1. The Molecular Nature of Matter and Change
Lecture PowerPoint
Chemistry
The Molecular Nature of Matter and Change
Fifth Edition
Martin S. Silberberg
Copyright  The McGraw-Hill Companies, Inc. Permission required for reproduction or display.

2. Chapter 18
Acid-Base Equilibria

3. 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 Relation 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

5. 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) H3O+(aq) + A-(aq)

6. The extent of dissociation for weak acids.
Figure 18.2
The extent of dissociation for weak acids.
Weak acid: HA(aq) + H2O(l) H3O+(aq) + A-(aq)

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

8. 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

9. ACID STRENGTH

10. 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.

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

12. 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 25oC
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-]

13. 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

14. 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 25oC and obtains a solution with [H3O+] = 3.0x10-4 M. Calculate [OH-]. Is the solution neutral, acidic, or basic?
PLAN:
Use the Kw at 25oC 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 =
3.3x10-11 M
[H3O+] is > [OH-] and the solution is acidic.

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

16. Table 18.3 The Relationship Between Ka and pKa
Acid Name (Formula)
Ka at 25oC
pKa
Hydrogen sulfate ion (HSO4-)
1.0x10-2
1.99
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

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

18. Sample Problem 18.3
Calculating [H3O+], pH, [OH-], and pOH
PROBLEM:
In an art restoration project, a conservator prepares copper-plate etching solutions by diluting concentrated HNO3 to 2.0 M, 0.30 M, and M HNO3. Calculate [H3O+], pH, [OH-], and pOH of the three solutions at 25oC.
PLAN:
HNO3 is a strong acid so [H3O+] = [HNO3]. Use Kw to find the [OH-] and then convert to pH and pOH.
SOLUTION:
For 2.0 M HNO3, [H3O+] = 2.0 M and -log [H3O+] = = pH
[OH-] = Kw/ [H3O+] = 1.0x10-14/2.0 = 5.0x10-15 M; pOH = 14.30
For 0.3 M HNO3, [H3O+] = 0.30 M and -log [H3O+] = 0.52 = pH
[OH-] = Kw/ [H3O+] = 1.0x10-14/0.30 = 3.3x10-14 M; pOH = 13.48
For M HNO3, [H3O+] = M and -log [H3O+] = 2.20 = pH
[OH-] = Kw/ [H3O+] = 1.0x10-14/6.3x10-3 = 1.6x10-12 M; pOH = 11.80

19. 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

20. Proton transfer as the essential feature of a Brønsted-Lowry acid-base reaction.
Figure 18.8
Lone pair binds H+
(acid, H+ donor)
(base, H+ acceptor)
Lone pair binds H+
(base, H+ acceptor)
(acid, H+ donor)

21. 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.

22. 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- +

23. 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

24. Sample Problem 18.5
Predicting the Net Direction of an Acid-Base Reaction
PROBLEM:
Predict the net direction and whether Kc 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 18.9 (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.

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

26. Sample Problem 18.6
Using Molecular Scenes to Predict the Net Direction of an Acid-Base Reaction
PROBLEM:
A 0.10 M solution of HX (blue and green) has a pH of 2.88 and a 0.10 M solution of HY (blue and orange) has a pH 3.52, which scene best represents the equilibrium mixture after equimolar solutions of HX and Y- (orange) are mixed?
PLAN:
Determine the relative acid strengths of HX and HY. The pH values are given at the same concentration of weak acid solutions, so pick the stronger acid directly from these values. The stronger acid will have fewer molecules of it in molecular scene than the weaker acid.
SOLUTION:
The HX solution has a lower pH than the HY solution and is the stronger acid. Y- is the stronger base. Therefore, the reaction of HX and Y- has a Kc > 1 and the mixture will have more HY than HX. Scene 3 is consistent with the relative acid strengths.

27. Sample Problem 18.7
Finding the Ka of a Weak Acid from the Solution pH
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 0.12 M 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]

28. Sample Problem 18.7
Finding the Ka of a Weak Acid from the Solution pH
Concentration (M)
HPAc(aq) + H2O(l) H3O+(aq) + PAc-(aq)
Initial
0.12 -
Change - -x +x
Equilibrium - x x
[H3O+] = 10-pH = 2.4x10-3 M which is >> 10-7 (the [H3O+] from water)
x ≈ 2.4x10-3 M ≈ [H3O+] ≈ [PAc-]
[HPAc]equilibrium = x ≈ 0.12 M
(2.4x10-3) (2.4x10-3)
0.12
So Ka =
= 4.8x10-5
[H3O+]from water;
1x10-7 M
2.4x10-3 M
x 100
Be sure to check for % error.
= 4x10-3 %
x 100
[HPAc]dissn;
2.4x10-3 M
0.12 M
= 2.0 %

