1. ADVANCED PLACEMENT CHEMISTRY ACIDS, BASES, AND AQUEOUS EQUILIBRIA

2. Acids- taste sour Bases(alkali)- tastes bitter and feels slippery

3. Lewis acid-base theory is no longer tested on the AP test.

4. hydronium ion (H 3 O + )- formed on reaction of a proton with a water molecule. H + and H 3 O + are used interchangeably in most situations.

5. HA(aq) + H 2 O(l)  H 3 O + (aq) + A - (aq) Acid Base Conjugate Conjugate Acid Base conjugate base- everything that remains of the acid molecule after a proton is lost conjugate acid- base plus a proton

6. Identify the conjugate bases of these substances. HCl HNO 3 NH 3 HC 2 H 3 O 2 CH 3 NH 3 + Cl - NO 3 - NH 2 - (amide ion) C2H3O2-C2H3O2- CH 3 NH 2 (methylamine)

7. Identify the conjugate acids of these substances. F - NH 3 H 2 O HF H3O+H3O+ NH 4 +

8. Acid dissociation constant (K a ) K a = [H 3 O + ][A - ] or K a = [H + ][A - ] [HA] [HA] Why isn’t water included in the K a expression?

9. Strong acid - mostly dissociated - equilibrium lies far to the right - a strong acid yields a weak conjugate base (much weaker than H 2 O) Weak acid- mostly undissociated - equilibrium lies far to the left - has a strong conjugate base (stronger than water, but not a “strong” base)

11. Sketch what five molecules of HF (weak) and five molecules of HCl (strong) would look like in a beaker of water. HF HCl

12. Sketch what five molecules of HF (weak) and five molecules of HCl (strong) would look like in a beaker of water. HF HCl

14. HC 2 H 3 O 2 1.8×10  5 HCN 6.2×10  10 HNO 2 4.0×10  4 Given the K a values for the acids above, rank these conjugate bases from weakest to strongest. H 2 O, Cl , C 2 H 3 O 2 , CN , NO 2  Cl  < H 2 O < NO 2  < C 2 H 3 O 2  < CN 

15. Common strong acids -all aqueous solutions (Know these!) H 2 SO 4 (sulfuric) HCl (hydrochloric) HNO 3 (nitric) HClO 4 (perchloric) HI (hydroiodic) HBr (hydrobromic)

16. Sulfuric acid is a diprotic acid which means that it has two acidic protons. The first (H 2 SO 4 ) is strong and the second (HSO 4  ) is weak.

17. Oxyacids- most acids are oxyacids - acidic proton is attached to O Examples of weak oxyacids: H 3 PO 4 (phosphoric) HNO 2 (nitrous) HOCl (hypochlorous)

18. Within a series, acid strength increases with increasing numbers of oxygen atoms. For example: HClO 4 > HClO 3 > HClO 2 > HClO and H 2 SO 4 > H 2 SO 3 (Electronegative O draws electrons away from O-H bond)

20. Acid strength increases with increasing electronegativity of oxyacids. For example: HOCl > HOBr > HOI > HOCH 3

21. Organic acids- -have carboxyl group -usually weak acids CH 3 COOH acetic C 6 H 5 COOH benzoic

22. Hydrohalic acids- H is attached to a halogen (HCl, HI, HBr, HF) HF is the only weak hydrohalic acid. Although the H-F bond is very polar, the bond is so strong (due to the small F atom) that the acid does not completely dissociate.

23. Weak acid strength is compared by the K a values of the acids. The smaller the K a, the weaker the acid. Strong acids do not usually have K a values because the [HA] is so small and can not be measured accurately.

24. Amphoteric substance- Substance that can act as an acid or as a base. Ex. H 2 O, NH 3, HSO 4  (anything that can both accept and donate a proton)

25. Autoionization of Water H 2 O + H 2 O  H 3 O + + OH - base acid conjugate conjugate acid base

26. Ion product constant for water (K w ) K w = [H 3 O + ][OH  ] K w = [H + ][OH  ] At 25 o C, K w = 1 × 10  14 because [H + ] = [OH  ] = 1 × 10  7 M

27. No matter what an aqueous solution contains, at 25 o C, [H + ][OH  ] = 1 × 10  14 Neutral solution [H + ] = [OH  ] Acidic solution [H + ] > [OH  ] Basic solution [H + ] < [OH  ] K w varies with temperature

28. pH =  log [H + ] If [H + ] = 1.0 × 10  7 M, pH = 7.00

29. You must have a general idea of the pH of a substance. What are reasonable pH values for strong acids, weak acids, weak bases, and strong bases??

