1. Chapter 18
Acid-Base Equilibria

2. Acid-Base Equilibria 18.1 Acids and Bases in Water
18.2 Autoionization of Water and the pH Scale
18.3 Proton Transfer and the Brønsted-Lowry Acid-Base Definition
18.4 Solving Problems Involving Weak-Acid Equilibria
18.5 Weak Bases and Their Relations to Weak Acids
18.6 Molecular Properties and Acid Strength
18.7 Acid-Base Properties of Salt Solutions
18.8 Generalizing the Brønsted-Lowry Concept: The Leveling Effect
18.9 Electron-Pair Donation and the Lewis Acid-Base Definition

3. The Nature of Acids and Bases:
Acids: Acids taste sour.
React with metals and produce H gas.
turns blue litmus red
pH < 7
* Bases: Bases taste bitter. They are slippery.
turns red litmus blue.
pH >7

5. Arrhenius concept Acids produce H+ ions. Bases produce OH- ions.
This concept is limited because it applies to only aqueous solution and defines only OH containing bases.

6. Neutralization:
acid + base _______salt + water.

7. Acid Dissociation Constant (Ka)
Write Ka expression for strong and weak acids.

8. The extent of dissociation for strong acids.
Figure 18.1
The extent of dissociation for strong acids.
Strong acid: HA(g or l) + H2O(l) H2O+(aq) + A-(aq)

9. The extent of dissociation for weak acids.
Figure 18.2
The extent of dissociation for weak acids.
Weak acid: HA(aq) + H2O(l) H2O+(aq) + A-(aq)

10. Reaction of zinc with a strong and a weak acid.
Figure 18.3
Reaction of zinc with a strong and a weak acid.
1M HCl(aq)
1M CH3COOH(aq)

11. The Acid-Dissociation Constant
Strong acids dissociate completely into ions in water.
HA(g or l) + H2O(l) H3O+(aq) + A-(aq)
Kc >> 1
Weak acids dissociate very slightly into ions in water.
HA(aq) + H2O(l) H3O+(aq) + A-(aq)
Kc << 1
The Acid-Dissociation Constant
Kc =
[H3O+][A-]
[H2O][HA]
stronger acid higher [H3O+]
larger Ka
Kc[H2O] = Ka =
[H3O+][A-]
[HA]
smaller Ka lower [H3O+]
weaker acid

12. Classifying the Relative Strengths of acids and Bases:
Strong acids:
HCl, HBr, HI
Oxo acids. HNO3, H2SO4, HClO4
Weak acids: HF
HCN , H2S (H not bonded to O or halogen)
Oxo acids. HClO, HNO2, H3PO4
Carboxylic acids. CH3COOH

13. Classifying the Relative Strengths of acids and Bases:
Strong bases:
M2O, MOH M= (group 1A metal)
MO , M(OH)2 M=group 2A metal
Weak bases: (N atom and lone pair of electrons)
NH3
Amines.

14. ACID STRENGTH

15. SAMPLE PROBLEM 18.1:
Classifying Acid and Base Strength from the
Chemical Formula
PROBLEM:
Classify each of the following compounds as a strong acid, weak acid, strong base, or weak base.
(a) H2SeO4
(b) (CH3)2CHCOOH
(c) KOH
(d) (CH3)2CHNH2
PLAN:
Pay attention to the text definitions of acids and bases. Look at O for acids as well as the -COOH group; watch for amine groups and cations in bases.
SOLUTION:
(a) Strong acid - H2SeO4 - the number of O atoms exceeds the number of ionizable protons by 2.
(b) Weak acid - (CH3)2CHCOOH is an organic acid having a -COOH group.
(c) Strong base - KOH is a Group 1A(1) hydroxide.
(d) Weak base - (CH3)2CHNH2 has a lone pair of electrons on the N and is an amine.

