CHM 1046: General Chemistry and
Qualitative Analysis
Unit 17:
Acid-Base Equilibria
Dr. Jorge L. Alonso
Miami-Dade College –
Kendall Campus
Miami, FL
Textbook Reference:
•Chapter 18
•Module # 6
Acids
and
Bases
Acid-Base Theories
Arrhenius
(1883)
NaOH
HCl
H2SO4
CO32BF3
NH3
H2O
Brønsted–Lowry (1923)
AlI3
Lewis (1923-38)
Acids
and
Bases
Acid-Base Definitions
• Svante Arrhenius (1883)
 Acid: Substance that, when dissolved in water,
{acid}
increases the concentration of hydrogen ions.
HX
H+ + X Base: Substance that, when dissolved in water,
increases the concentration of hydroxide ions.
MOH
M+ + OH Neutralization:
HX + MOH
acid
base
MX + HOH
salt
water
Acids
and
Bases
Acid-Base Definitions
• Brønsted–Lowry (1923)
 Acid: Proton (H+) donor.
… have a removable (acidic) proton which are donated to a bases.
HX
+
H2O
X-
H3O+
+
acid
base
conjugate b
 Base: Proton (H+) acceptor.
conjugate a.
…must have a pair of nonbonding electrons which accepts protons.
B
base
+
H2O
BH+ +
acid
conj. a
 Neutralization:
HX + B
acid
base
X+
conj. b
OH-
{base}
conj. b
BH+
conj. a
Acids
and
Bases
Strong Acids
HCl, HBr, HI, HNO3, H2SO4, HClO3, & HClO4
Are strong electrolytes and exist totally as ions in aqueous solution.
SOLUBILITY RULES: for Ionic Compounds (Salts)
• All salts containing the anions: NO3-, ClO3-, ClO4-, (C2H3O2-) are soluble.
• All Cl-, Br-, and I- are soluble except for Hg22+, Ag+, and Pb2+ salts.
• All SO42- are soluble except for Pb2+, Ba2+, and Sr2+.
Arrhenius:
Brønsted
–Lowry:
HX
HX
+
100%
H2O
H+(aq) +
100%
X-
X-(aq)
+
H3O+
acid
base
conj. b
conj. a
Strong acids have very weak conjugate bases
For the monoprotic strong acids, [acid]i = [H3O+] or [H+]
Acids
and
+
Bases
For strong acid: [1M HX]i = [1M H3O+] or [1M H ]
For weak acid: [1M HA]i > [0.1M H3O+] or [0.1M H+]
Strong Bases
SOLUBILITY RULES: for Ionic Compounds (Salts)
All OH- are insoluble except for IA metals, (NH4+), Ca2+, Ba2+ , & Sr2+ (heavy IIA).
Arrhenius:
these substances dissociate completely (100%) in aqueous solution.
MOH  M+ + OHBrønsted
–Lowry:
these substances accept protons from water to form very weak
conjugate acids.
Na+OH- + H2O  Na+ + HOH + OHbase
acid
conjugate acid
conjugate base
[OH-]
For the monobasic strong bases, [base]i =
[1M NaOH] = [1M OH-]
Acids
and
Bases
Review of Strong Acid & Bases
(according Brønsted–Lowry)
Strong acids have very weak conjugate bases
HX + H2O  H3O+ +
Xacid
base
conj. a
conj. b
Strong bases have very weak conjugate acids
MOH

M+
+ OHbase
conj. a
conj. b
Later on this course: Salts that are neutral
Salts of Strong acids & bases have very weak
conjugate acids & bases
MX + H2O  M+(aq)
+ X- (aq)
salt
conj. a
conj. b
Has no
affinity
for H+
Example:
HCl
Has no
affinity
for OH-
Example:
NaOH
Have no
affinity for
H+ or OH-
Acids
and
Bases
Example: NaCl
Concentrated 10M
Strong Acid
Strong Acid Dissociation
HX
HX
Concentrated 10M
Very Weak Acid
+
H3O
-
X
Weak Acid Dissociation
HA
Acids
and
Bases
Weak Acids
Reactions between acids and bases always yield their
corresponding conjugate bases and conjugate acids.
HA
+
H2O
↔
H3O+ +
A-
acid
base
conj. a
conj. b
Weak acids (HA) have very strong conjugate bases (A-)
H+
H+
Acids
and
Bases
•The weaker the acid, the stronger their conjugate base.
Weak Bases
B + HOH
BH+ + OH-
base
conj. a
acid
conj. b
Bases react with water to produce hydroxide ion.