29. Sample Problem 18.8
Determining Concentrations from Ka and Initial [HA]
PROBLEM:
Propanoic acid (CH3CH2COOH, which we simplify as HPr) is an organic acid whose salts are used to retard mold growth in foods. What is the [H3O+] of 0.10 M 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 - x x
Since Ka is small, we will assume that x << 0.10

30. Sample Problem 18.8
Determining Concentrations from Ka and Initial [HA]
1.3x10-5 =
[H3O+][Pr-]
[HPr] =
(x)(x)
0.10
= 1.1x10-3 M = [H3O+]
Check: [HPr]diss = 1.1x10-3 M/0.10 M x 100 = 1.1%

31. Percent HA dissociation = [HA]dissociated
[HA]initial
x 100
Polyprotic acids
acids with more than one 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

32. Sample Problem 18.9
Using Molecular Scenes to Determine the Extent of HA Dissociation
PROBLEM:
A 0.15 M solution of acid HA (blue and green) is 33% dissociated. Which scene best represents a sample of the solution after it is diluted with water?
PLAN:
Percent dissociation increase as solution is diluted. Calculate the percent dissociation of each sample.
% dissociation =
[H3O+]
[HA] + [H3O+]
x 100%
SOLUTION:
Solution 1. % dissociated = 4/(5 + 4) x 100 = 44%
Solution 2. % dissociated = 2/(7 + 2) x 100 = 22%
Solution 3. % dissociated = 3/(6 + 3) x 100 = 33%
Scene 1 represents the diluted solution.

33. ACID STRENGTH

34. Sample Problem 18.10
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 M 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]
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

35. Sample Problem 18.10
Calculating Equilibrium Concentrations for a Polyprotic Acid
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 -
7.1x10-4 M -x - +x
7.1x x - x
Ka2 = [Asc2-][H3O+]/[HAsc-] = 5x10-12 = (x)(7.1x x)/ (7.1x x) M = x

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

37. Abstraction of a proton from water by methylamine.
Figure 18.10
Abstraction of a proton from water by methylamine.
Lone pair binds H+ +
CH3NH2
H2O
methylamine +
CH3NH3+
OH-
methylammonium ion

38. Sample Problem 18.11
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 1.5 M (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 negligible
[(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

39. Sample Problem 18.11
Determining pH from Kb and Initial [B]
Kb = 5.9x10-4 =
[(CH3)2NH2+][OH-]
[(CH3)2NH]
5.9x10-4 =
(x)(x)
1.5 M
x = 3.0x10-2 M = [OH-]
Check assumption:
3.0x10-2 M/1.5 M x 100 = 2%
[H3O+] = Kw/[OH-] = 1.0x10-14/3.0x10-2 = 3.3x10-13 M
pH = -log 3.3x10-13 = 12.48

40. Sample Problem 18.12
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 0.25 M NaAc? Ka of acetic acid (HAc) is 1.8x10-5.
PLAN:
Sodium salts are soluble in water so [Ac-] = 0.25 M.
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
0.25 M -
Change -x +x -
Equilibrium -
0.25 M - x x
Kb =
[HAc][OH-]
[Ac-] = Kw Ka
Kb =
1.0x10-14
1.8x10-5
= 5.6x10-10 M

41. Sample Problem 18.12
Determining the pH of a Solution of A-
[Ac-] = 0.25 M - x ≈ 0.25 M
Kb =
[HAc][OH-]
[Ac-]
5.6x10-10 = x2/0.25 M
x = 1.2x10-5 M = [OH-]
Check assumption:
1.2x10-5 M/0.25 M x 100 = 4.8x10-3 %
[H3O+] = Kw/[OH-]
= 1.0x10-14/1.2x10-5
= 8.3x10-10 M
pH = -log 8.3x10-10 M = 9.08

42. The effect of atomic and molecular properties on
Figure 18.11
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

43. <<   The relative strengths of oxoacids. H O I Br Cl < 
Figure 18.12
The relative strengths of oxoacids. H O I Br Cl
<       H O Cl H O Cl  
<<  

44. Table 18.7 Ka Values of Some Hydrated Metal Ions at 25oC
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

45. 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

46. Sample Problem 18.13
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 (III) nitrate, Cr(NO3)3
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.
C6H5OO-(aq) + H2O(l) C6H5OOH(aq) + OH-(aq)
(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.
Cr(H2O)63+(aq) + H2O(l) Cr(H2O)5OH2+(aq) + H3O+(aq)

48. Sample Problem 18.14
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 relative 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.

49. Molecules as Lewis Acids
An acid is an electron-pair acceptor.
A base is an electron-pair donor.
acid
base
adduct

50. The Mg2+ ion as a Lewis acid in the chlorophyll molecule.
Figure 18.14
The Mg2+ ion as a Lewis acid in the chlorophyll molecule.

51. Sample Problem 18.15
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