Chapter 16
Aqueous Acidic Equilibrium
History of Acids and Bases
In the early days of chemistry chemists were organizing
physical and chemical properties of substances. They
discovered that many substances could be placed in two
different property categories:
Substance B
Substance A
1.
Bitter taste
1.
Sour taste
2.
Reacts with carbonates to make CO2 2.
3.
Reacts with metals to produce H2
3.
Do not react with metals
4.
Turns blue litmus pink
4.
Turns red litmus blue
5.
Reacts with B substances to make
salt water
5.
Reacts with A substances make
salt and water
Reacts with fats to make soaps
Arrhenius was the first person to suggest a reason why
substances are in A or B due to their ionization in water.
Arrhenius Acids
The Swedish chemist Svante Arrhenius
proposed the first definition of acids and
bases. (Substances A and B became
known as acids and bases)
According to the Arrhenius model:
“acids are substances that dissociate in water to
produce H+ ions and bases are substances that
dissociate in water to produce OH- ions”
NaOH (aq)  Na+ (aq) + OH- (aq)
Base
HCl (aq)  H+ (aq) + Cl- (aq)
Acid
Hydronium ion
Unknown to Arrhenius free H+ ions do not exist in water. They
covalently react with water to produce hydronium ions, H3O+.
or:
H+ (aq) + H2O (l)  H3O+ (aq)
This new bond is called a coordinate covalent bond since
both new bonding electrons come from the same atom
Acid Strength
What is the difference between a strong acid and a weak
acid?
Acid Strength
What is the difference between a strong acid and a weak
acid? Strong acids ionize 100% and weak ones do not!
Acid Strength
What is the difference between a strong acid and a weak
acid? Strong acids ionize 100% and weak ones do not!
A single arrow is used to represent the ionization of a strong
acid. Double arrows (Equilibrium) are used to represent
weak acids.
For example:
HCl (g)
H+ (aq) + Cl - (aq)
HF (g)
H+ (aq) + F -
Acid Strength
What is the difference between a strong acid and a weak
acid? Strong acids ionize 100% and weak ones do not!
A single arrow is used to represent the ionization of a strong
acid. Double arrows (Equilibrium) are used to represent
weak acids.
For example:
HCl (g)
H+ (aq) + Cl - (aq)
HF (g)
H+ (aq) + F -
According to Arrhenius, is water an acid or base?
HOH (l)
H+ (aq) + OH –
(aq)
Acid Strength
What is the difference between a strong acid and a weak
acid? Strong acids ionize 100% and weak ones do not!
A single arrow is used to represent the ionization of a strong
acid. Double arrows (Equilibrium) are used to represent
weak acids.
For example:
HCl (g)
H+ (aq) + Cl - (aq)
HF (g)
H+ (aq) + F -
According to Arrhenius, is water an acid or base?
HOH (l)
H+ (aq) + OH –
(aq)
Neither, he called it Neutral (same amount of OH- and H+
Acid Strength
How can we identify strong acids or bases?
Acid Strength
How can we identify strong acids or bases?
Easy memorize them!
Acid Strength
How can we identify strong acids or bases?
Easy, memorize them!
Memorized Strong Acids
1. HClO4
2. H2SO4
3. HI
4. HBr
5. HCl
6. HNO3
Memorized Strong Bases
Hydroxides of group 1 and 2
metals, excluding Be and Mg
Acid Strength
Bronsted
Lowry
Johannes Brønsted and Thomas Lowry revised Arrhenius’s
acid-base theory to include this behavior. They defined acids
and bases as follows:
“An acid is a hydrogen containing species that donates a
proton. A base is any substance that accepts a proton”
e.g.
HCl (aq) + H2O (l)  Cl- (aq) + H3O+ (aq)
In the above example what is the Brønsted acid? What is
the Brønsted base?
Bronsted Lowry Acids/Bases
In reality, the reaction of HCl with H2O is an equilibrium
and occurs in both directions, although in this case the
equilibrium lies far to the right.
HCl (aq) + H2O (l)  Cl - ( aq) + H3O+ (aq)
For the reverse reaction Cl - behaves as a Brønsted
base and H3O+ behaves as a Brønsted acid.
The Cl- is called the conjugate base of HCl. Brønsted
acids and bases always exist as conjugate acid-base
pairs.
Auto Ionization of Water
In pure water (no solute) water molecules behave as
both an acid and base!!
e.g.
H2O (l) + H2O (l)  H3O+ (aq) + OH- (aq)
This is called the self-ionization (autoionizaion) of
water. Although the equilibrium lies far to the left it is
very important to take into consideration, especially for
living systems.
Auto Ionization of Water
The self-ionization of water is described by the equation:
e.g.
H2O (l) + H2O (l)  H3O+ (aq) + OH- (aq)
The equilibrium constant for this reaction is given by:
[H3O  ][ OH ] [H3O  ][ OH ]
K

