solutions feel slippery and taste bitter Change color of vegetable dyes React with acids to form ionic salts neutralization 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

8. Common Bases 8

9. Indicators Chemicals that change color depending on acidity or basicity Many vegetable dyes are indicators Anthocyanin flavonoid (purple color of pansies, blueberries, blackberries, cranberries, red color of Fuji apples Litmus extracted from Lichens, e.g. Spanish moss Turns red in acid, blue in base Phenolphthalein found in laxatives red in base, colorless in acid 9

10. Arrhenius Theory Acids ionize in water to produce H + ions and anions molecular acids can’t dissociate because they contain no ions they must be ionized (pulled apart) by the water HCl(aq) → H + (aq) + Cl − (aq ) HC 2 H 3 O 2 (aq) → H + (aq) + C 2 H 3 O 2 − (aq) H + ions produced by an acid are so reactive they cannot exist as a distinct entity in water, they react with water to produce a complex ion, the hydronium ion, H 3 O + Bases, which do contain ions, dissociate in water to produce OH − ions and cations NaOH(aq) → Na + (aq) + OH − (aq) 10

11. Arrhenius Acid–Base Reactions The H + from an acid combines with the OH − from a base to make a molecule of H 2 O ( 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) 11 But the Arrhenius theory does not explain why molecular substances that do not contain OH – ions, dissolve in water to form basic solution  NH 3, Na 2 CO 3, Na 2 O why molecular substances that do not contain H + ions, such as CO 2, dissolve in water to form acidic solutions acid–base reactions that take place outside aqueous solution

12. Brønsted-Lowry Acid-Base Theory Brønsted and Lowry created a broader definition of acids and bases based on their reactivity. Any reaction that involves the transfer of H + from one molecule to another is an acid–base reaction  The acid is an H + donor  The base is an H + acceptor, that must contain an atom with an unshared pair of electrons H–A + :B  :A – + H–B +  does not require aqueous solution, or the presence of OH − When HCl dissolves in water, the HCl is the acid because HCl transfers an H + to H 2 O, forming H 3 O + ions 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) 12 HCl(aq) + H 2 O(l) → Cl – (aq) + H 3 O + (aq) acidbase NH 3 (aq) + H 2 O(l)  NH 4 + (aq) + OH – (aq) base acid

13. Brønsted-Lowry Acid-Base Reactions Brønsted-Lowry theory allows for reaction reversibility The original base, after rxn, forms an acid and the original acid, after rxn, forms a base The original base and the acid it becomes and the original acid and the base it becomes are called a conjugate pairs 13 H–A + :B  :A – + H–B + acid base conjugate base conjugate acid For 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 Acid base conjugate base conjugate acid

14. Write the formula for the conjugate acid…..the conjugate base 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 14 HCHO 2 + H 2 O  CHO 2 – +H 3 O + acid baseconjugate conjugate base acid H 2 O NH 3 CO 3 2− H 2 PO 4 1− H 2 OHO − NH 3 NH 2 − CO 3 2− No H, can’t be an acid H 2 PO 4 1− HPO 4 2−

15. Example 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 + H 2 SO 4 loses an H + so it is the acid and HSO 4  is its conjugate base H 2 O accepts an H + so H 2 O must be the base and H 3 O + is its conjugate acid 15

16. Example 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 – HCO 3  accepts an H + so it is the base and H 2 CO 3 is its conjugate acid H 2 O donates an H + so it is the acid and OH  is its conjugate base 16

17. 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) Accepts H + Base Conjugate Acid Donates H + Acid Conjugate Base 17

18. 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 − 18

19. 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−I− CO 3 2− + H 2 O  HCO 3 − + OH − Base Acid Conj. Conj. acid base 19 CO 3 2−

20. 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) 20

21. Strong Acids and Bases Strong acids and bases are strong electrolytes The stronger the acid, the more willing it is to donate H +  water is used as the standard base that accepts the H + the acid molecules are 100% ionized the base molecules are either dissociated into OH – or are reacted with water Ionization and dissociation favored Are strong electrolytes [H 3 O + ] = [strong acid] 0.1 M HCl  0.1 M H 3 O + 21

