1. The Molecular Nature of Matter and Change
Lecture PowerPoint
Chemistry
The Molecular Nature of Matter and Change
Seventh Edition
Martin S. Silberberg
and Patricia G. Amateis
Copyright  McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education

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

4. Table 18.1 Some Common Acids and Bases and their Household Uses.

5. Arrhenius Acid-Base Definition
This is the earliest acid-base definition, which classifies these substances in terms of their behavior in water.
An acid is a substance with H in its formula that dissociates to yield H3O+.
A base is a substance with OH in its formula that dissociates to yield OH-.
When an acid reacts with a base, they undergo neutralization:
H+(aq) + OH-(aq) → H2O(l) DH°rxn = kJ

6. Strong and Weak Acids
A strong acid dissociates completely into ions in water:
HA(g or l) + H2O(l) → H3O+(aq) + A-(aq)
A dilute solution of a strong acid contains no HA molecules.
A weak acid dissociates slightly to form ions in water:
HA(aq) + H2O(l) H3O+(aq) + A-(aq)
In a dilute solution of a weak acid, most HA molecules are undissociated.
[H3O+][A-]
[HA][H2O]
Kc =
has a very small value.

7. The extent of dissociation for strong acids.
Figure 18.1A
The extent of dissociation for strong acids.
Strong acid: HA(g or l) + H2O(l) → H3O+(aq) + A-(aq)
There are no HA molecules in solution.

8. Figure 18.1B
The extent of dissociation for weak acids.
Weak acid: HA(aq) + H2O(l) H3O+(aq) + A-(aq)
Most HA molecules are undissociated.

9. Reaction of zinc with a strong acid (left) and a weak acid (right).
Figure 18.2
Reaction of zinc with a strong acid (left) and a weak acid (right).
Zinc reacts rapidly with the strong acid, since [H3O+] is much higher.
1 M HCl(aq)
1 M CH3COOH(aq)

10. The Acid Dissociation Constant, Ka
HA(aq) + H2O(l) H3O+(aq) + A–(aq)
[H3O+][A–]
[HA][H2O]
Kc =
Kc[H2O] = Ka =
[H3O+][A–]
[HA]
The value of Ka is an indication of acid strength.
Stronger acid
larger Ka
higher [H3O+]
Weaker acid
smaller Ka
lower % dissociation of HA

11. Table 18.2 Ka Values for some Monoprotic Acids at 25°C

12. Classifying the Relative Strengths of Acids
Strong acids include
the hydrohalic acids (HCl, HBr, and HI) and
oxoacids in which the number of O atoms exceeds the number of ionizable protons by two or more (eg., HNO3, H2SO4, HClO4.)
Weak acids include
the hydrohalic acid HF,
acids in which H is not bonded to O or to a halogen (eg., HCN),
oxoacids in which the number of O atoms equals or exceeds the number of ionizable protons by one (eg., HClO, HNO2), and
carboxylic acids, which have the general formula RCOOH (eg., CH3COOH and C6H5COOH.)

13. Classifying the Relative Strengths of Bases
Strong bases include
water-soluble compounds containing O2- or OH- ions.
The cations are usually those of the most active metals:
M2O or MOH, where M = Group 1A(1) metal (Li, Na, K, Rb, Cs)
MO or M(OH)2 where M = group 2A(2) metal (Ca, Sr, Ba).
Weak bases include
ammonia (NH3),
amines, which have the general formula
The common structural feature is an N atom with a lone electron pair.

14. 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) KOH (b) (CH3)2CHCOOH
(c) H2SeO4 (d) (CH3)2CHNH2
PLAN:
We examine the formula and classify each acid or base, using the text descriptions. Particular points to note for acids are the numbers of O atoms relative to ionizable H atoms and the presence of the –COOH group. For bases, note the nature of the cation or the presence of an N atom that has a lone pair.
SOLUTION:
(a) Strong base: KOH is one of the group 1A(1) hydroxides.

15. Sample Problem 18.1
(b) Weak acid: (CH3)2CHCOOH is a carboxylic acid, as indicated by the –COOH group. The –COOH proton is the only ionizable proton in this compound.
(c) Strong acid: He2SO4 is an oxoacid in which the number of O atoms exceeds the number of ionizable protons by two.
(d) Weak base: (CH3)2CHNH2 has a lone pair of electrons on the N and is an amine.

16. Autoionization of Water
Water dissociates very slightly into ions in an equilibrium process known as autoionization or self-ionization.
2H2O (l) H3O+ (aq) + OH- (aq)

17. The Ion-Product Constant for Water (Kw)
2H2O (l) H3O+ (aq) + OH- (aq)
[H3O+][OH-]
[H2O]2
Kc =
Kc[H2O]2 = Kw = [H3O+][OH-] = 1.0x10-14 (at 25°C)
= 1.0x10-7 (at 25°C)
In pure water,
[H3O+] = [OH-] =
Both ions are present in all aqueous systems.