[H2O ][H2O ]
[H2O ]2
K[H2O ]2  [H3O  ][ OH ]
Kw = K[H2O]2 = 10-14 This equilibrium lies very much to the
left i.e. mostly water. For pure water [OH-] = [H+] = 1 x 10-7 M
Auto Ionization of Water
As [OH-] and [H+] are so small the [H2O] is not affected by their
formation. It is useful to define a new constant Kw such that:
1.00 g
ml
mole
ml
18.0 g
10-3
L
=
55.5 M
[H3O  ][ OH ] [H3O  ][ OH ]
K

[H2O ][H2O ]
[H2O ]2
K[H2O ]2  K w  [H3O  ][ OH ]
Kw is called the ion product of water.
What is the value for the ion product of water?
Auto Ionization of Water
As [OH-] and [H+] are so small the [H2O] is not affected by their
formation. It is useful to define a new constant Kw such that:
1.00 g
ml
mole
18.0 g
ml
10-3 L
=
55.5 M
[H3O  ][ OH ] [H3O  ][ OH ]
K

[H2O ][H2O ]
[H2O ]2
K[H2O ]2  K w  [H3O  ][ OH ]
Kw is called the ion product of water.
What is the value for the ion product of water?
[H+][OH-] = 10-14
Autoionization of Water
• An amphoteric substance is capable of
behaving either as an acid or as a base.
• Kw = [H+][OH-] = 1.00 x 10-14
Hydronium Ion in Water
Acids in Water
• When a Brønsted-Lowry acid donates a H+ ion, it
forms its conjugate base.
HF + H2O
H3O+ + Fa
b
ca
cb
• Conjugate acid-base pairs differ only from each
other only by the presence or absence of a proton.
• When a Brønsted-Lowry base accepts a proton, it
becomes it conjugate acid.
Acidic, Basic, or Neutral
We define an aqueous solution as being neutral when the
[H+] = [OH-].
We define an aqueous solution as being acidic when
[H+] > [OH-].
We define an aqueous solution as being basic when
[H+] H+] < [OH-].
However, in each case Kw = 1 x 10-14 M2, since [H+][OH-]=Kw
Acidic, Basic, or Neutral
We define an aqueous solution as being neutral when the
[H+] = [OH-].
We define an aqueous solution as being acidic when
[H+] > [OH-].
We define an aqueous solution as being basic when
[H+] H+] < [OH-].
However, in each case Kw = 1 x 10-14 M2, since [H+][OH-]=Kw
[H+] = 0.0000001 = 10-7 (how can this be abbreviated further?)
Acidic, Basic, or Neutral
We define an aqueous solution as being neutral when the
[H+] = [OH-].
We define an aqueous solution as being acidic when
[H+] > [OH-].
We define an aqueous solution as being basic when
[H+] H+] < [OH-].
However, in each case Kw = 1 x 10-14 M2, since [H+][OH-]=Kw
[H+] = 0.0000001 = 10-7 (how can this be abbreviated further?)
By just describing the power Called the power of H, or pH.
Acidic, Basic, or Neutral
We define an aqueous solution as being neutral when the
[H+] = [OH-].
We define an aqueous solution as being acidic when
[H+] > [OH-].
We define an aqueous solution as being basic when
[H+] H+] < [OH-].
However, in each case Kw = 1 x 10-14 M2, since [H+][OH-]=Kw
[H+] = 0.0000001 = 10-7 (how can this be abbreviated further?)
By just describing the power Called the power of H, or pH.
Our math departments tells us that log means
pH = 7
power too.
pH and Logs
The mathematical definition of pH using [H+] for
[H3O+] is listed below:
pH = -log [H+], or [H+]= 1x10-pH (both are
mathematically equivalent)
How about the power for the OH -, what should this
be called?
pH and Logs
The mathematical definition of pH using [H+] for
[H3O+] is listed below:
pH = -log [H+], or [H+]= 1x10-pH (both are
mathematically equivalent)
How about the power for the OH -, what should this
be called? Would you believe pOH?
A pH Number line
Number lines have been used in history and math classes,
so to keep up we use them in chemistry classes.
pH = 16
pH = 12
