1. 1 © 2009 Brooks/Cole - Cengage Acids and Bases Chapter 14

2. 2 © 2009 Brooks/Cole - Cengage Objectives Review acids and bases (Chem I & Ch. 4) Bronstead acids and bases Define autoionization of water, K w and pH Define K eq for acids (K a ) and bases (K b ) Types of reactions (qualitative) Calculations w/K eq ’s (K a, K b and K w ) Polyprotic acids Acid/base properties of salts Lewis acids and bases Molecular structure and acid strength

3. 3 © 2009 Brooks/Cole - Cengage Ch. 14.1 – The Nature of Acids and Bases

4. 4 © 2009 Brooks/Cole - Cengage Acid and Bases

5. 5 © 2009 Brooks/Cole - Cengage Acid and Bases

6. 6 © 2009 Brooks/Cole - Cengage Some Definitions Arrhenius acids and bases –Acid:Substance that, when dissolved in water, increases the concentration of hydrogen ions (protons, H + ). HCl(aq)  H + (aq) + Cl - (aq) –Base:Substance that, when dissolved in water, increases the concentration of hydroxide ions. NaOH(aq)  Na + (aq) + OH - (aq)

7. 7 © 2009 Brooks/Cole - Cengage ACID-BASE THEORIES The most general theory for common aqueous acids and bases is the BRØNSTED - LOWRY theoryThe most general theory for common aqueous acids and bases is the BRØNSTED - LOWRY theory ACIDS DONATE H + IONSACIDS DONATE H + IONS BASES ACCEPT H + IONSBASES ACCEPT H + IONS

8. 8 © 2009 Brooks/Cole - Cengage The Brønsted definition means NH 3 is a BASE in water — and water is itself an ACID ACID-BASE THEORIES

9. 9 © 2009 Brooks/Cole - Cengage ACID-BASE THEORIES NH 3 is a BASE in water — and water is itself an ACID NH 3 / NH 4 + is a conjugate pair — related by the gain or loss of H + Every acid has a conjugate base - and vice-versa.

10. 10 © 2009 Brooks/Cole - Cengage Conjugate Pairs

11. 11 © 2009 Brooks/Cole - Cengage Acid/Base Conjugate Pair Practice Write the formula for the conjugate acid of each of the following bases: BaseConj. Acid HSO 3 - H 2 SO 3 F - HF PO 4 3- HPO 4 2- COHCO +

12. 12 © 2009 Brooks/Cole - Cengage Conjugate Acid-Base Pair Examples H 2 S + NH 3  NH 4 + + HS - NH 3 + HCO 3 1-  NH 4 + + CO 3 2- H 3 PO 4 + H 2 O  H 2 PO 4 - + H 3 O +

13. 13 © 2009 Brooks/Cole - Cengage Ch. 14.2 – Acid Strength Generally divide acids and bases into STRONG or WEAK ones.Generally divide acids and bases into STRONG or WEAK ones. STRONG ACID: HNO 3 (aq) + H 2 O(liq)  H 3 O + (aq) + NO 3 - (aq) HNO 3 is about 100% dissociated in water.

14. 14 © 2009 Brooks/Cole - Cengage HNO 3, HCl, HI, HBr, H 2 SO 4, HClO 3 and HClO 4 are the strong acids. Strong and Weak Acids

15. 15 © 2009 Brooks/Cole - Cengage Weak acids are much less than 100% ionized in water.Weak acids are much less than 100% ionized in water. One of the best known is acetic acid = CH 3 CO 2 H Also written as HC 2 H 3 O 2 or HOAc Strong and Weak Acids

16. 16 © 2009 Brooks/Cole - Cengage Water as an Acid and a Base H 2 O can function as both an ACID and a BASE. In pure water there can be AUTOIONIZATION Dissociation Constant for Water= K w K w = [H 3 O + ] [OH - ] = 1.00 x 10 -14 at 25 o C

17. 17 © 2009 Brooks/Cole - Cengage Water as an Acid and as a Base K w = [H 3 O + ] [OH - ] = 1.00 x 10 -14 at 25 o C In a neutral solution [H 3 O + ] = [OH - ] so K w = [H 3 O + ] 2 = [OH - ] 2 and so [H 3 O + ] = [OH - ] = 1.00 x 10 -7 M