18. A change in [H3O+] causes an inverse change in [OH-], and vice versa.
Higher [H3O+]
lower [OH-]
Higher [OH-]
lower [H3O+]
We can define the terms “acidic” and “basic” in terms of the relative concentrations of H3O+ and OH– ions:
In an acidic solution, [H3O+] > [OH–]
In a neutral solution, [H3O+] = [OH–]
In a basic solution, [H3O+] < [OH–]

19. Figure 18.3
The relationship between [H3O+] and [OH-] and the relative acidity of solutions.

20. Sample Problem 18.2
Calculating [H3O+] or [OH–] in an Aqueous Solution
PROBLEM:
A research chemist adds a measured amount of HCl gas to pure water at 25°C and obtains a solution with [H3O+] = 3.0x10–4 M. Calculate [OH–]. Is the solution neutral, acidic, or basic?
PLAN:
We use the known value of Kw at 25°C (1.0x10–14) and the given [H3O+] to solve for [OH–]. We can then compare [H3O+] with [OH–] to determine whether the solution is acidic, basic, or neutral.
SOLUTION:
Kw = 1.0 x 10–14 = [H3O+] [OH–] so
[OH–] = Kw
[H3O+]
1.0 x 10–14
3.0 x 10–4 =
= 3.3x10–11 M
[H3O+] is > [OH–] and the solution is acidic.

21. The pH Scale pH = -log[H3O+]
The pH of a solution indicates its relative acidity:
In an acidic solution, pH < 7.00
In a neutral solution, pH = 7.00
In a basic solution, pH > 7.00
The higher the pH, the lower the [H3O+] and the less acidic the solution.

22. Copyright  The McGraw-Hill Companies, Inc
Copyright  The McGraw-Hill Companies, Inc. Permission required for reproduction or display.
Figure 18.4
The pH values of some familiar aqueous solutions.
pH = –log [H3O+]

23. A low pKa corresponds to a high Ka.
Table The Relationship between Ka and pKa
Acid Name (Formula)
Ka at 25°C
pKa
Hydrogen sulfate ion (HSO4-)
1.0x10–2
1.99
Nitrous acid (HNO2)
7.1x10–4
3.15
Acetic acid (CH3COOH)
1.8x10–5
4.75
Hypobromous acid (HBrO)
2.3x10–9
8.64
Phenol (C6H5OH)
1.0x10–10
10.00
pKa = –logKa
A low pKa corresponds to a high Ka.

24. pH, pOH, and pKw Kw = [H3O+][OH–] = 1.0x10–14 at 25°C pH = –log[H3O+]
pOH = –log[OH–]
pKw = pH + pOH = at 25°C
pH + pOH = pKw for any aqueous solution at any temperature.
Since Kw is a constant, the values of pH, pOH, [H3O+], and [OH–] are interrelated:
If [H3O+] increases, [OH–] decreases (and vice versa).
If pH increases, pOH decreases (and vice versa).

25. The relations among [H3O+], pH, [OH-], and pOH.
Figure 18.5
The relations among [H3O+], pH, [OH-], and pOH.
Copyright  The McGraw-Hill Companies, Inc. Permission required for reproduction or display.

26. Sample Problem 18.3
Calculating [H3O+], pH, [OH-], and pOH for Strong Acids and Bases
PROBLEM: Calculate [H3O+], pH, [OH–], and pOH for each solution at 25°C:
(a) 0.30 M HNO3, used for etching copper metal; and
(b) M Ca(OH)2, used in leather tanning to remove hair from hides.
PLAN:
HNO3 is a strong acid; so it dissociates completely, and [H3O+] = [HNO3]init. Likewise, Ca(OH)2 is a strong base that dissociates completely, and
[OH–] = 2[Ca(OH)2]init.
We use the given concentrations and the value of Kw at 25°C to find [OH–] and [H3O+]. We can then calculate pH and pOH.

27. Sample Problem 18.3
SOLUTION:
[H3O+] = 0.30 M
(a) Calculating the values for 0.30 M HNO3:
pH = –log[H3O+] = -log(0.30) = 0.52 Kw
[H3O+]
[OH–] = =
1.0 x 10–14
0.30
= 3.3 x M
pOH = –log[OH–] = –log(3.3 x 10–14) = 13.48
(b) Calculating the values for M Ca(OH)2:
pH = –log[H3O+] = –log(1.2 x 10–12) = 11.92
Ca(OH)2 is a strong electrolyte: Ca(OH)2(aq) → Ca2+(aq) + 2 OH–(aq)
[OH–] = 2 ( M) = M
pOH = –log[OH–] = -log = 2.08
1.0 x 10-14
0.0084
= 1.2 x 10–12 M
[H3O+] = Kw
[OH–] =

28. Methods for measuring the pH of an aqueous solution.
Figure 18.6
Methods for measuring the pH of an aqueous solution.
pH paper
pH meter
Copyright  The McGraw-Hill Companies, Inc. Permission required for reproduction or display.

