1. Acids and Bases

2. Acids & Bases The Bronsted-Lowry model defines an acid as a proton donor. A base is a proton acceptor. Note that this definition is based on the transfer of a proton from the acid to the base.

3. Acids & Bases H 2 O(l) + HCl(g)  H 3 O + (aq) + Cl 1- (aq) H 2 O(l) + HCl(g)  H 3 O + (aq) + Cl 1- (aq) In this reaction, water accepts a proton from HCl. Water is a base, and HCl is an acid. H 2 O(l) + HCl(g)  H 3 O + (aq) + Cl 1- (aq) proton proton acceptor donor B-L base B-L acid

4. Acids & Bases Since this reaction goes to completion (note the one-way arrow), we classify HCl(aq) as a strong acid. H 2 O(l) + HCl(g)  H 3 O + (aq) + Cl 1- (aq) proton proton acceptor donor

5. Acids & Bases Hydrochloric acid dissociates 100%, and exists as a solution of hydronium and chloride ions. H 2 O(l) + HCl(g)  H 3 O + (aq) + Cl 1- (aq) proton proton acceptor donor Although a bottle may be labeled 1.0M HCl, it really contains 1.0 M H 3 O + (aq) and 1.0 M Cl 1- (aq).

6. Acids & Bases Hydrochloric acid dissociates 100%, and exists as a solution of hydronium and chloride ions. H 2 O(l) + HCl(g)  H 3 O + (aq) + Cl 1- (aq) proton proton acceptor donor There is no reverse reaction because chloride has no tendency to accept a proton to form HCl.

7. Acids & Bases There are only a few common strong acids. They are: HCl(aq), HBr(aq), HI(aq), HNO 3 (aq), HClO 4 (aq) and H 2 SO 4 (aq)* *for the first proton only

8. Acids & Bases Most acids are weak acids in which only a small percentage of the molecules dissociate to protonate water. HA(aq) + H 2 O(aq) ↔ H 3 O + (aq) + A - (aq) HA is a generic monoprotic weak acid such as HF, HCN or CH 3 COOH.

9. Weak Acids Weak acids in water form an equilibrium with hydronium ion and the deprotonated anion of the acid. The reaction does not go to completion. H-A + H 2 O ↔ H 3 O + + A - H-A + H 2 O ↔ H 3 O + + A -

10. Acids The equilibrium of acids in water can be viewed as a competition between the forward reaction and the reverse reaction. H-A + H 2 O ↔ H 3 O + + A -

11. Acids The forward reaction involves HA, a generic acid, and water. Water acts as a base by accepting a proton from the acid. H-A + H 2 O ↔ H 3 O + + A -

12. Acids The forward reaction involves HA, a generic acid, and water. Water acts as a base by accepting a proton from the acid. Acid HA proton donor Base H 2 O proton acceptor H-A + H 2 O ↔ H 3 O + + A-

13. Acids The reverse reaction involves A - accepting a proton and acting as a base. H 3 O + donates a proton, and is an acid. Acid HA proton donor Base H 2 O proton acceptor Acid H 3 O + proton donor Base A - proton acceptor H-A + H 2 O ↔ H 3 O + + A - H-A + H 2 O ↔ H 3 O + + A -

14. K a Values Acid strength is determined by measuring the equilibrium constant for the following reaction: HA(aq) + H 2 O(l) ↔ H 3 O + (aq) + A - (aq) K a = [H 3 O + ][A - ] [HA] [HA]

15. K a Values HA(aq) + H 2 O(l) ↔ H 3 O + (aq) + A - (aq) K a = [H 3 O + ][A - ] [HA] [HA] Water is left out of the equilibrium constant expression because it is a pure liquid (with constant concentration). The “a” subscript stands for acid.

16. Conjugate Acids & Bases The deprotonated acid is a base, and the protonated base is an acid. These acids and bases are called conjugate acids and bases. Acid HA proton donor Base H 2 O proton acceptor Acid H 3 O + proton donor Base A - proton acceptor H-A + H 2 O ↔ H 3 O + + A - H-A + H 2 O ↔ H 3 O + + A -

17. Acids For weak acids, the equilibrium lies to the left, indicating that A - is a stronger base than water. Acid HA proton donor Base H 2 O proton acceptor Acid H 3 O + proton donor Base A - proton acceptor

18. Acids Acid HA Base H 2 O Acid H 3 O + Base A - Acid HA and base A - are related, and differ only by the addition or removal of H +. remove H + add H +