18. 18 © 2009 Brooks/Cole - Cengage Calculating [H 3 O + ] & [OH - ] You add 0.0010 mol of NaOH to 1.0 L of pure water. Calculate [HO + ] and [OH - ]. You add 0.0010 mol of NaOH to 1.0 L of pure water. Calculate [H 3 O + ] and [OH - ].Solution 2 H 2 O(liq)  HO + (aq) + OH - (aq) 2 H 2 O(liq)  H 3 O + (aq) + OH - (aq) Le Chatelier predicts equilibrium shifts to the __left____. [HO + ] < 10 -7 at equilibrium. [H 3 O + ] < 10 -7 at equilibrium. Set up a ICE table.

19. 19 © 2009 Brooks/Cole - Cengage Calculating [H 3 O + ] & [OH - ] You add 0.0010 mol of NaOH to 1.0 L of pure water. Calculate [HO + ] and [OH - ]. You add 0.0010 mol of NaOH to 1.0 L of pure water. Calculate [H 3 O + ] and [OH - ].Solution 2 H 2 O(liq)  H 3 O + (aq) + OH - (aq) initial00.0010 initial00.0010 change+x+x change+x+x equilibx0.0010 + x equilibx0.0010 + x K w = (x) (0.0010 + x) Because x << 0.0010 M, assume [OH - ] = 0.0010 M [H 3 O + ] = x = K w / 0.0010 = 1.0 x 10 -11 M

20. 20 © 2009 Brooks/Cole - Cengage Calculating [H 3 O + ] & [OH - ] You add 0.0010 mol of NaOH to 1.0 L of pure water. Calculate [HO + ] and [OH - ]. You add 0.0010 mol of NaOH to 1.0 L of pure water. Calculate [H 3 O + ] and [OH - ].Solution 2 H 2 O(liq)  HO + (aq) + OH - (aq) 2 H 2 O(liq)  H 3 O + (aq) + OH - (aq) [HO + ] = K w / 0.0010 = 1.0 x 10 -11 M [H 3 O + ] = K w / 0.0010 = 1.0 x 10 -11 M This solution is _________ because [H 3 O + ] < [OH - ]

21. 21 © 2009 Brooks/Cole - Cengage Ch. 14.3 – The pH Scale A common way to express acidity and basicity is with pH pH = - log [H 3 O + ] In a neutral solution, [HO + ] = [OH - ] = 1.00 x 10 -7 at 25 o C In a neutral solution, [H 3 O + ] = [OH - ] = 1.00 x 10 -7 at 25 o C pH = -log (1.00 x 10 -7 ) = - (-7) = 7

22. 22 © 2009 Brooks/Cole - Cengage The pH Scale What is the pH of the 0.0010 M NaOH solution? [HO + ] = 1.0 x 10 -11 M [H 3 O + ] = 1.0 x 10 -11 M pH = - log (1.0 x 10 -11 ) = 11.00 General conclusion — Basic solution pH > 7 Basic solution pH > 7 Neutral pH = 7 Neutral pH = 7 Acidic solutionpH < 7 Acidic solutionpH < 7

23. 23 © 2009 Brooks/Cole - Cengage The pH Scale See Active Figure 17.1

24. 24 © 2009 Brooks/Cole - Cengage pH Calculations If the pH of Coke is 3.12, it is _acidic__. Because pH = - log [HO + ] then Because pH = - log [H 3 O + ] then log [HO + ] = - pH log [H 3 O + ] = - pH Take antilog and get [HO + ] = 10 -pH [H 3 O + ] = 10 -pH [HO + ] = 10 -3.12 = 7.6 x 10 -4 M [H 3 O + ] = 10 -3.12 = 7.6 x 10 -4 M

25. 25 © 2009 Brooks/Cole - Cengage pH of Common Substances

26. 26 © 2009 Brooks/Cole - Cengage Other pX Scales In generalpX = -log X and so pOH = - log [OH - ] K w = [HO + ] [OH - ] = 1.00 x 10 -14 at 25 o C K w = [H 3 O + ] [OH - ] = 1.00 x 10 -14 at 25 o C Take the log of both sides -log (10 -14 ) = - log [HO + ] + (-log [OH - ]) -log (10 -14 ) = - log [H 3 O + ] + (-log [OH - ]) pK w = 14 = pH + pOH pK w = 14 = pH + pOH So if you know one, you can determine the others.