29. Brønsted-Lowry Acid-Base Definition
An acid is a proton donor, any species that donates an H+ ion.
An acid must contain H in its formula.
A base is a proton acceptor, any species that accepts an H+ ion.
A base must contain a lone pair of electrons to bond to H+.
An acid-base reaction is a proton-transfer process.

30. Figure 18.7
Dissolving of an acid or base in water as a Brønsted-Lowry acid-base reaction.
Copyright  The McGraw-Hill Companies, Inc. Permission required for reproduction or display.
Lone pair binds H+
(acid, H+ donor)
(base, H+ acceptor)
Lone pair binds H+
(base, H+ acceptor)
(acid, H+ donor)

31. Conjugate Acid-Base Pairs
In the forward reaction:
NH3 accepts an H+ to form NH4+.
H2S + NH HS– + NH4+
H2S donates an H+ to form HS-.
In the reverse reaction:
H2S + NH HS– + NH4+
NH4+ donates an H+ to form NH3.
HS– accepts an H+ to form H2S.

32. Conjugate Acid-Base Pairs
H2S + NH HS– + NH4+
H2S and HS– are a conjugate acid-base pair:
HS– is the conjugate base of the acid H2S.
NH3 and NH4+ are a conjugate acid-base pair:
NH4+ is the conjugate acid of the base NH3.
A Brønsted-Lowry acid-base reaction occurs when an acid and a base react to form their conjugate base and conjugate acid, respectively.
acid1 + base base1 + acid2

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

34. Sample Problem 18.4
Identifying Conjugate Acid-Base Pairs
PROBLEM:
The following reactions are important environmental processes. Identify the conjugate acid-base pairs.
(b) H2O(l) + SO32–(aq) OH–(aq) + HSO3–(aq)
(a) H2PO4–(aq) + CO32– (aq) HPO42-–(aq) + HCO3–(aq)
PLAN:
To find the conjugate pairs, we find the species that donated an H+ (acid) and the species that accepted it (base). The acid donates an H+ to become its conjugate base, and the base accepts an H+ to become its conjugate acid.
SOLUTION:
base2
acid2
acid1
base1
(a) H2PO4–(aq) + CO32–(aq) HPO42–(aq) + HCO3–(aq)
The conjugate acid-base pairs are H2PO4–/HPO42– and CO32–/HCO3–.

35. Sample Problem 18.4
acid2
base2
acid1
base1
(b) H2O(l) + SO32–(aq) OH–(aq) + HSO3–(aq)
The conjugate acid-base pairs are H2O/OH– and SO32–/HSO3–.

36. Net Direction of Reaction
The net direction of an acid-base reaction depends on the relative strength of the acids and bases involved.
A reaction will favor the formation of the weaker acid and base.
H2S NH HS– NH4+
stronger acid
weaker acid
stronger base
weaker base
This reaction favors the formation of the products.

37. Strengths of conjugate acid-base pairs.
Figure 18.8
Strengths of conjugate acid-base pairs.
Copyright  The McGraw-Hill Companies, Inc. Permission required for reproduction or display.
The stronger the acid is, the weaker its conjugate base. When an acid reacts with a base that is farther down the list, the reaction proceeds to the right (Kc > 1).

38. 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:
We identify the conjugate acid-base pairs and consult figure 18.8 to see which acid and base are stronger. The reaction favors the formation of the weaker acid and base.
SOLUTION:
(a) H2PO4–(aq) + NH3(aq) HPO42–(aq) + NH4+(aq)
stronger acid
weaker acid
stronger base
weaker base
The net direction for this reaction is to the right, so Kc > 1.

39. Sample Problem 18.5
(b) H2O(l) HS–(aq) OH–(aq) H2S(aq)
stronger acid
weaker acid
stronger base
weaker base
The net direction for this reaction is to the left, so Kc < 1.

40. Sample Problem 18.6
Using Molecular Scenes to Predict the Net Direction of an Acid-Base Reaction
PROBLEM:
Given that 0.10 M of HX (blue and green) has a pH of 2.88, and M HY (blue and orange) has a pH of 3.52, which scene best represents the final mixture after equimolar solutions of HX and Y- are mixed?
PLAN:
A stronger acid and base yield a weaker acid and base, so we have to determine the relative acid strengths of HX and HY to choose the correct molecular scene. The concentrations of the acid solutions are equal, so we can recognize the stronger acid by comparing the pH values of the two solutions.

