Chapter 17
Principles of
Chemical Reactivity:
The Chemistry of
Acids and Bases
Jeffrey Mack
California State University,
Sacramento
Acids & Bases: A Review
• In Chapter 3, you were introduced to two
definitions of acids and bases: the Arrhenius and
the Brønsted–Lowry definition.
• Arrhenius acid: Any substance that when
dissolved in water increases the concentration of
hydrogen ions, H+.
• Arrhenius base: Any substance that increases
the concentration of hydroxide ions, OH, when
dissolved in water.
• A Brønsted–Lowry acid is a proton (H+) donor.
• A Brønsted–Lowry base is a proton acceptor.
Strong & Weak Acids/Bases
• Generally divide acids and bases into STRONG or
WEAK ones.
STRONG ACID:
HNO3(aq) + H2O(liq)  H3O+(aq) + NO3-(aq)
HNO3 is about 100% dissociated in water.
Strong & Weak Acids/Bases
HNO3, HCl, H2SO4 and HClO4 are classified as
strong acids.
Strong & Weak Acids/Bases
• Strong Base: 100% dissociated in water.
NaOH(aq)  Na+(aq) + OH-(aq)
Other common strong
bases include KOH and
Ca(OH)2.
CaO (lime) + H2O 
CaO
Ca(OH)2 (slaked lime)
Strong & Weak Acids/Bases
• Weak base: less than 100% ionized in water
An example of a weak base is ammonia
NH3(aq) + H2O(liq)  NH4+(aq) + OH-(aq)
Strong & Weak Acids/Bases
Weak acids are much less than 100% ionized
in water.
Example: acetic acid = CH3CO2H
The Brønsted–Lowry Concept of
Acids & Bases Extended
• Proton donors may be molecular compounds,
cations or anions.
HNO3 (aq)  H2O(l)
NH4 (aq)  H2O(l)

HCO 3 (aq)  H2O(l)
NO3 (aq)  H3O  (aq)
NH3 (aq)  H3O  (aq)
2
3

CO (aq)  H3O (aq)
The Brønsted–Lowry Concept of
Acids & Bases Extended
• Proton acceptors may be molecular
compounds, cations or anions.
NH3 (aq)  OH (aq)
NH3 (aq)  H2O(l)
2
 Al H2O 5  OH  (aq)  H2O(l)
3
 Al H2O 6  (aq)  OH (aq)
CO32 (aq)  H2O(l)
HCO 3 (aq)  OH (aq)
The Brønsted–Lowry Concept of
Acids & Bases Extended
Using the Brønsted definition, NH3 is a BASE
in water and water is itself an ACID
Proton acceptor
NH3 (aq) + H2O(l)
NH+3 (aq) + OH- (aq)
Proton donor
The Brønsted–Lowry Concept of
Acids & Bases Extended
• Acids such as HF, HCl, HNO3, and CH3CO2H (acetic acid)
are all capable of donating one proton and so are called
monoprotic acids.
• Other acids, called polyprotic acids are capable of donating
two or more protons.
Conjugate Acid–Base Pairs
• A conjugate acid–base pair consists of two species that
differ from each other by the presence of one hydrogen ion.
• Every reaction between a Brønsted acid and a Brønsted
base involves two conjugate acid–base pairs
Conjugate Acid–Base Pairs
Water & the pH Scale
Water Autoionization and the Water Ionization
Constant, Kw:
H2O(l)  H2O(l)


