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.
H
H
H
–
O
O
Cl
+
H
Cl
H
H
HCl + H2O  H3O+ + Cl–
 Acid: proton (H+) donor
 Base: proton (H+) acceptor
NH3 + H2O  NH4+ + OH–
H
H
H
N
H
O
H
H
N
H
H
–
+
H
O
H
Acid: have a removable (acidic) proton.
HCl
NH4+
CH3COOH
H
Cl–
:NH3
CH3COO–
Base: have a pair of nonbonding electrons.
amphoteric: can be both acid & base
Amino Acids:
Tryptophan
HCO3−
−
HSO4
H 2O
Conjugate Acid-Base Pairs
• Reactions of acids and bases always yield
…conjugate bases and acids.
(a)
List the conjugate base of each of the
following species:
HClO4
ClO4–
HW p. 713
H2S
HS–
#18, 20
PH4+
PH3
HCO3–
CO32–
(b) List the conjugate acid of each of the
following species:
CN–
HCN
SO42–
HSO4–
H 2O
H 3O +
HCO3–
H2CO3
Strong Acids
• Recall the six Strong Acids are…
HCl, HBr, HI, HNO3 , H2SO4 , and HClO4 .
• strong electrolytes
(completely dissociated)
(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
Strong Bases
• Strong Bases are the soluble hydroxides
(OH–) of…
Group 1 (Li,Na,K,Rb,Cs) and Ca2+, Sr2+, Ba2+.
• strong electrolytes (dissociate completely
into ions in aqueous solution).
NaOH  Na+ + OH–
• and accept H’s easily in aqueous solution.
OH– + HA  H2O + A–
p. 674 Acid/Base
Strength
•Strong acids completely
dissociate in water.
conjugates do not
function as bases.
•Weak acids/bases only
partially dissociate.
conjugates are weak.
•Strong bases completely
dissociate & accept H’s.
conjugates do not
function as acids.
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,
HW p. 714
equilibrium favors left side reactants (K<1).
#26
Factors Affecting Acid Strength
p. 703
stronger acid
larger
X in
H–X
(bond is
longer,
weaker)
stronger acid
(greater ∆EN)
more polar H–X bond
(weaker, likely to break and lose H+)
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
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.
pH
pH = −log [H+]
H2O(l) + H2O(l)
or
pH = −log [H3O+]
 H3O+(aq) + OH−(aq)
In pure water, [H3O+] = [OH−]
Kw = [H3O+] [OH−] = 1.0  10−14
[H3O+] = √(1.0  10−14) = 1.0  10−7
•In pure water, pH = −log(1.0  10−7) = 7.00
•Acids have higher [H3O+], so pH <7.
•Bases have lower [H3O+], so pH >7.
[H3O+] & [OH–]
=
Acidic
Neutral
Basic
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
pOH = −log [OH–]
−log [H3O+] + −log [OH−] = −log Kw = 14.00
pH
+
pOH
=
pKw = 14
[H+]
Neutral
pH
pOH
[OH–]
1 x 10–0
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
7.0
1 x 10–7
1 x 10–8
8.0
6.0
1 x 10–6
1 x 10–9
9.0
5.0
1 x 10–5
1 x 10–10
10.0
4.0
1 x 10–4
1 x 10–11
11.0
3.0
1 x 10–3
1 x 10–12
12.0
2.0
1 x 10–2
1 x 10–13
13.0
1.0
1 x 10–1
1 x 10–14
14.0
0.0
1 x 10–0
Kw = [H+]x[OH–]
1
1
1
1
x
x
x
x
10–14
10–14
10–14
10–14
1
1
1
1
1
1
x
x
x
x
x
x
10–14
10–14
10–14
10–14
10–14
10–14
1
1
1
1
1
x
x
x
x
x
10–14
10–14
10–14
10–14
10–14
[H+]
Neutral
pH
pOH
[OH–]
1 x 10–0
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
7.0
1 x 10–7
1 x 10–8
8.0
6.0
1 x 10–6
1 x 10–9
9.0?
5.0?