• Weak bases have strong conjugate acids
H+
H+
•The weaker the base, the stronger their
conjugate acid.
Acids
and
Bases
conjugate
Acid / Base
Strength
HA +
acid
H2O ↔
base
H3O+ +
conj. a
Aconj. b
The stronger an
acid, the weaker its
conjugate base.
B + HOH
base acid
↔ BH+ + OHconj. a
conj. b
The stronger an
base, the weaker
Acids
and
its conjugate acid.
Bases
+
The Leveling
Effect of Water
Strong acids react with
water to produce H3O+,
the strongest acid that
can exist in an aqueous
solution.
Strong bases react with
water to produce OH-, the
strongest base that can
exist in an aqueous
solution.
+
Acids
and
Bases
Review of Weak Acid & Bases
Weak acids (HA) have very strong conjugate bases (A-)
HA + H2O
↔ H 3 O+ +
AHas
affinity
acid
base
conj. a conj. b
Example:
HF
Weak bases (B) have very strong conjugate acids (BH+)
B + HOH ↔
BH+ + OHHas
affinity
base
acids
conj. a
conj. b
Example:
NH3
for H+
Later on this course: Salts that are acidic or basic
Salts: (1) of strong base & weak acid:
MA  M+(aq) + A-(aq)
A- + HOH  HA + OH- (hydrolysis)
(2) of a weak base & strong acid:
BHX  BH+(aq) + X-(aq)
BH+ + HOH  B + H3O+ (hydrolysis)
(3) of a weak base & weak acid:
BHA ↔ BH+(aq) + A-(aq)
for OH-
Example:
NaF
Example:
NH4Cl
Acids
Example:and
Bases
NH4F
Lewis Acids and Bases (1923-38)
• Lewis acids are electron-pair acceptors (have empty valance orbitals).
+
+
• Lewis bases are electron-pair donors.
+
+
• Lewis acid-base neutralization reactions
+
{Lewis A-B}
..
..
+
:
:
H + O
.. H  H:O
..:H
Acids
and
Bases
Is Water an Acid or a Base?
The Autoionization of Water
• In pure water, a few water molecules act as
acids and a few act as bases.
H2O(l)
H2O(l) + H2O(l)
H+(aq) + OH−(aq)
H3O+(aq) + OH−(aq)
1 H3O+ and 1 OH- for every 10 million (107) H2O molecules
• This is referred to as autoionization.
Acids
and
Bases
{Movie}
Autoionization of Water
H2O(l) + H2O(l)
H3O+(aq) + OH−(aq)
• The equilibrium expression for this process is
Kw = [H3O+] [OH−] = 1.0  10−14
−14
Kw =
=
1.0

10
[H2O]2
• Kw is referred to as the ion-product constant
Acids
for water (@ 25°C).
and
Bases
What proportion of H2O molecules
+
dissociate into H3O & OH ?
• In pure water,
Kw = [H3O+] [OH−] = 1.0  10−14
• Because in pure water [H3O+] = [OH−],
Kw = [1.0  10−7 ] [1.0  10−7 ] = 1.0  10−14
1 H3O+ and 1 OH- for every 10 million (107) H2O molecules
Acids
and
Bases
pH
…..defined as the negative base-10
logarithm of the hydronium ion
concentration.
pH = −log [H3O+]
[1.0  10−7 ] = 7.0
Problem: Calculate pH when [H3O+] = 2.3 x 10-3 M
For pure H2O:
10x
log
base 10 log
log - 2.3 * 10^- 3
= 2.64
Problem: Calculate [H3O+] when pH = 2.3 ?
antilog
10x
log
-3
10^ -2.3 = 5.0 * 10
Acids
and
Bases
Other “p” Scales
• The “p” in pH tells us to take the negative log
of the quantity (in this case, hydrogen ions).