pH = 7
[H+] = 10-2
pH = 2
A pH Number line
Number lines have been used in history and math classes,
so to keep up we use them in chemistry classes.
pH = 16
pH = 12
pH = 7
pH = 2
[H+] = 10-2
[OH -] = 10-12
A pH Number line
Number lines have been used in history and math classes,
so to keep up we use them in chemistry classes.
pH = 16
pH = 12
pH = 7
pH = 2
[H+] = 10-2
[OH -] = 10-12
[H+] > [OH -]
acidic
A pH Number line
Number lines have been used in history and math classes,
so to keep up we use them in chemistry classes.
pH = 16
pH = 12
pH = 7
[H+] = 10-7
[OH -] = 10-7
pH = 2
[H+] = 10-2
[OH -] = 10-12
[H+] > [OH -]
acidic
A pH Number line
Number lines have been used in history and math classes,
so to keep up we use them in chemistry classes.
pH = 16
pH = 12
pH = 7
[H+] = 10-7
[OH -] = 10-7
[H+] = [OH -]
neutral
pH = 2
[H+] = 10-2
[OH -] = 10-12
[H+] > [OH -]
acidic
A pH Number line
Number lines have been used in history and math classes,
so to keep up we use them in chemistry classes.
pH = 16
pH = 12
[H+] =10-12
[OH -] = 10-2
pH = 7
[H+] = 10-7
[OH -] = 10-7
[H+] = [OH -]
neutral
pH = 2
[H+] = 10-2
[OH -] = 10-12
[H+] > [OH -]
acidic
A pH Number line
Number lines have been used in history and math classes,
so to keep up we use them in chemistry classes.
pH = 16
pH = 12
[H+] =10-12
[OH -] = 10-2
pH = 7
[H+] = 10-7
[OH -] = 10-7
[H+] = [OH -]
neutral
pH = 2
[H+] = 10-2
[OH -] = 10-12
[H+] > [OH -]
acidic
[H+] < [OH -]
basic
A pH Number line
Number lines have been used in history and math classes,
so to keep up we use them in chemistry classes.
[H+] =10-16
pH = 16
[OH -] =
pH = 12
[H+] =10-12
[OH -] = 10-2
pH = 7
[H+] = 10-7
[OH -] = 10-7
[H+] = [OH -]
neutral
pH = 2
[H+] = 10-2
[OH -] = 10-12
[H+] > [OH -]
acidic
[H+] < [OH -]
basic
A pH Number line
Number lines have been used in history and math classes,
so to keep up we use them in chemistry classes.
[H+] =10-16
pH = 16
[OH -] = 102
pH = 12
[H+] =10-12
[OH -] = 10-2
pH = 7
[H+] = 10-7
[OH -] = 10-7
[H+] = [OH -]
neutral
pH = 2
[H+] = 10-2
[OH -] = 10-12
[H+] > [OH -]
acidic
[H+] < [OH -]
basic
A pH Number line
Number lines have been used in history and math classes,
so to keep up we use them in chemistry classes.
[H+] =10-1
pH = 16
[OH -] = 102
[H+] < [OH -]
basic
pH = 12
[H+] =10-12
[OH -] = 10-2
pH = 7
[H+] = 10-7
[OH -] = 10-7
[H+] = [OH -]
neutral
pH = 2
[H+] = 10-2
[OH -] = 10-12
[H+] > [OH -]
acidic
[H+] < [OH -]
basic
A pH Number line
Number lines have been used in history and math classes,
so to keep up we use them in chemistry classes.
[H+] =10-16
pH = 16
[OH -] = 102
[H+] < [OH -]
basic
pH = 12
[H+] =10-12
[OH -] = 10-2
pH = 7
[H+] = 10-7
[OH -] = 10-7
[H+] = [OH -]
neutral
pH = 2
[H+] = 10-2
[OH -] = 10-12
[H+] > [OH -]
acidic
acidic
[H+] < [OH -]
basic
A pH Number line
Number lines have been used in history and math classes,
so to keep up we use them in chemistry classes.
[H+] =10-16
pH = 16
[OH -] = 102
basic
[H+] < [OH -]
basic
pH = 12
[H+] =10-12
[OH -] = 10-2
pH = 7
[H+] = 10-7
[OH -] = 10-7
[H+] = [OH -]
neutral
pH = 2
[H+] = 10-2
[OH -] = 10-12
[H+] > [OH -]
acidic
acidic
[H+] < [OH -]
basic
Relative Strengths of Acids and Bases
The leveling effect of
water refers to the
observation that
strong acids all have
the same strength in
water and are
completely converted
into a solution of
hydronium ions.
The pH Scale
• pH = -log[H+]
• pOH = -log[OH-]
• pKw = pH + pOH
Comparison of Strong and Weak Acids
Weak Acids
H3O+(aq) + NO2-(aq)
HNO2(aq) + H2O(l)
K=








