1. Acid-Base Equilibria 熊同銘 tmhsiung@gmail.com
Chapter 16
Acid-Base Equilibria
熊同銘

2. Contents
Acids and Bases: A Brief Review
Bronsted–Lowry Acids and Bases
The Autoionization of Water
The pH Scale
Strong Acids and Bases
Weak Acids
Weak Bases
Relationship between Ka and Kb
Acid–Base Properties of Salt Solutions
Acid–Base Behavior and Chemical Structure
Lewis Acids and Bases

3. 1. Acids and Bases: A Brief Review
Arrhenius concept:
Acid: A substance that provides H+ in aqueous solution, e.g., HCl
Base (alkalis): A substance that provides OH– in aqueous solution, e.g., NaOH
H2O
HCl(g) → H+(aq) + Cl-(aq)
NaOH(s) → Na+(aq) + OH-(aq)
H2O
Na+(aq) + OH-(aq) + H+(aq) + Cl-(aq) → H2O(l) + Na+(aq) + Cl-(aq)
H+(aq) + OH-(aq) → H2O(l)
Arrhenius theory did not handle non OH– bases such as ammonia (NH3).

4. 2. Bronsted–Lowry Acids and Bases
Definition:
Acid: the substance act as H+ donor
Base: the substance act as H+ acceptor

5. Conjugate Acid–Base Pairs
* H2O is an amphiprotic substance
* H3O+(aq): hydronium ion, OH-(aq): hydroxyl ion

6. Relative Strengths of Acids and Bases
The stronger an acid, the weaker its conjugate base, and the stronger a base, the weaker its conjugate acid.
Acids above the line with H2O as a base are strong acids; their conjugate bases do not act as acids in water.
Bases below the line with H2O as an acid are strong bases; their conjugate acids do not act as acids in water.
The substances between the lines with H2O are conjugate acid–base pairs in water.

7. HCl(aq) + H2O(l) → H3O+(aq) + Cl(aq)
Continued
Stronger acids react with water to produce H3O+(aq) ions, and stronger bases react with water to produce OH-(aq) ions, a phenomenon known as the leveling effect.
For example, e.g., HI and HBr are leveled to the same acidic strength in H2O.
in every acid–base reaction, equilibrium favors transfer of the proton from the stronger acid to the stronger base to form the weaker acid and the weaker base.
HCl(aq) + H2O(l) → H3O+(aq) + Cl(aq)
H2O is a much stronger base than Cl, so the equilibrium lies far to the right (K >> 1)
CH3COOH(aq) + H2O(l) ⇌ H3O+(aq) + CH3COO–(aq)
CH3COO– is a stronger base than H2O, so the equilibrium favors the left side (K < 1).

8. Sample Exercise 16.1 Identifying Conjugate Acids and Bases
(a) What is the conjugate base of HClO4, H2S, PH4+, HCO3−?
(b) What is the conjugate acid of CN−, SO42− , H2O, HCO3−?
Solution
(a) ClO4−, HS−, PH3, and CO32−.
(b) HCN, HSO4−, H3O+, and H2CO3.

9. Sample Exercise 16.2 Writing Equations for Proton-Transfer Reactions
The hydrogen sulfite ion (HSO3−) is amphiprotic. Write an equation for the reaction of HSO3– with water (a) in which the ion acts as an acid and (b) in which the ion acts as a base. In both cases identify the conjugate acid–base pairs.
Solution
(a) HSO3− (aq) + H2O(l) SO32 − (aq) + H3O+(aq)
The conjugate pairs in this equation are HSO3− (acid) and SO32 − (conjugate base), and H2O (base) and H3O+ (conjugate acid).
(b) HSO3− (aq) + H2O(l) H2SO3(aq) + OH− (aq)
The conjugate pairs in this equation are H2O (acid) and OH− (conjugate base), and HSO3− (base) and H2SO3 (conjugate acid).