19. Acids Acid HA Base H 2 O Acid H 3 O + Base A - HA and A - are called conjugate acid-base pairs. A - is the conjugate base of the acid HA. remove H + add H +

20. Acids Acid HA Base H 2 O Acid H 3 O + Base A - Likewise, H 2 O and H 3 O + are related, and differ only by the addition or removal of H +. remove H + add H +

21. Acids Acid HA Base H 2 O Acid H 3 O + Base A - H 2 O and H 3 O + are conjugate acid-base pairs. H 2 O is the conjugate base of H 3 O +. remove H + add H +

22. Conjugate Acid & Bases Provide the formulas for the conjugate bases of: Provide the formulas for the conjugate bases of: H 2 SO 4 HCN H 2 O CH 3 COOH

23. Conjugate Acid & Bases Provide the formulas for the conjugate acids of: Provide the formulas for the conjugate acids of: NH 3 OH 1- HCO 3 1- H 2 O

24. Structure and Acid Strength For binary acids of the general formula H n X (where X is a non-metal), the acidity reflects the H-X bond strength and polarity. If the bond is strong and non-polar, such as with carbon (in CH 4 ), the compound is non- acidic, and the C-H bonds remain intact.

25. Structure and Acids & Bases The effects of bond polarity can be seen in the hydrogen halides: HF, HCl, HBr and HI. HF has the most polar bond, yet it is the weakest acid of the group. The HF bond is the strongest of the group, and this very high bond strength results in a lower tendency for dissociation in water.

26. Binary acid strength depends upon the electronegativity of the non-metal, and also the bond strength of the bond with hydrogen.

27. The bond strength is greatest when there is a good match in energy between the 1s orbital on hydrogen, and the bonding orbital on the non-metal.

28. Structure and Acids & Bases The acids HCl, HBr and HI are all strong acids. In all cases, the bonds are polar, but they are also weaker than the H-F bond, and are more easily broken in aqueous solution.

29. Structure and Acids & Bases

30. The Oxyacids The oxyacids typically have one or more oxygen atoms attached to a central non-metal. Attached to one or more of the oxygen atoms are acidic hydrogens. Examples include H 2 SO 4, HNO 3, HNO 2, H 3 PO 4 and HClO 4. It is important to remember that in all cases, the acidic hydrogen is attached to oxygen (X-O-H).

31. The Oxyacids

32. Oxyacids For a given central atom, the greater the number of oxygen atoms attached, the more acidic the acid. The larger number of oxygen atoms polarizes and weakens the O-H bonds.

33. Oxyacids The greater the number of oxygen atoms, the weaker the O-H bonds, and the stronger the acid.

34. Oxyacids The nature of the central atom also affects the acidity of the acid. The more electronegative X is (in X-O-H), the weaker the O-H bond, and the stronger the acid.

35. Oxyacids

36. Strong and Weak Acids HX is a weak acid and forms only a small amount of H 3 O + and X -

37. Strong and Weak Acids HY is a strong acid and dissociates completely to form H 3 O + and Y -

38. Strong and Weak Acids

39. Acid Strength The conjugate bases of strong acids have no tendency to pick up a proton. There is no reverse reaction. The conjugate bases of infinitely strong acids will not act as a base and accept a proton.

40. K a Values

41. Bases Strong bases are very effective at accepting protons. The most common strong bases are the soluble group IA and IIA metal hydroxides. In general, the metal ion is non-reactive, and serves as a spectator ion. Another strong base is oxide ion, O 2-. Oxide reacts with water to become fully protonated. O 2- (aq) + H 2 O(l)  2 OH - (aq)

42. Bases In this reaction, water is donating a proton, and hence acting as an acid. In previous reactions, water accepted a proton and served as a base. Substances that can behave as either an acid or a base are called amphoteric.

43. Amphoteric Nature of Water Depending upon its environment, water may donate a proton, acting as an acid, or accept a proton, and act as a base. This behavior is characteristic of amphoteric substances. Pure water molecules can react with each other, to a very small extent, to form hydronium and hydroxide ions.

44. Autoionization of Water This process is called autoionization or selfionization. One water molecule donates a proton to another. The result is the formation of equal amounts of hydronium and hydroxide ions. H 2 O(l) + H 2 O(l) ↔ H 3 O + (aq) + OH - (aq)

45. Autoionization of Water H 2 O(l) + H 2 O(l) ↔ H 3 O + (aq) + OH - (aq) K w = [H 3 O + ][OH - ] = 1.0 x 10 -14 at 25 o C In pure water, the concentration of hydronium ion equals the concentration of hydroxide. Both ions have a concentration of 1.0 x 10 -7 M.