41. Sample Problem 18.6
SOLUTION:
The HX solution has a lower pH than the HY solution, so HX is the stronger acid and Y– is the stronger base. Therefore the reaction of HX and Y– has a Kc > 1, which means the equilibrium mixture will contain more HY than HX.
Scene 1 has equal numbers of HX and HY, which could occur if the acids were of equal strength. Scene 2 shows fewer HY than HX, which would occur if HY were the stronger acid.
Scene 3 is consistent with the relative acid strengths, because it contains more HY than HX.

42. Solving Problems Involving Weak-Acid Equilibria
Problem-solving approach
Write a balanced equation.
Write an expression for Ka.
Define x as the change in concentration that occurs during the reaction.
Construct a reaction table in terms of x.
Make assumptions that simplify the calculation.
Substitute values into the Ka expression and solve for x.
Check that the assumptions are justified.

43. Solving Problems Involving Weak-Acid Equilibria
The notation system
Molar concentrations are indicated by [ ].
A bracketed formula with no subscript indicates an equilibrium concentration.
The assumptions
[H3O+] from the autoionization of H2O is negligible.
A weak acid has a small Ka and its dissociation is negligible. [HA] ≈ [HA]init.

44. Sample Problem 18.7
Finding 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:
We start with the balanced dissociation equation and write the expression for Ka. We assume that [H3O+] from H2O is negligible and use the given pH to find [H3O+], which equals [PAc–] and [HPAc]dissoc. We assume that [HPAc] ≈ [HPAc]init because HPAc is a weak acid.
SOLUTION:
HPAc(aq) + H2O(l) H3O+(aq) + PAc–(aq)
Ka =
[H3O+][PAc–]
[HPAc]

45. HPAc(aq) + H2O(l) H3O+(aq) + PAc–(aq)
Sample Problem 18.7
Concentration (M)
HPAc(aq) + H2O(l) H3O+(aq) + PAc–(aq)
Initial
0.12 -
Change -
– x
+ x +x
Equilibrium -
0.12 – x x
[H3O+] = 10–pH = 2.4 x 10–3 M which is >> 10–7 (the [H3O+] from water)
x ≈ 2.4 x 10–3 M ≈ [H3O+] ≈ [PAc–]
[HPAc] = x ≈ 0.12 M
(2.4 x 10–3) (2.4 x 10–3)
0.12
So Ka =
= 4.8 x 10–5
Checking the assumptions by finding the percent error in concentration:
= 4 x 10–3 % (< 5%; assumption is justified).
[H3O+] =
1 x 10–7 M
2.4 x 10–3 M
x 100
from H2O
x 100
[HPAc]dissoc =
2.4 x 10–3 M
0.12 M
= 2.0 % (< 5%; assumption is justified).

46. Sample Problem 18.8
Determining Concentration from Ka and Initial [HA]
PROBLEM:
Propanoic acid (CH3CH2COOH, which we simplify as HPr) is a carboxylic 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:
We write a balanced equation and the expression for Ka. We know [HPr]init but not [HPr] (i.e., the concentration at equilibrium). We define x as [HPr]dissoc and set up a reaction table. We assume that, since HPr has a small Ka value, it dissociates very little and therefore [HPr] ≈ [HPr]init.
SOLUTION:
HPr(aq) + H2O(l) H3O+(aq) + Pr−(aq)
Ka =
[H3O+][Pr−]
[HPr]

47. Sample Problem 18.8
HPr(aq) + H2O(l) H3O+(aq) + Pr–(aq)
Concentration (M)
Initial
0.10 -
Change - -x +x
Equilibrium - x x
Since Ka is small, we will assume that x << 0.10 and [HPr] ≈ 0.10 M.
Ka = 1.3x10–5 =
[H3O+][Pr–]
[HPr] ≈ x2
0.10
x ≈ √(0.10)(1.3 x 10–5)
= 1.1 x 10–3 M = [H3O+]
Check: [HPr]diss =
1.1 x 10–3 M
0.10 M
x 100 = 1.1% (< 5%; assumption is justified.)

48. Concentration and Extent of Dissociation
Percent HA dissociated =
[HA]dissoc
[HA]init
x 100
As the initial acid concentration decreases, the percent dissociation of the acid increases.
HA(aq) + H2O(l) H3O+(aq) + A–(aq)
A decrease in [HA]init means a decrease in [HA]dissoc = [H3O+] = [A–], causing a shift toward the products.
The fraction of ions present increases, even though the actual [HA]dissoc decreases.

