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

2. Copyright  2011 Pearson Education, Inc. Polyprotic Acids Acid molecules often have more than one ionizable H – these are called polyprotic acids the ionizable H’s may have different acid strengths or be equal 1 H = monoprotic, 2 H = diprotic, 3 H = triprotic HCl = monoprotic, H 2 SO 4 = diprotic, H 3 PO 4 = triprotic Polyprotic acids ionize in steps each ionizable H is removed sequentially Removing of the first H automatically makes removal of the second H harder H 2 SO 4 is a stronger acid than HSO 4  2 Tro, Chemistry: A Molecular Approach, 2/e

3. Copyright  2011 Pearson Education, Inc. 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 3 Tro, Chemistry: A Molecular Approach, 2/e

4. Copyright  2011 Pearson Education, Inc. Example 15.9: What is the percent ionization of a 2.5 M HNO 2 solution? 4 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 Tro, Chemistry: A Molecular Approach, 2/e

5. Copyright  2011 Pearson Education, Inc. Example 15.9: What is the percent ionization of a 2.5 M HNO 2 solution? 5 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 Tro, Chemistry: A Molecular Approach, 2/e

6. Copyright  2011 Pearson Education, Inc. Example 15.9: What is the percent ionization of a 2.5 M HNO 2 solution? 6 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 Tro, Chemistry: A Molecular Approach, 2/e

7. Copyright  2011 Pearson Education, Inc. Example 15.9: What is the percent ionization of a 2.5 M HNO 2 solution? 7 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 Tro, Chemistry: A Molecular Approach, 2/e

8. Copyright  2011 Pearson Education, Inc. 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 @ 25 °C) 8 Tro, Chemistry: A Molecular Approach, 2/e

9. Copyright  2011 Pearson Education, Inc. Practice – What is the percent ionization of a 0.012 M solution of nicotinic acid, HC 6 H 4 NO 2 ? 9 write the reaction for the acid with water construct an ICE table for the reaction enter the initial concentrations – assuming the [H 3 O + ] from water is ≈ 0 HC 6 H 4 NO 2 + H 2 O  C 6 H 4 NO 2  + H 3 O + [HA][A − ][H 3 O + ] initial 0.012 0 ≈ 0 change equilibrium Tro, Chemistry: A Molecular Approach, 2/e

10. Copyright  2011 Pearson Education, Inc. [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 ? 10 represent the change in the concentrations in terms of x sum the columns to find the equilibrium concentrations in terms of x substitute into the equilibrium constant expression +x+x +x+x xx 0.012  x xx HC 6 H 4 NO 2 + H 2 O  C 6 H 4 NO 2  + H 3 O + Tro, Chemistry: A Molecular Approach, 2/e

11. Copyright  2011 Pearson Education, Inc. 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 11 determine the value of K a because K a is very small, approximate the [HA] eq = [HA] init and solve for x [HA][A − ][H 3 O + ] initial 0.012 0 ≈ 0 change −x−x+x+x+x+x equilibrium 0.012xx 0.012  x HC 6 H 4 NO 2 + H 2 O  C 6 H 4 NO 2  + H 3 O + Tro, Chemistry: A Molecular Approach, 2/e

12. Copyright  2011 Pearson Education, Inc. Practice – What is the percent ionization of a 0.012 M solution of nicotinic acid, HC 6 H 4 NO 2 ? 12 substitute x into the equilibrium concentration definitions and solve x = 4.1 x 10 −4 [HA][A − ][H 3 O + ] initial 0.012 0 ≈ 0 change −x−x+x+x+x+x equilibrium 0.012−x xx Tro, Chemistry: A Molecular Approach, 2/e

13. Copyright  2011 Pearson Education, Inc. Practice – What is the percent ionization of a 0.012 M solution of nicotinic acid, HC 6 H 4 NO 2 ? 13 apply the definition and compute the percent ionization because the percent ionization is < 5%, the “x is small” approximation is valid [HA][A − ][H 3 O + ] initial 0.012 0 ≈ 0 change −x−x+x+x+x+x equilibrium 0.0124.1E-04 Tro, Chemistry: A Molecular Approach, 2/e

14. Copyright  2011 Pearson Education, Inc. Relationship Between [H 3 O + ] equilibrium & [HA] initial Increasing the initial concentration of acid results in increased [H 3 O + ] at equilibrium Increasing the initial concentration of acid results in decreased percent ionization This means that the increase in [H 3 O + ] concentration is slower than the increase in acid concentration 14 Tro, Chemistry: A Molecular Approach, 2/e

15. Copyright  2011 Pearson Education, Inc. 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 15 Tro, Chemistry: A Molecular Approach, 2/e

16. Copyright  2011 Pearson Education, Inc. 16 Finding the pH of Mixtures of Acids Generally, you can ignore the contribution of the weaker acid to the [H 3 O + ] equil For a mixture of a strong acid with a weak acid, the complete ionization of the strong acid provides more than enough [H 3 O + ] to shift the weak acid equilibrium to the left so far that the weak acid’s added [H 3 O + ] is negligible For mixtures of weak acids, generally only need to consider the stronger for the same reasons as long as one is significantly stronger than the other, and their concentrations are similar Tro, Chemistry: A Molecular Approach, 2/e