10. HSO4− (aq) + CO32 − (aq) SO42 − (aq) + HCO3− (aq)
Sample Exercise Predicting the Position of a Proton-Transfer Equilibrium
For the following proton-transfer reaction use Figure 16.4 to predict whether the equilibrium lies to the left (Kc < 1) or to the right (Kc > 1):
HSO4− (aq) + CO32 − (aq) SO42 − (aq) + HCO3− (aq)
Solution
The CO32 − ion appears lower in the right-hand column in Figure 16.4 and is therefore a stronger base than SO42 −. Therefore, CO32− will get the proton preferentially to become HCO3−, while SO42 − will remain mostly unprotonated. The resulting equilibrium lies to the right, favoring products (that is, Kc > 1):

11. 3. The Autoionization of Water
Water is amphoteric.
In pure water, a few molecules act as bases and a few act as acids.
This is referred to as autoionization.

12. Ion-product of water
Ion-product of water, KW:
KW = [H3O+][OH–]
At 25oC, KW = 1x10–14
KW applies to all aqueous solutions—acids, bases, salts, and nonelectrolytes—not just to pure water.

13. Sample Exercise 16.4 Calculating [H+] for Pure Water
Calculate the values of [H+] and [OH−] in a neutral aqueous solution at 25 ℃.
Solution
[H+][OH−] = (x)(x) = 1.0 × 10−14
x2 = 1.0 × 10−14
x = 1.0 × 10−7M = [H+] = [OH−]
It is a neutral aqueous solution.

14. Sample Exercise 16.5 Calculating [H+] from [OH−]
Calculate the concentration of H+(aq) in (a) a solution in which [OH−] is M, (b) a solution in which [OH−] is 1.8 × 10−9M. Note: In this problem and all that follow, we assume, unless stated otherwise, that the temperature is 25 ℃.
Solution
(a)
(b)
It is a basic solution.
It is a acidic solution.

15. p function: –log
pH = –log [H3O+] [H3O+] = 10–pH
pOH = –log [OH–] [OH–] = 10–pOH
Acidic solution: [H3O+] > [OH–] pH < pOH
Basic solution: [H3O+] < [OH–] pH > pOH
Aqueous solution at 25oC:
pKW = pH + pOH = 14.00
acidic solution: [H3O+] > 1.0×10–7 pH < 7.00
basic solution: [H3O+] < 1.0×10–7 pH > 7.00
neutral solution: [H3O+] = 1.0×10–7 pH = 7.00

16. Sample Exercise 16.6 Calculating pH from [H+]
Calculate the pH values for the two solutions of Sample Exercise 16.5.
Solution
(a) [H+] = 1.0 × 10−12 M
pH = −log(1.0 × 10−12) = −(−12.00) = 12.00
(b) [H+] = 5.6 × 10−6 M
pH = −log(5.6 × 10−6) = 5.25

17. [H+] = antilog (−3.76) = 10−3.76 = 1.7 × 10−4 M
Sample Exercise Calculating [H+] from pOH
A sample of freshly pressed apple juice has a pOH of Calculate [H+].
Solution
pH = − pOH
pH = − = 3.76
[H+] = antilog (−3.76) = 10−3.76 = 1.7 × 10−4 M

18. Measuring pH
Indicators, including litmus paper, are used for less accurate measurements; an indicator is one color in its acid form and another color in its basic form.
pH meters are used for accurate measurement of pH; electrodes indicate small changes in voltage to detect pH.

19. 5. Strong Acids and Strong Bases
The contribution due to the self-ionization of water can generally be ignored (unless the solution is extremely dilute), i.e., for strong acids and bases, dissociated completely. Therefore,
[H3O+]  CHCl CHCl: initial concentration of HCl
[OH–]  CNaOH CNaOH: initial concentration of NaOH.

20. Sample Exercise 16.8 Calculating the pH of a Strong Acid
What is the pH of a M solution of HClO4?
Solution
HClO4 is a strong acid, it is completely ionized, giving [H+] = [ClO4−] = M.
pH = −log(0.040) = 1.40

21. Sample Exercise 16.9 Calculating the pH of a Strong Base
What is the pH of (a) a M solution of NaOH, (b) a M solution of Ca(OH)2?
Solution
(a) [OH−] = M.
pOH = −log(0.028) = 1.55 pH = − pOH = 12.45
(b) [OH−] = 2 × ( M) = M.
pOH = −log(0.0022) = 2.66 pH = − pOH = 11.34

22. For extremely dilute solution of a strong acid and strong base, 1
For extremely dilute solution of a strong acid and strong base, 1.0 x 10–8 M HCl for example:

23. 6. Weak Acids
Comparing Strong and Weak Acids
Strong acids completely dissociate to ions.
Weak acids only partially dissociate to ions.