22. Weak Acids or Bases 22 Weak acids or bases are weak electrolytes only a small percentage of the acid molecules ionize only a small percentage of the base molecules form OH – ions, either through dissociation or reaction with water reactants favored, equilibrium situation Weak acids donate a small fraction of their H + most of the weak acid molecules do not donate H + to water much less than 1% ionized in water [weak acid] << [H 3 O + ] 0.1 M H 3 O + ≠ 0.1 M HF [H 3 O + ] << [HF]

24. Strengths of Acids & Bases acid or base strength can be measured by finding their equilibrium constant in their reaction with water HA + H 2 O  A − + H 3 O + B: + H 2 O  HB + + OH − The larger the equilibrium constant, the more products are formed, the stronger the acid or base The position of equilibrium depends on the strength of attraction between the base and the H + stronger attraction means a stronger base or a weaker acid 24

25. General Trends in Acidity Stronger acid  weaker conjugate base Higher oxidation number = stronger oxyacid Stronger: H 2 SO 4 > H 2 SO 3 ; HNO 3 > HNO 2 ; Weaker Cations are stronger acids than neutral molecules H 3 O + > H 2 O neutral species are stronger acids than anions NH 3 > NH 2 − Opposite trend for base strength 25

26. Acid Ionization Constant, K a HA + H 2 O  A − + H 3 O + The equilibrium constant for this reaction is called the acid ionization constant, K a larger K a = stronger acid 26

27. 27

28. Auto-ionization of Water Water is an extremely weak electrolyte Approx. 2 out of every 1 billion water molecules form ions through a process called auto-ionization H 2 O  H + + OH – or H 2 O + H 2 O  H 3 O + + OH – All aqueous solutions contain both H 3 O + and OH – In water: [H 3 O + ] = [OH – ] = 10 −7 M @ 25 °C [H 3 O + ] x [OH – ] always = K w, = 1.00 x 10 −14 @ 25 °C K w = Dissociation Constant of Water measure one of the concentrations, you can calculate the other When [H 3 O + ] increases, the [OH – ] must decrease their product stays constant 28

29. Acidic and Basic Solutions Neutral solutions, [H 3 O + ] = [OH – ] = 1.00 x 10 −7 Acidic solutions, [H 3 O + ] > [OH – ] [H 3 O + ] > 1.00 x 10 −7 ; [OH – ] < 1.00 x 10 −7 Basic solutions, [OH – ] > [H 3 O + ] [H 3 O + ] 1.00 x 10 −7 29

30. [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 Acid Base 30  The divisions above are not to scale

31. 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 31 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: Conceptual Plan: Relationships: Given: Find: [H 3 O + ][OH  ]

32. Practice – Determine the [H 3 O + ] when the [OH − ] = 2.5 x 10 −9 M 32 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 + ]

33. Measuring Acidity: pH 33 Solution acidity or basicity is often expressed as pH = −log[H 3 O + ] pH has positive sign Reverse calculation [H 3 O + ] = 10 −pH pH water = −log[10 −7 ] = 7 (neutral) pH 7 is basic 1 pH unit is a factor of 10 acidity difference pH 0 is [H 3 O + ] = 1 M, pH 14 is [OH – ] = 1 M pH can extend beyond the 0 to 14 range

34. Significant Figures (sf) & Logs For 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... The significant figures are determined by the decimal part, the sf’s are the digits after the decimal point in the log 34 log(2.0 x 10 6 ) = 6.30 (2 sf) log(2.0 x 10 6 ) = log(10 6 )+log(2.0) = 6+0.30303… = 6.30303... (2 sf’s)

35. 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 35 pH is unitless; the fact that the pH > 7 means the solution is basic [OH  ] = 1.3 x 10 −2 M pH Check: Solution: Conceptual Plan: Relationships: Given: Find: [H 3 O + ][OH  ]pH