17. Copyright  2011 Pearson Education, Inc. Example 15.10: Find the pH of a mixture of 0.150 M HF(aq) solution and 0.100 M HClO 2 (aq) 17 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 Tro, Chemistry: A Molecular Approach, 2/e

18. Copyright  2011 Pearson Education, Inc. Example 15.10: Find the pH of a mixture of 0.150 M HF(aq) solution and 0.100 M HClO 2 (aq) 18 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 Tro, Chemistry: A Molecular Approach, 2/e

19. Copyright  2011 Pearson Education, Inc. Example 15.10: Find the pH of a mixture of 0.150 M HF(aq) solution and 0.100 M HClO 2 (aq) 19 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 Tro, Chemistry: A Molecular Approach, 2/e

20. Copyright  2011 Pearson Education, Inc. Example 15.10: Find the pH of a mixture of 0.150 M HF(aq) solution and 0.100 M HClO 2 (aq) 20 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 Tro, Chemistry: A Molecular Approach, 2/e

21. Copyright  2011 Pearson Education, Inc. Example 15.10: Find the pH of a mixture of 0.150 M HF(aq) solution and 0.100 M HClO 2 (aq) 21 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 Tro, Chemistry: A Molecular Approach, 2/e

22. Copyright  2011 Pearson Education, Inc. Example 15.10: Find the pH of a mixture of 0.150 M HF(aq) solution and 0.100 M HClO 2 (aq) 22 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 Tro, Chemistry: A Molecular Approach, 2/e

23. Copyright  2011 Pearson Education, Inc. 23 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 Tro, Chemistry: A Molecular Approach, 2/e

24. Copyright  2011 Pearson Education, Inc. Practice – Determine the pH @ 25 ºC of a solution that is a mixture of 0.045 M HCl and 0.15 M HF 24 Tro, Chemistry: A Molecular Approach, 2/e

25. Copyright  2011 Pearson Education, Inc. Practice – Determine the pH @ 25 ºC of a solution that is a mixture of 0.045 M HCl and 0.15 M HF 25 pH is unitless; the fact that the pH < 7 means the solution is acidic [HCl] = 4.5 x 10 −2 M, [HF] = 0.15 M pH Check: Solution: Conceptual Plan: Relationships: Given: Find: [H 3 O + ][HCl]pH Because HCl is a strong acid and HF is a weak acid, [H 3 O + ] = [HCl] Tro, Chemistry: A Molecular Approach, 2/e

26. Copyright  2011 Pearson Education, Inc. Strong Bases 26 NaOH  Na + + OH − The stronger the base, the more willing it is to accept H use water as the standard acid For ionic bases, practically all units are dissociated into OH – or accept H’s strong electrolyte multi-OH strong bases completely dissociated [HO – ] = [strong base] x (# OH) Tro, Chemistry: A Molecular Approach, 2/e

27. Copyright  2011 Pearson Education, Inc. 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 27 Tro, Chemistry: A Molecular Approach, 2/e

28. Copyright  2011 Pearson Education, Inc. 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 28 Tro, Chemistry: A Molecular Approach, 2/e

29. Copyright  2011 Pearson Education, Inc. Practice – Calculate the pH of the following strong acid or base solutions 0.0020 M HCl 0.0015 M Ca(OH) 2 [H 3 O + ] = [HCl] = 2.0 x 10 −3 M pH = −log(2.0 x 10 −3 ) = 2.70 [OH − ] = 2 x [Ca(OH) 2 ] = 3.0 x 10 −3 M pOH = −log(3.0 x 10 −3 ) = 2.52 pH = 14.00 − pOH = 14.00 − 2.52 pH = 11.48 29 Tro, Chemistry: A Molecular Approach, 2/e

30. Copyright  2011 Pearson Education, Inc. Weak Bases 30 NH 3 + H 2 O  NH 4 + + OH − In weak bases, only a small fraction of molecules accept H’s weak electrolyte most of the weak base molecules do not take H from water much less than 1% ionization in water [HO – ] << [weak base] Finding the pH of a weak base solution is similar to finding the pH of a weak acid Tro, Chemistry: A Molecular Approach, 2/e

31. Copyright  2011 Pearson Education, Inc. Base Ionization Constant, K b Base strength measured by the size of the equilibrium constant when react with H 2 O :Base + H 2 O  OH − + H:Base + The equilibrium constant is called the base ionization constant, K b larger K b = stronger base 31 Tro, Chemistry: A Molecular Approach, 2/e

32. Copyright  2011 Pearson Education, Inc. 32 Tro, Chemistry: A Molecular Approach, 2/e

33. Copyright  2011 Pearson Education, Inc. Structure of Amines 33 Tro, Chemistry: A Molecular Approach, 2/e

34. Copyright  2011 Pearson Education, Inc. [NH 3 ][NH 4 + ][OH  ] initial change equilibrium Example 15.12:Find the pH of 0.100 M NH 3 (aq) 34 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 Tro, Chemistry: A Molecular Approach, 2/e

35. Copyright  2011 Pearson Education, Inc. [NH 3 ][NH 4 + ][OH  ] initial 0.100 00 change equilibrium Example 15.12: Find the pH of 0.100 M NH 3 (aq) 35 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 Tro, Chemistry: A Molecular Approach, 2/e