16. Autoionization of water and the pH scale
Water dissociates into its ions.

17. Autoionization of Water and the pH Scale
H2O(l)
H2O(l) +
OH-(aq)
H3O+(aq) +

18. H2O(l) + H2O(l) H3O+(aq) + OH-(aq)
Kc =
[H3O+][OH-]
[H2O]2
The Ion-Product Constant for Water
Kc[H2O]2 =
Kw =
[H3O+][OH-]
= 1.0 x at 250C
A change in [H3O+] causes an inverse change in [OH-].
In an acidic solution, [H3O+] > [OH-]
In a basic solution, [H3O+] < [OH-]
In a neutral solution, [H3O+] = [OH-]

19. The relationship between [H3O+] and [OH-] and the
relative acidity of solutions.
Figure 18.4
Divide into Kw
[H3O+]
[OH-]
[H3O+] > [OH-]
[H3O+] = [OH-]
[H3O+] < [OH-]
ACIDIC SOLUTION
NEUTRAL SOLUTION
BASIC SOLUTION

20. SAMPLE PROBLEM 18.2:
Calculating [H3O+] and [OH-] in an Aqueous
Solution
PROBLEM:
A research chemist adds a measured amount of HCl gas to pure water at 250C and obtains a solution with [H3O+] = 3.0x10-4M. Calculate [OH-]. Is the solution neutral, acidic, or basic?
PLAN:
Use the Kw at 250C and the [H3O+] to find the corresponding [OH-].
SOLUTION:
Kw = 1.0x10-14 = [H3O+] [OH-] so
[OH-] = Kw/ [H3O+] = 1.0x10-14/3.0x10-4 =
3.3x10-11M
[H3O+] is > [OH-] and the solution is acidic.

21. pH scale pH  log[H+] pH in water ranges from 0 to 14.
Kw = 1.00  1014 = [H+] [OH]
pKw = = pH + pOH
As pH rises, pOH falls (sum = 14.00).
pH=7-neutral, pH>7-basic, pH<7-acidic

22. pH scale Acidic solns have a higher pOH. pK=-log K
equation reaches equ, mostly products present, low pK (high K)
Reverse of the above statement is also true.
pH is measure using pH meter, pH paper or acid-base indicator.

23. Methods for measuring the pH of an aqueous solution
Figure 18.7
Methods for measuring the pH of an aqueous solution
pH (indicator) paper
pH meter

24. The pH values of some familiar aqueous solutions
Figure 18.5
The pH values of some familiar aqueous solutions
pH = -log [H3O+]

25. Table 18.3 The Relationship Between Ka and pKa
Acid Name (Formula)
Ka at 250C
pKa
Hydrogen sulfate ion (HSO4-)
1.02x10-2
1.991
3.15
Nitrous acid (HNO2)
7.1x10-4
4.74
Acetic acid (CH3COOH)
1.8x10-5
2.3x10-9
8.64
Hypobromous acid (HBrO)
Phenol (C6H5OH)
1.0x10-10
10.00

26. The relations among [H3O+], pH, [OH-], and pOH.
Figure 18.6
The relations among [H3O+], pH, [OH-], and pOH.

27. Problems P.3.Calculate pH and pOH at 25 C for: 1.0 M H+
P.4.pH=6.88, Calculate [ H+] and [ OH-] for this sample.
P.5.Calculate pH of 0.10 M HNO3
P.6.Calculate pH of 1.0 x HCl.

28. Bronsted-Lowry model:
An acid is a proton (H+ ) donor. A base is a proton acceptor.
* A conjugate acid-base pair only differs by one H.
HCl H2O _________ H3O+ Cl-
Acid base conj acid conj base
HA(aq) + H2O (l)  H3O+ (aq) +A-(aq)
CA CB CA CB1

29. Bronsted-Lowry model:
conjugate base: everything that remains of the acid molecule after a proton is lost.
Has one H less and one more minus charge than the acid.
conjugate acid: formed when the proton is transferred to the base.
Has one more H and one less charge than the base.
Do Follow up problem-18.4-Page.779

30. Brønsted-Lowry Acid-Base Definition
An acid is a proton donor, any species which donates a H+.
A base is a proton acceptor, any species which accepts a H+.
An acid-base reaction can now be viewed from the standpoint of the reactants AND the products.
An acid reactant will produce a base product and the two will constitute an acid-base conjugate pair.