36. Practice – Determine the pH @ 25 ºC of a solution that has [OH − ] = 2.5 x 10 −9 M 36 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

37. Practice – Determine the [OH − ] of a solution with a pH of 5.40 37 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  ]

38. pOH 38 Another way of expressing the acidity/basicity of a solution is pOH = −log[OH  ] [OH  ] = 10 −pOH pOH water = −log[10 −7 ] = 7 (neutral) at 25 °C pOH 7 acidic pH + pOH = 14.0 = (−log[H 3 O + ] ) + (−log[OH  ] )

39. Example: 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 39 pH is unitless; the fact that the pH > 7 means the solution is basic [OH  ] = 1.3 x 10 −2 M pH Check: Solution: Conceptual Plan: Relationships: Given: Find: pOH[OH  ]pH

40. Determine the pOH @ 25 ºC of a solution that has [H 3 O + ] = 2.5 x 10 −9 M 40 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

41. pKpK A way of expressing the strength of an acid or base is the -log of the equilibrium constant, 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 41

42. [H 3 O + ] and [OH − ] in a Strong Acid or Strong Base Solution There are 2 sources of H 3 O + in an aqueous solution of a strong acid one source is the strong acid itself and the second source is from the water, whose contribution is negligible There are 2 sources of OH − in an aqueous solution of a strong base one source is the strong base itself and the second source is from the water, whose contribution is negligible the strong [acid] or [base] shifts the K w equilibrium so far to the right that the contribution from water is too small to be significant (not true in dilute solutions) 42

43. Finding pH of a Strong Acid or Strong Base Solution For a strong acid containing only one proton (monoprotic) [H 3 O + ] = [HAcid] 0.10 M HCl has [H 3 O + ] = 0.10 M and pH = 1.00 For a strong ionic base, [OH − ] = [Base] x (the # of OH − groups in the base) 0.10 M Ca(OH) 2 has [OH − ] = 2 x 0.10M = 0.20 M and pH = 13.30 for molecular bases with more than one lone pair of e -, only one lone pair accepts an H + rxn with the other lone pair(s) can generally be ignored 43

44. Polyprotic Acids Acid molecules can have more than one ionizable H – these are called polyprotic acids 2 H = diprotic = H 2 SO 4, 3 H = triprotic = H 3 PO 4 Each H can have different acid strengths or be equal for polyprotic acids, 2 nd & 3 rd ionizations can generally be ignored, however for H 2 SO 4, the second ionization cannot be ignored Polyprotic acids ionize in sequentially in steps Removing the first H makes removal of the second harder H 2 SO 4 is a stronger acid than HSO 4  44

45. 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 finding the [H 3 O + ] is complicated by the fact that the weak 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 + 45

46. [HNO 2 ][NO 2 − ][H 3 O + ] initial change equilibrium Example 15.6: Find the pH of 0.200 M HNO 2 (aq) solution @ 25 °C 46 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 because 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

47. [HNO 2 ][NO 2 − ][H 3 O + ] initial 0.200 00 change equilibrium Example 15.6: Find the pH of 0.200 M HNO 2 (aq) solution @ 25 °C 47 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 HNO 2 + H 2 O (l)  NO 2 − (aq) + H 3 O + (aq)

48. Example 15.6: Find the pH of 0.200 M HNO 2 (aq) solution @ 25 °C 48 determine the value of K a from Table 15.5 because 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

49. Example 15.6: Find the pH of 0.200 M HNO 2 (aq) solution @ 25 °C 49 check if the approximation is valid by seeing if x < 5% of [HNO 2 ] init 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 the approximation is valid x = 9.6 x 10 −3

50. Example 15.6: Find the pH of 0.200 M HNO 2 (aq) solution @ 25 °C 50 substitute x into the equilibrium concentration definitions and solve 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 [HNO 2 ][NO 2 − ][H 3 O + ] initial 0.200 0 ≈ 0 change −x−x+x+x+x+x equilibrium 0.1900.0096