36. Copyright  2011 Pearson Education, Inc. Example 15.12: Find the pH of 0.100 M NH 3 (aq) 36 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 Tro, Chemistry: A Molecular Approach, 2/e

37. Copyright  2011 Pearson Education, Inc. Example 15.12: Find the pH of 0.100 M NH 3 (aq) 37 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 Tro, Chemistry: A Molecular Approach, 2/e

38. Copyright  2011 Pearson Education, Inc. Example 15.12: Find the pH of 0.100 M NH 3 (aq) 38 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 Tro, Chemistry: A Molecular Approach, 2/e

39. Copyright  2011 Pearson Education, Inc. Example 15.12: Find the pH of 0.100 M NH 3 (aq) 39 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 Tro, Chemistry: A Molecular Approach, 2/e

40. Copyright  2011 Pearson Education, Inc. Example 15.12: Find the pH of 0.100 M NH 3 (aq) 40 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 Tro, Chemistry: A Molecular Approach, 2/e

41. Copyright  2011 Pearson Education, Inc. Example 15.12: Find the pH of 0.100 M NH 3 (aq) 41 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 Tro, Chemistry: A Molecular Approach, 2/e

42. Copyright  2011 Pearson Education, Inc. Practice – Find the pH of a 0.0015 M morphine solution, K b = 1.6 x 10 −6 42 Tro, Chemistry: A Molecular Approach, 2/e

43. Copyright  2011 Pearson Education, Inc. Practice – Find the pH of a 0.0015 M morphine solution 43 write the reaction for the base with water construct an ICE table for the reaction enter the initial concentrations – assuming the [OH  ] from water is ≈ 0 because no products initially, Q c = 0, and the reaction is proceeding forward B + H 2 O  BH + + OH  [B][BH + ][OH  ] initial 0.0015 0 ≈ 0 change equilibrium Tro, Chemistry: A Molecular Approach, 2/e

44. Copyright  2011 Pearson Education, Inc. [B][BH + ][OH  ] initial 0.0015 00 change equilibrium Practice – Find the pH of a 0.0015 M morphine solution 44 represent the change in the concentrations in terms of x sum the columns to find the equilibrium concentrations in terms of x substitute into the equilibrium constant expression +x+x +x+x xx 0.0015  x xx B + H 2 O  BH + + OH  Tro, Chemistry: A Molecular Approach, 2/e

45. Copyright  2011 Pearson Education, Inc. Practice – Find the pH of a 0.0015 M morphine solution 45 determine the value of K b because K b is very small, approximate the [B] eq = [B] init and solve for x K b for morphine = 1.6 x 10 −6 [B][BH + ][OH  ] initial 0.0015 0 ≈ 0 change −x−x+x+x+x+x equilibrium 0.0015xx 0.0015  x Tro, Chemistry: A Molecular Approach, 2/e

46. Copyright  2011 Pearson Education, Inc. Practice – Find the pH of a 0.0015 M morphine solution 46 check if the approximation is valid by seeing if x < 5% of [B] init the approximation is valid x = 4.9 x 10 −5 K b for morphine = 1.6 x 10 −6 [B][BH + ][OH  ] initial 0.0015 0 ≈ 0 change −x−x+x+x+x+x equilibrium 0.0015 xx Tro, Chemistry: A Molecular Approach, 2/e

47. Copyright  2011 Pearson Education, Inc. Practice – Find the pH of a 0.0015 M morphine solution 47 substitute x into the equilibrium concentration definitions and solve x = 4.9 x 10 −5 K b for morphine = 1.6 x 10 −6 [B][BH + ][OH  ] initial 0.0015 0 ≈ 0 change −x−x+x+x+x+x equilibrium 0.0015  x xx [B][BH + ][OH  ] initial 0.0015 0 ≈ 0 change −x−x+x+x+x+x equilibrium 0.00154.9E−5 Tro, Chemistry: A Molecular Approach, 2/e

48. Copyright  2011 Pearson Education, Inc. Practice – Find the pH of a 0.0015 M morphine solution 48 use the [OH − ] to find the [H 3 O + ] using K w substitute [H 3 O + ] into the formula for pH and solve K b for morphine = 1.6 x 10 −6 [B][BH + ][OH  ] initial 0.0015 0 ≈ 0 change −x−x+x+x+x+x equilibrium 0.00154.9E−5 Tro, Chemistry: A Molecular Approach, 2/e

49. Copyright  2011 Pearson Education, Inc. Practice – Find the pH of a 0.0015 M morphine solution 49 use the [OH − ] to find the pOH use pOH to find pH K b for morphine = 1.6 x 10 −6 [B][BH + ][OH  ] initial 0.0015 0 ≈ 0 change −x−x+x+x+x+x equilibrium 0.00154.9E−5 Tro, Chemistry: A Molecular Approach, 2/e

50. Copyright  2011 Pearson Education, Inc. Practice – Find the pH of a 0.0015 M morphine solution 50 check by substituting the equilibrium concentrations back into the equilibrium constant expression and comparing the calculated K b to the given K b the answer matches the given K b K b for morphine = 1.6 x 10 −6 [B][BH + ][OH  ] initial 0.0015 0 ≈ 0 change −x−x+x+x+x+x equilibrium 0.00154.9E−5 Tro, Chemistry: A Molecular Approach, 2/e