31. Table 18.4 The Conjugate Pairs in Some Acid-Base Reactions+
Base
Acid +
Conjugate Pair
Reaction 1 HF
H2O + F-
H3O+ +
Reaction 2
HCOOH
CN- +
HCOO-
HCN +
Reaction 3
NH4+
CO32- +
NH3
HCO3- +
Reaction 4
H2PO4-
OH- +
HPO42-
H2O +
Reaction 5
H2SO4
N2H5+ +
HSO4-
N2H62+ +
Reaction 6
HPO42-
SO32- +
PO43-
HSO3- +

32. Proton transfer as the essential feature of a Brønsted-
Lowry acid-base reaction.
Figure 18.8
HCl
H2O +
Lone pair binds H+
Cl-
H3O+ +
(acid, H+ donor)
(base, H+ acceptor)
NH3
H2O +
Lone pair binds H+
NH4+
OH- +
(base, H+ acceptor)
(acid, H+ donor)

33. SAMPLE PROBLEM 18.4:
Identifying Conjugate Acid-Base Pairs
PROBLEM:
The following reactions are important environmental processes. Identify the conjugate acid-base pairs.
(a) H2PO4-(aq) + CO32-(aq) HPO42-(aq) + HCO3-(aq)
(b) H2O(l) + SO32-(aq) OH-(aq) + HSO3-(aq)
PLAN:
Identify proton donors (acids) and proton acceptors (bases).
conjugate pair2
conjugate pair1
SOLUTION:
(a) H2PO4-(aq) + CO32-(aq) HPO42-(aq) + HCO3-(aq)
proton donor
proton acceptor
proton acceptor
proton donor
conjugate pair2
conjugate pair1
(b) H2O(l) + SO32-(aq) OH-(aq) + HSO3-(aq)
proton donor
proton acceptor
proton acceptor
proton donor

34. SAMPLE PROBLEM 18.5:
Predicting the Net Direction of an Acid-Base
Reaction
PROBLEM:
Predict the net direction and whether Ka is greater or less than 1 for each of the following reactions (assume equal initial concentrations of all species):
(a) H2PO4-(aq) + NH3(aq) HPO42-(aq) + NH4+(aq)
(b) H2O(l) + HS-(aq) OH-(aq) + H2S(aq)
PLAN:
Identify the conjugate acid-base pairs and then consult Figure (button) to determine the relative strength of each. The stronger the species, the more preponderant its conjugate.
SOLUTION:
(a) H2PO4-(aq) + NH3(aq) HPO42-(aq) + NH4+(aq)
stronger acid
weaker acid
stronger base
weaker base
Net direction is to the right with Kc > 1.
(b) H2O(l) + HS-(aq) OH-(aq) + H2S(aq)
stronger acid
weaker acid
stronger base
weaker base
Net direction is to the left with Kc < 1.

35. Strengths of conjugate acid-base pairs
Figure 18.9
Strengths of conjugate acid-base pairs

36. Solving Problems Involving Weak-Acid Equilibria.
Write balanced equation and Ka expression.
Make an ICE table.
Make required assumptions.
Substitute values and solve for x.
Verify assumptions by calculating % error.

