1. Unit 9 (Chp 16): Acid-Base Equilibria (Ka, Kb, Kw, pH, pOH)
Chemistry, The Central Science, 10th edition
Theodore L. Brown; H. Eugene LeMay, Jr.; and Bruce E. Bursten
Unit 9 (Chp 16): Acid-Base Equilibria (Ka, Kb, Kw, pH, pOH)
John D. Bookstaver
St. Charles Community College
St. Peters, MO
 2006, Prentice Hall, Inc.

2. HCl + H2O  H3O+ + Cl– NH3 + H2O  NH4+ + OH– + – + – O O Cl Cl
Acid: proton (H+) donor
Base: proton (H+) acceptor
NH3 + H2O  NH OH– H + – H N O N O H H H H H H H H

3. H Acid: have a removable (acidic) proton. HCl NH4+ CH3COOH
Base:
have a pair of nonbonding electrons.

4. HCO3−
HSO4−
H2O
amphoteric:
can be both acid & base
Amino Acids:
Tryptophan

5. Conjugate Acid-Base Pairs
Reactions of acids and bases always yield …conjugate acids & bases.
(differ by only a proton,H+)

6. (a)  List the conjugate base of each of the
following species:
HClO4
H2S
PH4+
HCO3–
(b) List the conjugate acid of each of the
CN–
SO42–
H2O
ClO4–
HS–
PH3
CO32–
HW p. 713 #18, 20
HCN
HSO4–
H3O+
H2CO3

7. Strong Acids Recall the six Strong Acids are…
HCl, HBr, HI, HNO3 , H2SO4 , and HClO4 .
strong electrolytes
(completely ionized)
(exist totally as ions in aqueous solution)
HNO3 + H2O  H3O+ + NO3–
For the monoprotic strong acids,
[acid] = [H3O+]
0.50 M HCl means… [H3O+] = 0.50 M

8. Strong Bases Strong Bases are the soluble hydroxides (OH–) of…
Group 1 (Li,Na,K,Rb,Cs) and Ca2+, Sr2+, Ba2+.
strong electrolytes (completely dissociated into ions in aqueous solution).
NaOH  Na+ + OH–
and accept H’s easily in aqueous solution.
OH– + HA  H2O + A–

9. Acid/Base Strength p. 674 Strong acids completely ionize in water.
conjugates do not function as bases.
Weak acids/bases only partially ionize.
conjugates are weak.
Strong bases completely dissociate & accept H+.
conjugates do not function as acids.

10. Acid/Base Strength The equilibrium will favor the direction that
moves the proton away from stronger acid.
(OR…the stronger base gets the proton)
HCl(aq) + H2O(l)  H3O+(aq) + Cl−(aq)
HCl is stronger acid than H3O+, and equilibrium
lies so far to the right K is not measured (K>>>1).
Which side is favored?
HC2H3O2(aq) + H2O(l) 
H3O+(aq) + C2H3O2−(aq)
H3O+ is a stronger acid than HC2H3O2,
equilibrium favors left side reactants (K<1).
HW p. 714 #26

11. Factors Affecting Acid Strength
larger X in H–X
p. 703
stronger acid
stronger acid
(bond is longer, weaker)
(greater ∆EN)
more polar H–X bond
(weaker, likely to break and lose H+)

12. Factors Affecting Acid Strength
p. 705
The more O’s the more polar (weaker) the H-O bond the more acidic (more likely to lose H+)
HW p. 716 #93

13. Autoionization of Water
As we have seen, water is amphoteric.
In pure water, a few molecules act as bases and a few act as acids.
Write the K exp. for the autoionization of water:
H2O(l) + H2O(l) 
H3O+(aq) + OH−(aq)
Kw = [H3O+] [OH−]
= 1.0  10–14 (at 25oC)
This special equilibrium constant is called the ion-product constant for water, Kw.