51. Copyright  2011 Pearson Education, Inc. Acid–Base Properties of Salts Salts are water-soluble ionic compounds Salts that contain the cation of a strong base and an anion that is the conjugate base of a weak acid are basic NaHCO 3 solutions are basic  Na + is the cation of the strong base NaOH  HCO 3 − is the conjugate base of the weak acid H 2 CO 3 Salts that contain cations that are the conjugate acid of a weak base and an anion of a strong acid are acidic NH 4 Cl solutions are acidic  NH 4 + is the conjugate acid of the weak base NH 3  Cl − is the anion of the strong acid HCl 51 Tro, Chemistry: A Molecular Approach, 2/e

52. Copyright  2011 Pearson Education, Inc. Anions as Weak Bases Every anion can be thought of as the conjugate base of an acid Therefore, every anion can potentially be a base A − (aq) + H 2 O(l)  HA(aq) + OH − (aq) The stronger the acid is, the weaker the conjugate base is An anion that is the conjugate base of a strong acid is pH neutral Cl − (aq) + H 2 O(l)  HCl(aq) + OH − (aq) An anion that is the conjugate base of a weak acid is basic F − (aq) + H 2 O(l)  HF(aq) + OH − (aq) 52 Tro, Chemistry: A Molecular Approach, 2/e

53. Copyright  2011 Pearson Education, Inc. 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 53 Tro, Chemistry: A Molecular Approach, 2/e

54. Copyright  2011 Pearson Education, Inc. 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 54 Tro, Chemistry: A Molecular Approach, 2/e

55. Copyright  2011 Pearson Education, Inc. 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

56. Copyright  2011 Pearson Education, Inc. [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 56 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 Tro, Chemistry: A Molecular Approach, 2/e

57. Copyright  2011 Pearson Education, Inc. Example 15.14: Find the pH of 0.100 M NaCHO 2 (aq) 57 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 Tro, Chemistry: A Molecular Approach, 2/e

58. Copyright  2011 Pearson Education, Inc. Example 15.14: Find the pH of 0.100 M NaCHO 2 (aq) 58 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 Tro, Chemistry: A Molecular Approach, 2/e

59. Copyright  2011 Pearson Education, Inc. [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) 59 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 Tro, Chemistry: A Molecular Approach, 2/e

60. Copyright  2011 Pearson Education, Inc. Example 15.14: Find the pH of 0.100 M NaCHO 2 (aq) 60 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 Tro, Chemistry: A Molecular Approach, 2/e

61. Copyright  2011 Pearson Education, Inc. Example 15.14: Find the pH of 0.100 M NaCHO 2 (aq) 61 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 Tro, Chemistry: A Molecular Approach, 2/e

62. Copyright  2011 Pearson Education, Inc. Example 15.14: Find the pH of 0.100 M NaCHO 2 (aq) 62 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 Tro, Chemistry: A Molecular Approach, 2/e

63. Copyright  2011 Pearson Education, Inc. Practice – If a 0.15 M NaA solution has a pOH of 5.45, what is the K a of HA? 63 Tro, Chemistry: A Molecular Approach, 2/e

64. Copyright  2011 Pearson Education, Inc. If a 0.0015 M NaA solution has a pOH of 5.45, what is the K a of HA? Na + is the cation of a strong base – pOH neutral. Because pOH is < 7, the solutinon is basic. A − is basic. write the reaction for the anion with water construct an ICE table for the reaction enter the initial concentrations – assuming the [OH  ] from water is ≈ 0 A − + H 2 O  HA + OH  [A − ][HA][OH  ] initial 0.1000≈ 0 change equilibrium 64

65. Copyright  2011 Pearson Education, Inc. 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 65 Tro, Chemistry: A Molecular Approach, 2/e

66. Copyright  2011 Pearson Education, Inc. 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 66 Practice – If a 0.15 M NaA solution has a pOH of 5.45, what is the K a of HA? Tro, Chemistry: A Molecular Approach, 2/e

67. Copyright  2011 Pearson Education, Inc. use K b of A − to find K a of HA 67 Practice – If a 0.15 M NaA solution has a pOH of 5.45, what is the K a of HA? [A − ][HA][OH  ] initial 0.150≈ 0 change − 3.6E−6 + 3.6E−6 equilibrium 0.153.6E-6 Tro, Chemistry: A Molecular Approach, 2/e

68. Copyright  2011 Pearson Education, Inc. Polyatomic Cations as Weak Acids Some cations can be thought of as the conjugate acid of a weak base others are the counter-ions of a strong base Therefore, some cations can potentially be acidic MH + (aq) + H 2 O(l)  MOH(aq) + H 3 O + (aq) The stronger the base is, the weaker the conjugate acid is a cation that is the counter-ion of a strong base is pH neutral a cation that is the conjugate acid of a weak base is acidic NH 4 + (aq) + H 2 O(l)  NH 3 (aq) + H 3 O + (aq)  because NH 3 is a weak base, the position of this equilibrium favors the right 68 Tro, Chemistry: A Molecular Approach, 2/e