27. 27 © 2009 Brooks/Cole - Cengage pH, pOH, [H+], [OH-] Practice pHpOH[H + ][OH-]Acidic/Basic? 6.217.796.2 x 10 -7 1.6 x 10 -8 acidic 3.8710.131.3 x 10 -4 7.4 x 10 -11 acidic 2.4611.543.5 x 10 -3 2.9 x 10 -12 acidic 10.753.251.8 x 10 -11 5.6 x 10 -4 basic

28. 28 © 2009 Brooks/Cole - Cengage Ch. 14.4 – Calculating the pH of Strong Acid Solutions What are the major species in solution? What is the dominant reaction that will take place? –Is it an equilibrium reaction or a reaction that will go essentially to completion? –React all major species until you are left with an equilibrium reaction. Solve for the pH if needed.

29. 29 © 2009 Brooks/Cole - Cengage pH of Strong Acid Solutions Consider an aqueous solution of 2.0 × 10 –3 M HCl. What are the major species in solution? H +, Cl –, H 2 O What is the pH? pH = 2.70

30. 30 © 2009 Brooks/Cole - Cengage pH of Strong Acid Solutions Calculate the pH of a 1.5 × 10 –11 M solution of HCl. pH = 7.00

31. 31 © 2009 Brooks/Cole - Cengage pH of Strong Acid Solutions Calculate the pH of a 1.5 × 10 –2 M solution of HNO 3. When HNO 3 is added to water, a reaction takes place immediately:  HNO 3 + H 2 O  H 3 O + + NO 3 – The reaction controlling the pH is: H 2 O + H 2 O  H 3 O + (aq) + OH - (aq)

32. 32 © 2009 Brooks/Cole - Cengage pH of Strong Acid Solutions Let’s Think About It… What reaction controls the pH? H 2 O(l) + H 2 O(l)  H 3 O + (aq) + OH – (aq) In aqueous solutions, this reaction is always taking place. But is water the major contributor of H + (H 3 O + )? pH = - log (1.5x10 -2 ) = 1.82

33. 33 © 2009 Brooks/Cole - Cengage Ch. 14.5 – Calculating pH of Weak Acid Solutions Aspirin is a good example of a weak acid, where most of the species exist as the neutral molecule and not as ions.

34. 34 © 2009 Brooks/Cole - Cengage Calculating pH of Weak Acid Solutions AcidConjugate Base AcidConjugate Base acetic, CH 3 CO 2 HCH 3 CO 2 -, acetate ammonium, NH 4 + NH 3, ammonia bicarbonate, HCO 3 - CO 3 2-, carbonate A weak acid (or base) is one that ionizes to a VERY small extent (< 5%).

35. 35 © 2009 Brooks/Cole - Cengage Calculating pH of Weak Acid Solutions Consider acetic acid, CH 3 CO 2 H (HOAc) HOAc + H 2 O  H 3 O + + OAc - Acid Conj. base (K is designated K a for ACID) Because [H 3 O + ] and [OAc - ] are SMALL, K a << 1.

36. 36 © 2009 Brooks/Cole - Cengage Equilibrium Constants for Weak Acids Weak acid has K a < 1 Leads to small [H 3 O + ] and a pH of 2 - 7 pK a = -log K a

37. 37 © 2009 Brooks/Cole - Cengage Ionization Constants for Acids/Bases Acids ConjugateBases Increase strength

38. 38 © 2009 Brooks/Cole - Cengage Calculating pH of Weak Acid Solutions You have 1.00 M HOAc. Calculate the equilibrium concentrations of HOAc, H 3 O +, OAc -, and the pH. Step 1. Define equilibrium concs. in ICE table. [HOAc][H 3 O + ][OAc - ] [HOAc][H 3 O + ][OAc - ]initialchangeequilib

39. 39 © 2009 Brooks/Cole - Cengage Calculating pH of Weak Acid Solutions You have 1.00 M HOAc. Calc. the equilibrium concs. of HOAc, H 3 O +, OAc -, and the pH. Step 1. Define equilibrium concs. in ICE table. HOAC(aq) + H 2 O  OAC - + H 3 O + [HOAc][H 3 O + ][OAc - ] initial1.0000 change-x+x+x equilib1.00-xxx Note that we neglect [H 3 O + ] from H 2 O.