• Some examples of other “p” scales are:
 pH = −log [H3O+]
[H3O+] = 1 x 10-7
pH = 7
 pOH = −log [OH−]
[OH-] = 1 x 10-7
pOH = 7
 pKw = −log Kw
Kw = 1 x 10-14
pKw = 14
 pKa = −log Ka
Ka = 6.8 x 10-4
 pKb = −log Kb
-5
Kb = 1.8 x 10
pKa = 3.2
Acids
and
Bases
pKb = 4.7
pH and pOH equilibrium
in pure Water
• In pure water,
[H3O+] [OH−] = Kw
[1.0  10−7 ] [1.0  10−7 ] = 1.0  10−14
−log [H3O+] + −log [OH−] = −log Kw
pH + pOH = pKw
• Because in pure water [H3O+] = [OH−],
Kw = [1.0  10−7 ] [1.0  10−7 ] = 1.0  10−14
7
+
7
=
14
Acids
and
Bases
pH and pOH equilibrium
in Water to which
Acids & Bases are Added
Add acid
H3O
+
H3O+(aq) + OH−(aq)
H2O(l) + H2O(l)
H2O
[H3O+]
[OH−]
Kw = [1.0  10−7 ] [1.0  10−7 ] = 1.0  10−14
Kw = [1.0  10−6 ] [1.0  10−8 ] = 1.0  10−14
pH + pOH = pKw
Acids
6 + 8 = 14
and
Bases
pH and pOH equilibrium
in Water to which
Acids & Bases are Added
Add base
-
OH
H3O+(aq) + OH−(aq)
H2O(l) + H2O(l)
H2O
Kw = [1.0  10−7 ]
[H3O+]
[OH−]
[1.0  10−7 ]
= 1.0  10−14
Kw = [1.0  10−8 ] [1.0  10−6 ]
pH + pOH = pKw
= 1.0  10−14
8 + 6 = 14
Acids
and
Bases
Notice the
relationship
between
[H3
O+]
and
[OH−]
It is an inverse
relationship!
∝
[H3O+]
1
[OH−]
pH + pOH = 14
pH
[H3O+]
0
1.0 (10 )
1
[OH−]
0
0.00000000000001
1.0 (10 )
-1
0.0000000000001
2
0.01
0.000000000001
3
0.001
0.00000000001
4
0.0001
0.0000000001
5
0.00001
0.000000001
6
0.000001
0.00000001
7
0.0000001 (10 )
0.0000001 (10 )
8
0.00000001
0.0000001
9
0.000000001
0.00001
10
0.0000000001
0.0001
11
0.00000000001
0.001
12
0.000000000001
0.01
-7
-7
-1
13
0.0000000000001
1.0 (10 )
14
0.00000000000001
1.0 (10 )
0
Acids
and
Bases
pH
These are
the pH
values for
several
common
substances.
Acids
and
Bases
pH + pOH = 14
Dissociation Constants for Weak
Acids
• For a generalized acid dissociation,
HA(aq) + H2O(l)
A−(aq) + H3O+(aq)
the equilibrium expression would be
[H3O+] [A−]
Ka =
[HA]
• This equilibrium constant is called the aciddissociation (ionization) constant, Ka.
Acids
and
Bases
Dissociation Constants
The greater the value of Ka, the stronger the acid.
pKaa
3.2
3.3
4.2
4.7
7.5
9.3
Acids
9.9
and
Bases
Salts of weak acids are strong bases
Calculating Ka from the pH
Problem: (to determine Ka, simply dissolve a known [HA]i then measure pH)
A 0.10 M solution of formic acid, HCOOH, at 25°C has a
pH = 2.38. Calculate Ka for formic acid at this temperature.
Also calculate % ionization.
HCOO−(aq) + H3O+(aq)
HCOOH (aq) + H2O(l)
What is the Ka expression?
[H3O+] [A−]
Ka =
[HA]
pH = −log [H3O+]
2.38 = −log [H3O+]
−2.38 = log [H3O+]
10−2.38 = 10log [H3O+] = [H3O+]
Acids
and −]
4.2  10−3 = [H3O+] = [HCOO
Bases
Problem:
Calculating Ka from pH
• The pH of a 0.10 M solution of formic acid, HCOOH, at
25°C is 2.38. Calculate Ka for formic acid at this
temperature. Also calculate % ionization.
HCOO−(aq) + H3O+(aq)
HCOOH (aq) + H2O(l)
[H3O+] = [HCOO−] = 4.2  10−3
[H3O+] [A−] [4.2  10−3] [4.2  10−3]
Ka =
= 1.8  10−4
=
[HA]
[0.10]
[HCOOH], M
Initially
Change
Equilibrium
0.10
−4.2  10-3
0.0958 = 0.10
[H3O+], M [HCOO−], M
0
0
+4.2  10-3 +4.2  10−3Acids
and
−3
−3
4.2  10
4.2  10 Bases
Calculating Percent Ionization
HCOO−(aq) + H3O+(aq)
HCOOH(aq) + H2O(l)
Whole
(Initial)
Part
(@ Equi)
[H3O+]eq
• Percent Ionization =
 100
[HA]initial
• In this example
[HCOOH], M
[H3O+], M
[HCOO−], M
I
0.10
0
0
C
−4.2  10-3
+4.2  10-3
+4.2  10−3
E
0.0958 = 0.10
4.2  10−3
4.2  10−3
4.2  10−3
Percent Ionization = 0.10  100
= 4.2%
Acids
and
Bases
Calculating pH from Ka
Problem: Calculate the pH of a 0.30 M solution of acetic acid,
HC2H3O2, at 25°C. For acetic acid at 25°C is Ka =1.8  10−5
HC2H3O2(aq) + H2O(l)
H3O+(aq) + C2H3O2−(aq)
[H3O+] [C2H3O2−]
Ka =
[HC2H3O2]
=
[x] [x]
[0.3]
[C2H3O2], M
[H3O+], M
[C2H3O2−], M
Initially
0.30
0
0
Change
−x
+x
+x
0.30 − x  0.30
x
x
At Equilibrium
We are assuming that x will be very small compared to 0.30 and
can, therefore, be ignored.