H3O (aq)  OH (aq)
The water autoionization equilibrium lies far to the left
side. In fact, in pure water at 25 °C, only about two
out of a billion (109) water molecules are ionized at
any instant.
+
-
K w = éëH3O ùû éëOH ùû = 1.00 ´ 10
-14
Even in pure water, there is a small concentration of
ions present at all times. [H3O+] = [OH] = 1.00  107
Water & the pH Scale
H2O can function as both an ACID and a BASE.
In pure water there can be AUTOIONIZATION.
Equilibrium constant for autoionization = Kw
Kw = [H3O+] [OH-] = 1.00 x 10-14 at 25 °C
Water & the pH Scale
In a neutral solution, [H3O+] = [OH]
Both are equal to 1.00  10 7 M
In an acidic solution, [H3O+] > [OH]
[H3O+] > 1.00  10 7 M and [OH] < 1.00  10
7 M
• In a basic solution, [H3O+] < [OH]
• [H3O+] < 1.00  10 7 M and [OH] > 1.00  10
7 M
•
•
•
•
The pH Scale
The pH Scale
• The pH of a solution is defined as the
negative of the base (10) logarithm (log) of
the hydronium ion concentration.
pH =  log[H3O+]
• In a similar way, we can define the pOH of a
solution as the negative of the base - 10
logarithm of the hydroxide ion concentration.
pOH =  log[OH]
pH + pOH = pKw = 14
The pH Scale
• The concentration of acid, [H3O+] is found by
taking the antilog of the solutions pH.
+
éëH3O ùû = 10
- pH
• In a similar way, [OH] can be found from:
-
éëOH ùû = 10
-pOH
The pH Scale
Once [H3O+] is known, [OH] can be found
from:
Kw
[OH ] =
+
[H3O ]
-
And vice versa.
Kw
[H3O ] =
[OH ]
+
Equilibrium Constants for Acids &
Bases
• In Chapter 3, it was stated that acids and bases
can be divided roughly into those that are strong
electrolytes (such as HCl, HNO3, and NaOH)
and those that are weak electrolytes (such as
CH3CO2H and NH3)
• In this chapter we will discuss the quantitative
aspects of dissociation of weak acids and bases.
• The relative strengths of weak acids and bases
can be ranked based on the magnitude of
individual equilibrium constants.
Equilibrium Constants for Acids &
Bases
• Strong acids and bases almost completely
ionize in water (~100%):
Kstrong >> 1
(product favored)
• Weak acids and bases almost completely
ionize in water (<<100%):
Kweak << 1
(Reactant favored)
Equilibrium Constants for Acids &
Bases
• The relative strength of an acid or base can also be
expressed quantitatively with an equilibrium
constant, often called an ionization constant. For
the general acid HA, we can write:
HA(aq) + H2O(l)
Conjugate
acid
+
-
H3O (aq) + A (aq)
éëH3O+ ùû éë A - ùû
Ka =
[HA ]
Conjugate
base
Equilibrium Constants for Acids &
Bases
• The relative strength of an acid or base can
also be expressed quantitatively with an
equilibrium constant, often called an
ionization constant. For the general base B,
we can write:
B(aq) + H2O(l)
Conjugate
base
+
-
BH (aq) + OH (aq)
éëBH+ ùû éëOH- ùû
Kb =
[B]
Conjugate
Acid
Ionization Constants for
Acids/Bases
Acids
Conjugate
Bases
Increase
strength
Increase
strength
Equilibrium Constants for Acids &
Bases
• The strongest acids are at the upper left.
They have the largest Ka values.
• Ka values become smaller on descending the
chart as the acid strength declines.
• The strongest bases are at the lower right.
They have the largest Kb values.
• Kb values become larger on descending the
chart as base strength increases.
Equilibrium Constants for Acids &
Bases
• The weaker the acid, the stronger its
conjugate base: The smaller the value of Ka,
the larger the value of Kb.
• Aqueous acids that are stronger than H3O+
are completely ionized.
• Their conjugate bases (such as NO3) do not
produce meaningful concentrations of OH
ions, their Kb values are “very small.”
• Similar arguments follow for strong bases
and their conjugate acids.
Equilibrium Constants for Acids &
Bases
¾¾¾¾¾¾¾¾¾¾¾
®
Increasing Acid Strength
Acid
HCO3
Ka
4.8  1011
HClO
HF
3.5  108 7.2  104
¬¾¾¾¾¾¾¾¾¾¾
¾
Increasing Base Strength
Base
CO32
Kb
2.1  104
ClO
F
2.9  107 1.4  1011
Equilibrium Constants for Acids &
Bases
Equilibrium Constants for Acids &
Bases
Ka Values for Polyprotic Acids
H2S(aq)  H2O(l)


H3O (aq)  HS (aq)
K a(1)  1 10 7

HS (aq)  H2O(l)

2
H3O (aq)  S (aq)
K a(2)  1 10 19
In general, each successive dissociation
produces a weaker acid.
Equilibrium Constants for Acids &
Bases
Logarithmic Scale of Relative Acid
Strength, pKa
• Many chemists use a logarithmic scale to
report and compare relative acid strengths.
pKa =  log(Ka)
¾¾¾¾¾¾¾¾¾¾¾
®
Increasing Acid Strength
Acid
HCO3
HClO
HF
pKa
10.32
7.46
3.14
The lower the pKa, the stronger the acid.
Equilibrium Constants for Acids &
Bases
Relating the Ionization Constants for an
Acid and Its Conjugate Base
Ka
¾¾®
H2S(aq) + H2O(l) ¬¾¾ H3O+ (aq) + HS- (aq)
Kb
¾¾®
HS (aq) + H2O(l) ¬¾¾ H2S(aq) + OH- (aq)
-
Equilibrium Constants for Acids &
Bases
Relating the Ionization Constants for an
Acid and Its Conjugate Base
Ka
¾¾®
H2S(aq) + H2O(l) ¬¾¾ H3O+ (aq) + HS- (aq)
Kb
¾¾®
HS (aq) + H2O(l) ¬¾¾ H2S(aq) + OH- (aq)
-
Equilibrium Constants for Acids &
Bases
Relating the Ionization Constants for an
Acid and Its Conjugate Base
Ka
¾¾®
H2S(aq) + H2O(l) ¬¾¾ H3O+ (aq) + HS- (aq)
Kb
¾¾®
HS (aq) + H2O(l) ¬¾¾ H2S(aq) + OH- (aq)
2H2O(l)
H3O+ (aq) + OH- (aq)
-
Equilibrium Constants for Acids &
Bases
Relating the Ionization Constants for an
Acid and Its Conjugate Base
Ka
¾¾®
H2S(aq) + H2O(l) ¬¾¾ H3O+ (aq) + HS- (aq)
Kb
¾¾®
HS (aq) + H2O(l) ¬¾¾ H2S(aq) + OH- (aq)
2H2O(l)
H3O+ (aq) + OH- (aq)
-
H3O  HS  H2S OH 