69. Copyright  2011 Pearson Education, Inc. Metal Cations as Weak Acids Cations of small, highly charged metals are weakly acidic alkali metal cations and alkali earth metal cations are pH neutral cations are hydrated Al(H 2 O) 6 3+ (aq) + H 2 O(l)  Al(H 2 O) 5 (OH) 2+ (aq) + H 3 O + (aq) 69 Tro, Chemistry: A Molecular Approach, 2/e

70. Copyright  2011 Pearson Education, Inc. Example 15.15: Determine if the given cation Is acidic or neutral a) C 5 N 5 NH 2 + the conjugate acid of the weak base pyridine, therefore acidic b) Ca 2+ the counter-ion of the strong base Ca(OH) 2, therefore neutral c) Cr 3+ a highly charged metal ion, therefore acidic 70 Tro, Chemistry: A Molecular Approach, 2/e

71. Copyright  2011 Pearson Education, Inc. Classifying Salt Solutions as Acidic, Basic, or Neutral If the salt cation is the counter-ion of a strong base and the anion is the conjugate base of a strong acid, it will form a neutral solution NaCl Ca(NO 3 ) 2 KBr If the salt cation is the counter-ion of a strong base and the anion is the conjugate base of a weak acid, it will form a basic solution NaF Ca(C 2 H 3 O 2 ) 2 KNO 2 71 Tro, Chemistry: A Molecular Approach, 2/e

72. Copyright  2011 Pearson Education, Inc. Classifying Salt Solutions as Acidic, Basic, or Neutral If the salt cation is the conjugate acid of a weak base and the anion is the conjugate base of a strong acid, it will form an acidic solution NH 4 Cl If the salt cation is a highly charged metal ion and the anion is the conjugate base of a strong acid, it will form an acidic solution Al(NO 3 ) 3 72 Tro, Chemistry: A Molecular Approach, 2/e

73. Copyright  2011 Pearson Education, Inc. Classifying Salt Solutions as Acidic, Basic, or Neutral If the salt cation is the conjugate acid of a weak base and the anion is the conjugate base of a weak acid, the pH of the solution depends on the relative strengths of the acid and base NH 4 F because HF is a stronger acid than NH 4 +, K a of NH 4 + is larger than K b of the F − ; therefore the solution will be acidic 73 Tro, Chemistry: A Molecular Approach, 2/e

74. Copyright  2011 Pearson Education, Inc. Example 15.16: Determine whether a solution of the following salts is acidic, basic, or neutral a) SrCl 2 Sr 2+ is the counter-ion of a strong base, pH neutral Cl − is the conjugate base of a strong acid, pH neutral solution will be pH neutral b) AlBr 3 Al 3+ is a small, highly charged metal ion, weak acid Cl − is the conjugate base of a strong acid, pH neutral solution will be acidic c) CH 3 NH 3 NO 3 CH 3 NH 3 + is the conjugate acid of a weak base, acidic NO 3 − is the conjugate base of a strong acid, pH neutral solution will be acidic 74 Tro, Chemistry: A Molecular Approach, 2/e

75. Copyright  2011 Pearson Education, Inc. Example 15.16: Determine whether a solution of the following salts is acidic, basic, or neutral d) NaCHO 2 Na + is the counter-ion of a strong base, pH neutral CHO 2 − is the conjugate base of a weak acid, basic solution will be basic e) NH 4 F NH 4 + is the conjugate acid of a weak base, acidic F − is the conjugate base of a weak acid, basic K a (NH 4 + ) > K b (F − ); solution will be acidic 75 Tro, Chemistry: A Molecular Approach, 2/e

76. Copyright  2011 Pearson Education, Inc. the solution is pH neutral Co 3+ is a highly charged cation, pH acidic Practice – Determine whether a solution of the following salts is acidic, basic, or neutral 76 KNO 3 CoCl 3 Ba(HCO 3 ) 2 CH 3 NH 3 NO 3 K + is the counter-ion of a strong base, pH neutral NO 3 − is the counter-ion of a strong acid, pH neutral Cl − is the counter-ion of a strong acid, pH neutral the solution is pH acidic Ba 2+ is the counter-ion of a strong base, pH neutral HCO 3 − is the conjugate of a weak acid, pH basic the solution is pH basic CH 3 NH 3 + is the conjugate of a weak base, pH acidic NO 3 − is the counter-ion of a strong acid, pH neutral the solution is pH acidic Tro, Chemistry: A Molecular Approach, 2/e

77. Copyright  2011 Pearson Education, Inc. Ionization in Polyprotic Acids Because polyprotic acids ionize in steps, each H has a separate K a K a1 > K a2 > K a3 Generally, the difference in K a values is great enough so that the second ionization does not happen to a large enough extent to affect the pH most pH problems just do first ionization except H 2 SO 4  use [H 2 SO 4 ] as the [H 3 O + ] for the second ionization [A 2− ] = K a2 as long as the second ionization is negligible 77 Tro, Chemistry: A Molecular Approach, 2/e

78. Copyright  2011 Pearson Education, Inc. 78 Tro, Chemistry: A Molecular Approach, 2/e

79. Copyright  2011 Pearson Education, Inc. Practice – 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 ) 79 Tro, Chemistry: A Molecular Approach, 2/e