14. pH pH = −log [H+] or pH = −log [H3O+] H2O(l) + H2O(l) 
H3O+(aq) + OH−(aq)
In pure water, [H3O+] = [OH−]
Kw = [H3O+] [OH−] = 1.0  10−14
x2 = 1.0  10−14
[H3O+] = 1.0  10−7 M
pH = −log(1.0  10−7) = 7.00
Acids pH <7 b/c higher [H3O+], 1 x 10–(<7)
Bases pH >7 b/c lower [H3O+], 1 x 10–(>7)
(in pure water)

15. [H3O+] & [OH–] =
Acidic
Neutral
Basic

16. Other “p” Scales
The “p” in pH means to take the negative log of the quantity (in this case, hydrogen ions).
Some similar examples are
−log [H3O+] + −log [OH−] = −log Kw = 14.00
pOH = −log [OH–]
[H3O+] [OH−] = Kw = 1.0  10−14
pH pOH = pKw = 14
(given on exam)

17. Neutral [H+] pH pOH [OH–] Kw = [H+]x[OH–] 1 x 10–14 1 x 10–14
0.0
14.0
1 x 10–14
1 x 10–1
1.0
13.0
1 x 10–13
1 x 10–2
2.0
12.0
1 x 10–12
1 x 10–3
3.0
11.0
1 x 10–11
1 x 10–4
4.0
10.0
1 x 10–10
1 x 10–5
5.0
9.0
1 x 10–9
1 x 10–6
6.0
8.0
1 x 10–8
1 x 10–7
7.0
Kw = [H+]x[OH–]
1 x 10–14
1 x 10–14
1 x 10–14
1 x 10–14
1 x 10–14
1 x 10–14
1 x 10–14
Neutral
1 x 10–14
1 x 10–14
1 x 10–14
1 x 10–14
1 x 10–14
1 x 10–14
1 x 10–14
1 x 10–14

18. Neutral [H+] pH pOH [OH–] Kw = [H+]x[OH–] 1 x 10–14 1 x 10–14
0.0
14.0
1 x 10–14
1 x 10–1
1.0
13.0
1 x 10–13
1 x 10–2
2.0
12.0
1 x 10–12
1 x 10–3
3.0
11.0
1 x 10–11
1 x 10–4
4.0
10.0
1 x 10–10
1 x 10–5
5.0
9.0
1 x 10–9
1 x 10–6
6.0
8.0
1 x 10–8
1 x 10–7
7.0
Kw = [H+]x[OH–]
1 x 10–14
1 x 10–14
1 x 10–14
1 x 10–14
1 x 10–14
1 x 10–14
1 x 10–14
Neutral
1 x 10–14
1 x 10–14 ? ? ?
1 x 10–14
1 x 10–14
1 x 10–14
1 x 10–14
1 x 10–14
1 x 10–14

19. Neutral [H+] pH pOH [OH–] Kw = [H+]x[OH–] 1 x 10–14 1 x 10–14
0.0
14.0
1 x 10–14
1 x 10–1
1.0
13.0
1 x 10–13
1 x 10–2
2.0
12.0
1 x 10–12
1 x 10–3
3.0
11.0
1 x 10–11
1 x 10–4
4.0
10.0
1 x 10–10
1 x 10–5
5.0
9.0
1 x 10–9
1 x 10–6
6.0
8.0
1 x 10–8
1 x 10–7
7.0
Kw = [H+]x[OH–]
1 x 10–14
1 x 10–14
1 x 10–14
1 x 10–14 ? ? ?
1 x 10–14
1 x 10–14
1 x 10–14
Neutral
1 x 10–14
1 x 10–14
1 x 10–14
1 x 10–14
1 x 10–14
1 x 10–14
1 x 10–14
1 x 10–14

20. pH pOH [H+] [OH–] Calculations
on equation sheet
on equation sheet
pH = –log[H+]
pOH = –log[OH–]
10–pH = [H+]
10–pOH = [OH–]
NOT
NOT
on equation sheet
Kw = [H+] [OH–] = 1.0 x 10–14
pH + pOH = 14
on equation sheet

21. OH– H+ pH pOH pH & pOH Calculations [OH–] = 1 x 10–14 [H+]
[OH–] = 10–pOH
pOH = –log[OH–]
pH = –log[H+]
pOH = 14 – pH pH
pOH
pH = 14 – pOH

22. How Do We Measure pH? Demo Indicators Litmus paper (less accurate)
Demo – rainbow reaction (aDaD)
Litmus paper (less accurate)
blue pH > 8 (basic) red pH < 5 (acidic)

23. How Do We Measure pH?
For more accurate measurements, one uses a pH meter, which measures the voltage in the solution by tracking the number of H+ particles that contact the device.