  K w
Ka  Kb 


H
O
OH
3




HS 
H2S
Equilibrium Constants for Acids &
Bases
Relating the Ionization Constants for an
Acid and Its Conjugate Base
Ka
¾¾®
H2S(aq) + H2O(l) ¬¾¾ H3O+ (aq) + HS- (aq)
Kb
¾¾®
HS (aq) + H2O(l) ¬¾¾ H2S(aq) + OH- (aq)
2H2O(l)
H3O+ (aq) + OH- (aq)
-
éëH3O+ ùû éëHS- ùû [H2S]éëOH- ùû
+
é
ù
é
ùû = K w
Ka ´ Kb =
´
=
H
O
OH
3
ë
û
ë
éëHS ùû
[H2S]
Ka ´ Kb = K w
When adding equilibria,
multiply the K values.
Acid–Base Properties of Salts
Acid–Base Properties of Salts
Anions that are conjugate bases of strong
acids (for examples, Cl or NO3.

3
NO (aq)  H2O(l)
No Reaction
These species are such weak bases that they
have no effect on solution pH.
Acid–Base Properties of Salts
Anions such as CO3 that are the conjugate
bases of weak acids will raise the pH of a
solution.
CO32 (aq)  H2O(l)
HCO3 (aq)  OH (aq)
Hydroxide ions are produced via “Hydrolysis”.
Acid–Base Properties of Salts
Anions such as CO3 that are the conjugate
bases of weak acids will raise the pH of a
solution.
2
3
CO (aq)  H2O(l)

3

HCO (aq)  OH (aq)
Hydroxide ions are produced via “Hydrolysis”.
A partially deprotonated anion (such as HCO3)
is amphiprotic. Its behavior will depend on the
other species in the reaction.
Acid–Base Properties of Salts
Alkali metal and alkaline earth cations have no
measurable effect on solution pH.

Na (aq)  H2O(l)
No Reaction
Since these cations are conjugate acids of
strong bases, hydrolysis does not occur.
Acid–Base Properties of Salts
Basic cations are conjugate bases of acidic cations
such as [Al(H2O)6]3+.
Acidic cations fall into two categories: (a) metal
cations with 2+ and 3+ charges and (b) ammonium
ions (and their organic derivatives).
All metal cations are hydrated in water, forming ions
such as [M(H2O)6]n+.
3
 Al H2O 6  (aq)  H2O(l)
2
 Al H2O 5 (OH )  (aq)  H3O (aq)

Ka  7.9  10 4
Acid–Base Properties of Salts:
Practice
Salt
CaCl2
NH4Br
NH4F
KNO3
KHCO3
pH of (aq) solution
Acid–Base Properties of Salts:
Practice
Salt
pH of (aq) solution
CaCl2
Neutral
NH4Br
Acidic
NH4F
Basic
KNO3
Neutral
KHCO3
Basic
Predicting the Direction of Acid–
Base Reactions
• According to the Brønsted–Lowry theory, all acid–
base reactions can be written as equilibria involving
the acid and base and their conjugates.
Acid  Base
Conjugate base of the acid + Conjugate acid of the base
• All proton transfer reactions proceed from the
stronger acid and base to the weaker acid and
base.
Predicting the Direction of Acid–
Base Reactions
• When a weak acid is in solution, the products are a
stronger conjugate acid and base. Therefore
equilibrium lies to the left.
• All proton transfer reactions proceed from the
stronger acid and base to the weaker acid and
base.
Predicting the Direction of Acid–
Base Reactions
• Will the following acid/base reaction occur
spontaneously?
H3PO4 (aq)  CH3CO2 (aq)
H2PO3 (aq)  CH3CO2H(aq)
Predicting the Direction of Acid–
Base Reactions
• Will the following acid/base reaction occur
spontaneously?
Kb = 5.6  1010
H3PO4 (aq)  CH3CO2 (aq)
Ka = 7.5  105
Kb = 1.3  1012
H2PO3 (aq)  CH3CO2H(aq)
Ka = 1.8  105
Predicting the Direction of Acid–
Base Reactions
• Will the following acid/base reaction occur
spontaneously?
Kb = 5.6  1010
H3PO4 (aq)  CH3CO2 (aq)
Ka = 7.5  105
Stronger Acid + Stronger Base
Kb = 1.3  1012
H2PO3 (aq)  CH3CO2H(aq)
Ka = 1.8  105
Weaker Base + Weaker Acid
• Equilibrium lies to the right since all proton
transfer reactions proceed from the stronger
acid and base to the weaker acid and base.
Types Acids–Base Reactions
Strong acid (HCl) + Strong base (NaOH)
HCl (aq)  NaOH (aq)