H3O+ NO-2 






2 






HNO H2O
Ka =








H3O+ NO-2 




HNO


2 

• Degree of ionization isthe ratio of the quantity of
a substance that is ionized to the concentration of
the substance before ionization.
• When expressed as a percentage, it is called
percent ionization.
Practice
Calculate and compare [H+] for a 0.100 M
solution of HClO4 and a 0.100 M solution of
HClO (Ka = 2.9 x 10-8 for HClO).
Practice
Calculate and compare [H+] for a 0.100 M solution of
HClO4 and a 0.100 M solution of HClO (Ka = 2.9 x 10-8
for HClO).
0.100
HClO4
0
[H+]=0.100 = 10-1
0.100
HClO
0.100 - x
0
0
H+ + ClO4-
0.100
0.100
pH = 1
0
0
H+ + ClOx
x
initial
final
initial
equilibrium
Continued
Ka =
[H+][ClO-]
[HClO]
=
X2
0.100 - x
Continued
Ka =
[H+][ClO-]
[HClO]
=
X2
0.100 - x
= 2.9 x 10-8
Continued
Ka =
[H+][ClO-]
[HClO]
=
X2
0.100 - x
= 2.9 x 10-8
Assume X small
X2 = (0.100)(2.9X10-8)
X = [H+] = 5.4 X 10-5
pH = 4.27
Continued
Ka =
[H+][ClO-]
[HClO]
=
X2
0.100 - x
= 2.9 x 10-8
Assume X small
X2 = (0.100)(2.9X10-8)
X = [H+] = 5.4 X 10-5
pH = 4.27
% ionization =
[H+]final
[HClO]final
X 100 = 5.4 X 10-2%
Practice
Calculate the degree of ionization for 1.28 M HNO2 and
0.015 M HNO2 solutions (Ka = 4.0 x 10-4).
0.0150
HNO2
0.0150 - x
0
0
NO2- + H+
X
x
Initial
equilibrium
Practice
Calculate the degree of ionization for 1.28 M HNO2 and
0.015 M HNO2 solutions (Ka = 4.0 x 10-4).
0.0150
HNO2
0.0150 - x
Ka =
[NO2-][ H+]
[HNO2]
0
0
NO2- + H+
X
x
Initial
Practice
Calculate the degree of ionization for 1.28 M HNO2 and
0.015 M HNO2 solutions (Ka = 4.0 x 10-4).
0.0150
HNO2
0.0150 - x
Ka =