37. Problems
P.7. Calculate the pH of 1.00 M aqueous solution of HF. Ka=7.2 x 10-4
P.8 Calculate the pH of M aq solution of HOCl. Ka=3.5 x 10-8

38. SAMPLE PROBLEM 18.6:
Finding the Ka of a Weak Acid from the pH of
Its Solution
PROBLEM:
Phenylacetic acid (C6H5CH2COOH, simplified here as HPAc) builds up in the blood of persons with phenylketonuria, an inherited disorder that, if untreated, causes mental retardation and death. A study of the acid shows that the pH of 0.12M HPAc is What is the Ka of phenylacetic acid?
PLAN:
Write out the dissociation equation. Use pH and solution concentration to find the Ka.
Assumptions:
With a pH of 2.62, the [H3O+]HPAc >> [H3O+]water.
[PAc-] ≈ [H3O+]; since HPAc is weak, [HPAc]initial ≈ [HPAc]initial - [HPAc]dissociation
SOLUTION:
HPAc(aq) + H2O(l) H3O+(aq) + PAc-(aq)
Ka =
[H3O+][PAc-]
[HPAc]

39. SAMPLE PROBLEM 18.6:
Finding the Ka of a Weak Acid from the pH of
Its Solution
continued
Concentration(M)
HPAc(aq) + H2O(l) H3O+(aq) + PAc-(aq)
Initial
0.12 -
1x10-7
Change - -x +x
Equilibrium -
0.12-x x
x +(<1x10-7)
[H3O+] = 10-pH = 2.4x10-3 M which is >> 10-7 (the [H3O+] from water)
x ≈ 2.4x10-3 M ≈ [H3O+] ≈ [PAc-]
[HPAc]equilibrium = 0.12-x ≈ 0.12 M
(2.4x10-3) (2.4x10-3)
0.12
So Ka =
= 4.8 x 10-5
[H3O+]from water;
1x10-7M
2.4x10-3M
x100
Be sure to check for % error.
= 4x10-3 %
x100
[HPAc]dissn;
2.4x10-3M
0.12M
= 2.0 %

40. SAMPLE PROBLEM 18.7:
Determining Concentrations from Ka and
Initial [HA]
PROBLEM:
Propanoic acid (CH3CH2COOH, which we simplify and HPr) is an organic acid whose salts are used to retard mold growth in foods. What is the [H3O+] of 0.10M HPr (Ka = 1.3x10-5)?
PLAN:
Write out the dissociation equation and expression; make whatever assumptions about concentration which are necessary; substitute.
Assumptions:
For HPr(aq) + H2O(l) H3O+(aq) + Pr-(aq)
x = [HPr]diss = [H3O+]from HPr= [Pr-]
Ka =
[H3O+][Pr-]
[HPr]
SOLUTION:
Concentration(M)
HPr(aq) + H2O(l) H3O+(aq) + Pr-(aq)
Initial
0.10 -
Change - -x +x
Equilibrium -
0.10-x x
Since Ka is small, we will assume that x << 0.10

41. SAMPLE PROBLEM 18.7:
Determining Concentrations from Ka and
Initial [HA]
continued
1.3x10-5 =
[H3O+][Pr-]
[HPr] =
(x)(x)
0.10
= 1.1x10-3 M = [H3O+]
Check: [HPr]diss = 1.1x10-3M/0.10 M x 100 = 1.1%

42. Percent dissociation:
For a weak acid % dissociation increases as the acid becomes more dilute.
For solution of any weak acid, [ H+] decreases as [HA]0 decreases, but % dissociation increases as [HA]0 increases.

43. Problems
P.11.Calculate the % dissociation of acetic acid (Ka= 1.8 x 10-5 ) 1.00 M and M solutions.
P.12.In a M HC3H5O3 aqueous solution, lactic acid is 3.7 % dissociated. Calculate ka for the acid.