25. Equilibrium of monoprotic acid, HA
Assume the [H3O+] from autoionization of H2O is negligible, [A-]from HA ≈ [H3O+]equilibrium:
HA H2O  H3O+ + A–
CHA – x x x
Acid-dissociation constant
Therefore
5% rule:
CHA: [HA]originally
Assume CHA – x  CHA
* CHA – x  CHA, using:

26. Percent Ionization (A weak acid HA for example)
HA + H2O H3O+ + A-
Percent ionization =
[A-]from HA
[HA]originally
 100%
If the [H3O+] from autoionization of H2O is negligible,
[A-]from HA ≈ [H3O+]equilibrium:
Percent ionization =
[H+]equilibrium
[HA]originally
 100%

27. Effect of concentration on percent ionization

28. Sample Exercise 16.10 Calculating Ka from Measured pH
A student prepared a 0.10 M solution of formic acid (HCOOH) and found its pH at 25 ℃ to be Calculate Ka for formic acid at this temperature.
Solution
[H+] = 10–2.38 = 4.2 × 10–3 M
[A-]from HA ≈
[H3O+]equilibrium
(0.10 – 4.2 × 10–3)
≃ 0.10

29. Sample Exercise 16.11 Calculating Percent Ionization
As calculated in Sample Exercise 16.10, a 0.10 M solution of formic acid (HCOOH) contains 4.2 × 10–3 M H+(aq). Calculate the percentage of the acid that is ionized.
Solution
HA + H2O H3O+ + A-

30. HCN + H2O H3O+ + CN- Sample Exercise 16.12 Using Ka to Calculate pH
Calculate the pH of a 0.20 M solution of HCN. (Ka for HCN is 4.9 × 10–10)
Solution
HCN H2O H3O+ + CN-

31. Sample Exercise 16.13 Using the Quadratic Equation to Calculate pH and Percent Ionization
Calculate the pH and percentage of HF molecules ionized in a 0.10 M HF solution. (Ka for HF is 6.8 × 10–4)
Solution
Assume 0.10 – x ≃ 0.10
x = 8.2 × 10–3 M
(8.2 × 10–3/0.10) x 100% = 8.2% (>5%)
we should work the problem in standard quadratic form!

32. Continued
x2 = (0.10 − x)(6.8 × 10−4)
= 6.8 × 10−5 – (6.8 × 10−4)x
x2 + (6.8 × 10−4)x – 6.8 × 10−5 = 0
x = [H+] = [F–] = 7.9 × 10–3 M
pH = –log[H+] = 2.10

33. Polyprotic Acids

34. Diprotic acid Triprotic acid H2A + H2O  HA– + H3O+
HA– + H2O  A2– + H3O+
Triprotic acid
Phosphoric acid for example:
H3PO4 + H2O  H2PO4– + H3O+
H2PO4– + H2O  HPO42– + H3O+
HPO42– + H2O  PO43– + H3O+
Ionization constants for polyprotic acid progressively decrease:
Ka1 > Ka2 > Ka3 > …..
Except in very dilute solutions, essentially all of the H3O+ ions come from the first ionization step alone.

35. Sample Exercise 16.14 Calculating the pH of a Solution of a Polyprotic Acid
The solubility of CO2 in water at 25 ℃ and 0.1 atm is M. The common practice is to assume that all the dissolved CO2 is in the form of carbonic acid (H2CO3), which is produced in the reaction
CO2(aq) + H2O(l) H2CO3(aq)
What is the pH of a M solution of H2CO3? (H2CO3: Ka1 = 4.3 × 10–7 ; Ka2 = 5.6 × 10–11)
Solution

36. Continued
Assume
0.037 – x ≃

37. 7. Weak Bases

38. Equilibrium of monobasic base, B
Assume the [OH-] from autoionization of H2O is negligible, [HB+]from B ≈ [OH-]equilibrium:
B H2O  HB+ + OH–
CB – x x x
Base-dissociation constant
Therefore
5% rule:
CB: [B]originally
Assume CB – x  CB
* CB – x  CB, using:

39. Sample Exercise 16.15 Using Kb to Calculate OH−
Calculate the concentration of OH− in a 0.15 M solution of NH3.
(Kb for NH3 is 1.8 × 10–5)
Solution

40. Sample Exercise 16.16 Using pH to Determine the Concentration of a Salt
A solution made by adding solid sodium hypochlorite (NaClO) to enough water to make 2.00 L of solution has a pH of Calculate the number of moles of NaClO added to the water. (Kb for ClO– is 3.3 × 10–7)
Solution
pOH = – pH = – = 3.50 [OH–] = 10–3.50 = 3.2 × 10–4 M
0.62 mol of NaClO is the amount of the salt that was added to the 2L water.