40. 40 © 2009 Brooks/Cole - Cengage Calculating pH of Weak Acid Solutions Step 2. Write K a expression You have 1.00 M HOAc. Calc. the equilibrium concs. of HOAc, H 3 O +, OAc -, and the pH. This is a quadratic. However…

41. 41 © 2009 Brooks/Cole - Cengage Calculating pH of Weak Acid Solutions Step 3. Solve K a expression You have 1.00 M HOAc. Calc. the equilibrium concs. of HOAc, H 3 O +, OAc -, and the pH. First assume x is very small because K a is so small. Therefore, Now we can more easily solve this approximate expression.

42. 42 © 2009 Brooks/Cole - Cengage Calculating pH of Weak Acid Solutions Step 3. Solve K a approximate expression You have 1.00 M HOAc. Calc. the equilibrium concs. of HOAc, H 3 O +, OAc -, and the pH. x = [ H 3 O + ] = [ OAc - ] = [K a 1.00] 1/2 x = [ H 3 O + ] = [ OAc - ] = 4.2 x 10 -3 M pH = - log [ H 3 O + ] = -log (4.2 x 10 -3 ) = 2.37

43. 43 © 2009 Brooks/Cole - Cengage Calculating pH of Weak Acid Solutions Consider the approximate expression For many weak acids [H 3 O + ] = [conj. base] = [K a C o ] 1/2 [H 3 O + ] = [conj. base] = [K a C o ] 1/2 where C 0 = initial conc. of acid Useful Rule of Thumb: If C o > 100*K a then [H 3 O + ] = [K a C o ] 1/2 If C o > 100*K a then [H 3 O + ] = [K a C o ] 1/2 In other words, the amount of ionization is small relative to the initial amount. 

44. 44 © 2009 Brooks/Cole - Cengage Calculating pH of Weak Acid Solutions Calculate the pH of a 0.0010 M solution of formic acid, HCO 2 H. HCO 2 H + H 2 O  HCO 2 - + H 3 O + HCO 2 H + H 2 O  HCO 2 - + H 3 O + K a = 1.8 x 10 -4 Approximate solution [H 3 O + ] = [K a C o ] 1/2 = 4.2 x 10 -4 M, pH = 3.37 [H 3 O + ] = [K a C o ] 1/2 = 4.2 x 10 -4 M, pH = 3.37 Exact Solution [H 3 O + ] = [HCO 2 - ] = 3.4 x 10 -4 M [H 3 O + ] = [HCO 2 - ] = 3.4 x 10 -4 M [HCO 2 H] = 0.0010 - 3.4 x 10 -4 = 0.0007 M [HCO 2 H] = 0.0010 - 3.4 x 10 -4 = 0.0007 M pH = 3.47 pH = 3.47

45. 45 © 2009 Brooks/Cole - Cengage Calculating pH of Weak Acid Solutions Calculate the pH of a 0.50 M aqueous solution of HF. (K a = 7.2 x 10 -4 ) What are the major species in solution? HF, H 2 O Why aren’t H + and F – major species? Only a few are formed from the dissociation of HF.

46. 46 © 2009 Brooks/Cole - Cengage Calculating pH of Weak Acid Solutions What are the possibilities for the dominant reaction?  HF(aq) + H 2 O(l)  H 3 O + (aq) + F – (aq) K a =7.2 × 10 -4  H 2 O(l) + H 2 O(l)  H 3 O + (aq) + OH – (aq) K w =1.0 × 10 -14 Which reaction controls the pH? Why?