51. Example 15.6: Find the pH of 0.200 M HNO 2 (aq) solution @ 25 °C 51 substitute [H 3 O + ] into the formula for pH and solve 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

52. Example 15.6: Find the pH of 0.200 M HNO 2 (aq) solution @ 25 °C 52 check by substituting the equilibrium concentrations back into the equilibrium constant expression and comparing the calculated K a to the given K a 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 though not exact, the answer is reasonably close

53. What is the pH of a 0.012M solution of nicotinic acid, HC 6 H 4 NO 2 ? (K a = 1.4 x 10 −5 @ 25°C) 53 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

54. [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 ? 54 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 +

55. 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 +

56. 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 56 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

57. 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 57 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

58. 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 58 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

59. 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 59 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

60. [HClO 2 ][ClO 2 − ][H 3 O + ] initial 0.100 0≈ 0 change equilibrium Example 15.7: Find the pH of 0.100 M HClO 2 (aq) solution @ 25 °C 60 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 +

61. Example 15.7: Find the pH of 0.100 M HClO 2 (aq) solution @ 25 °C 61 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

62. Example 15.7: Find the pH of 0.100 M HClO 2 (aq) solution @ 25 °C 62 determine the value of K a from Table 15.5 because 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

63. Example 15.7: Find the pH of 0.100 M HClO 2 (aq) solution @ 25 °C 63 check if the approximation is valid by seeing if x < 5% of [HNO 2 ] init K a for HClO 2 = 1.1 x 10 −2 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

64. Example 15.7: Find the pH of 0.100 M HClO 2 (aq) solution @ 25 °C 64 if the approximation is invalid, solve for x using the quadratic formula K a for HClO 2 = 1.1 x 10 −2

65. Example 15.7: Find the pH of 0.100 M HClO 2 (aq) solution @ 25 °C 65 substitute x into the equilibrium concentration definitions and solve K a for HClO 2 = 1.1 x 10 −2 x = 0.028 [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

66. Example 15.7: Find the pH of 0.100 M HClO 2 (aq) solution @ 25 °C 66 substitute [H 3 O + ] into the formula for pH and solve 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 0.0720.028

67. Example 15.7: Find the pH of 0.100 M HClO 2 (aq) solution @ 25°C 67 check by substituting the equilibrium concentrations back into the equilibrium constant expression and comparing the calculated K a to the given K a K a for HClO 2 = 1.1 x 10 −2 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

68. Example 15.8: What is the K a of a weak acid if a 0.100 M solution has a pH of 4.25? 68start 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

69. [HA][A − ][H 3 O + ] initial 0.100 00 change equilibrium Example 15.8: What is the K a of a weak acid if a 0.100 M solution has a pH of 4.25? 69 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

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? 70 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

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? 71 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 +

72. 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 72

73. Example 15.9: What is the percent ionization of a 2.5 M HNO 2 solution? 73 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 2.50≈0 change equilibrium [HNO 2 ][NO 2 − ][H 3 O + ] initial 2.50≈ 0 change equilibrium +x+x +x+x xx 2.5  x xx

74. Example 15.9: What is the percent ionization of a 2.5 M HNO 2 solution? 74 determine the value of K a from Table 15.5 because 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

75. Example 15.9: What is the percent ionization of a 2.5 M HNO 2 solution? 75 substitute x into the equilibrium concentration definitions and solve 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 x = 3.4 x 10 −2

76. Example 15.9: What is the percent ionization of a 2.5 M HNO 2 solution? 76 apply the definition and compute the percent ionization 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 because the percent ionization is < 5%, the “x is small” approximation is valid

77. 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 @ 25 °C) 77 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

78. [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 ? K a = 1.4 x 10 −5 78 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 +

79. 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 79 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 +