41. 8. Relationship Between Ka and Kb For an conjugate acid-base pair:
pKa + pKb = pKW = at 25oC
* Many reference books only give tables of Ka values because Kb values can be found from them.

43. Sample Exercise 16.17 Calculating Ka or Kb for a Conjugate Acid–Base Pair
Calculate (a) Kb for the fluoride ion. (Ka for HF is 6.8 × 10−4)
(b) Ka for the ammonium ion. (Kb for NH3 is 1.8 ×10−5)
Solution
(a)
(b)

44. Diprotic acid H2A + H2O  HA– + H3O+ HA– + H2O  A2– + H3O+
For conjugate base:
A2– + H2O  HA– + OH–
HA– + H2O  H2A + OH–
Ka1 x Kb2 = Kw Ka2 x Kb1 = Kw

45. Triprotic acid Phosphoric acid for example:
H3PO4 + H2O  H2PO4– + H3O+
H2PO4– + H2O  HPO42– + H3O+
HPO42– + H2O  PO43– + H3O+
Ka1 x Kb3 = Kw Ka2 x Kb2 = Kw Ka3 x Kb1 = Kw

46. 9. Acid–Base Properties of Salt Solutions
Hydrolysis: The reaction between an ion and water to create H+ or OH–..
(1) NaCl(aq)  Na+(aq) + Cl–(aq) Neutral
Na+(aq) + H2O 
Cl–(aq) + H2O 
(2) NH4Cl(aq)  NH4+(aq) + Cl–(aq) Acidic
NH4+(aq) + H2O  NH3(aq) + H3O+(aq)
(3) CH3COONa(aq)  Na+(aq) + CH3COO–(aq) Basic
CH3COO–(aq) + H2O  CH3COOH(aq) + OH–(aq)
(4) CH3COONH4(aq)  NH4+(aq) + CH3COO– (aq) ????? X X X X

47. Anions
Anions of strong acids are neutral. For example, Cl– will not react with water, so OH– can’t be formed.
Anions of weak acids are conjugate bases, so they create OH– in water;
e.g., C2H3O2– + H2O ⇌ HC2H3O2 + OH–
Protonated anions from polyprotic acids can be acids or bases: If Ka > Kb, the anion will be acidic; if Kb > Ka, the anion will be basic.

48. Cations
Group I or Group II (Ca2+, Sr2+, or Ba2+) metal cations are neutral.
Polyatomic cations are typically the conjugate acids of a weak base; e.g., NH4+.
Transition and post-transition metal cations are acidic.

49. Hydrolysis of the hydrated metal ion
* Cations of small, highly charged metals are weakly acidic. Fe(H2O)63+(aq) + H2O(l)  Fe(H2O)5(OH)2+(aq) + H3O+(aq)

51. The pH of Salt Solutions
Type of Salt
Example
Ion That
Undergo
Hydrolysis
pH of
Solution
Cation from strong base/
Anion from strong acid
NaCl, KI, KNO3,
RbBr, BaCl2
None 7
Anion from weak acid
CH3COONa,
KNO2, NaOCl
Anion
>7
Cation from weak base/
NH4Cl, NH4NO3
Cation
<7
NH4NO2, NH4CN,
CH3COONH4
Anion and
<7 if Kb<Ka
7 if KbK a
>7 if Kb>Ka

52. Continued Type of Salt Example Ion That Undergo Hydrolysis pH of
Solution
Cation is small, highly charged/
Anion from strong acid
AlCl3, Fe(NO3)3
Hydrated
Cation
<7
Anion from weak acid
Al(CH3COO)3, CrF3
Anion and
Hydrated cation
<7 if Kb<Ka
7 if KbK a
>7 if Kb>Ka