[NO2-][ H+]
[HNO2]
0
0
NO2- + H+
X
x
=
(X)(X)
0.015-X
Initial
Practice
Calculate the degree of ionization for 1.28 M HNO2 and
0.015 M HNO2 solutions (Ka = 4.0 x 10-4).
0.0150
HNO2
0.0150 - x
Ka =
[NO2-][ H+]
[HNO2]
0
0
NO2- + H+
X
x
=
(X)(X)
0.015-X
small
Initial
= 4.0 x 10-4
Practice
Calculate the degree of ionization for 1.28 M HNO2 and
0.015 M HNO2 solutions (Ka = 4.0 x 10-4).
0.0150
HNO2
0
NO2- + H+
0.0150 - x
Ka =
0
[NO2-][ H+]
[HNO2]
X
x
=
(X)(X)
0.015-X
Initial
= 4.0 x 10-4
small
X2 = (0.0150)(4.0 x 10-4) = 6.00 X 10-6
X = 2.45 x 10-3
Practice
Calculate the degree of ionization for 1.28 M HNO2 and
0.015 M HNO2 solutions (Ka = 4.0 x 10-4).
0.0150
HNO2
0.0150 - x
Ka =
[NO2-][ H+]
[HNO2]
0
0
NO2- + H+
X
x
=
(X)(X)
0.015-X
Initial
= 4.0 x 10-4
small
X2 = (0.0150)(4.0 x 10-4) = 6.00 X 10-6
X = 2.45 x 10-3
[H+]
X 100
% ionization =
[HNO2] Iinitial
Practice
Calculate the degree of ionization for 1.28 M HNO2 and
0.015 M HNO2 solutions (Ka = 4.0 x 10-4).
0.0150
HNO2
0.0150 - x
Ka =
[NO2-][ H+]
[HNO2]
0
0
NO2- + H+
X
x
=
(X)(X)
0.015-X
Initial
= 4.0 x 10-4
small
X2 = (0.0150)(4.0 x 10-4) = 6.00 X 10-6
X = 2.45 x 10-3
[H+]
2.45 x 10-3
X 100 =
% ionization =
X 100
[HNO2] Iinitial
0.015
= 16 %
Practice
Calculate the pH for a 0.0500 M HCl solution.
Practice
Calculate the pH of a 1.22 M HNO2 solution.
Strong Bases
• Strong bases completely dissociate in
water producing hydroxide ions.
• All soluble metal hydroxides are strong
bases.
Weak Bases
• Weak bases accept hydrogen ions from
water to form hydroxide ions.
Weak Base Equilibria
NH4+ +
NH3 + HOH
Kb =
NH OH 
NH 
OH-