47. 47 © 2009 Brooks/Cole - Cengage Calculating pH of Weak Acid Solutions Steps Toward Solving for pH HF(aq) +H2OH2O H 3 O + (aq) + F – (aq) Initial0.50 M~ 0 Change–x+x Equilibrium0.50–xxx K a = 7.2 × 10 –4 [H 3 O + ] =(K a *C o ) 1/2 = (7.2x10 -4 *0.05) 1/2 =0.0190 pH = -log(0.0190) = 1.72

48. 48 © 2009 Brooks/Cole - Cengage Using K a Example

49. 49 © 2009 Brooks/Cole - Cengage Using K a Practice A solution prepared from 0.55 mol of butanoic acid is dissolved in water to give 1.0 L of solution with a pH of 2.21. Determine the Ka for butanoic acid. The acid ionizes according to the equation below. CH 3 (CH 2 ) 2 COOH(aq) + H 2 O(l)  H 3 O + (aq) + CH 3 (CH 2 ) 2 COO -

50. 50 © 2009 Brooks/Cole - Cengage Percent Dissociation pH of an acetic acid solution. What are your observations? pH of an acetic acid solution. What are your observations? a pH meter 2.0 M 0.0001 M 0.003 M 0.06 M

51. 51 © 2009 Brooks/Cole - Cengage Percent Dissociation (Ionization) For a given weak acid, the percent dissociation increases as the acid becomes more dilute.

52. 52 © 2009 Brooks/Cole - Cengage Percent Dissociation Example

53. 53 © 2009 Brooks/Cole - Cengage Percent Dissociation The value of K a for a 4.00 M formic acid solution should be: higher than lower thanthe same as the value of K a of an 8.00 M formic acid solution. Explain.

54. 54 © 2009 Brooks/Cole - Cengage Percent Dissociation higher!! However, the % Dissociation and the pH are higher!! Dominant rxn: HCHO 2 (aq) + H 2 O  H 3 O + (aq) + CHO 2 - (aq) HCHO 2 (aq) + H 2 O  H 3 O + (aq) + CHO 2 - (aq) I 4.00 0 0 C -x +x +x E 4.00-x x x K a = 1.8 x 10 -4 = (x) 2 /(4.00-x) x = 0.027; (0.027/4.00) x 100 = 0.67% pH=-log(0.027) = 1.57

55. 55 © 2009 Brooks/Cole - Cengage Strong Base: 100% dissociated in water.Strong Base: 100% dissociated in water. NaOH(aq)  Na + (aq) + OH - (aq) NaOH(aq)  Na + (aq) + OH - (aq) Ch. 14.6 - Bases Other common strong bases include KOH and Ca(OH) 2. CaO (lime) + H 2 O  Ca(OH) 2 (slaked lime) Ca(OH) 2 (slaked lime) CaO

56. 56 © 2009 Brooks/Cole - Cengage Weak base: less than 100% ionized in waterWeak base: less than 100% ionized in water One of the best known weak bases is ammonia NH 3 (aq) + H 2 O(liq)  NH 4 + (aq) + OH - (aq) BasesBases

57. 57 © 2009 Brooks/Cole - Cengage Weak Bases Weak base has K b < 1 Leads to small [OH - ] and a pH of 12 - 7 pK b = -log K b

58. 58 © 2009 Brooks/Cole - Cengage Weak Bases

59. 59 © 2009 Brooks/Cole - Cengage Equilibria Involving A Weak Base You have 0.010 M NH 3. Calc. the pH. NH 3 + H 2 O  NH 4 + + OH - NH 3 + H 2 O  NH 4 + + OH - K b = 1.8 x 10 -5 Step 1. Define equilibrium concs. in ICE table [NH 3 ][NH 4 + ][OH - ] [NH 3 ][NH 4 + ][OH - ]initialchangeequilib

60. 60 © 2009 Brooks/Cole - Cengage Equilibria Involving A Weak Base You have 0.010 M NH 3. Calc. the pH. NH 3 + H 2 O  NH 4 + + OH - NH 3 + H 2 O  NH 4 + + OH - K b = 1.8 x 10 -5 Step 1. Define equilibrium concs. in ICE table [NH 3 ][NH 4 + ][OH - ] [NH 3 ][NH 4 + ][OH - ] initial0.01000 change-x+x+x equilib0.010 - xx x

61. 61 © 2009 Brooks/Cole - Cengage Equilibria Involving A Weak Base You have 0.010 M NH 3. Calc. the pH. NH 3 + H 2 O  NH 4 + + OH - NH 3 + H 2 O  NH 4 + + OH - K b = 1.8 x 10 -5 Step 2. Solve the equilibrium expression Assume x is small (C o >100K b ), so x = [OH - ] = [NH 4 + ] = 4.2 x 10 -4 M x = [OH - ] = [NH 4 + ] = 4.2 x 10 -4 M and [NH 3 ] = 0.010 - 4.2 x 10 -4 ≈ 0.010 M The approximation is valid!