49. Sample Problem 18.9
Finding the Percent Dissociation of a Weak Acid
PROBLEM: In 2011, researchers showed that hypochlorous acid (HClO)
generated by white blood cells kills bacteria. Calculate the percent dissociation of (a) 0.40 M HClO; (b) M HClO (Ka = 2.9 x 10–8).
PLAN: Percent dissociation of HClO =
[HClO]dissoc
[HClO]init
x 100
The value of [HClO]init is given. To calculate [HClO]dissoc, we write the balanced equation for the dissociation of HClO, set up a reaction table with x = [HClO]dissoc = [ClO–] = [H3O+], and use the expression for Ka. We assume that HClO dissociates to a very small extent since Ka is small (2.9 x 10–8).
SOLUTION: (a) Writing the balanced equation and the expression for Ka:
HClO(aq) + H2O(l) H3O+(aq) + ClO– (aq)
Ka =
[H3O+][ClO–]
[HClO]
= 2.9 x 10–8

50. HClO(aq) + H2O(l) H3O+(aq) + ClO–(aq) Concentration (M)
Sample Problem 18.9
HClO(aq) + H2O(l) H3O+(aq) + ClO–(aq)
Concentration (M)
Initial
0.40 -
Change - –x +x
Equilibrium -
0.40 – x x
Since Ka is small, we will assume that x << 0.40 and [HClO] ≈ 0.40 M. x2
0.40
Ka =
[H3O+][ClO–]
[HClO]
= 2.9 x 10–8 =
% Dissociation =
1.1 x 10–4 M
0.40 M
x 100 = 0.028% (< 5%; assumption is justified.)
= 1.1 x 10–4 M = [HClO]dissoc
x ≈ √
(0.40)(2.9 x 10–8)

51. (b) Performing the same calculations using [HClO]init = 0.035 M:
Sample Problem 18.9
(b) Performing the same calculations using [HClO]init = M: x2
0.035
Ka =
[H3O+][ClO–]
[HClO]
= 2.9 x 10–8 =
% Dissociation =
3.2 x 10–5 M
0.035 M
x 100 = 0.091% (< 5%; assumption is justified.)
= 3.2 x 10–5 M = [HClO]dissoc
x ≈ √
(0.035)(2.9 x 10–8)

52. Polyprotic Acids
A polyprotic acid is an acid with more than one ionizable proton. In solution, each dissociation step has a different value for Ka:
Ka1 =
[H3O+][H2PO4–]
[H3PO4]
= 7.2x10–3
H3PO4(aq) + H2O(l) H2PO4–(aq) + H3O+(aq)
Ka2 =
[H3O+][HPO42–]
[H2PO4–]
= 6.3x10–8
H2PO4–(aq) + H2O(l) HPO42–(aq) + H3O+(aq)
Ka3 =
[H3O+][PO43–]
[HPO42–]
= 4.2x10–13
HPO42–(aq) + H2O(l) PO43–(aq) + H3O+(aq)
Ka1 >> Ka2 >> Ka3
We usually neglect [H3O+] produced after the first dissociation.

53. Table 18.5 Successive Ka values for Some Polyprotic Acids at 25°C

54. Calculating Equilibrium Concentrations for a Polyprotic Acid
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:
We first write the dissociation equations and the associated Ka expressions. Since Ka1 >> Ka2, we can assume that the first dissociation produces almost all the H3O+. Also, since Ka1 is small, the amount of H2ASc that dissociates can be neglected. We set up a reaction table for the first dissociation, with x equal to [H2Asc]dissoc, and solve for [H3O+] and [HAsc-].
SOLUTION:
H2Asc(aq) + H2O(l) Hasc–(aq) + H3O+(aq)
Ka1 =
[HAsc–][H3O+]
[H2Asc]
= 1.0x10-5
Ka2 =
[Asc2–][H3O+]
[HAsc–]
= 5x10-12
HAsc–(aq) + H2O(l) Asc2–(aq) + H3O+(aq)

55. Sample Problem 18.10
Concentration (M)
H2Asc(aq) + H2O(l) HAsc–(aq) + H3O+(aq)
Initial
0.050 -
Change
– x - +x
Equilibrium
0.050 – x - x
Ka1 =
[HAsc–][H3O+]
[H2Asc]
= 1.0x10–5 = x2
0.050 – x ≈ x2
0.050
x = [H3O+] = [Asc–] = √(0.050)(1.0 x 10–5)
= 7.1x10–4 M
pH = –log[H3O+] = –log(7.1x10–4) = 3.15

56. Checking assumptions:
Sample Problem 18.10
Checking assumptions:
1. [H3O+]
from HAsc-
<< [H3O+]
from H2Asc
: For any second dissociation,
= √ (7.1 x 10–4)(5 x 10–12)
= 6 x 10–8 M
[H3O+]
from HAsc-
≈ √ [HAsc–](Ka2)
This is even less than [H3O+]
from H2O
, so the assumption is justifed.
2. [H2Asc]dissoc << [H2Asc]init:
7.1 x 10–4 M
0.050 M
x 100 = 1.4% (< 5%; assumption is justified).