80. Practice – What is the percent ionization of a 0.012 M solution of nicotinic acid, HC 6 H 4 NO 2 ? 80 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

81. Practice – What is the percent ionization of a 0.012 M solution of nicotinic acid, HC 6 H 4 NO 2 ? 81 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

82. Relationship Between [H 3 O + ] equilibrium & [HA] initial Increasing the [acid] initial eventually results in increased [H 3 O + ] at equilibrium Increasing the [acid] initial results, at first, in decreased % ionization This means that the increase in [H 3 O + ] concentration is slower than the increase in initial acid concentration 82

83. 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 aqueous components, the equilibrium should shift to the right to increase the total number of dissolved particles we can reduce all the aqueous component concentrations by using an initial acid concentration that is more dilute 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 83

84. 84 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 weak acid mixtures, you generally need to consider the stronger of the two for the same reason as long as one is significantly stronger than the other, and their concentrations are similar

85. Example 15.10: Find the pH of a mixture of 0.150 M HF(aq) solution and 0.100 M HClO 2 (aq) 85 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

86. Example 15.10: Find the pH of a mixture of 0.150 M HF(aq) solution and 0.100 M HClO 2 (aq) 86 represent the change in the concentrations in term of x sum the columns to find the equilibrium concentrations in terms of x substitute into the equilibrium constant expression [HF][F - ][H 3 O + ] initial 0.150 00 change equilibrium +x+x+x+x xx 0.150  x xx

87. Example 15.10: Find the pH of a mixture of 0.150 M HF(aq) solution and 0.100 M HClO 2 (aq) 87 because 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

88. Example 15.10: Find the pH of a mixture of 0.150 M HF(aq) solution and 0.100 M HClO 2 (aq) 88 check if the approximation is valid by seeing if x < 5% of [HF] init 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 the approximation is valid x = 7.2 x 10 −3

89. Example 15.10: Find the pH of a mixture of 0.150 M HF(aq) solution and 0.100 M HClO 2 (aq) 89 substitute x into the equilibrium concentration definitions and solve 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 [HF][F - ][H 3 O + ] initial 0.150 0 ≈ 0 change −x−x+x+x+x+x equilibrium 0.1430.0072

90. Example 15.10: Find the pH of a mixture of 0.150 M HF(aq) solution and 0.100 M HClO 2 (aq) 90 substitute [H 3 O + ] into the formula for pH and solve 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 0.1430.0072

91. 91 Example 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

92. Practice – Determine the pH @ 25 ºC of a solution that is a mixture of 0.045 M HCl and 0.15 M HF 92 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]

93. Strong Bases 93 NaOH  Na + + OH − The stronger the base, the stronger the drive to accept H + For ionic bases, practically all units dissociate into OH – or accept H’s strong electrolyte multi-OH strong bases completely dissociated [HO – ] = [strong base] x (# OH)

94. Example 15.11: 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: Conceptual Plan: Relationships: Given: Find: [H 3 O + ][OH  ]pH[Sr(OH) 2 ] [OH  ]=2[Sr(OH) 2 ] [OH  ] = 2(0.0015) = 0.0030 M 94

95. Example 15.11: 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: Conceptual Plan: Relationships: Given: Find: pOH[OH  ]pH[Sr(OH) 2 ] [OH  ]=2[Sr(OH) 2 ] [OH  ] = 2(0.0015) = 0.0030 M pH + pOH = 14.00 pH = 14.00 – 2.52 = 11.48 95

96. Calculating 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 96

97. Weak Bases 97 NH 3 + H 2 O  NH 4 + + OH − For weak bases, only a small fraction of molecules accept H + are weak electrolytes most 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

98. Base Ionization Constant, K b Base strength measured by the size of the equilibrium constant when reacting with H 2 O :Base + H 2 O  OH − + H:Base + The equilibrium constant is called the base ionization constant, K b a non-ionic base ionizes water larger K b = stronger base 98