H3O (aq)  NaCl (aq)
Net ionic equation
H3O (aq)  OH (aq)
2H2O(l)
Mixing equal molar quantities of a strong acid
and strong base produces a neutral solution.
Types Acids–Base Reactions
Weak acid (HCN) + Strong base (NaOH)

HCN (aq) + OH (aq)

CN (aq) + H2O (l)
Mixing equal amounts (moles) of a strong base
and a weak acid produces a salt whose anion
is the conjugate base of the weak acid. The
solution is basic, with the pH depending on Kb
for the anion.

CN (aq) + H2O (l)

HCN (aq) + OH (aq)
Types Acids–Base Reactions
Strong acid (HCl) + Weak base (NH3)
H3O (aq) + NH3 (aq)
H2O (l) + NH4 (aq)
Mixing equal amounts (moles) of a weak base
and a strong acid produces a conjugate acid of
the weak base. The solution is basic, with the
pH depending on Ka for the acid.
NH4 (aq) + H2O (l)
H3O (aq) + NH3 (aq)
Types Acids–Base Reactions
Weak acid (CH3CO2H) + Weak base (NH3)
CH3CO2H (aq) + NH3 (aq)

2

4
CH3CO + NH (aq)
Mixing equal amounts (moles) of a weak acid
and a weak base produces a salt whose cation
is the conjugate acid of the weak base and
whose anion is the conjugate base of the weak
acid. The solution pH depends on the relative
Ka and Kb values.
Types Acids–Base Reactions
Weak acid + Weak base
• Product cation = conjugate acid of weak base.
• Product anion = conjugate base of weak acid.
• pH of solution depends on relative strengths of
cation and anion.
Types Acids–Base Reactions
Summary
Calculations with Equilibrium
Constants
Determining K from Initial Concentrations and
pH
HNO2 (aq) + H2O(l)
H3O+ (aq) + NO2- (aq)
0.10 M HNO2 (aq) pH = 2.17
[H2S]
Initial
Change
Equilibrium
0.10
[H3O+]
[HS]
Calculations with Equilibrium
Constants
Determining K from Initial Concentrations and
pH
HNO2 (aq) + H2O(l)
H3O+ (aq) + NO2- (aq)
0.10 M HNO2 (aq) pH = 2.17
[H2S]
[H3O+]
[HS]
0.10
0
0
Change
0.10 - x
+x
+x
Equilibrium
0.10 - x
x
x
Initial
Calculations with Equilibrium
Constants
Determining K from Initial Concentrations and
pH
HNO2 (aq) + H2O(l)
H3O+ (aq) + NO2- (aq)
0.10 M HNO2 (aq) pH = 2.17
[H2S]
[H3O+]
[HS]
0.10
0
0
Change
0.10 - x
+x
+x
Equilibrium
0.10 - x
x
x
Initial
H3O  NO2 
xx
x2
Ka 


0.10  x 0.10  x
HNO2 
Calculations with Equilibrium
Constants
Determining K from Initial Concentrations and
pH
+
HNO2 (aq) + H2O(l)
H3O (aq) + NO2 (aq)
0.10 M HNO2 (aq) pH = 2.17
H3O  NO2 
xx
x2
Ka 


0.10  x 0.10  x
HNO2 
[H3O+] = [NO2] = 6.76  103
 6.76  10 
3 2
Ka 
0.10  6.76  103
 4.9  104
Calculations with Equilibrium
Constants
Determining K from Initial Concentrations and
pH
H2S(aq) + H2O(l)
1.00 M H2S(aq)
[H2S]
Initial
Change
Equilibrium
1.00
H3O+ (aq) + HS- (aq)
K a = 1.0 ´ 10-7
[H3O+]
[HS]
Calculations with Equilibrium
Constants
Determining K from Initial Concentrations and
pH
H2S(aq) + H2O(l)
H3O+ (aq) + HS- (aq)
1.00 M H2S(aq)
Initial
Change
Equilibrium
K a = 1.0 ´ 10-7
[H2S]
[H3O+]
[HS]
1.00
0
0
-x
+x
+x
1.00 - x
x
x
H3O  HS 
xx
x2
Ka 


 1.0  107
1.00  x 1.00  x
H2S
Calculations with Equilibrium
Constants
Determining K from Initial Concentrations and
pH
H3O  HS 
xx
x2
Ka 