1 x 10
?–5
1 x 10–10
10.0
4.0
1 x 10–4
1 x 10–11
11.0
3.0
1 x 10–3
1 x 10–12
12.0
2.0
1 x 10–2
1 x 10–13
13.0
1.0
1 x 10–1
1 x 10–14
14.0
0.0
1 x 10–0
Kw = [H+]x[OH–]
1
1
1
1
x
x
x
x
10–14
10–14
10–14
10–14
1
1
1
1
1
1
x
x
x
x
x
x
10–14
10–14
10–14
10–14
10–14
10–14
1
1
1
1
1
x
x
x
x
x
10–14
10–14
10–14
10–14
10–14
[H+]
Neutral
pH
pOH
[OH–]
1 x 10–0
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
7.0
1 x 10–7
1 x 10–8
8.0
6.0
1 x 10–6
1 x 10–9
9.0
5.0
1 x 10–5
1 x 10–10
10.0
4.0
1 x 10–4
1 x 10–11
11.0
3.0
1 x 10–3
1 x 10–12
12.0
2.0
1 x 10–2
1 x 10–13
13.0
1.0
1 x 10–1
1 x 10–14
14.0
0.0
1 x 10–0
Kw = [H+]x[OH–]
1
1
1
1
x
x
x
x
10–14
10–14
10–14
10–14
1
1
1
1
1
1
x
x
x
x
x
x
10–14
10–14
10–14
10–14
10–14
10–14
1
1
1
1
1
x
x
x
x
x
10–14
10–14
10–14
10–14
10–14
–
[OH ]
+
[H ]
pH pOH
Calculations
on equation sheet
pH = –log[H+]
–pH
10
=
[H+]
Kw =
NOT
on equation sheet
pOH = –log[OH–]
–pOH
10
= [OH–]
on equation sheet
[H+] [OH–] = 1.0
pH + pOH = 14
on equation sheet
x 10
–14
NOT
pH & pOH Calculations
[H+] = 1 x 10–14
[OH–]
pOH = 14 – pH
pH
pH = 14 – pOH
–
OH
pOH = –log[OH–]
pH = –log[H+]
[H+] = 10–pH
+
H
[OH–] = 10–pOH
[OH–] = 1 x 10–14
[H+]
pOH
How Do We Measure pH?
Demo
 Litmus paper (less accurate)
• blue pH > 8 (basic)
red pH < 5 (acidic)
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.
I Think I Know This Stuff.
1) Which one of the following
solutions is the most basic?
A)
soap
pH ≈ 10
B)
milk
pH ≈ 6.5
C)
vinegar
pH ≈ 2.5
D)
ammonia
pH ≈ 11
E)
tomato juice
F)
blood
pH ≈ 4.0
pH ≈ 7.4
I Do Know This Stuff!
2) The [H3O+] for four solutions is
given below. Which one of the
solutions is the most acidic?
–3
10 M
A)
1x
B)
1 x 10–7M
C)
1x
–9
10 M
D)
1 x 10–14M
Can I Calculate This Stuff?
3) What is the pH of a solution with a [H+]
concentration of 2.50 x 10–11M ?
A)
12.93
B)
10.60
C)
8.92
D)
5.50
I Can Calculate This Stuff!
4) What is the pH of a solution with a
[OH–] concentration of 1.50 x 10–3M ?
A)
B)
C)
2.82
11.18
6.67 x 10–12
D) 1.50 x 10–12
WS Acids & Bases 1 #1-3
pOH = –log[OH–]
pH = 14 – pOH
OR
[H+] = 1 x 10–14
[OH–]
pH = –log[H+]
Dissociation Constants (Ka , Kb)
• For a generalized acid dissociation,
HA(aq) + H2O(l)
 H3O+(aq) + A−(aq)
the equilibrium expression would be
[H3O+] [A−] on equation
Ka =
sheet
[HA]
• called the acid-dissociation constant, Ka.
pKa = −log Ka
on equation
sheet
The greater the Ka, the stronger the acid.
(reactant dissociates more to products)
Weak (p. 685)
Acids
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.
HCOOH(aq) + H2O(l)  H3O+(aq) + HCOO–(aq)
+] [HCOO−]
[H
O
• We know: Ka = 3
[HCOOH]
• We need: all 3 concentrations at equilibrium.