99. 99

100. [NH 3 ][NH 4 + ][OH  ] initial change equilibrium Example 15.12:Find the pH of 0.100 M NH 3 (aq) 100 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 NH 3 + H 2 O  NH 4 + + OH  [NH 3 ][NH 4 + ][OH  ] initial 0.100 0 ≈ 0 change equilibrium

101. [NH 3 ][NH 4 + ][OH  ] initial 0.100 00 change equilibrium Example 15.12: Find the pH of 0.100 M NH 3 (aq) 101 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

102. Example 15.12: Find the pH of 0.100 M NH 3 (aq) 102 determine the value of K b from Table 15.8 because 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

103. Example 15.12: Find the pH of 0.100 M NH 3 (aq) 103 check if the approximation is valid by seeing if x < 5% of [NH 3 ] init 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 the approximation is valid x = 1.33 x 10 −3

104. Example 15.12: Find the pH of 0.100 M NH 3 (aq) 104 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

105. Example 15.12: Find the pH of 0.100 M NH 3 (aq) 105 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

106. Example 15.12: Find the pH of 0.100 M NH 3 (aq) 106 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

107. Example 15.12: Find the pH of 0.100 M NH 3 (aq) 107 check by substituting the equilibrium concentrations back into the equilibrium constant expression and comparing the calculated K b to the given K b 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 though not exact, the answer is reasonably close

108. Find the pH of a 0.0015 M morphine solution, K b = 1.6 x 10 −6 108 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

109. [B][BH + ][OH  ] initial 0.0015 00 change equilibrium Find the pH of a 0.0015 M morphine solution, K b = 1.6 x 10 −6 109 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 

110. Find the pH of a 0.0015 M morphine solution, K b = 1.6 x 10 −6 110 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

111. Find the pH of a 0.0015 M morphine solution, K b = 1.6 x 10 −6 111 check if the approximation is valid by seeing if x < 5% of [B] init the approximation is valid x = 4.9 x 10 −5 [B][BH + ][OH  ] initial 0.0015 0 ≈ 0 change −x−x+x+x+x+x equilibrium 0.0015 xx

112. Find the pH of a 0.0015 M morphine solution, K b = 1.6 x 10 −6 112 substitute x into the equilibrium concentration definitions and solve x = 4.9 x 10 −5 [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

113. Find the pH of a 0.0015 M morphine solution, K b = 1.6 x 10 −6 113 use the [OH − ] to find the [H 3 O + ] using K w substitute [H 3 O + ] into the formula for pH and solve [B][BH + ][OH  ] initial 0.0015 0 ≈ 0 change −x−x+x+x+x+x equilibrium 0.00154.9E−5

114. Find the pH of a 0.0015 M morphine solution, K b = 1.6 x 10 −6 114 use the [OH − ] to find the pOH use pOH to find pH [B][BH + ][OH  ] initial 0.0015 0 ≈ 0 change −x−x+x+x+x+x equilibrium 0.00154.9E−5

115. Find the pH of a 0.0015 M morphine solution, K b = 1.6 x 10 −6 115 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 [B][BH + ][OH  ] initial 0.0015 0 ≈ 0 change −x−x+x+x+x+x equilibrium 0.00154.9E−5

116. Ionization in Polyprotic Acids Because polyprotic acids ionize in steps, each H has its own K a K a1 > K a2 > K a3 Generally, the difference between K a1 and K a2 is large enough that the second ionization will not affect the pH except H 2 SO 4 116

117. 117

118. What is the pH of a 0.12 M solution of carbonic acid, H 2 CO 3 ? (K a1 = 4.3 x 10 −7, K a2 = 5.6 x 10 −11 ) 118 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 +

119. [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 ? 119 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 +

120. 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 −11 120

121. Practice – What is the pH of a 0.12 M solution of carbonic acid, H 2 CO 3 ? 121 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

122. Practice – What is the pH of a 0.12 M solution of carbonic acid, H 2 CO 3 ? 122 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

123. Practice – What is the pH of a 0.12 M solution of carbonic acid, H 2 CO 3 ? 123 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