44. Percent HA dissociation = [HA]dissociated
[HA]initial
x 100
Polyprotic acids
acids with more than more ionizable proton
Ka1 =
[H3O+][H2PO4-]
[H3PO4]
H3PO4(aq) + H2O(l) H2PO4-(aq) + H3O+(aq)
= 7.2x10-3
Ka2 =
[H3O+][HPO42-]
[H2PO4-]
H2PO4-(aq) + H2O(l) HPO42-(aq) + H3O+(aq)
= 6.3x10-8
Ka3 =
[H3O+][PO43-]
[HPO42-]
HPO42-(aq) + H2O(l) PO43-(aq) + H3O+(aq)
= 4.2x10-13
Ka1 > Ka2 > Ka3

45. ACID STRENGTH

46. Polyprotic Acids
They can furnish more than one proton (H+) to the solution.
Characteristics: 1. Ka values are much smaller than the first value , so only the first dissociation step makes a significant contribution to equilibrium concentration of H+ .
2. For sulfuric acid, it behaves as a strong acid in the first dissociation step and a weak acid in the second step, where its contribution of H+ ions can be neglected. Use quadratic equation to solve such kind of a problem.

47. Problems
P.13.Calculate the pH of a 5.0 M H3PO4 solution and the equilibrium concentration of the species H3PO4 , H2PO4-, HPO42-, PO43-
P.14.Calculate the pH of 1.0 M sulfuric acid solution.

48. SAMPLE PROBLEM 18.8:
Calculating Equilibrium Concentrations for a
Polyprotic Acid
PROBLEM:
Ascorbic acid (H2C6H6O6; H2Asc for this problem), known as vitamin C, is a diprotic acid (Ka1 = 1.0x10-5 and Ka2 = 5x10-12) found in citrus fruit. Calculate [H2Asc], [HAsc-], [Asc2-], and the pH of 0.050M H2Asc.
PLAN:
Write out expressions for both dissociations and make assumptions.
Ka1 >> Ka2 so the first dissociation produces virtually all of the H3O+.
Ka1 is small so [H2Asc]initial ≈ [H2Asc]diss
After finding the concentrations of various species for the first dissociation, we can use them as initial concentrations for the second dissociation.
SOLUTION:
Ka1 =
[HAsc-][H3O+]
[H2Asc]
H2Asc(aq) + H2O(l) HAsc-(aq) + H3O+(aq)
= 1.0x10-5
Ka2 =
[Asc2-][H3O+]
[HAsc-]
HAsc-(aq) + H2O(l) Asc2-(aq) + H3O+(aq)
= 5x10-12

49. SAMPLE PROBLEM 18.8:
Calculating Equilibrium Concentrations for a
Polyprotic Acid
continued
Concentration(M)
H2Asc(aq) + H2O(l) HAsc-(aq) + H3O+(aq)
Initial
0.050 -
Change
- x -
+ x
Equilibrium x - x
Ka1 = [HAsc-][H3O+]/[H2Asc] = 1.0x10-5 = (x)(x)/0.050 M x
x = 7.1x10-4 M
pH = -log(7.1x10-4) = 3.15
HAsc-(aq) + H2O(l) Asc2-(aq) + H3O+(aq)
Concentration(M)
Equilibrium
Change
Initial
7.1x10-4M -
- x -
+ x
7.1x x - x x
= 6x10-8 M

50. Bases
“Strong” and “weak” are used in the same sense for bases as for acids.
strong = complete dissociation (hydroxide ion supplied to solution)
NaOH(s)  Na+(aq) + OH(aq)
weak = very little dissociation (or reaction with water)
H3CNH2(aq) + H2O(l)  H3CNH3+(aq) + OH(aq)

51. BASE STRENGTH
[BH+][OH-]
[B]
Kb =

52. Problems P.15.Calculate pH of 5.0 x 10-2 M NaOH solution.
P.16.Calculate the pH for a 1.50 M solution of ammonia
(Kb=1.8 x )

53. Abstraction of a proton from water by methylamine.
Figure 18.11
Abstraction of a proton from water by methylamine.
Lone pair binds H+ +
CH3NH2
H2O
Methylamine
Base in water +
CH3NH3+
OH-
methylammonium ion