46. [H 3 O + ] and [OH - ] in Aqueous Solution The product of the hydroxide and hydronium concentration in any aqueous solution must equal K w. As a result, when a solution is acidic, the hydronium concentration increases, and the hydroxide concentration decreases. K w = [H 3 O + ][OH - ] = 1.0 x 10 -14 at 25 o C

47. [H 3 O + ] and [OH - ] in Aqueous Solution Likewise, in basic solutions, the hydroxide ion concentration is greater than the hydronium ion concentration. Likewise, in basic solutions, the hydroxide ion concentration is greater than the hydronium ion concentration. There is always some hydroxide ion and some hydronium ion present in any aqueous solution.

48. [H 3 O + ] and [OH - ] in Aqueous Solution

49. The pH Scale A scale of acidity, the pH scale, is used to indicate the degree of acidity of aqueous solutions. pH = -log[H 3 O + ] or –log[H + ] The scale generally runs from 0-14, though negative pH values are possible. A neutral solution will have a pH = 7.00

50. The pH Scale A one unit change in pH is a ten-fold change in the concentration of hydronium ion. Acidic solutions have pH values less than 7.00, and basic solutions have pH values greater than 7.00

51. pH Values Most foods have pH values in the acidic range.

52. Determination of pH Universal Indicator

53. Determination of pH A portable pH meter

54. Problem: Calculation of pH Calculate the pH of 0.10M HCl. Calculate the pH of 0.10M HCl. - Is it an acid or a base? - Is it strong or weak? - Write the appropriate chemical reaction(s). HCl(aq) + H 2 O(l)  H 3 O + (aq) + Cl - (aq) or HCl(aq)  H + (aq) + Cl - (aq)

55. Problem: Calculation of pH Calculate the pH of 0.10M HCl. Calculate the pH of 0.10M HCl. HCl(aq) + H 2 O(l)  H 3 O + (aq) + Cl - (aq) [H 3 O + ] = 0.10M pH = - log [H 3 O + ] = - log( 0.10) = 1.00 For every significant digit in the concentration, there is a place after the decimal in the pH.

56. Problem: Calculation of pH Calculate the pH of 0.10M HCl. Calculate the pH of 0.10M HCl. pH = - log [H 3 O + ] = - log( 0.10) = 1.00 - Does your answer make sense? The pH of a fairly concentrated strong acid should be much less than 7. Yes, the answer makes sense.

57. Problem: pH of 0.50M HCN Calculate the pH of 0.50M HCN. Calculate the pH of 0.50M HCN. - Acid or base? - Weak or strong? HCN is a weak acid, with K a = 6.2 x 10 -10. HCN(aq) + H 2 O(l) ↔ H 3 O + (aq) + CN - (aq)

58. Problem: pH of 0.50M HCN Calculate the pH of 0.50M HCN. Calculate the pH of 0.50M HCN. HCN is a weak acid, with K a = 6.2 x 10 -10. HCN(aq) + H 2 O(l) ↔ H 3 O + (aq) + CN - (aq) - Write the equilibrium constant expression. K a = 6.2 x 10 -10 = [H 3 O + ][CN - ] [HCN] [HCN]

59. Problem: pH of 0.50M HCN - Make a table of initial, change and equilibrium concentrations. HCN(aq) + H 2 O(aq) ↔ H 3 O + (aq) + CN - (aq) HCN H3O+H3O+H3O+H3O+ CN - initial 0.5000 change -x+x+x equil. 0.50-xxx

60. Problem: pH of 0.50M HCN - Substitute and solve. HCN H3O+H3O+H3O+H3O+ CN - initial 0.5000 change -x+x+x equil. 0.50-xxx K a = 6.2 x 10 -10 = [H 3 O + ][CN - ] [HCN] [HCN] K a = 6.2 x 10 -10 = [x] [x] [.50-x] [.50-x]

61. Problem: pH of 0.50M HCN - State any assumptions. - Solve the problem. - Check your assumptions. - Answer the question. - Does your answer make sense? - Check your results, if possible.

62. pH of Weak Acids If the same question were asked concerning a 0.50 M solution of chlorous acid, HClO 2 (K a =1.2 x 10 -2 ), you would need to use the quadratic formula to answer the question. Generally, if the acid concentration is.10M or larger, and K is 10 -5 or smaller, you won’t need to use the quadratic formula.