124. Practice – What is the pH of a 0.12 M solution of carbonic acid, H 2 CO 3 ? 124 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

125. What is the [CO 3 2− ] in a 0.12 M solution of carbonic acid, H 2 CO 3 ? (K a1 = 4.3 x 10 −7, K a2 = 5.6 x 10 −11 ) 125 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 +

126. [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 ? 126 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

127. 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 127

128. Ionization of 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 (100%) 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 taken into account Because the first ionization is 100% complete, use the given [H 2 SO 4 ] = [HSO 4 − ] initial = [H 3 O + ] initial 128

129. Example 15.18: Find the pH of 0.0100 M H 2 SO 4 (aq) solution @ 25 °C 129 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 +

130. 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 130

131. 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 − = 0.012 131

132. 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 132

133. 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 133

134. Example 15.18: Find the pH of 0.0100 M H 2 SO 4 (aq) solution @ 25 °C 134stop 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

135. Strengths of Binary Acids The more polarized the bond, the more acidic the bond  + H─X  − 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 135

136. Relationship between Bond Strength and Acidity Acid Bond Energy kJ/mol Type of Acid HF565weak HCl431strong HBr364strong 136

137. Strengths of Oxyacids, H–O–Y The more electronegative atom Y is, the stronger the oxyacid HClO > HIO (Cl is more EN than I)  helps weakens the H–O bond acidity of oxyacids decreases down a group 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 The more oxygens attached to Y, the stronger the oxyacid further weakens and polarizes the H–O bond HClO 3 > HClO 2 137

138. 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 −11 138

139. Relationship Between Number of Oxygens on the Central Atom and Acidity 139

140. 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 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− 140

141. Lewis Acid–Base Theory Lewis Acid–Base theory focuses on donating/accepting an electron pair The Lewis Acid is an electron acceptor electron deficient electrophile doesn’t having a full valence shell electron deficient due to attached electronegative atom(s) Must have empty orbital able to accept an electron pair H + has empty 1s orbital Boron in BF 3 has an 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 141

142. Lewis Bases The Lewis Base is an electron donor An electron rich nucleophile Anions are better Lewis Bases than neutral species N: < N: − The more electronegative the atom, the less willing it is to be a Lewis Base O: < S: 142 The Lewis base donates an electron pair to the Lewis acid, resulting in covalent bond formation H 3 N: + BF 3  H 3 N─BF 3 (an adduct) Arrhenius and Brønsted-Lowry acid–base reactions are also Lewis acid–base reactions

143. Examples of Lewis Acid–Base Reactions 143 Ag + (aq) + 2 :NH 3(aq)  Ag(NH 3 ) 2 + (aq) Lewis Acid Lewis Base Adduct

144. Identify the Lewis Acid and Lewis Base in Each Reaction 144 Lewis Base Lewis Base Lewis Acid Lewis Acid

145. Acid–Base Properties of Salts Salts that form Acidic solutions: (example = NH 4 Cl) are composed of a cation that is the conjugate acid of a weak base (NH 4 + is the conjugate acid of the weak base, NH 3 ) and an anion from a strong acid (Cl − is the anion of the strong acid HCl) 145 Salts that form Basic solutions: (example = NaHCO 3 ) are composed of a cation of a strong base (Na + is the cation of the strong base NaOH), and an anion that is the conjugate base of a weak acid (HCO 3 − is the conjugate base of the weak acid H 2 CO 3 ) Salts that form neutral solutions are composed of cations from a strong base and anions from a strong acid (e.g. NaCl). Dissolved salts (water-soluble ionic compounds), can create solutions that are acidic, neutral or basic  Anions that are the conjugate base of a weak acid are basic F − (aq) + H 2 O (l)  HF (aq) + OH − (aq)  Anions that are the conjugate base of a strong acid are pH neutral  Cl − (aq) + H 2 O (l)  HCl (aq) + OH − (aq)