57. Weak Bases
A Brønsted-Lowry base is a species that accepts an H+. For a weak base that dissolves in water:
B(aq) + H2O(l) BH+(aq) + OH–(aq)
The base-dissociation or base-ionization constant is given by:
[BH+][OH–]
[B]
Kb =
Note that no base actually dissociates in solution, but ions are produced when the base reacts with H2O.

58. Figure 18.9 Abstraction of a proton from water by the base methylamine.
Lone pair of N pair binds H+
Copyright  The McGraw-Hill Companies, Inc. Permission required for reproduction or display.

59. Table 18.6 Kb Values for Some Molecular (Amine) Bases at 25°C

60. 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 5.9x10–4. What is the pH of 1.5 M (CH3)2NH?
PLAN:
We start with the balanced equation for the reaction of the amine with H2O, remembering that it is a weak base. We then write the expression for Kb, set up a reaction table and solve for [OH–]. From [OH–] we can calculate [H3O+] and pH.
We make similar assumptions to those made for weak acids. Since Kb >> Kw, the [OH–] from H2O is negligible. Since Kb is small, we can assume that the amount of amine reacting is also small, so [(CH3)2NH] ≈ [(CH3)2NH]init.
SOLUTION:
(CH3)2NH(aq) + H2O(l) (CH3)2NH2+(aq) + OH–(aq)
Kb =
[(CH3)2NH2+][OH–]
[(CH3)2NH]

61. Sample Problem 18.11
(CH3)2NH(aq) + H2O(l) (CH3)2NH2+(aq) + OH–(aq)
Concentration (M)
Initial
1.5 -
Change -x - +x
Equilibrium
1.5 - x - x
Since Kb is small, x << 1.5 and 1.5 – x ≈ 1.5
Kb =
[(CH3)2NH2+][OH–]
[(CH3)2NH]
= 5.9 x 10–4 ≈ x2
1.5
x = [OH–] = 3.0 x 10–2 M
Check assumption:
3.0 x 10–2 M
1.5 M
x 100 = 2.0% (< 5%; assumption is justified).

62. Sample Problem 18.11
[H3O+] = Kw
[OH–] =
1.0x10–14
3.0x10–2
= 3.3 x 10–13 M
pH = –log (3.3 x 10–13) = 12.48

63. Anions of Weak Acids as Weak Bases
The anions of weak acids often function as weak bases.
A-(aq) + H2O(l) HA(aq) + OH-(aq)
Kb =
[HA][OH–]
[A–]
A solution of HA is acidic, while a solution of A- is basic.
HF(aq) + H2O(l) H3O+(aq) + F–(aq)
HF is a weak acid (much weaker than H3O+), so this equilibrium lies to the left.
[HF] >> [F–], but [H3O+]from HF >> [OH–] ;
the solution is therefore acidic.
from H2O

64. F–(aq) + H2O(l) HF(aq) + OH–(aq)
If NaF is dissolved in H2O, it dissolves completely, and F- can act as a weak base:
F–(aq) + H2O(l) HF(aq) + OH–(aq)
Equilibrium favors the formation of the weaker acid and base. Since HF (although a weak acid) is a stronger acid than H2O, this equilibrium also lies to the left.
[F–] >> [HF], but [OH–] >> [H3O+ ] ;
the solution is therefore basic.
from H2O
from F-

65. Ka and Kb for a Conjugate Acid-Base Pair
HA + H2O H3O+ + A-
A- + H2O HA + OH–
2H2O H3O+ + OH–
Kc for the overall equation = K1 x K2, so
[H3O+][A–]
[HA]
[HA][OH–]
[A–] x
= [H3O+][OH–]
Ka x Kb = Kw
This relationship is true for any conjugate acid-base pair.

66. 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 at 25ºC? Ka of acetic acid (HAc) is 1.8 x 10–5.
PLAN:
Sodium salts are soluble in water and acetate is the anion of HAc so it acts as a weak base. We write the base dissociation equation and the expression for Kb, and solve for [OH-]. We recall that any soluble ionic salt dissociates completely in solution, so [Ac–]init = 0.25 M.
SOLUTION:
Ac–(aq) + H2O(l) HAc(aq) + OH–(aq)
Kb =
[HAc][OH–]
[Ac–]