30. Significant figures in pH and other log values: The number of decimal places in the log value should equal the number of significant digits in the original number (concentration). pH = 4.42 How many significant figures?? 2

31. pOH =  log [OH  ] pK =  log K pH and pOH are logarithmic functions. The pH changes by 1 for every power of 10 change in [H + ]. pH decreases as [H + ] increases. A substance with a pH of 3.00 has ten times the [H + ] as a substance with a pH of 4.00.

32. pH + pOH = 14 [H + ] = 10 (  pH) [OH  ] = 10 (  pOH)

33. Calculating pH of Strong Acid Solutions Calculating pH of strong acid solutions is generally very simple. The pH is simply calculated by taking the negative logarithm of concentration of a monoprotic strong acid. For example, the pH of 0.1 M HCl is 1.0. However, if the acid concentration is less than 1.0 × 10  7, the water becomes the important source of [H + ] and the pH is 7.00. The pH of an acidic solution can not be greater than 7 at 25 o C!!!!!

34. Another exception is calculating the pH of an H 2 SO 4 solution that is more dilute than 1.0 M. At this concentration, the [H + ] of the HSO 4  must also be calculated. You won’t be asked to do this on the AP test because it involves a quadratic equation. However, you should be able to predict that a 0.01 M solution of H 2 SO 4 has a pH of less than 2, while the pH of 0.01 M HCl = 2.0.

35. Ex. Calculate the [H + ] and pH in a 1.0 M solution of HCl. HCl is a strong monoprotic acid, therefore its concentration is equal to the hydrogen ion concentration. [H + ] = 1.0 M pH =  log (1.0) = 0.00

36. Ex. Calculate the pH of 1.0 × 10  10 M HCl. Since the [H + ] is less than 1.0 x 10  7, the [H + ] from the acid is negligible and the pH = 7.00 If you calculated the answer, using the 1.0 × 10  10 as the [H + ], you would get a pH of 10.00, which would mean the solution is basic. Impossible!

37. Calculating pH of Weak Acid Solutions Calculating pH of weak acids involves setting up an equilibrium. Always start by writing the equation, setting up the acid equilibrium expression (K a ), defining initial concentrations, changes, and final concentrations in terms of x, substituting values and variables into the K a expression and solving for x.

38. Ex. Calculate the pH of a 0.100 M solution of acetic acid. The K a of acetic acid is 1.8 × 10  5. HC 2 H 3 O 2  H + + C 2 H 3 O 2  You should know the formula for acetic acid. It may be written as either HC 2 H 3 O 2 or as CH 3 COOH. Acetic acid is also called ethanoic acid.

39. Reaction HC 2 H 3 O 2  H + + C 2 H 3 O 2  Initial 0.100 0 0 Change -x +x +x Equilibrium 0.100- x x x Often, the -x in a K a expression can be treated as negligible. x = 1.34 × 10  3 Remember the 5% rule? Was the x negligible?

40. The question asked for the pH. pH =  log [H + ] [H + ] = x = 1.34×10  3 pH =  log 1.34×10  3 pH = 2.87 The answer has two significant figures because the K a had only two significant figures. The answer is a reasonable pH for a solution of a weak acid.

41. To be valid, x must be less than 5% of the number that it was to be subtracted from. In this example 1.34×10  3 is less than 5% of 0.100. This means that the assumption that x was negligible was valid and the quadratic equation or the method of successive approximation does not need to be used.