54. SAMPLE PROBLEM 18.9:
Determining pH from Kb and Initial [B]
PROBLEM:
Dimethylamine, (CH3)2NH, a key intermediate in detergent manufacture, has a Kb of 5.9x What is the pH of 1.5M (CH3)2NH?
PLAN:
Perform this calculation as you did those for acids. Keep in mind that you are working with Kb and a base.
(CH3)2NH(aq) + H2O(l) (CH3)2NH2+(aq) + OH-(aq)
Assumptions:
Kb >> Kw so [OH-]from water is neglible
[(CH3)2NH2+] = [OH-] = x ; [(CH3)2NH2+] - x ≈ [(CH3)2NH]initial
SOLUTION:
(CH3)2NH(aq) + H2O(l) (CH3)2NH2+(aq) + OH-(aq)
Concentration
Initial
1.50M -
Change
- x -
+ x
Equilibrium x - x

55. SAMPLE PROBLEM 18.9:
Determining pH from Kb and Initial [B]
continued
Kb = 5.9x10-4 =
[(CH3)2NH2+][OH-]
[(CH3)2NH]
5.9x10-4 =
(x) (x)
1.5M
x = 3.0x10-2M = [OH-]
Check assumption:
3.0x10-2M/1.5M x 100 = 2%
[H3O+] = Kw/[OH-] = 1.0x10-14/3.0x10-2 = 3.3x10-13M
pH = -log 3.3x10-13 = 12.48

56. SAMPLE PROBLEM 18.10:
Determining the pH of a Solution of A-
PROBLEM:
Sodium acetate (CH3COONa, or NaAc for this problem) has applications in photographic development and textile dyeing. What is the pH of 0.25M NaAc? Ka of acetic acid (HAc) is 1.8x10-5.
PLAN:
Sodium salts are soluble in water so [Ac-] = 0.25M.
Write the association equation for acetic acid; use the Ka to find the Kb.
SOLUTION:
Ac-(aq) + H2O(l) HAc(aq) + OH-(aq)
Concentration
Initial
0.25M -
Change -x +x -
Equilibrium -
0.25M-x x
Kb =
[HAc][OH-]
[Ac-] = Kw Ka
Kb =
1.0x10-14
1.8x10-5
= 5.6x10-10M

57. SAMPLE PROBLEM 18.10:
Determining the pH of a Solution of A-
continued
[Ac-] = 0.25M-x ≈ 0.25M
Kb =
[HAc][OH-]
[Ac-]
5.6x10-10 = x2/0.25M
x = 1.2x10-5M = [OH-]
Check assumption:
1.2x10-5M/0.25M x 100 = 4.8x10-3 %
[H3O+] = Kw/[OH-]
= 1.0x10-14/1.2x10-5
= 8.3x10-10M
pH = -log 8.3x10-10M = 9.08

58. Molecular Properties and Acid strength:
Trends in acid strength of Nonmetal Hydrides:
Across a period nonmetal hydride acid strength increases.
Down a group nonmetal hydride acid strength increases.

59. The effect of atomic and molecular properties on
Figure 18.12
The effect of atomic and molecular properties on
nonmetal hydride acidity.
6A(16)
H2O
H2S
H2Se
H2Te
7A(17) HF
HCl
HBr HI
Electronegativity increases, acidity increases
Bond strength decreases, acidity increases

60. Trends in Acid strength of Oxoacids
1. For oxoacids with same number of O around E , acid strength increases with electronegativity of E.
2. For oxoacids with different number of O around E, acid strength increases with number of O atoms.