67. Sample Problem 18.12
Ac–(aq) + H2O(l) HAc(aq) + OH–(aq)
Concentration (M)
Initial
0.25 -
Change -x +x -
Equilibrium - x x
Kb of Ac- = Kw Ka =
1.0x10–14
1.8x10–5
= 5.6x10–10 M
Kb = 5.6x10-10 =
[HAc][OH–]
[Ac–] ≈ x2
0.25
so x = [OH–] = 1.2x10–5 M
Checking the assumption:
1.2x10–5
0.25
x 100 = 4.8x10-3% (< 5%; assumption is justified)

68. Sample Problem 18.12
[H3O+] = Kw
[OH–] =
1.0x10–14
1.2x10–5
= 8.3x10–10 M
pH = – log (8.3x10–10) = 9.08

69. Acid Strength of Nonmetal Hydrides
For nonmetal hydrides (E-H), acid strength depends on:
the electronegativity of the central nonmetal (E), and
the strength of the E-H bond.
Across a period, acid strength increases.
Electronegativity increases across a period, so the acidity of E-H increases.
Down a group, acid strength increases.
The length of the E-H bond increases down a group and its bond strength therefore decreases.

70. The effect of atomic and molecular properties on
Figure 18.10
The effect of atomic and molecular properties on
nonmetal hydride acidity.
Copyright  The McGraw-Hill Companies, Inc. Permission required for reproduction or display.
Electronegativity increases, so acidity increases
Bond strength decreases, so acidity increases
6A(16)
H2O
H2S
H2Se
H2Te
7A(17) HF
HCl
HBr HI

71. Acid Strength of Oxoacids
All oxoacids have the acidic H bonded to an O atom.
Acid strength of oxoacids depends on:
the electronegativity of the central nonmetal (E), and
the number of O atoms around E.
For oxoacids with the same number of O atoms, acid strength increases as the electronegativity of E increases.
For oxoacids with different numbers of O atoms, acid strength increases with the number of O atoms.

72. Figure 18.11
The relative strengths of oxoacids.

73. Hydrated Metal Ions
Some hydrated metal ions are able to transfer an H+ to H2O. These metal ions will form acidic solutions.
Consider a metal ion in solution, Mn+:
Mn+(aq) + H2O(l) → M(H2O)xn+(aq)
If Mn+ is small and highly charged, it will withdraw enough e- density from the O-H bonds of the bound H2O molecules to release H+:
M(H2O)xn+(aq) + H2O(l) M(H2O)x-1OH(n-1)+(aq) + H3O+(aq)

74. Figure 18.12
The acidic behavior of the hydrated Al3+ ion.

75. Salts that Yield Neutral Solutions
A salt that consists of the cation of a strong base and the anion of a strong acid yields a neutral solution.
NaNO3
Na+ is the cation of NaOH, a strong base.
NO3– is the anion of HNO3, a strong acid.
This solution will be neutral, because neither Na+ nor NO3– will react with H2O to any great extent.

76. Salts that Yield Acidic Solutions
A salt that consists of the cation of a weak base and the anion of a strong acid yields an acidic solution.
NH4Cl
NH4 + is the cation of NH3, a weak base.
Cl– is the anion of HCl, a strong acid.
This solution will be acidic, because NH4+ will react with H2O to produce H3O+:
NH4+(aq) + H2O(l) NH3(aq) + H3O+(aq)

77. Salts that Yield Acidic Solutions
A salt that consists of a small, highly charged metal cation and the anion of a strong acid yields an acidic solution.
Fe(NO3)3
Fe3+ is a small, highly charged metal cation.
NO3 - is the anion of HNO3, a strong acid.
This solution will be acidic, because the hydrated Fe3+ ion will react with H2O to produce H3O+:
Fe(H2O)63+(aq) + H2O(l) Fe(H2O)5OH2+(aq) + H3O+(aq)

78. Salts that Yield Basic Solutions
A salt that consists of the anion of a weak acid and the cation of a strong base yields a basic solution.
CH3COONa
CH3COO– is the anion of CH3COOH, a weak acid.
Na+ is the cation of NaOH, a strong base.
This solution will be basic, because CH3COO– will react with H2O to produce OH–:
CH3COO–(aq) + H2O(l) CH3COOH(aq) + OH–(aq)

79. Sample Problem 18.13
Predicting Relative Acidity of Salt Solutions from Reactions of the Ions with Water
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:
We identify the cation and anion from the formula for each salt. Depending on an ion’s ability to react with water, the solution will be neutral (strong-acid anion with strong-base cation or small, highly charged metal cation), acidic (weak-base cation with strong-acid anion), or basic (weak-acid anion and strong-base cation).
SOLUTION:
(a) K+ is the cation of a strong base (KOH) while ClO4– is the anion of a strong acid (HClO4). This solution will be neutral.