62. 62 © 2009 Brooks/Cole - Cengage Equilibria Involving A Weak Base You have 0.010 M NH 3. Calc. the pH. NH 3 + H 2 O  NH 4 + + OH - NH 3 + H 2 O  NH 4 + + OH - K b = 1.8 x 10 -5 Step 3. Calculate pH [OH - ] = 4.2 x 10 -4 M so pOH = - log [OH - ] = 3.37 Because pH + pOH = 14, pH = 10.63

63. 63 © 2009 Brooks/Cole - Cengage K a and K b are linked: Combined reaction = ? Relationship between K a and K b

64. 64 © 2009 Brooks/Cole - Cengage Relationship between K a and K b K a and K b are related in this way: K a  K b = K w Therefore, if you know one of them, you can calculate the other. 64

65. 65 © 2009 Brooks/Cole - Cengage K a and K b are linked: Combined reaction = ? Relationship between K a and K b

66. 66 © 2009 Brooks/Cole - Cengage K a and K b Comparison Several acids are listed here with their respective Ka’s.  C 6 H 6 OH(aq) + H 2 O(l)  H 3 O + (aq) + C 6 H 5 O - (aq) K a = 1.3 x 10 -10  HCO 2 H(aq) + H 2 O(l)  H 3 O + (aq) + HCO 2 - (aq) K a = 1.8 x 10 -4  HC 2 O 4 - (aq) + H 2 O(l)  H 3 O + (aq) + C 2 O 4 2- (aq) K a = 6.4 x 10 -5 a)Which acid is the strongest? Weakest? b)Which acid has the weakest conjugate base? Strongest? c)What are the pKa values of each acid?

67. 67 © 2009 Brooks/Cole - Cengage Ch. 14.7 - Polyprotic Acids Have more than one acidic proton. If the difference between the K a for the first dissociation and subsequent K a values is 10 3 or more, the pH generally depends only on the first dissociation. 67

68. 68 © 2009 Brooks/Cole - Cengage Polyprotic Acids Acids that can furnish more than one proton. Always dissociates in a stepwise manner, one proton at a time. The conjugate base of the first dissociation equilibrium becomes the acid in the second step. Example: Citric Acid Dissociation: H 3 C 6 H 5 O 7 + H 2 O  H 2 C 6 H 5 O 7 - + H 3 O + K a1 = 7.4x10 -4 H 2 C 6 H 5 O 7 - + H 2 O  HC 6 H 5 O 7 2- + H 3 O + K a2 = 1.7x10 -5 HC 6 H 5 O 7 2- + H 2 O  C 6 H 5 O 7 3- + H 3 O + K a3 = 4.0x10 -7 68

69. 69 © 2009 Brooks/Cole - Cengage Polyprotic Acids For a typical weak polyprotic acid: K a1 > K a2 > K a3 For a typical polyprotic acid in water, only the first dissociation step is important to pH, especially if K a1 >1000K a2 and so on. Example Ascorbic Acid: H 2 C 6 H 6 O 6 + H 2 O  HC 6 H 6 O 6 - + H 3 O + K a1 = 8.0 x10 -5 HC 6 H 6 O 6 - + H 2 O  C 6 H 6 O 6 2- + H 3 O + K a2 = 1.6x10 -12 In other words, the amount of H 3 O + that HC 6 H 6 O 6 - is contributing to the solution is negligible compared to the H 2 C 6 H 6 O 6. (It’s negligible compared to water too!) 69

70. 70 © 2009 Brooks/Cole - Cengage Ch. 14.8 - Acid-Base Properties of Salts Salts are ionic compounds. Salts of the cation of strong bases and the anion of strong acids are neutral. For example: NaCl, KNO 3 There is no equilibrium for strong acids and bases. We ignore the reverse reaction. So…