80. Copyright  2011 Pearson Education, Inc. Practice – What is the pH of a 0.12 M solution of carbonic acid, H 2 CO 3 ? 80 write the reactions for the acid with water one H at a time construct an ICE table for the reaction enter the initial concentrations – assuming the second ionization is negligible H 2 CO 3 + H 2 O  HCO 3  + H 3 O + [HA][A − ][H 3 O + ] initial 0.12 0 ≈ 0 change equilibrium HCO 3 − + H 2 O  CO 3 2− + H 3 O + Tro, Chemistry: A Molecular Approach, 2/e

81. Copyright  2011 Pearson Education, Inc. [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 ? 81 represent the change in the concentrations in terms of x sum the columns to find the equilibrium concentrations in terms of x substitute into the equilibrium constant expression +x+x +x+x xx 0.12  x xx H 2 CO 3 + H 2 O  HCO 3  + H 3 O + Tro, Chemistry: A Molecular Approach, 2/e

82. Copyright  2011 Pearson Education, Inc. 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 82 Tro, Chemistry: A Molecular Approach, 2/e

83. Copyright  2011 Pearson Education, Inc. Practice – What is the pH of a 0.12 M solution of carbonic acid, H 2 CO 3 ? 83 check if the approximation is valid by seeing if x < 5% of [H 2 CO 3 ] init K a1 for H 2 CO 3 = 4.3 x 10 −7 the approximation is valid x = 2.27 x 10 −4 [HA][A − ][H 3 O + ] initial 0.12 0 ≈ 0 change −x−x+x+x+x+x equilibrium 0.12 xx Tro, Chemistry: A Molecular Approach, 2/e

84. Copyright  2011 Pearson Education, Inc. Practice – What is the pH of a 0.12 M solution of carbonic acid, H 2 CO 3 ? 84 substitute x into the equilibrium concentration definitions and solve x = 2.3 x 10 −4 [HA][A − ][H 3 O + ] initial 0.12 0 ≈ 0 change −x−x+x+x+x+x equilibrium 0.12−x xx Tro, Chemistry: A Molecular Approach, 2/e

85. Copyright  2011 Pearson Education, Inc. Practice – What is the pH of a 0.12 M solution of carbonic acid, H 2 CO 3 ? 85 substitute [H 3 O + ] into the formula for pH and solve [HA][A − ][H 3 O + ] initial 0.12 0 ≈ 0 change −x−x+x+x+x+x equilibrium 0.120.00023 Tro, Chemistry: A Molecular Approach, 2/e

86. Copyright  2011 Pearson Education, Inc. Practice – What is the pH of a 0.12 M solution of carbonic acid, H 2 CO 3 ? 86 check by substituting the equilibrium concentrations back into the equilibrium constant expression and comparing the calculated K a to the given K a the values match within sig figs [HA][A − ][H 3 O + ] initial 0.12 0 ≈ 0 change −x−x+x+x+x+x equilibrium 0.120.00023 K a1 for H 2 CO 3 = 4.3 x 10 −7 Tro, Chemistry: A Molecular Approach, 2/e

87. Copyright  2011 Pearson Education, Inc. Practice – 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 ) 87 Tro, Chemistry: A Molecular Approach, 2/e

88. Copyright  2011 Pearson Education, Inc. Practice – What is the [CO 3 2− ] in a 0.12 M solution of carbonic acid, H 2 CO 3 ? 88 write the reactions for the acid with water one H at a time construct an ICE table for the reaction enter the initial concentrations for the second ionization using the equilibrium concentrations from first ionization H 2 CO 3 + H 2 O  HCO 3  + H 3 O + [HCO 3 − ][CO 3 2− ][H 3 O + ] initial 0.00023 0 change equilibrium HCO 3 − + H 2 O  CO 3 2− + H 3 O + Tro, Chemistry: A Molecular Approach, 2/e

89. Copyright  2011 Pearson Education, Inc. [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 ? 89 represent the change in the concentrations in terms of x sum the columns to find the equilibrium concentrations in terms of x substitute into the equilibrium constant expression +x+x +x+x xx 2.3E−4  x x HCO 3 − + H 2 O  CO 3 2− + H 3 O + 2.3E−4 +x Tro, Chemistry: A Molecular Approach, 2/e

90. Copyright  2011 Pearson Education, Inc. 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 90 Tro, Chemistry: A Molecular Approach, 2/e

91. Copyright  2011 Pearson Education, Inc. Ionization in H 2 SO 4 The ionization constants for H 2 SO 4 are H 2 SO 4 + H 2 O  HSO 4  + H 3 O + strong HSO 4  + H 2 O  SO 4 2  + H 3 O + K a2 = 1.2 x 10 −2 For most sulfuric acid solutions, the second ionization is significant and must be accounted for Because the first ionization is complete, use the given [H 2 SO 4 ] = [HSO 4 − ] initial = [H 3 O + ] initial 91 Tro, Chemistry: A Molecular Approach, 2/e