63. Problem A 0.0875M solution of an unknown monoprotic acid has a pH of 2.28. Calculate the K a of the acid. A 0.0875M solution of an unknown monoprotic acid has a pH of 2.28. Calculate the K a of the acid. - Write the chemical reaction. HA(aq) + H 2 O(l) ↔ H 3 O + (aq) + A - (aq) - Write the equilibrium constant expression. K a = [H 3 O + ][A - ] [HA] [HA]

64. Problem A 0.0875M solution of an unknown monoprotic acid has a pH of 2.28. Calculate the K a of the acid. A 0.0875M solution of an unknown monoprotic acid has a pH of 2.28. Calculate the K a of the acid. - Make a table of concentrations.

65. Problem A 0.0875M solution of an unknown monoprotic acid has a pH of 2.28. Calculate the K a of the acid. A 0.0875M solution of an unknown monoprotic acid has a pH of 2.28. Calculate the K a of the acid. [HA] [H 3 O + ] [A - ] init.087500 chg equil

66. Problem A 0.0875M solution of an unknown monoprotic acid has a pH of 2.28. Calculate the K a of the acid. A 0.0875M solution of an unknown monoprotic acid has a pH of 2.28. Calculate the K a of the acid. [HA] [H 3 O + ] [A - ] init.087500 chg equil The pH can be used to calculate how much H 3 O + and A - form.

67. Problem A 0.0875M solution of an unknown monoprotic acid has a pH of 2.28. Calculate the K a of the acid. A 0.0875M solution of an unknown monoprotic acid has a pH of 2.28. Calculate the K a of the acid. To calculate [H 3 O + ] from pH, enter the pH, change its sign, and take the inverse log. You will get a [H 3 O + ] = 5.2 x10 -3. This is the equilibrium concentration of H 3 O + (and A - ).

68. Problem A 0.0875M solution of an unknown monoprotic acid has a pH of 2.28. Calculate the K a of the acid. A 0.0875M solution of an unknown monoprotic acid has a pH of 2.28. Calculate the K a of the acid. [HA] [H 3 O + ] [A - ] init.087500 chg equil 5.2 x10 -3

69. Problem A 0.0875M solution of an unknown monoprotic acid has a pH of 2.28. Calculate the K a of the acid. A 0.0875M solution of an unknown monoprotic acid has a pH of 2.28. Calculate the K a of the acid. [HA] [H 3 O + ] [A - ] init.087500 chg -.0052 + 5.2 x 10 -3 equil 5.2 x10 -3

70. Problem A 0.0875M solution of an unknown monoprotic acid has a pH of 2.28. Calculate the K a of the acid. A 0.0875M solution of an unknown monoprotic acid has a pH of 2.28. Calculate the K a of the acid. [HA] [H 3 O + ] [A - ] init.087500 chg -.0052 + 5.2 x 10 -3 equil.0823 5.2 x10 -3

71. Problem A 0.0875M solution of an unknown monoprotic acid has a pH of 2.28. Calculate the K a of the acid. A 0.0875M solution of an unknown monoprotic acid has a pH of 2.28. Calculate the K a of the acid. [HA] [H 3 O + ] [A - ] equil.0823 5.2 x10 -3 Substitute into the equilibrium constant expression and solve for K a.

72. Problem A 0.0875M solution of an unknown monoprotic acid has a pH of 2.28. Calculate the K a of the acid. A 0.0875M solution of an unknown monoprotic acid has a pH of 2.28. Calculate the K a of the acid. [HA] [H 3 O + ] [A - ] equil.0823 5.2 x10 -3 K a = [H 3 O + ] [A - ]/ [HA] = (5.2 x10 -3 ) (5.2 x10 -3 ) (0.0823) K a = 3.3 x 10 -4

73. Percent Ionization Sometimes the ionization of a weak acid solution is expressed in % ionization. For a given solution, % ionization = (100%)[H 3 O + ] eq /[HA] init

74. Problem: pH Calculate the pH of 0.0050M Ca(OH) 2. Calculate the pH of 0.0050M Ca(OH) 2. - Acid or base? - Strong or weak? - Write the reaction(s). Ca(OH) 2 in water is a strong base. Ca(OH) 2 (aq)  Ca 2+ (aq) + 2 OH - (aq)

75. Problem: pH Calculate the pH of 0.0050M Ca(OH) 2. Calculate the pH of 0.0050M Ca(OH) 2. Ca(OH) 2 (aq)  Ca 2+ (aq) + 2 OH - (aq) 0.0050M0.0050M 2(0.0050M) 0.0050M0.0050M 2(0.0050M) [OH - ] = 2(0.0050M) = 0.0100M Since [OH - ] and [H 3 O + ] are related by K w, the pH can be calculated.