71. 71 © 2009 Brooks/Cole - Cengage MX + H 2 O  acidic or basic solution? Consider NH 4 Cl NH 4 Cl(aq)  NH 4 + (aq) + Cl - (aq) (a)Reaction of Cl - with H 2 O Cl - + H 2 O  HCl + OH - Cl - + H 2 O  HCl + OH - baseacidacidbase baseacidacidbase Cl - ion is a VERY weak base because its conjugate acid is strong. Therefore, [Cl - ] does not affect pH and above reaction lies completely to the left. Acid-Base Properties of Salts

72. 72 © 2009 Brooks/Cole - Cengage NH 4 Cl(aq)  NH 4 + (aq) + Cl - (aq) (b)Reaction of NH 4 + with H 2 O NH 4 + + H 2 O  NH 3 + H 3 O + NH 4 + + H 2 O  NH 3 + H 3 O + acidbasebaseacid acidbasebaseacid NH 4 + ion is a moderate acid because its conjugate base is weak. Therefore, NH 4 + is an acidic solution See TABLE 14.6 for a summary of acid-base properties of ions. Acid-Base Properties of Salts

73. 73 © 2009 Brooks/Cole - Cengage Acid-Base Properties of Salts

74. 74 © 2009 Brooks/Cole - Cengage What is the pH of a 0.015 M solution of sodium acetate? C 2 H 3 O 2 - (aq) + H 2 O(l)  C 2 H 3 O 2 H(aq) + OH-(aq) I 0.015 0 0 C -x +x +x E 0.015-x x +x Since the acetate ion is the conjugate base of a weak acid, it’s going to act as a base and take H + away from the ‘acid’ water. Acid-Base Properties of Salts Example

75. 75 © 2009 Brooks/Cole - Cengage Acid-Base Properties of Salts Example

76. 76 © 2009 Brooks/Cole - Cengage Acid-Base Properties of Salts Example pH = 14 – 5.54 = 8.46

77. 77 © 2009 Brooks/Cole - Cengage Acid-Base Properties of Salts Practice

78. 78 © 2009 Brooks/Cole - Cengage Step 1.Set up ICE table [CO 3 2- ][HCO 3 - ][OH - ] [CO 3 2- ][HCO 3 - ][OH - ] initial0.1000 initial0.1000 change-x+x+x equilib0.10 - xxx equilib0.10 - xxx Acid-Base Example of a Polyprotic Salt Calculate the pH of a 0.10 M solution of Na 2 CO 3. Na + + H 2 O  neutral CO 3 2- +H 2 O  HCO 3 - +OH - baseacid acid base K b = 2.1 x 10 -4 K b = 2.1 x 10 -4

79. 79 © 2009 Brooks/Cole - Cengage Calculate the pH of a 0.10 M solution of Na 2 CO 3. Na + + H 2 O is neutral CO 3 2- +H 2 O  HCO 3 - +OH - baseacid acidbase K b = 2.1 x 10 -4 K b = 2.1 x 10 -4 Acid-Base Example of a Polyprotic Salt Assume 0.10 - x ≈ 0.10, because 100K b < C o x = [HCO 3 - ] = [OH - ] = 0.0046 M Step 2.Solve the equilibrium expression

80. 80 © 2009 Brooks/Cole - Cengage Calculate the pH of a 0.10 M solution of Na 2 CO 3. Na + + H 2 O is neutral CO 3 2- +H 2 O  HCO 3 - +OH - baseacid acidbase K b = 2.1 x 10 -4 K b = 2.1 x 10 -4 Acid-Base Example of a Polyprotic Salt Step 3.Calculate the pH [OH - ] = 0.0046 M pOH = - log [OH - ] = 2.34 pH + pOH = 14, so pH = 11.66, and the solution is ___basic_.