80. Sample Problem 18.13
(b) Na+ is the cation of a strong base (NaOH) while the benzoate anion (C6H5COO–) is the anion of a weak acid (benzoic acid). The benzoate ion will react with H2O to produce OH– ions:
C6H5COO–(aq) + H2O(l) C6H5COOH(aq) + OH– (aq)
This solution will be basic.
(c) NO3– is the anion of a strong acid (HNO3) and will not react with H2O to any great extent. Cr3+ is a small metal cation with a fairly high charge density. It will become hydrated and the hydrated ion will react with H2O to form H3O+ ions:
Cr(H2O)63+(aq) + H2O(l) Cr(H2O)5OH2+(aq) + H3O+(aq)
This solution will be acidic.

81. Salts of Weakly Acidic Cations and Weakly Basic Anions
If a salt that consists of the cation of a weak base and the anion of a weak acid, the pH of the solution will depend on the relative acid strength or base strength of the ions.
NH4CN
NH4+ is the cation of a weak base, NH3.
CN– is the anion of a weak acid, HCN.
NH4+(aq) + H2O(l) NH3(aq) + H3O+(aq)
CN–(aq) + H2O(l) HCN(aq) + OH– (aq)

82. The reaction that proceeds farther to the right determines the pH of the solution, so we need to compare the Ka of NH4+ with the Kb of CN–.
Ka of NH4+ = Kw
Kb of NH3
1.0x10–14
1.76x10–5 =
= 5.7x10–10
Kb of CN– = Kw
Ka of HCN
1.0x10–14
6.2x10–10 =
= 1.6x10–5
Since Kb of CN– > Ka of NH4+, CN– is a stronger base than NH4+ is an acid. A solution of NH4CN will be basic.

83. Table 18.7 The Acid-Base Behavior of Salts in Water

84. 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, at 25°C is acidic, basic, or neutral.
PLAN:
Zn2+ is a small, highly charged metal cation, while HCOO– is the anion of a weak acid. Both will react with H2O, so to determine the acidity of the solution we must compare the Ka of the hydrated Zn2+ ion with the Kb of the HCOO– ion.
SOLUTION:
Zn(H2O)62+(aq) + H2O(l) Zn(H2O)5OH+(aq) + H3O+(aq)
HCOO–(aq) + H2O(l) HCOOH(aq) + OH–(aq)
Ka of Zn(H2O)62+ = 1x10–9 (from Appendix C.)
Kb of HCOO– = Kw
Ka of HCOOH
1.0x10–14
1.8x10–4 =
= 5.6x10–11
Ka for Zn(H2O)62+ >> Kb HCOO–, therefore the solution is acidic.

85. The Leveling Effect
All strong acids and bases are equally strong in water.
All strong acids dissociate completely to form H3O+, while all strong bases dissociate completely to form OH–.
In water, the strongest acid possible is H3O+ and the strongest base possible is OH–.
H2O exerts a leveling effect on any strong acid or base.

86. The Lewis Acid-Base Definition
A Lewis base is any species that donates an electron pair to form a bond.
A Lewis acid is any species that accepts an electron pair to form a bond.
The Lewis definition views an acid-base reaction as the donation and acceptance of an electron pair to form a covalent bond.

87. Lewis Acids and Bases
A Lewis base must have a lone pair of electrons to donate.
Any substance that is a Brønsted-Lowry base is also a Lewis base.
A Lewis acid must have a vacant orbital (or be able to rearrange its bonds to form one) to accept a lone pair and form a new bond.
Many substances that are not Brønsted-Lowry acids are Lewis acids.
The Lewis definition expands the classes of acids.

88. Electron-Deficient Molecules as Lewis Acids
B and Al often form electron-deficient molecules, and these atoms have an unoccupied p orbital that can accept a pair of electrons:
BF3 accepts an electron pair from ammonia to form a covalent bond.

89. Lewis Acids with Polar Multiple Bonds
Molecules that contain a polar multiple bond often function as Lewis acids:
The O atom of an H2O molecule donates a lone pair to the S of SO2, forming a new S‒O σ bond and breaking one of the S‒O p bonds.

90. Metal Cations as Lewis Acids
A metal cation acts as a Lewis acid when it dissolves in water to form a hydrated ion:
The O atom of an H2O molecule donates a lone pair to an available orbital on the metal cation.

91. Figure 18.13
The Mg2+ ion as a Lewis acid in chlorophyll.

92. 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:
We examine the formulas to see which species accepts the electron pair (Lewis acid) and which donates it (Lewis base) in forming the adduct.
SOLUTION:
(a) The H+ ion accepts the electron pair from OH–. H+ is the Lewis acid and OH– is the Lewis base.
(b) BCl3 accepts an electron pair from Cl–. Cl– is the Lewis base and BCl3 is the Lewis acid.
An O atom from each H2O molecule donates an electron pair to
K+. H2O is therefore the Lewis base, and K+ is the Lewis acid.