 1.0  107
1.00  x 1.00  x
H2S
Since x << 1.00, the equation reduces to:
x2
 1.0  10 7
1.00
x = 3.2  104
pH = 3.50
Calculations with Equilibrium
Constants
Determining K from Initial Concentrations and
pH
In general, the approximation that
[HA]equilibrium = [HA]initial  x  [HA]initial
is valid whenever [HA]initial is greater than or
equal to 100  Ka.
If this is not the case, the quadratic equation
must by used.
Calculations with Equilibrium
Constants
Determining pH after an acid/base reaction:
Calculate the hydronium ion concentration and
pH of the solution that results when 22.0 mL of
0.15 M acetic acid, CH3CO2H, is mixed with 22.0
mL of 0.15 M NaOH.
Calculations with Equilibrium
Constants
Determining pH after an acid/base reaction:
Calculate the hydronium ion concentration and
pH of the solution that results when 22.0 mL of
0.15 M acetic acid, CH3CO2H, is mixed with 22.0
mL of 0.15 M NaOH.
Solution: From the volume and concentration of
each solution, the moles of acid and base can be
calculated. Knowing the moles after the reaction
and the equilibrium constants, the concentration
of H3O+ and pH can be calculated.
Calculate the hydronium ion concentration and pH of the
solution that results when 22.0 mL of 0.15 M acetic acid,
CH3CO2H, is mixed with 22.0 mL of 0.15 M NaOH.
CH3CO2H (aq)  OH (aq)
22.0 mL 
CH3CO2 (aq)  H2O (l)
1L
0.15 mols



0.0033
mols
of
CH
CO
H
&
OH
3
2
103 mL
1L
All of the acetic acid is converted to acetate ion.
Calculate the hydronium ion concentration and pH of the
solution that results when 22.0 mL of 0.15 M acetic acid,
CH3CO2H, is mixed with 22.0 mL of 0.15 M NaOH.
CH3CO2H (aq)  OH (aq)
22.0 mL 
CH3CO2 (aq)  H2O (l)
1L
0.15 mols



0.0033
mols
of
CH
CO
H
&
OH
3
2
103 mL
1L
All of the acetic acid is converted to acetate ion.
3
0.0033
mols
10
mL
CH3CO2  

 0.075 M
44.0 mL
1L
Calculate the hydronium ion concentration and pH of the
solution that results when 22.0 mL of 0.15 M acetic acid,
CH3CO2H, is mixed with 22.0 mL of 0.15 M NaOH.
3
0.0033
mols
10
mL
éëCH3CO2 ùû =
´
= 0.075 M
44.0 mL
1L
CH3CO2 (aq)  H2O (l)
Initial
Change
Equilibrium
CH3CO2H (aq)  OH (aq)
[CH3CO2-]
[CH3CO2H]
[OH-]
0.075
0
0
Calculate the hydronium ion concentration and pH of the
solution that results when 22.0 mL of 0.15 M acetic acid,
CH3CO2H, is mixed with 22.0 mL of 0.15 M NaOH.
3
0.0033
mols
10
mL
éëCH3CO2 ùû =
´
= 0.075 M
44.0 mL
1L
CH3CO2 (aq)  H2O (l)
Initial
Change
Equilibrium
CH3CO2H (aq)  OH (aq)
[CH3CO2-]
[CH3CO2H]
[OH-]
0.075
0
0
-x
+x
+x
0.075 - x
x
x
Calculate the hydronium ion concentration and pH of the
solution that results when 22.0 mL of 0.15 M acetic acid,
CH3CO2H, is mixed with 22.0 mL of 0.15 M NaOH.
Initial
Change
Equilibrium
[CH3CO2H]éëOH- ùû
éëCH3CO2- ùû
[CH3CO2-]
[CH3CO2H]
[OH-]
0.075
0
0
-x
+x
+x
0.075 - x
x
x
K w 1.00 ´ 10-14
x2
-10
=
= Kb =
=
=
5.6
´
10
0.075 - x
Ka
1.8 ´ 10-5
Calculate the hydronium ion concentration and pH of the
solution that results when 22.0 mL of 0.15 M acetic acid,
CH3CO2H, is mixed with 22.0 mL of 0.15 M NaOH.
Initial
Change
Equilibrium
[CH3CO2H]éëOH- ùû
éëCH3CO2- ùû
[CH3CO2-]
[CH3CO2H]
[OH-]
0.075
0
0
-x
+x
+x
0.075 - x
x
x
K w 1.00 ´ 10-14
x2
-10
=
= Kb =
=
=
5.6
´
10
0.075 - x
Ka
1.8 ´ 10-5
Since Kb100 > [CH3CO2-]initial, the quadratic equation
is not needed.
Calculate the hydronium ion concentration and pH of the
solution that results when 22.0 mL of 0.15 M acetic acid,
CH3CO2H, is mixed with 22.0 mL of 0.15 M NaOH.
x2
= 5.6 ´ 10 -10
0.075
x = éëOH- ùû = 6.4 ´ 10 -6
pOH = - log(6.4 ´ 10 -6 ) = 5.19
pH = 14 - pOH = 8.81
éëH3O+ ùû = 10 -pH = 1.5 ´ 10-9 M
Polyprotic Acuids & Bases
• Because polyprotic acids are capable of donating
more than one proton they present us with
additional challenges when predicting the pH of
their solutions.
• For many inorganic polyprotic acids, the ionization
constant for each successive loss of a proton is
about 104 to 106 smaller than the previous step.
• This implies that the pH of many inorganic
polyprotic acids depends primarily on the hydronium
ion generated in the first ionization step.
• The hydronium ion produced in the second step can
be neglected.
Polyprotic Acids & Bases
Sulfurous acid, H2SO3, is a weak acid capable of providing two
H+ ions.
(a) What is the pH of a 0.45 M solution of H2SO3?
(b) What is the equilibrium concentration of the sulfite ion,
SO32- in the 0.45 M solution of H2SO3?
Polyprotic Acids & Bases
Sulfurous acid, H2SO3, is a weak acid capable of providing two
H+ ions.
(a) What is the pH of a 0.45 M solution of H2SO3?
(b) What is the equilibrium concentration of the sulfite ion,
SO32- in the 0.45 M solution of H2SO3?
1.2  10–2
[HSO3 ][H3O ]
x2