61. <<   The relative strengths of oxoacids. H O I Br Cl > 
Figure 18.13
The relative strengths of oxoacids. H O I Br Cl
>       H O Cl H O Cl  
<<  

62. Table 18.7 Ka Values of Some Hydrated Metal Ions at 250C
Free Ion
Hydrated Ion Ka
ACID STRENGTH
Fe3+
Fe(H2O)63+(aq)
6 x 10-3
Sn2+
Sn(H2O)62+(aq)
4 x 10-4
Cr3+
Cr(H2O)63+(aq)
1 x 10-4
Al3+
Al(H2O)63+(aq)
1 x 10-5
Cu2+
Cu(H2O)62+(aq)
3 x 10-8
Pb2+
Pb(H2O)62+(aq)
3 x 10-8
Zn2+
Zn(H2O)62+(aq)
1 x 10-9
Co2+
Co(H2O)62+(aq)
2 x 10-10
Ni2+
Ni(H2O)62+(aq)
1 x 10-10

63. The acidic behavior of the hydrated Al3+ ion.
Figure 18.13
The acidic behavior of the hydrated Al3+ ion.
Electron density drawn toward Al3+
Nearby H2O acts as base
Al(H2O)63+
Al(H2O)5OH2+
H3O+
H2O

64. Acid-Base properties of Salt:
Salt is an ionic compound.
Ka xKb=Kw

66. SAMPLE PROBLEM 18.11:
Predicting Relative Acidity of Salt Solutions
PROBLEM:
Predict whether aqueous solutions of the following are acidic, basic, or neutral, and write an equation for the reaction of any ion with water:
(a) Potassium perchlorate, KClO4
(b) Sodium benzoate, C6H5COONa
(c) Chromium trichloride, CrCl3
(d) Sodium hydrogen sulfate, NaHSO4
PLAN:
Consider the acid-base nature of the anions and cations. Strong acid-strong base combinations produce a neutral solution; strong acid-weak base, acidic; weak acid-strong base, basic.
SOLUTION:
(a) The ions are K+ and ClO4- , both of which come from a strong base(KOH) and a strong acid(HClO4). Therefore the solution will be neutral.
(b) Na+ comes from the strong base NaOH while C6H5COO- is the anion of a weak organic acid. The salt solution will be basic.
(c) Cr3+ is a small cation with a large + charge, so it’s hydrated form will react with water to produce H3O+. Cl- comes from the strong acid HCl. Acidic solution.
(d) Na+ comes from a strong base. HSO4- can react with water to form H3O+. So the salt solution will be acidic.

67. SAMPLE PROBLEM 18.12:
Predicting the Relative Acidity of Salt
Solutions from Ka and Kb of the Ions
PROBLEM:
Determine whether an aqueous solution of zinc formate, Zn(HCOO)2, is acidic, basic, or neutral.
PLAN:
Both Zn2+ and HCOO- come from weak conjugates. In order to find the relatively acidity, write out the dissociation reactions and use the information in Tables and 18.7.
SOLUTION:
Zn(H2O)62+(aq) + H2O(l) Zn(H2O)5OH+(aq) + H3O+(aq)
HCOO-(aq) + H2O(l) HCOOH(aq) + OH-(aq)
Ka Zn(H2O)62+ = 1x10-9
Ka HCOO- = 1.8x10-4 ; Kb = Kw/Ka = 1.0x10-14/1.8x10-4 = 5.6x10-11
Ka for Zn(H2O)62+ >>> Kb HCOO-, therefore the solution is acidic.

68. Molecules as Lewis Acids
An acid is an electron-pair acceptor.
A base is an electron-pair donor.
acid
base
adduct
H2O(l)
M(H2O)42+(aq)
M2+
adduct

69. The Mg2+ ion as a Lewis acid in the chlorophyll molecule.
Figure 18.15
The Mg2+ ion as a Lewis acid in the chlorophyll molecule.

70. SAMPLE PROBLEM 18.13:
Identifying Lewis Acids and Bases
PROBLEM:
Identify the Lewis acids and Lewis bases in the following reactions:
(a) H+ + OH H2O
(b) Cl- + BCl BCl4-
(c) K+ + 6H2O K(H2O)6+
PLAN:
Look for electron pair acceptors (acids) and donors (bases).
SOLUTION:
acceptor
(a) H+ + OH H2O
donor
donor
(b) Cl- + BCl BCl4-
acceptor
acceptor
(c) K+ + 6H2O K(H2O)6+
donor