146. 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 146 Every anion is the conjugate base of an acid A − (aq) +H 2 O (l)  HA (aq) +OH − (aq) The stronger the acid, the weaker the conjugate base HNO 2 NO 2 ‾

147. 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 147

148. Example 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 K b for CHO 2 − = 5.6 x 10 −11

149. [CHO 2 − ][HCHO 2 ][OH  ] initial 0.1000≈ 0 change equilibrium 0.100  x Example 15.14: Find the pH of 0.100 M NaCHO 2 (aq) solution 149 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

150. Example 15.14: Find the pH of 0.100 M NaCHO 2 (aq) 150 because 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

151. Example 15.14: Find the pH of 0.100 M NaCHO 2 (aq) 151 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

152. [CHO 2 − ][HCHO 2 ][OH  ] initial 0.1000≈ 0 change −x−x+x+x+x+x equilibrium 0.1002.4E-6 Example 15.14: Find the pH of 0.100 M NaCHO 2 (aq) 152 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

153. Example 15.14: Find the pH of 0.100 M NaCHO 2 (aq) 153 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

154. Example 15.14: Find the pH of 0.100 M NaCHO 2 (aq) 154 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

155. Example 15.14: Find the pH of 0.100 M NaCHO 2 (aq) 155 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

156. If a 0.15 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 0.150≈ 0 change equilibrium 156

157. 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 157

158. 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-6 158 Practice – If a 0.15 M NaA solution has a pOH of 5.45, what is the K a of HA?

159. use K b of A − to find K a of HA 159 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

160. Polyatomic Cations as Weak Acids (reference = H 2 O) The stronger the base, the weaker the conjugate acid Cations that are conjugate acids of a weak base can potentially be 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 products (  ) Cations that are the counter-ion of a strong base are pH neutral Group IA and IIA cations (alkali & alkali earth metal cations) are pH neutral 160

161. Metal Cations as Weak Acids (reference = H 2 O) Cations of small, highly charged metals (hydrated) are weakly acidic Al(H 2 O) 6 3+ (aq) + H 2 O (l)  Al(H 2 O) 5 (OH) 2+ (aq) + H 3 O + (aq) 161

162. Example 15.15: Determine if the given cation is acidic or neutral (reference = H 2 O) a) C 5 H 5 NH + the conjugate acid of the weak base pyridine, C 5 H 5 N, 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 162

163. Classifying Salt Solutions as Acidic, Basic, or Neutral Anion Conjugate base of strong acid Conjugate base of weak acid Cation Conjugate acid of a weak base AcidicDepends on relative strength Small, highly charged metal ion AcidicDepends on relative strength Counterion of strong base NeutralBasic 163 Conjugate acid of a weak base Small, highly charged metal ion Counterion of strong base Na + Cl - Neutral Ca 2+ NO 3 - K+ Br - Neutral Na + F-F- Ca 2+ C2H3O2-C2H3O2- K+K+ NO 2 - Basic NH 4 + Cl - Acidic Al 3+ NO 3 - HF is a stronger acid than NH 4 + (K a of NH 4 + > K b of the F − ) NH 4 + F-F- Depends on relative strength Acidic

164. 164 Anion Conjugate base of strong acid Conjugate base of weak acid Cation Conjugate acid of a weak base AcidicDepends on relative strength Small, highly charged metal ion AcidicDepends on relative strength Counterion of strong base NeutralBasic Conjugate acid of a weak base Small, highly charged metal ion Counterion of strong base Sr 2+ Cl − Neutral Al 3+ Acidic CH 3 NH 3 + NO 3 − Acidic Na + CHO 2 − Basic HCO 3 - K+K+ Co 3+ Ba 2+

165. 165 STOP

166. 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) 166

167. 167 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

168. 168 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

169. pH of Rain in Different Regions 169

170. 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 170

171. Sources of SO 2 from Utilities 171

172. 172 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

173. 173 Damage from Acid Rain

174. 174 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