• We can find [H3O+] from pH, and…
[H3O+] = [HCOO−] (due to 1:1 mol ratio)
Calculating Ka from the pH
[H3O+]eq = 0.0042 M
Given:
I
pH = −log [H3O+]
2.38 = −log [H3O+]
−2.38 = log [H3O+]
10−2.38 = [H3O+]
[HCOOH]
[H3O+]
[HCOO−]
0.10 M
0
0
0.0042 M
0.0042 M
C
E
Calculating Ka from the pH
Ka =
[H3
O +]
[HCOO−]
[HCOOH]
2
(0.0042)
Ka =
(0.10)
Ka = 1.8  10−4
[HCOOH]
[H3O+]
[HCOO−]
I
0.10 M
0M
0M
C
−0.0042
+0.0042
+0.0042
0.0042 M
0.0042 M
0.10 − 4.2  10−3
E
= 0.0958 = 0.10 M
Calculating Percent Ionization
[H3O+]eq
% ionization =
 100
[HA]in
HCOOH(aq) + H2O(l)  H3O+(aq) + HCOO–(aq)
I
C
E
[HCOOH]
0.10 M
−0.0042
0.0958 = 0.10 M
[H3O+]
0M
+0.0042
0.0042 M
[HCOO−]
0M
+0.0042
0.0042 M
0.0042
% ionization =
 100 = 4.2%
0.10
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−]
Ka =
[HC2H3O2]
I
[HC2H3O2]
[H3O+]
[C2H3O2−]
0.30 M
0
0
+x
+x
C
–x
E
0.30 – x
x
x
 0.30 M (assume x is negligible
compared to 0.30 b/c K <<<1)
Calculating pH from Ka
[H3O+] [C2H3O2−]
Ka =
[HC2H3O2]
2
(x)
1.8  10−5 =
(0.30)
(1.8  10−5) (0.30) = x2
5.4  10−6 = x2
2.3  10−3 = x
pH = −log [H3O+]
pH = −log (2.3  10−3)
WS Acids & Bases 1 #7-9pH = 2.64
Weak Bases
Bases accept H+ or produce OH– ions with H2O.
Dissociation Constants
• For a generalized base dissociation,
B(aq) + H2O(l)
 BH+(aq) + OH−(aq)
the equilibrium expression would be
[BH+] [OH−]
Kb =
[B]
where Kb is the base-dissociation constant.
OR
B–(aq) + H2O(l)
 HB(aq) + OH−(aq)
Kb can be used to find [OH−] and, through it, pH.
Weak (p. 694)
Bases
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−]
Kb =
[NH3]
Kb = 1.8  10−5
Calculating pH of Basic Solutions
Given:
Initial
Change
Equilibrium
Kb = 1.8  10−5
[NH3]
[NH4+]
[OH−]
0.15 M
0M
0M
Calculating pH of Basic Solutions
ICE:
Initial
Change
Equilibrium
Kb = 1.8  10−5
[NH3]
[NH4+]
[OH−]
0.15 M
0M
0M
–x
+x
+x
0.15 – x
x
x
 0.15 M
b/c K <<<1
Calculating pH of Basic Solutions
[NH4+] [OH−]
Kb =
[NH3]
Kb = 1.8  10−5
2
(x)
1.8  10−5 =
(0.15)
(1.8  10−5) (0.15) = x2
2.7  10−6 = x2
1.6  10−3 = x2
[OH−] = 1.6  10−3 M
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 + 2.80 = 14.00
pH = 11.20
HW p. 715
#56, 62ab
Ka and Kb
Ka and Kb are related in this way:
Ka  Kb = Kw
Therefore:
HW p. 716
If you know one of them,
#82, 84
you can calculate the other.
Reactions of 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)
anion base
 HX(aq) + OH−(aq)
conj. acid
• CATIONS with acidic protons
(like NH4+) will lower pH.
NH4+(aq) + H2O(l)  NH3(aq) + H3O+(aq)
Reactions of Ions with Water
• Most metal CATIONS (Fe3+, Cu2+) lower
_____ pH of H2O.
• Makes O-H bond (of H2O)
more polar making water
more acidic (lose H+).
Cu+(aq) + H2O(l)
 CuOH(aq) + H+(aq)
• Greater charge and
smaller size make a
cation more acidic.
Effect of Cations and Anions
Na+
Ca2+
Zn2+
Consider the size and charge of each metal
ion and its effect on the pH of solution.
Al3+
(donates H+)
Cations:
NH4+(aq) + H2O(l)
(acidic, ↓pH) Cu+(aq) + H O(l)
2
 NH3(aq) + H3O+(aq)

CuOH(aq) + H+(aq)
1. Most metal cations will lower pH (acidic).
(Fe3+, Zn2+, …) (take OH– from H2O to lose H+)
2. 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)
3. Anions from conj. base of a weak acid will increase
pH (basic). (F–, HCO3–, NO2–)
(take H+ from H2O)
4. Anions from conj. base of a strong acid CAN’T
affect pH. (Cl–, HSO4–, NO3–) (can’t take H+ from H2O)