76. Problem: pH from [OH - ] [OH - ] = 2(0.0050M) = 0.0100M You can use: [H 3 O + ] [OH - ] = 1.0 x 10 -14 to calculate hydronium concentration and the pH, or: pH + pOH = -log K w =14.00 ( @ 25 o C)

77. Problem: What is the pH of 0.15M NH 3 ? NH 3 is a weak base, and NH 4 + is its conjugate acid. K b = [OH - ][NH 4 + ] [NH 3 ]

78. Values of Kb for Some Common Weak Bases

79. What is the pH of 0.15M NH 3 ? K b = [OH - ][NH 4 + ] = 1.8 x 10 -5 [NH 3 ] [NH 3 ] NH 3 (aq) + H 2 O(l) ↔ OH - (aq) + NH 4 + (aq) NH 3 OH - NH 4 + initial0.1500 change-x+x+x equil..15-xxx

80. pH of Salt Solutions When salts (ionic compounds) dissolve in water, they dissociate into separate hydrated ions. Since many ions are conjugate acids or bases, they may react to form hydronium or hydroxide ions.

81. Acid Strength The conjugate bases of strong acids have no tendency to pick up a proton. There is no reverse reaction. The conjugate bases of infinitely strong acids are infinitely weak.

82. Conjugate Acid- Base Strength The weaker the acid, the stronger its conjugate base. As the acid gets weaker, the reverse reaction with water becomes more significant.

83. Conjugate Acid-Base Strength

84. pH of Salt Solutions Since the conjugate bases of strong acids have no tendency to accept protons, these ions will have no effect on pH. The “neutral” anions are: Cl -, Br -, I -, NO 3 - and ClO 4 - (Although HSO 4 1- isn’t basic, it is acidic, and will donate a proton to form hydronium ion.)

85. pH of Salt Solutions The cations of strong bases also have no effect on pH when dissolved in water. As a result, the group IA and group IIA metal ions will produce neutral solutions. Some small, highly positively charged metal ions, such as Al 3+, Cr 3+ or Fe 3+ produce slightly acidic solutions.

86. pH of Salt Solutions Are the following aqueous solutions acidic, basic or neutral? Are the following aqueous solutions acidic, basic or neutral? NaNO 3, KCN, NH 4 Cl, Ca(HSO 4 ) 2, LiCH 3 CO 2

87. pH of Salt Solutions Calculate the pH of 0.10M NaCH 3 CO 2. Calculate the pH of 0.10M NaCH 3 CO 2. NaCH 3 CO 2 (aq)  Na + (aq) + CH 3 CO 2 - (aq) Acetate ion is the conjugate base of the weak acid, acetic acid. So, acetate will act as a base and accept protons from water. CH 3 CO 2 - (aq) + H 2 O(l) ↔ CH 3 CO 2 H(aq) + OH - (aq) The solution should be basic with the pH>7.0

88. pH of Salt Solutions Calculate the pH of 0.10M NaCH 3 CO 2. Calculate the pH of 0.10M NaCH 3 CO 2. CH 3 CO 2 - (aq) + H 2 O(l) ↔ CH 3 CO 2 H(aq) + OH - (aq) K b = [CH 3 CO 2 H][OH - ] [CH 3 CO 2 - ] The value of K b for acetate ion can be calculated from the value of K a for acetic acid.

89. pH of Salt Solutions For any weak conjugate acid-base pairs, K a (K b ) = K w (K a for acetic acid) (K b for acetate ion)= 1.0 x 10 -14 K b for acetate ion=1.0 x 10 -14 /1.8 x 10 -5 = 5.6 x 10 -10

90. pH of Salt Solutions Calculate the pH of 0.10M NaCH 3 CO 2. Calculate the pH of 0.10M NaCH 3 CO 2. CH 3 CO 2 - (aq) + H 2 O(l) ↔ CH 3 CO 2 H(aq) + OH - (aq) K b = [CH 3 CO 2 H][OH - ] = 5.6 x 10 -10 [CH 3 CO 2 - ] [CH 3 CO 2 - ] [CH 3 CO 2 - ] [CH 3 CO 2 H] [OH - ] initial 0.1000 change -x+x+x equil..10-xxx

91. Polyprotic Acids Acids that have more than one proton that can be donated are called polyprotic acids. Examples are H 2 SO 4, H 3 PO 4, H 2 S and H 2 CO 3. The loss of protons can be viewed as a step- wise process.

