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

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

3. Arrhenius concept- acids produce hydrogen ions in aqueous solution while bases produce hydroxide ions

4. Bronsted-Lowry model- acids are proton (H + ) donors and bases are proton acceptors

5. Lewis model- acids are electron pair acceptors while bases are electron pair donors

6. 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.

7. 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

8. Acid dissociation constant (K a ) K a = [H 3 O + ][A - ] or K a = [H + ][A - ] [HA] [HA]

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)

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

13. 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.

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

15. 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)

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

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

19. Hydrohalic acids- H is attached to a halogen (HCl, HI, etc.) 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.

20. 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 have K a values because the [HA] is so small and can not be measured accurately.

21. 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)

22. autoionization of water H 2 O + H 2 O  H 3 O + + OH - base acid conjugate conjugate acid base

24. Ion product constant for water (K w ) K w = [H 3 O + ][OH - ] K w = [H + ][OH - ] At 25 o C, K w = 1 x 10 -14 mol 2 /L 2 because [H + ] = [OH - ] = 1 x 10 -7 M

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

26. pH = -log [H + ] If [H + ] = 1.0 x 10 -7 M, pH = 7.00

28. 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).

29. 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.

30. pH + pOH = 14 [H + ] = antilog(-pH) [OH - ] = antilog(-pOH)

31. 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 x 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!!!!!

32. Another exception is calculating the pH of a 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.

33. 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

34. Ex. Calculate the pH of 1.0 x 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

35. 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.

36. Ex. Calculate the pH of a 1.00 x 10 -4 M solution of acetic acid. The K a of acetic acid is 1.8 x 10 -5 HC 2 H 3 O 2  H + + C 2 H 3 O 2 - K a = [H + ][C 2 H 3 O 2 - ] = 1.8 x 10 -5 [HC 2 H 3 O 2 ]

37. Reaction HC 2 H 3 O 2  H + + C 2 H 3 O 2 - Initial 1.00 x 10 -4 0 0 Change -x +x +x Equilibrium 1.00 x 10 -4 - x x x Often, the -x in a K a expression 1.8 x 10 -5 = (x)(x) can be treated as negligible. 1.00x10 -4 - x 1.8 x 10 -5  (x)(x) x = 4.2 x 10 -5 1.00 x 10 -4

38. When you assume that x is negligible, you must check the validity of this assumption. To be valid, x must be less than 5% of the number that it was to be subtracted from. In this example 4.2 x 10 -5 is greater than 5% of 1.00 x 10 -4. This means that the assumption that x was negligible was invalid and x must be solved for using the quadratic equation or the method of successive approximation.

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

40. Another method which some people prefer is the method of successive approximations. In this method, you start out assuming that x is negligible, solve for x, and repeatedly plug your value of x into the equation again until you get the same value of x two successive times.

41. Using successive approximation for the previous example would go as follows: x = 4.2 x 10 -5 x = 3.2 x 10 -5 x = 3.5 x 10 -5 x = 3.4 x 10 -5 x = 3.4 x 10 -5 [H + ] = 3.4 x 10 -5 pH = 4.47

42. or/ with a graphing calculator:  (1.8 x 10 -5 x 1.00 x 10 -4 ) = 4.2 x 10 -5 (not negl)  ((1.8 x 10 -5 )( 1.00 x 10 -4 -ans)) = 3.2 x 10 -5 =3.5 x 10 -5 =3.4 x 10 -5 =3.4 x 10 -5 Use answer key on calculator for this! Press Enter key repeatedly until you get the same answer each time

43. 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.

44. 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.

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

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

47. 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.

48. Determination of the Percent Dissociation of a Weak Acid % dissociation = amt. dissociated (mol/L) x100 initial concentration (mol/L) = final [H + ] x 100 initial [HA]

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

51. 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.

52. 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 x 10 -14 /1 x 10 -2 = 1.00 x 10 -12 M pH = - log 1.00 x 10 -12 = 12.000

53. 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 ).

54. 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.

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

56. 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!

57. Ex. Calculate the [OH - ] and the pH for a 15.0 M NH 3 solution. The K b for NH 3 is 1.8 x 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 x 10 -5 = [NH 4 + ][OH - ] = x 2  x 2 [NH 3 ] 15.0-x 15.0 x = 1.6 x 10 -2 = [OH - ] pOH = -log 1.6 x 10 -2 = 1.78 pH = 14-1.78 = 12.22

58. Determination of the pH of Salts

59. 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

60. 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. Ex. NH 4 Cl The cation hydrolyzes the water molecule to produce hydronium ions and thus an acidic solution. NH 4 + + H 2 O  H 3 O + + NH 3 strong acid weak base

61. 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. Ex. NaC 2 H 3 O 2, KNO 2 The anion hydrolyzes the water molecule to produce hydroxide ions and thus a basic solution. C 2 H 3 O 2 - + H 2 O  OH - + HC 2 H 3 O 2 strong base weak acid

62. 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 x K b = K w

63. Ex. Calculate the pH of a 0.15 M solution of sodium acetate. Sodium acetate is the salt of a strong base (NaOH) and a weak acid (acetic acid) and thus 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

64. K b = [HC 2 H 3 O 2 ][OH - ] [C 2 H 3 O 2 - ] K b = K w = 1 x 10 -14 = 5.6 x 10 -10 K a 1.8 x 10 -5 5.6 x 10 -10 = x 2  x 2 x = 9.2 x 10 -6 0.15 - x 0.15 [OH - ] = 9.2 x 10 -6 pOH = - log 9.2 x 10 -6 = 5.04 pH = 14.00-5.04 = 8.96

65. 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

66. 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 +

67. 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.