92. Copyright  2011 Pearson Education, Inc. Example 15.18: Find the pH of 0.0100 M H 2 SO 4 (aq) solution @ 25 °C 92 write the reactions for the acid with water construct an ICE table for the second ionization reaction enter the initial concentrations – assuming the [HSO 4 − ] and [H 3 O + ] is ≈ [H 2 SO 4 ] [HSO 4  ][SO 4 2  ][H 3 O + ] initial 0.01000 change equilibrium HSO 4  + H 2 O  SO 4 2  + H 3 O + H 2 SO 4 + H 2 O  HSO 4  + H 3 O + Tro, Chemistry: A Molecular Approach, 2/e

93. Copyright  2011 Pearson Education, Inc. Example 15.18: Find the pH of 0.0100 M H 2 SO 4 (aq) solution @ 25 °C represent the change in the concentrations in terms of x sum the columns to find the equilibrium concentrations in terms of x substitute into the equilibrium constant expression [HSO 4  ][SO 4 2  ][H 3 O + ] initial 0.01000 change −x−x+x+x+x+x equilibrium 0.0100 −x x Tro, Chemistry: A Molecular Approach, 2/e93

94. Copyright  2011 Pearson Education, Inc. Example 15.18: Find the pH of 0.0100 M H 2 SO 4 (aq) solution @ 25 °C expand and solve for x using the quadratic formula K a for HSO 4 − = 0.012 Tro, Chemistry: A Molecular Approach, 2/e94

95. Copyright  2011 Pearson Education, Inc. Example 15.18: Find the pH of 0.0100 M H 2 SO 4 (aq) solution @ 25 °C substitute x into the equilibrium concentration definitions and solve x = 0.0045 [HSO 4  ][SO 4 2  ][H 3 O + ] initial 0.01000 change −x−x+x+x+x+x equilibrium 0.0100 −xx K a for HSO 4 − = 0.012 [HSO 4  ][SO 4 2  ][H 3 O + ] initial 0.01000 change −x−x+x+x+x+x equilibrium 0.00550.00450.0145 Tro, Chemistry: A Molecular Approach, 2/e95

96. Copyright  2011 Pearson Education, Inc. Example 15.18: Find the pH of 0.0100 M H 2 SO 4 (aq) solution @ 25 °C substitute [H 3 O + ] into the formula for pH and solve K a for HSO 4 − = 0.012 [HSO 4  ][SO 4 2  ][H 3 O + ] initial 0.01000 change −x−x+x+x+x+x equilibrium 0.00550.00450.0145 Tro, Chemistry: A Molecular Approach, 2/e96

97. Copyright  2011 Pearson Education, Inc. Example 15.18: Find the pH of 0.0100 M H 2 SO 4 (aq) solution @ 25 °C 97 check by substituting the equilibrium concentrations back into the equilibrium constant expression and comparing the calculated K a to the given K a the answer matches K a for HSO 4 − = 0.012 [HSO 4  ][SO 4 2  ][H 3 O + ] initial 0.01000 change −x−x+x+x+x+x equilibrium 0.00550.00450.0145 Tro, Chemistry: A Molecular Approach, 2/e

98. Copyright  2011 Pearson Education, Inc. Strengths of Binary Acids The more  + H─X  − polarized the bond, the more acidic the bond The stronger the H─X bond, the weaker the acid Binary acid strength increases to the right across a period acidity: H─C < H─N < H─O < H─F Binary acid strength increases down the column acidity: H─F < H─Cl < H─Br < H─I 98 Tro, Chemistry: A Molecular Approach, 2/e

99. Copyright  2011 Pearson Education, Inc. Relationship between Bond Strength and Acidity Acid Bond Energy kJ/mol Type of Acid HF565weak HCl431strong HBr364strong 99 Tro, Chemistry: A Molecular Approach, 2/e

100. Copyright  2011 Pearson Education, Inc. Strengths of Oxyacids, H–O–Y The more electronegative the Y atom, the stronger the oxyacid HClO > HIO acidity of oxyacids decreases down a group  same trend as binary acids helps weakens the H–O bond The larger the oxidation number of the central atom, the stronger the oxyacid H 2 CO 3 > H 3 BO 3 acidity of oxyacids increases to the right across a period  opposite trend of binary acids The more oxygens attached to Y, the stronger the oxyacid further weakens and polarizes the H–O bond HClO 3 > HClO 2 100 Tro, Chemistry: A Molecular Approach, 2/e

101. Copyright  2011 Pearson Education, Inc. 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 −11 101 Tro, Chemistry: A Molecular Approach, 2/e

102. Copyright  2011 Pearson Education, Inc. Relationship Between Number of Oxygens on the Central Atom and Acidity 102 Tro, Chemistry: A Molecular Approach, 2/e

103. Copyright  2011 Pearson Education, Inc. Practice – Order the Following By Acidity (Least to Most) H 3 PO 4 HNO 3 H 3 PO 3 H 3 AsO 3 By Acidity (Least to Most) HClHBrH 2 SHS − By Basicity (Least to Most) CO 3 2− NO 3 − HCO 3 − BO 3 3− Tro, Chemistry: A Molecular Approach, 2/e103

104. Copyright  2011 Pearson Education, Inc. Practice – Order the Following By Acidity (Least to Most) H 3 PO 4 HNO 3 H 3 PO 3 H 3 AsO 3 H 3 AsO 3 < H 3 PO 3 < H 3 PO 4 < HNO 3 By Acidity (Least to Most) HClHBrH 2 SHS − HS − < H 2 S < HCl < HBr By Basicity (Least to Most) CO 3 2− NO 3 − HCO 3 − BO 3 3− NO 3 − < HCO 3 − < CO 3 2− < BO 3 3− Tro, Chemistry: A Molecular Approach, 2/e104