24. I Think I Know This Stuff.
1) Which one of the following
solutions is the most basic?
A) soap
B) milk
C) vinegar
D) ammonia
E) tomato juice
F) blood
pH ≈ 10
pH ≈ 6.5
pH ≈ 2.5
pH ≈ 11
pH ≈ 4.0
pH ≈ 7.4 24

25. I Do Know This Stuff! given below. Which one of the
2) The [H3O+] for four solutions is
given below. Which one of the
solutions is the most acidic?
A) 1 x 10–3M
B) 1 x 10–7M
C) 1 x 10–9M
D) 1 x 10–14M 25

26. Can I Calculate This Stuff?
3) What is the pH of a solution with a [H+] concentration of x 10–11M ?
A) 12.93
B) 10.60
C) 8.92
D) 5.50 26

27. I Can Calculate This Stuff!
4) What is the pH of a solution with a
[OH–] concentration of x 10–3M ?
A) 2.82
B) 11.18
C) 6.67 x 10–12
D) 1.50 x 10–12
pOH = –log[OH–]
pH = 14 – pOH OR
[H+] = 1 x 10–14
[OH–]
WS Acids & Bases 1 #1-3
pH = –log[H+] 27

28. Dissociation Constants (Ka , Kb)
For a generalized acid dissociation,
the equilibrium expression would be
called the acid-dissociation constant, Ka.
HA(aq) + H2O(l) 
H3O+(aq) + A−(aq)
[H3O+] [A−]
[HA]
Ka =
on equation sheet
pKa = −log Ka
on equation sheet

29. (reactant dissociates more to products)
The greater the Ka, the stronger the acid.
(reactant dissociates more to products)
Weak
(p. 685)
Acids

30. Calculating Ka from the pH
The pH of a 0.10 M solution of formic acid, HCOOH, at 25oC is 2.38.
Calculate Ka for formic acid at 25oC.
We know:
HCOOH(aq) + H2O(l)  H3O+(aq) + HCOO–(aq)
[H3O+] [HCOO−]
[HCOOH]
Ka =
We need: all 3 concentrations at equilibrium.
We are given:
[HCOOH]initial = 0.10 M
get [H3O+]eq from pH.

31. Calculating Ka from the pH+
HCOOH
H2O
H3O+
HCOO− I
0.10 M
0 M C E M
pH = −log [H3O+]
2.38 = −log [H3O+]
−2.38 = log [H3O+]
10−2.38 = [H3O+]
[H3O+] = [HCOO−]
(due to 1:1 mol ratio)
[H3O+]eq = M

32. Calculating Ka from the pH+
HCOOH
H2O
H3O+
HCOO− I
0.10 M
0 M C
−0.0042 E
0.10 − = M
= 0.10 M
(0.0042)2
(0.10)
Ka =
[H3O+] [HCOO−]
[HCOOH]
Ka =
Ka = 1.8  10−4

33. Calculating Percent Ionization
[H3O+]eq
[HA]in
NOT on equation sheet
% ionization =  100%
HCOOH(aq) + H2O(l)  H3O+(aq) + HCOO–(aq) +
HCOOH
H2O
H3O+
HCOO− I
0.10 M
0 M C
−0.0042 E
0.10 − = M
= 0.10 M
0.0042
0.10
% ionization =  100% =
4.2%

34. HC2H3O2(aq) + H2O(l)  H3O+(aq) + C2H3O2–(aq)
Calculating pH from Ka
Calculate the pH of a 0.30 M solution of acetic acid, HC2H3O2 (Ka = 1.8  10−5 at 25oC).
HC2H3O2(aq) + H2O(l)  H3O+(aq) + C2H3O2–(aq)
[H3O+] [C2H3O2−]
[HC2H3O2]
Ka = +
HC2H3O2
H2O
H3O+
C2H3O2− I
0.30 M C E –x +x +x
0.30 – x x x
 0.30 M
(x is negligible compared to b/c K <<<1)