53. Sample Exercise 16.18 Determining Whether Salt Solutions Are Acidic, Basic, or Neutral
Determine whether aqueous solutions of each of these salts are acidic, basic, or neutral: (a) Ba(CH3COO)2, (b) NH4Cl, (c) CH3NH3Br, (d) KNO3, (e) Al(ClO4)3.
Solution
(a) Ba2+ will not affect the pH, CH3COO− will hydrolyze to produce OH− ions, the solution is basic.
NH4+ will hydrolyze to produce H3O+, Cl− will not affect the pH, the solution is acidic.
CH3NH3+ will hydrolyze to produce H3O+, Br− will not affect the pH, the solution is acidic.
K+ will not affect the pH, NO3− will not affect the pH, the solution is neutral.
(e) Hydrated Al3+ will hydrolyze to produce H3O+, ClO4− will not affect the pH, the solution is acidic.

54. Sample Exercise 16.19 Predicting Whether the Solution of an Amphiprotic Anion Is Acidic or Basic
Predict whether the salt Na2HPO4 forms an acidic solution or a basic solution when dissolved in water.
Solution
H3PO4 + H2O  H2PO4– + H3O+ Ka1 = 7.5 x 10 –3
H2PO4– + H2O  HPO42– + H3O+ Ka2 = 6.2 x 10 –8
HPO42– + H2O  PO43– + H3O+ Ka3 = 4.2 x 10 –13
PO43–+ H2O  HPO42– + OH– Kb1 = KW/Ka3 =2.4 x 10 –2
HPO42– + H2O  H2PO4– + OH– Kb2 = KW/Ka2 =1.6 x 10 –7
H2PO4– + H2O  H3PO4 + OH– Kb3 = KW/Ka1 =1.3 x 10 –12
For the amphiprotic anion, HPO42–, Kb2 > Ka3, the solution is basic

55. 10. Acid–Base Behavior and Chemical Structure
Factors That Affect Acid Strength
H—A bond must be polarized with δ+ on the H atom and δ– on the A atom.
Bond strength: Weaker bonds can be broken more easily, making the acid stronger.
Stability of A–: More stable conjugate base means stronger acid.

56. Comparing binary acids of X the same group
Binary Acids HX
Comparing binary acids of X the same group
The smaller bond energy (the weaker H—X bond) the stronger acid.
*****
Molecule HF HCl HBr HI
BE (kJ/mol)
Anion radius (pm)
Ka 6.6x10–4 ~106 ~108 ~109
Acidity: HF < HCl < HBr < HI
*** HF(aq) is a weak acid
Other example:
Acidity: H2O < H2S < H2Se < H2Te

57. Comparing binary acids of X in the same period
The higher polarity of the bond (the larger ΔEN (electronegativity difference)), the stronger acid.
Small ΔEN has more covalent character
Large ΔEN has more ionic character
*****
Molecule CH4 NH3 H2O HF
ΔEN
Acidity: CH4 < NH3 < H2O < HF

58. Strengths of Oxoacids (H–O–EOn)
Comparing the EN of E
The larger EN (electronegativity) of E, the weaker H–O bond, the stronger acid.
Molecule H–OI H–OBr H–OCl EN
Ka 2.3x10–11 2.5x10–9 2.9x10–8
Acid strength HOI < HOBr < HOCl
More examples:
Acid strength: H2SeO3 < H2SO3
HBrO4 < HClO4

59. *O is the element has second higher electronegativity.
Comparing the n of On
The more number (n) of terminal O, the weaker protonated H–O bond, the stronger acid.
ClO– : hypoclorite (次氯酸根)
ClO2–: chlorite ion (亞氯酸根)
ClO3–: chlorate ion (氯酸根)
ClO4–: perchlorate ion (過氯酸根)
不要和Binary Acid弄混了。
*O is the element has second higher electronegativity.
More examples:
Ka1(H2SO3) < Ka1(H2SO4)

60. Carboxylic acids (R–COOH)
Carboxylic acids are organic acids containing the —COOH group.
Factors contributing to their acidic behavior:
The additional O attached to C draws electron density from O—H bond, increasing polarity.
Its conjugate base (carboxylate anion) has resonance forms to stabilize the anion.

61. Glossary and Definition
11. Lewis Acids and Bases
Glossary and Definition
Lewis acid: The species act as electron-pair acceptor.
Lewis base: The species act as electron-pair donor. 61

62. End of Chapter 16