105. Copyright  2011 Pearson Education, Inc. Lewis Acid–Base Theory Lewis Acid–Base theory focuses on transferring an electron pair lone pair  bond bond  lone pair Does NOT require H atoms The electron donor is called the Lewis Base electron rich, therefore nucleophile The electron acceptor is called the Lewis Acid electron deficient, therefore electrophile 105 Tro, Chemistry: A Molecular Approach, 2/e

106. Copyright  2011 Pearson Education, Inc. Lewis Bases Lewis Base has electrons it is willing to give away to or share with another atom Lewis Base must have lone pair of electrons on it that it can donate Anions are better Lewis Bases than neutral atoms or molecules N: < N: − Generally, the more electronegative an atom, the less willing it is to be a Lewis Base O: < S: 106 Tro, Chemistry: A Molecular Approach, 2/e

107. Copyright  2011 Pearson Education, Inc. Lewis Acids Electron deficient, either from being attached to electronegative atom(s) not having an octet Must have empty orbital willing to accept the electron pair H + has empty 1s orbital B in BF 3 has empty 2p orbital and an incomplete octet Many small, highly charged metal cations have empty orbitals they can use to accept electrons Atoms that are attached to highly electronegative atoms and have multiple bonds can be Lewis Acids 107 Tro, Chemistry: A Molecular Approach, 2/e

108. Copyright  2011 Pearson Education, Inc. Lewis Acid–Base Reactions The base donates a pair of electrons to the acid Generally results in a covalent bond forming H 3 N: + BF 3  H 3 N─BF 3 The product that forms is called an adduct Arrhenius and Brønsted-Lowry acid–base reactions are also Lewis 108 Tro, Chemistry: A Molecular Approach, 2/e

109. Copyright  2011 Pearson Education, Inc. Examples of Lewis Acid–Base Reactions 109 Tro, Chemistry: A Molecular Approach, 2/e

110. Copyright  2011 Pearson Education, Inc. Examples of Lewis Acid–Base Reactions Ag + (aq) + 2 :NH 3(aq)  Ag(NH 3 ) 2 + (aq) Lewis Acid Lewis Base Adduct 110 Tro, Chemistry: A Molecular Approach, 2/e

111. Copyright  2011 Pearson Education, Inc. Practice – Identify the Lewis Acid and Lewis Base in Each Reaction 111 Lewis Base Lewis Base Lewis Acid Lewis Acid Tro, Chemistry: A Molecular Approach, 2/e

112. Copyright  2011 Pearson Education, Inc. 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) 112 Tro, Chemistry: A Molecular Approach, 2/e

113. Copyright  2011 Pearson Education, Inc. 113 What Is Acid Rain? Natural rain water has a pH of 5.6 naturally slightly acidic due mainly to CO 2 Rain water with a pH lower than 5.6 is called acid rain Acid rain is linked to damage in ecosystems and structures Tro, Chemistry: A Molecular Approach, 2/e

114. Copyright  2011 Pearson Education, Inc. 114 What Causes Acid Rain? Many natural and pollutant gases dissolved in the air are nonmetal oxides CO 2, SO 2, NO 2 Nonmetal oxides are acidic CO 2 (g) + H 2 O(l)  H 2 CO 3 (aq) 2 SO 2 (g) + O 2 (g) + 2 H 2 O(l)  2 H 2 SO 4 (aq) 4 NO 2 (g) + O 2 (g) + 2 H 2 O(l)  4 HNO 3 (aq) Processes that produce nonmetal oxide gases as waste increase the acidity of the rain natural – volcanoes and some bacterial action man-made – combustion of fuel Tro, Chemistry: A Molecular Approach, 2/e

115. Copyright  2011 Pearson Education, Inc. pH of Rain in Different Regions Tro, Chemistry: A Molecular Approach, 2/e115

116. Copyright  2011 Pearson Education, Inc. 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 116 Tro, Chemistry: A Molecular Approach, 2/e

117. Copyright  2011 Pearson Education, Inc. Sources of SO 2 from Utilities Tro, Chemistry: A Molecular Approach, 2/e 117

118. Copyright  2011 Pearson Education, Inc. 118 Damage from Acid Rain Acids react with metals, and materials that contain carbonates Acid rain damages bridges, cars, and other metallic structures Acid rain damages buildings and other structures made of limestone or cement Acidifying lakes affects aquatic life Soil acidity causes more dissolving of minerals and leaching more minerals from soil making it difficult for trees Tro, Chemistry: A Molecular Approach, 2/e

119. Copyright  2011 Pearson Education, Inc. 119 Damage from Acid Rain Tro, Chemistry: A Molecular Approach, 2/e

120. Copyright  2011 Pearson Education, Inc. 120 Acid Rain Legislation 1990 Clean Air Act attacks acid rain forces utilities to reduce SO 2 Result is acid rain in the Northeast stabilized and beginning to be reduced Tro, Chemistry: A Molecular Approach, 2/e