42. Quadratic Formula Problem (not tested on AP Test!) Ex. Calculate the pH of a 1.00×10 -4 M solution of acetic acid. The K a of acetic acid is 1.8×10  5. Reaction HC 2 H 3 O 2  H + + C 2 H 3 O 2  Initial 1.00×10 -4 M 0 0 Change -x +x +x Equilibrium 1.00×10 -4 - x x x x=4.2×10 -5

43. Use of the quadratic equation: x 2 + 1.8×10  5 x  1.8×10  9 = 0 x = 3.5×10  5 and  5.2×10  5 Since a concentration can not be negative, x = 3.5×10  5 M x = [H + ] = 3.5×10  5 pH =  log 3.5×10  5 = 4.46

44. Calculating pH of polyprotic acids All polyprotic acids dissociate stepwise. Each dissociation has its own K a value. As each H is removed, the remaining acid gets weaker and therefore has a smaller K a. As the negative charge on the acid increases it becomes more difficult to remove the positively charged proton.

45. Except for H 2 SO 4, polyprotic acids have K a2 and K a3 values so much weaker than their K a1 value that the 2nd and 3rd (if applicable) dissociation can be ignored. The [H + ] obtained from this 2nd and 3rd dissociation is negligible compared to the [H + ] from the 1st dissociation. Because H 2 SO 4 is a strong acid in its first dissociation and a weak acid in its second, we need to consider both if the concentration is more dilute than 1.0 M. The quadratic equation is needed to work this type of problem and it will not be asked on the AP test.

46. Ex. Calculate the pH of a 1.00 × 10  2 M H 2 SO 4 solution. The K a of HSO 4  is 1.2 × 10  2 H 2 SO 4  H + + HSO 4  Before 1.00×10  2 0 0 Change  1.00×10  2 +1.00×10  2 +1.00×10  2 After 0 1.00×10  2 1.00×10  2 Reaction HSO 4   H + + SO 4  Initial 1×10  2 1×10  2 0 Change -x +x +x Equil. 1×10  2 -x 1×10  2 +x x

47. K a = [H + ][SO 4  ]= 1.2×10  2 [HSO 4  ] 1.2×10  2 = (1×10  2 + x)(x) (1×10  2  x) Using the quadratic equation, x = 4.52×10  3 [H + ]= 1×10  2 + (4.52×10  3 ) = 1.45×10  2 pH = 1.84

48. Determination of the pH of a Mixture of Weak Acids Only the acid with the largest K a value will contribute an appreciable [H + ]. Determine the pH based on this acid and ignore any others.

49. Determination of the Percent Dissociation (or Ionization) of a Weak Acid % dissociation = amt. dissociated (mol/L) ×100 initial concentration (mol/L) = final [H + ] ×100 initial [HA]

50. For a weak acid, percent dissociation (or ionization) increases as the acid becomes more dilute. Equilibrium shifts to the right.

52. BASES The hydroxides of Group I and IIA metals are all strong bases. The Group IIA hydroxides are not very soluble. This property allows some of them to be used effectively as stomach antacids.

53. Ex. Calculate the [OH  ], [H + ], and pH of a 0.0100 M solution of NaOH. NaOH is a strong base. [OH  ] = 0.0100 M [H + ] = 1×10  14 /1×10  2 = 1.00×10  12 M pH =  log 1.00×10  12 = 12.000 Watch out for problems with Ca(OH) 2 or Ba(OH) 2. The concentration must be doubled to get the [OH  ].

54. Weak bases (bases without OH  ) react with water to produce a hydroxide ion. Common examples of weak bases are ammonia (NH 3 ), methylamine (CH 3 NH 2 ), and ethylamine (C 2 H 5 NH 2 ).

55. B(aq) + H 2 O(l)  BH + (aq) + OH  (aq) base acid conjugate conjugate acid base NH 3 + H 2 O  NH 4 + + OH  base acid conjugate conjugate acid base The lone pair on N forms a bond with a H +. Most weak bases involve N.

56. Base dissociation constant (K b ) K b = [BH + ][OH  ] K b = [NH 4 + ][OH  ] [B] [NH 3 ]

57. Determination of the pH of a weak base is very similar to the determination of the pH of a weak acid. Follow the same steps. Remember, however, that x is the [OH  ] and taking the negative log of x will give you the pOH and not the pH!