[H2SO3 ]
0.45 – x
Polyprotic Acids & Bases
Sulfurous acid, H2SO3, is a weak acid capable of providing two
H+ ions.
(a) What is the pH of a 0.45 M solution of H2SO3?
(b) What is the equilibrium concentration of the sulfite ion,
SO32- in the 0.45 M solution of H2SO3?
1.2  10–2
[HSO3 ][H3O ]
x2


[H2SO3 ]
0.45 – x
Since 100  Ka is not << 0.45M, the quadratic equation must
be used
Polyprotic Acids & Bases
Sulfurous acid, H2SO3, is a weak acid capable of providing two
H+ ions.
(a) What is the pH of a 0.45 M solution of H2SO3?
(b) What is the equilibrium concentration of the sulfite ion,
SO32- in the 0.45 M solution of H2SO3?
(a)
x = [H3O+] = 0.0677 M
pH = 1.17
Polyprotic Acids & Bases
Sulfurous acid, H2SO3, is a weak acid capable of providing two
H+ ions.
(a) What is the pH of a 0.45 M solution of H2SO3?
(b) What is the equilibrium concentration of the sulfite ion,
SO32- in the 0.45 M solution of H2SO3?
(a) in part a we found that x = [H3O+] = 0.0677 M
(b)
HSO3- (aq) + H2O(l)
K a2 = 6.2 ´ 10 -8
SO32- (aq) + H3O+ (aq)
éëSO32- ùû éëH3O+ ùû éëSO32- ùû [0.0677]
=
=
éëHSO3 ùû
[0.0677]
éëSO32- ùû = K a2 = 6.2 ´ 10-8 M
Molecular Structure, Bonding, &
Acid–Base Behavior
Halide Acid Strengths
• Experiments show that the acid strength increases in the
order:
HF << HCl < HBr < HI.
• Stronger acids result when the HX bond is readily broken
(as signaled by a smaller, positive value of H for bond
dissociation) and a more negative value for the electron
attachment enthalpy of X.
Molecular Structure, Bonding, &
Acid–Base Behavior
Comparing Oxoacids: HNO2 and HNO3
• In all the series of related oxoacid compounds, the
acid strength increases as the number of oxygen
atoms bonded to the central element increases.
• Thus, nitric acid (HNO3) is a stronger acid than
nitrous acid (HNO2).
Molecular Structure, Bonding, &
Acid–Base Behavior
Why Are Carboxylic Acids Brønsted Acids?
• There is a large class of organic acids, all like acetic
acid (CH3CO2H) have the carboxylic acid group,
CO2H
• They are collectively called carboxylic acids.
Molecular Structure, Bonding, &
Acid–Base Behavior
Why Are Carboxylic Acids Brønsted Acids?
• The carboxylate anion is stabilized by
resonance.
Molecular Structure, Bonding, &
Acid–Base Behavior
Why Are Carboxylic Acids Brønsted Acids?
• The acidity of carboxylic acids is enhanced if
electronegative substituents replace the hydrogen
atoms in the alkyl (–CH3 or –C2H5) groups.
• Compare, for example, the pKa values of a series of
acetic acids in which hydrogen is replaced sequentially
by the more electronegative element chlorine.
Acetic acid
Trichloroacetic acid
Ka = 1.8 x 10-5
Ka = 0.3
• Trichloroacetic acid is a much stronger acid
owing to the high electronegativity of Cl.
• Cl withdraws electrons from the rest of the
molecule.
• This makes the O—H bond highly polar. The H
of O—H is very positive.
The Lewis Concept of Acids &
Bases
• The concept of acid–base behavior advanced by
Brønsted and Lowry in the 1920’s works well for
reactions involving proton transfer.
• However, a more general acid– base concept,
was developed by Gilbert N. Lewis in the
1930’s.
• A Lewis acid is a substance that can accept a
pair of electrons from another atom to form a
new bond.
• A Lewis base is a substance that can donate a
pair of electrons to another atom to form a new
bond.
The Lewis Concept of Acids &
Bases
A
Acid
+
B:
Base