92. Sulfuric Acid Sulfuric acid, H 2 SO 4, can lose two protons. It is called a diprotic acid. Sulfuric acid is also one of the common strong acids. This applies to loss of the first proton only. H 2 SO 4 (aq) + H 2 O(l)  H 3 O + (aq) + HSO 4 - (aq) This reaction goes 100% to the right.

93. Sulfuric Acid H 2 SO 4 (aq) + H 2 O(l)  H 3 O + (aq) + HSO 4 - (aq) HSO 4 - can react with water to lose an additional proton. HSO 4 - can react with water to lose an additional proton.

94. Sulfuric Acid HSO 4 - (aq) + H 2 O(l) ↔ H 3 O + (aq) + SO 4 2- (aq) The double arrows indicate an equilibrium is established because HSO 4 - is a weak acid. K a for HSO 4 - = [H 3 O + ][SO 4 2- ] = 1.2 x 10 -2 [HSO 4 - ] [HSO 4 - ]

95. Sulfuric Acid HSO 4 - (aq) + H 2 O(l) ↔ H 3 O + (aq) + SO 4 2- (aq) K a for HSO 4 - = [H 3 O + ][SO 4 2- ] = 1.2 x 10 -2 [HSO 4 - ] [HSO 4 - ] The value of K a indicates that hydrogen sulfate ion is a relatively strong weak acid. It is stronger than most weak acids, but weaker than the strong acids HCl, HNO 3, H 2 SO 4 or HClO 4.

96. Polyprotic Acids H 3 A(aq) + H 2 O(l) ↔ H 3 O + (aq) + H 2 A - (aq) K a1 = [H 3 O + ][H 2 A - ]/[H 3 A] H 2 A - (aq) + H 2 O(l) ↔ H 3 O + (aq) + HA 2- (aq) K a2 = [H 3 O + ][HA 2- ]/[H 2 A - ] HA 2- (aq) + H 2 O(l) ↔ H 3 O + (aq) + A 3- (aq) K a3 = [H 3 O + ][A 3- ]/[HA 2- ]

97. Polyprotic Acids For all polyprotic acids, K a1 >K a2 >K a3. It becomes progressively more difficult to remove protons as the acid becomes more negative in charge. Since the first dissociation usually predominates, it can often be used to determine the pH of a solution of the polyprotic acid. The subsequent steps are usually negligible.

98. Polyprotic Acids

99. Problem: Polyprotic Acids Calculate the pH and the concentration of all species present in 0.40M H 2 CO 3. (K a1 = 4.3 x 10 -7, K a2 =5.6 x 10 -11 ). Calculate the pH and the concentration of all species present in 0.40M H 2 CO 3. (K a1 = 4.3 x 10 -7, K a2 =5.6 x 10 -11 ). Since K a1 >>K a2, the pH can be obtained by considering only the first dissociation.

100. Problem: Polyprotic Acids Calculate the pH and concentration of all species present in 0.40M H 2 CO 3. (K a1 = 4.3 x 10 -7, K a2 =5.6 x 10 -11 ). Calculate the pH and concentration of all species present in 0.40M H 2 CO 3. (K a1 = 4.3 x 10 -7, K a2 =5.6 x 10 -11 ). H 2 CO 3 (aq) + H 2 O(l) ↔ H 3 O + (aq) + HCO 3 1- (aq) K a1 =[H 3 O + ][HCO 3 1- ] = 4.3 x 10 -7 [H 2 CO 3 ]

101. Lewis Acids & Bases The Lewis definition of acids and bases is broader than the Bronsted-Lowry definition. It defines a Lewis base as an electron pair donor. Thus, all Lewis bases have at least one lone pair of electrons. A Lewis acid is an electron pair acceptor. H + (aq) + H 2 O(l)  H 3 O + (aq)

102. Lewis Acids & Bases A Lewis acid had an empty orbital than can accept an electron pair. Metal ions can act as Lewis acids: Ag + (aq) + 2 :NH 3 (aq)  Ag(NH 3 ) 2 + (aq) The product is often referred to as an adduct.