58. Ex. Calculate the [OH  ] and the pH for a 15.0 M NH 3 solution. The K b for NH 3 is 1.8×10  5. Reaction NH 3 + H 2 O  NH 4 + + OH  Initial 15.0 --- 0 0 Change  x --- +x +x Equil 15.0  x --- x x K b = 1.8×10 -5 = [NH 4 + ][OH  ] = x 2  x 2 [NH 3 ] 15.0  x 15.0 x = 1.6×10 -2 = [OH - ] pOH =  log 1.6×10  2 = 1.78 pH = 14  1.78 = 12.22

59. Determination of the pH of Salts

60. Neutral Salts- Salts that are formed from the cation of a strong base and the anion from a strong acid form neutral solutions when dissolved in water. Ex. NaCl, KNO 3 Think of this as a “tug-of-war”. Since both the acid and base were strong, no one wins and the solution is neutral. The real reason that the salt is neutral is that neither ion reacts with water. They are simply spectator ions.

61. Acid Salts- Salts that are formed from the cation of a weak base and the anion from a strong acid form acidic solutions when dissolved in water. The acid was strong (tug-of-war) so the salt is acidic. This is not the real reason, only a quick trick. Ex. NH 4 Cl Real reason: The cation (weak part) hydrolyzes the water molecule to produce hydronium ions and thus an acidic solution The Cl  was only a spectator. NH 4 + + H 2 O  H 3 O + + NH 3 strong acid weak base

62. Basic Salts- Salts that are formed from the cation of a strong base and the anion from a weak acid form basic solutions when dissolved in water. The base was strong (tug-of-war) so the salt is basic. This is not the real reason, only a quick trick. Ex. NaC 2 H 3 O 2 Real reason: The anion hydrolyzes the water molecule to produce hydroxide ions and thus a basic solution. The Na + was a spectator ion. C 2 H 3 O 2  + H 2 O  OH  + HC 2 H 3 O 2 strong base weak acid

63. When determining the exact pH of salt solutions, we can use the K a of the weak acid formed to find the K b of the salt or we can use the K b of the weak base formed to find the K a of the salt. K a × K b = K w

64. General Order of pH Strong Bases (SB) Weak Bases (WB) Basic Salts (BS) Neutral Salts (NS) Acidic Salts (AS) Weak Acids (WA) Strong Acids (SA) When asked to rank a mixture of compounds by pH, label each as shown above and then arrange them. High pH Low pH

65. Arrange the following 0.1 M solutions in order of decreasing pH. NaOH NaCN KCl CH 3 NH 2 H 3 PO 4 NH 4 NO 3 HBr NaOH>CH 3 NH 2 >NaCN>KCl>NH 4 NO 3 >H 3 PO 4 > HBr

66. Ex. Calculate the pH of a 0.15 M solution of sodium acetate. First, determine what NaC 2 H 3 O 2 is. Sodium acetate is the salt of a strong base (NaOH) and a weak acid (acetic acid) and thus is a basic salt and forms a basic solution. The acetate ion hydrolyzes to produce acetic acid and hydroxide ions. Reaction C 2 H 3 O 2  + H 2 O  HC 2 H 3 O 2 + OH  Initial 0.15M ---- 0 0 Change  x +x +x Equil. 0.15  x x x

67. [OH  ] = 9.2×10  6 pOH = - log 9.2×10  6 = 5.04 pH = 14.00  5.04 = 8.96 The pH is slightly basic, which is reasonable for a basic salt.

68. Acidic and Basic Oxides When metallic (ionic) oxides dissolve in water they produce a metallic hydroxide (basic solution). When nonmetallic (covalent) oxides dissolve in water they produce a weak acid (acidic solution). CaO + H 2 O  Ca(OH) 2 CO 2 + H 2 O  H 2 CO 3

69. Salts of Highly Charged Metals Salts that contain a highly charged metal ion produce an acidic solution. AlCl 3 + 6H 2 O  Al(H 2 O) 6 3+ + 3Cl  Al(H 2 O) 6 3+  Al(H 2 O) 5 (OH) 2+ + H +

71. The higher the charge on the metal ion the stronger the acidity of the hydrated ion. The electrons are pulled away from the O-H bond and toward the positively charged metal ion. FeCl 3 and Al(NO 3 ) 3 also behave this way.