Acids
and
Bases
Calculating pH from Ka
Ka =
[H3O+]
2
(x)
1.8  10−5 =
(0.30)
[C2H3O2−]
[HC2H3O2]
(1.8  10−5) (0.30) = x2
5.4  10−6 = x2
[H3O+] = 2.3  10−3 = x
What is the pH?
+
• pH = −log [H3O ] = − log x
• pH = −log (2.3  10−3)
• pH = 2.64
Acids
and
Bases
2002B Q1
(b) Calculate the pH of 0.50 M Lactic Acid.
Acids
and
Bases
Polyprotic Acids
• Have more than one acidic proton.
• If the difference between the Ka for the first
dissociation and subsequent Ka values is
103 or more, the pH generally depends only
on the first dissociation.
Acids
and
Bases
Dissociation Constants for Weak Bases
Bases react with water to produce hydroxide ion.
The generalized equilibrium constant expression for
these reactions is
[HB+] [OH−]
Kb =
[B]
where Kb is the base-dissociation constant.
Acids
and
Bases
Weak Bases
Kb can be used to find [OH−] and, through it, pH.
pKb
pKb
4.7
8.8
8.0
3.4
6.7
3.7
Acids
and
Bases
6.5
Salts of weak bases are strong acids
pH & pOH for Weak Acids & Bases
What is the pH of a 0.15 M solution of NH3?
Kb = 1.8 x 10-5
NH4+(aq) + OH−(aq)
NH3(aq) + H2O(l)
[NH3], M
Initially
Change
At Equilibrium
[NH4+], M [OH−], M
0.15
-x
0
+x
0
+x
0.15 - x  0.15
x
x
2
[NH4+] [OH−]
(x)
Kb =
= 1.8  10−5 =
[NH3]
(0.15)
Acids
and
Bases
pH of Basic Solutions
2
[NH4+] [OH−]
(x)
Kb =
= 1.8  10−5 =
[NH3]
(0.15)
(1.8  10−5) (0.15) = x2
−6 = x2
2.7

10
What is pH of soln?
What is x?
1.6  10−3 = x2
[OH−] = 1.6  10−3 M
pOH = −log (1.6  10−3)
pOH = 2.80
pH = 14.00 − 2.80
pH = 11.20
Acids
and
Bases
2002A Q1(a-b)
Acids
and
Bases
2005 A acid base equilibria
Acids
and
Bases
2005 A acid base equilibria ANSWERS
Acids
and
Bases
c) ii
Acids
and
Bases
2005 B acid base equilibria
Acids
and
Bases
2005 B acid base equilibria ANSWERS
a)
b)
c)
Acids
and
Bases
d)i
d)ii
Acids
and
Bases
d)iii
Acids
and
Bases
2001 a/b equilibria titration
Acids
and
Bases
2001 answers
Acids
and
Bases
Acids
and
Bases
2002
Acids
and
Bases
a)
Acids
and
Bases
Acids
and
Bases
Acids
and
Bases
Acids
and
Bases
Acids
and
Bases
Acids
and
Bases
2002 B
Acids
and
Bases
Acids
and
Bases
Acids
and
Bases
Acids
and
Bases
Acids
and
Bases
2003 A
Acids
and
Bases
Acids
and
Bases
Acids
and
Bases
Acids
and
Bases
Acids
and
Bases
2003 B
Acids
and
Bases
Acids
and
Bases
Acids
and
Bases
Acids
and
Bases
2005 A
Acids
and
Bases
Acids
and
Bases
Acids
and
Bases
H2O(l) + H2O(l)
H3O+(aq) + OH−(aq)
Hydroxide is a
much stronger
base than H2O,
so the equilibrium
favors the left
side (K<<<1).
Acids
and
Bases