81. 81 © 2009 Brooks/Cole - Cengage Acid-Base Example of a Polyprotic Salt Notice now that HCO 3 - can act as a base as well. Does the K b for this base affect pH?Notice now that HCO 3 - can act as a base as well. Does the K b for this base affect pH? HCO 3 - + H 2 O  H 2 CO 3 + OH - HCO 3 - + H 2 O  H 2 CO 3 + OH - base acid acid base base acid acid base K b = 2.4 x 10 -8 K b = 2.4 x 10 -8 Step 1.Set up ICE table as before only… [HCO 3 1- ][H 2 CO 3 ][OH - ] [HCO 3 1- ][H 2 CO 3 ][OH - ] initial0.0046 00.0046 initial0.0046 00.0046 change-x+x+x equilib0.0046 - xx0.0046 + x equilib0.0046 - xx0.0046 + x

82. 82 © 2009 Brooks/Cole - Cengage Acid-Base Example of a Polyprotic Salt

83. 83 © 2009 Brooks/Cole - Cengage Ch. 14.9 – Effect of Structure on Acid/Base Properties Why are some compounds acids?Why are some compounds acids? Why are some compounds bases?Why are some compounds bases? Why do acids and bases vary in strength?Why do acids and bases vary in strength? Can we predict variations in acidity or basicity?Can we predict variations in acidity or basicity?

84. 84 © 2009 Brooks/Cole - Cengage Why is CH 3 CO 2 H an Acid? 1. The electronegativity of the O atoms causes the H attached to O to be highly positive. 2. The O—H bond is highly polar. 3. The H atom of O—H is readily attracted to polar H 2 O. –0.32 +0.24 +0.12 See Figure 17.12

85. 85 © 2009 Brooks/Cole - Cengage Trichloroacetic acid is a much stronger acid owing to the high electronegativity of Cl.Trichloroacetic acid is a much stronger acid owing to the high electronegativity of Cl. Cl withdraws electrons from the rest of the molecule.Cl withdraws electrons from the rest of the molecule. This makes the O—H bond highly polar. The H of O—H is very positive.This makes the O—H bond highly polar. The H of O—H is very positive. Acetic acid Trichloroacetic acid K a = 1.8 x 10 -5 K a = 0.3

86. 86 © 2009 Brooks/Cole - Cengage Factors Affecting Acid Strength For a series of oxyacids, acidity increases with the number of oxygens. 86

87. 87 © 2009 Brooks/Cole - Cengage These ions are BASES. They become more and more basic as the negative charge increases. As the charge goes up, they interact more strongly with polar water molecules. NO 3 - CO 3 2- Basicity of Oxoanions PO 4 3-

88. 88 © 2009 Brooks/Cole - Cengage Electron withdrawing explains why water solutions of Fe 3+, Al 3+, Cu 2+, Pb 2+, etc. are acidic. Acid Properties of Metal Ion Solutions This interaction weakens this bond Another H 2 O pulls this H away as H +

89. 89 © 2009 Brooks/Cole - Cengage Ch. 14.10 - Acid/Base Properties of Oxides Acidic Oxides (Acid Anhydrides): –O—X bond is strong and covalent. SO 2, NO 2, CO 2 When H—O—X grouping is dissolved in water, the O—X bond will remain intact. It will be the polar and relatively weak H—O bond that will tend to break, releasing a proton.

90. 90 © 2009 Brooks/Cole - Cengage Acid/Base Properties of Oxides Basic Oxides (Basic Anhydrides): –O—X bond is ionic. K 2 O, CaO If X has a very low electronegativity (i.e. metal), the O—X bond will be ionic and subject to being broken in polar water, producing a basic solution.

91. 91 © 2009 Brooks/Cole - Cengage Lewis acid a substance that accepts an electron pair Ch. 14.11 - Lewis Acid-Base Model Lewis base a substance that donates an electron pair

92. 92 © 2009 Brooks/Cole - Cengage Reaction of a Lewis Acid and Lewis Base New bond formed using electron pair from the Lewis base.New bond formed using electron pair from the Lewis base. Coordinate covalent bondCoordinate covalent bond Notice geometry change on reaction.Notice geometry change on reaction.

93. 93 © 2009 Brooks/Cole - Cengage Formation of hydronium ion is also an excellent example. Lewis Acids & Bases Electron pair of the new O-H bond originates on the Lewis base.Electron pair of the new O-H bond originates on the Lewis base.

94. 94 © 2009 Brooks/Cole - Cengage Lewis Acid/Base Reaction

95. 95 © 2009 Brooks/Cole - Cengage AP Exam Practice 2009B AP Exam #1 2009 AP Exam #1 2007 AP Exam #1 2005 AP Exam #1(a,b,d,e)