BA
Adduct
• The product is often called an acid–base adduct.
In Section 8.3, this type of chemical bond was
called a coordinate covalent bond.
• Lewis acid-base reactions are very common. In
general, they involve Lewis acids that are cations or
neutral molecules with an available, empty valence
orbital and bases that are anions or neutral
molecules with a lone electron pair.
The Lewis Concept of Acids &
Bases
Lewis acid
a substance that accepts
an electron pair
Lewis base
a substance that donates
an electron pair
Reaction of a Lewis Acid & Lewis
Base
• New bond formed
using electron pair
from the Lewis base.
• Coordinate covalent
bond
• Notice geometry
change on reaction.
The Lewis Concept of Acids &
Bases
The formation of a hydronium ion is an
example of a Lewis acid / base reaction
H+
ACID
•• ••
O—H
H
BASE
••
H O—H
H
The H+ is an electron pair acceptor.
Water with it’s lone pairs is a Lewis acid
donor.
Lewis AcidBase Reactions
Lewis Acids & Bases
Metal cations often act as Lewis acids because
of open d-orbitals.
Lewis Acids & Bases
The combination of metal ions (Lewis acids) with
Lewis bases such as H2O and NH3 leads to
Coordinate Complex ions.
Lewis Acids & Bases
Aqueous solutions of Fe3+, Al3+, Cu2+, Pb2+,
etc. are acidic through hydrolysis.
This interaction weakens this bond
Another H2O pulls
this H away as H+
[Al(H2O)6]3+(aq) + H2O(l)  [Al(H2O)5(OH)]2+(aq) + H3O+(aq)
Molecular Lewis Acids
• Because oxygen is more electronegative than C,
the CO bonding electrons in CO2 are polarized
away from carbon and toward oxygen.
• This causes the carbon atom to be slightly positive,
and it is this atom that the negatively charged Lewis
base OH can attack to give, ultimately, the
bicarbonate ion.
Molecular Lewis Acids
• Ammonia is the parent compound of an enormous
number of compounds that behave as Lewis and
Brønsted bases. These molecules all have an
electronegative N atom with a partial negative
charge surrounded by three bonds and a lone pair
of electrons.
• This partially negative N atom can extract a proton
from water.
Lewis Acids & Bases
Many complex ions containing water undergo
HYDROLYSIS to give acidic solutions.
[Cu(H2O)4 ]2+ (aq) + H2O(l)
2+
[Cu(H2O)3 (OH)]+ (aq) + H3O+ (aq)
+
+
Reaction of NH3 with Cu2+(aq)
Lewis Acid–Base Interactions in
Biology
• The heme group in
hemoglobin can interact
with O2 and CO.
• The Fe ion in
hemoglobin is a Lewis
acid
• O2 and CO can act as
Lewis bases
Heme group
pH
–
A
Measure
of
Acidity
Slides from Chang Book
pH = -log [H+]
Solution Is
At 250C
neutral
[H+] = [OH-] [H+] = 1 x 10-7
acidic
[H+] > [OH-] [H+] > 1 x 10-7
basic
[H+] < [OH-] [H+] < 1 x 10-7
pH
[H+]
99
pH = 7
pH < 7
pH > 7
percent ionization =
% Ionization =
Ionized acid concentration at equilibrium
x 100%
Initial concentration of acid
For a monoprotic acid HA
[H+]
Percent ionization = x 100% [HA]0 = initial concentration
[HA]0
100
What is the pH of a 2 x 10-3 M HNO3 solution?
HNO3 is a strong acid – 100% dissociation.
0.0 M
0.0 M
Start 0.002 M
HNO3 (aq) + H2O (l)
H3O+ (aq) + NO3- (aq)
0.002 M 0.002 M
End 0.0 M
pH = -log [H+] = -log [H3O+] = -log(0.002) = 2.7
What is the pH of a 1.8 x 10-2 M Ba(OH)2 solution?
Ba(OH)2 is a strong base – 100% dissociation.
0.0 M
0.0 M
Start0.018 M
Ba(OH)2 (s)
Ba2+ (aq) + 2OH- (aq)
0.018 M 0.036 M
End 0.0 M
pH = 14.00 – pOH = 14.00 + log(0.036) = 12.6
101
What is the pH of a 0.5 M HF solution (at 250C)?
HF (aq)
+][F-]
[H
H+ (aq) + F- (aq) Ka = [HF] = 7.1 x 10-4
HF (aq)
Initial (M)
Change (M)
H+ (aq) + F- (aq)
0.50
0.00
0.00
-x
+x