35. Calculating pH from Ka [H3O+] [C2H3O2−] [HC2H3O2] Ka = (x)2 (0.30)
1.8  10−5 =
(1.8  10−5) (0.30) = x2
5.4  10−6 = x2
2.3  10−3 = x
HW p. 715 #56, 62ab
pH = −log [H3O+]
pH = −log (2.3  10−3)
pH =
WS Acids & Bases 1 #7-9
2.64

36. Weak Bases
Bases accept H+ or produce OH– ions with H2O.

37. Dissociation Constants
For a generalized base dissociation,
the equilibrium expression would be OR
B(aq) + H2O(l) 
BH+(aq) + OH−(aq)
[BH+] [OH−]
[B]
Kb =
where Kb is the base-dissociation constant.
B–(aq) + H2O(l) 
HB(aq) + OH−(aq)

38. Kb can be used to find [OH−], then pOH, & pH.
Weak
(p. 694)
Bases

39. Calculating pH of Basic Solutions
What is the pH of a 0.15 M solution of NH3?
NH3(aq) + H2O(l) 
NH4+(aq) + OH−(aq)
[NH4+] [OH−]
[NH3]
Kb =
Kb = 1.8  10−5

40. Calculating pH of Basic Solutions
Given:
Kb = 1.8  10−5
NH3
H2O
NH4+
OH− I
0.15 M
0 M C E +

41. Calculating pH of Basic Solutions
ICE:
Kb = 1.8  10−5
NH3
H2O
NH4+
OH− I
0.15 M
0 M C –x +x E
0.15 – x x +
 0.15 M
b/c K <<1

42. Calculating pH of Basic Solutions
[NH4+] [OH−]
[NH3]
Kb =
Kb = 1.8  10−5
(x)2
(0.15)
1.8  10−5 =
(1.8  10−5) (0.15) = x2
2.7  10−6 = x2
1.6  10−3 = x2
[OH−] = 1.6  10−3 M

43. Calculating pH of Basic Solutions
[OH−] = 1.6  10−3 M
Therefore,
pOH = −log [OH−]
pOH = −log (1.6  10−3)
pOH = 2.80
pH + pOH = 14
pH =
HW p. 716 #76, 78
pH = 11.20

44. Ka and Kb Ka and Kb are related in this way: Ka  Kb = Kw Therefore:
If you know one of them,
you can calculate the other.
HW p. 716 #82, 84a

45. Acidic/Basic Ions with Water
ANIONS are bases (raise pH).
react with water in hydrolysis reaction to form OH− and the conjugate acid:
X−(aq) + H2O(l) 
HX(aq) + OH−(aq)
anion base
conj. acid
CATIONS with acidic protons (like NH4+) will lower pH.
NH4+(aq) + H2O(l) 
NH3(aq) + H3O+(aq)

46. Acidic/Basic Ions with Water
lower
Most metal CATIONS (Fe3+, Cu2+) _____ pH of H2O.
Makes O-H bond (of H2O) more polar making water more acidic (lose H+).
Cu+(aq) + H2O(l) 
CuOH(s) + H+(aq)
Greater charge and smaller size make a cation more acidic.

47. Effect of Cations and Anions
Na+
Zn2+
Al3+
Ca2+
Consider the size and charge of each metal ion and its effect on the pH of solution.

48. Cations: Anions: NH4+(aq) + H2O(l)  NH3(aq) + H3O+(aq) (acidic, ↓pH)
(donates H+)
NH4+(aq) + H2O(l)  NH3(aq) + H3O+(aq)
Cations:
(acidic, ↓pH)
Cu+(aq) + H2O(l)  CuOH(s) + H+(aq)
Most metal cations will lower pH (acidic).
(Fe3+, Zn2+, …) (take OH– from H2O to lose H+)
But, metal cations of strong base CAN’T affect pH.
(Li+,Na+,K+,Ca2+,Ba2+) (can’t take OH– from H2O)
Anions:
(basic, ↑pH)
F−(aq) + H2O(l)  HF(aq) + OH−(aq)
Most anions will increase pH (basic) (F–, HCO3–, NO2–) (take H+ from H2O)
But, anions from conj. base of a strong acid CAN’T affect pH. (Cl–, HSO4–, NO3–) (can’t take H+ from H2O)