+x
x
x
Equilibrium (M) 0.50 - x
x2
= 7.1 x 10-4 Ka << 1 0.50 – x  0.50
Ka =
0.50 - x
x2
Ka 
= 7.1 x 10-4 x2 = 3.55 x 10-4 x = 0.019 M
0.50
[H+] = [F-] = 0.019 M
[HF] = 0.50 – x = 0.48 M
pH = -log [H+] = 1.72
102
When can I use the approximation?
Ka << 1
Ka *100 << [HA]0
0.50 – x  0.50
When x is less than 5% of the value from which it is
subtracted. 0.019 M
Less than 5%
x 100% = 3.8%
x = 0.019
0.50 M
Approximation
ok.
What is the pH of a 0.05 M HF solution (at 250C)?
x2
Ka 
= 7.1 x 10-4 x = 0.006 M
0.05
More than 5%
0.006 M x 100% = 12%
0.05 M
Approximation not
ok.
Must solve for x exactly using quadratic equation or
method of successive approximations. 103
What is the pH of a 0.122 M monoprotic acid whose
Ka is 5.7 x 10-4?
HA (aq)
Initial (M)
Change (M)
H+ (aq) + A- (aq)
0.122
0.00
0.00
-x
+x
+x
x
x
Equilibrium (M) 0.122 - x
x2
= 5.7 x 10-4 Ka << 1 0.122 – x  0.122
Ka =
0.122 - x
x2
Ka 
= 5.7 x 10-4 x2 = 6.95 x 10-5 x = 0.0083 M
0.122
More than 5%
0.0083 Mx 100% = 6.8%
0.122 M
Approximation not
ok.
104
x2
= 5.7 x 10-4 x2 + 0.00057x – 6.95 x 10-5 = 0
Ka =
0.122 - x
ax2 + bx + c =0
x = 0.0081
HA (aq)
Initial (M)
Change (M)
 b2 – 4ac
-b
±
x=
2a
x = - 0.0081
H+ (aq) + A- (aq)
0.122
0.00
0.00
-x
+x
+x
x
x
Equilibrium (M) 0.122 - x
[H+] = x = 0.0081 M
pH = -log[H+] = 2.09
105
Acid-Base Properties of Salts
Neutral Solutions:
Salts containing an alkali metal or alkaline earth
metal ion (except Be2+) and the conjugate base
of a strong acid (e.g. Cl-, Br-, and NO3-).
NaCl (s)
H2O
Na+ (aq) + Cl- (aq)
Basic Solutions:
Salts derived from a strong base and a weak
acid.
H 2O
NaCH3COOH (s)
Na+ (aq) + CH3COO- (aq)
CH3COO- (aq) + H2O (l)
CH3COOH (aq) + OH- (aq)
106
Acid-Base Properties of Salts
Acid Solutions:
Salts derived from a strong acid and a weak
base.
H2O
NH4Cl (s)
NH4+ (aq) + Cl- (aq)
NH4+ (aq)
NH3 (aq) + H+ (aq)
Salts with small, highly charged metal cations
(e.g. Al3+, Cr3+, and Be2+) and the conjugate
base of a strong
acid.
3+
2+
Al(H2O)6 (aq)
Al(OH)(H2O)5 (aq) + H+ (aq)
107
Acid-Base Properties of Salts
Solutions in which both the cation and the anion hydrolyze:
• Kb for the anion > Ka for the cation, solution will be basic
•
Kb for the anion < Ka for the cation, solution will be acidic
•
Kb for the anion  Ka for the cation, solution will be neutral
108
What is the molarity of an NH4NO3(aq) solution that has
a pH = 4.80?
Strategy
Ammonium nitrate is the salt of a strong acid (HNO3) and a weak base
(NH3). In NH4NO3(aq), NH4+ hydrolyzes and NO3– does not. The ICE format
must be based on the hydrolysis equilibrium for NH4+(aq). In that format
[H3O+], derived from the pH, will be a known quantity, and the initial
concentration of NH4+ will be the unknown.
Solution
We begin by writing the equation for the hydrolysis equilibrium and the
equation for Ka in terms of Kw and Kb.
As usual, we can calculate [H3O+] from the pH of the solution.
log[H3O+] = –pH = –4.80
[H3O+] = 10–4.80 = 1.6 x 10–5 M
109
Example 15.15 continued
Solution continued
If we assume that all the hydronium ion comes from the hydrolysis reaction,
we can set up an ICE format in which x represents the unknown initial
concentration of NH4+.
We now substitute equilibrium concentrations into the ionization constant
expression for the hydrolysis reaction.
We can assume that the ammonium ion is mostly nonhydrolyzed and that
the change in [NH4+] is much smaller than the initial [NH4+], so that 1.6 x 10–
5 << x and we can replace (x – 1.6 x 10–5) by x. Then we can solve for x.
The solution is 0.46 M NH4NO3.
110