+
4
3
= 1.76 x 10
-5
Practice
Calculate and contrast [OH-] in 0.200 M
LiOH and 0.200 M NH3 (Kb = 1.76 x 10-5).
Calculation of pH for Bases
• pOH = -log[OH-]
• Kw = [H3O+][OH-] = 1.0 x 10-14
• -log(Kw) = -log[OH-] - log[H3O+] = 14.00
• pKw = pOH + pH = 14.00
Practice
Calculate the pH of a 0.200 M KOH
solution.
Polyprotic Acids
• Monoprotic acids have one ionizable
hydrogen atom per molecule, whereas
polyprotic have two or more.
• Carbonic Acid
H2CO3(aq)
HCO3-(aq)
H+(aq) + HCO3-(aq)
H+(aq) + CO32-(aq)
Practice
Calculate the pH of 0.100 M carbonic acid
solution (Ka1 = 4.3 x 10-7 and Ka2 = 4.7 x 10-11).
0.100
H2CO3
0.100 - x
[H+][HCO3]
Ka1 =
[H2CO3]
0
0
initial
H+ + HCO3x
x
Equilibrium
Practice
Calculate the pH of 0.100 M carbonic acid
solution (Ka1 = 4.3 x 10-7 and Ka2 = 4.7 x 10-11).
0.100
H2CO3
0.100 - x
[H+][HCO3]
Ka1 =
=
[H2CO3]
0
0
initial
H+ + HCO3x
x2
0.100-x
small
x
Equilibrium
Practice
Calculate the pH of 0.100 M carbonic acid
solution (Ka1 = 4.3 x 10-7 and Ka2 = 4.7 x 10-11).
0
0.100
H+ + HCO3-
H2CO3
0.100 - x
[H+][HCO3]
Ka1 =
[H2CO3]
0
x
=
x
initial
Equilibrium
x2
-7 X=2.07x10-4
=
4.3
x
10
0.100-x
small
Practice
Calculate the pH of 0.100 M carbonic acid
solution (Ka1 = 4.3 x 10-7 and Ka2 = 4.7 x 10-11).
0
0.100
0.100 - x
2.07x10-4
HCO3-
2.07x10-4 - y
initial
H+ + HCO3-
H2CO3
[H+][HCO3]
Ka1 =
[H2CO3]
0
x
=
x
Equilibrium
x2
4.3 x 10-7 X=2.07x10-4
=
0.100-x
0
2.07x10-4
initial
H+ + CO32.07x10-4
+y
y
Equilibrium
Practice
Calculate the pH of 0.100 M carbonic acid
solution (Ka1 = 4.3 x 10-7 and Ka2 = 4.7 x 10-11).
0.100
0
0.100 - x
[H+][HCO3]
Ka1 =
=
[H2CO3]
2.07x10-4
x
Equilibrium
-7 X=2.07x10-4
x2
4.3
x
10
0.100-x =
0
initial
H+ + CO3-
2.07x10-4 + y
2.07x10-4 - y
3
x
2.07x10-4
HCO3-
Ka2 =
initial
H+ + HCO3-
H2CO3
[H+][CO32-]
[ HCO -]
0
=
[2.07x10-4 + y][y]
2.07x10-4 - y
small
=
y
4.7 x 10-11
Summary of Results
[H+] = 2.07 x 10-4
[HCO3-] = 2.07 x 10-4
[CO32-] = 4.7 x10-11
pH = -log [2.07 x 10-4] = 3.684
pOH = 10.32
Acid Strength and Structure
The greater the ratio of O to H the stronger the acid, since the
electronegative nature of the oxygen withdraws elections away
from the hydrogen weakening the H-O bond.
Structure
and Acid
Strength
Periodic Trends and Acid Strengths
Oxy Acid Strength
Oxo acid is an acid where the proton is directly
attached to the oxygen.
Examples: sulfuric, perchloric, nitric
• Since the oxygen is attached to another atom (Z) then that
atom defines the strength
• The strength depends on the electronegativity of Z, which
draws electrons away from the H-O bond
• The greater the electronegativity, the weaker the bond and the
greater the strength of the acid, thus HOCl is stronger than
HOBr
Oxy Acid Strength
Oxygen atoms attached to Z is important, as well.
The more the better, thus removing more
electrons and weakening the bond, thus HClO4
should be the strongest acid.
Oxygen proton ratio; the greater the stronger
1. oxygen to Proton ratio is most important
2. The larger oxygen to proton ratio, the stronger the
acid.
Binary Acid Strength
Molecular structure and acid strength (Bond
polarity and strength)
Binary acids
In group VIIA (The halogens)
1. Since acids loose protons, then bond strength
ought to be a factor in proton loss
2. Generally speaking the stronger the H-X bond the
weaker the acid
3. Bond strength is related to bond length
4. The shorter the bond the stronger it is
Binary Acid Strength
HI
HBr
Bond Length (pm)
160.9 141.2
Bond Dissociation energy (Kj/molr) 297 368
Acid Ionization Constant K
109
108
HCl
127.4
431
HF
91.7
569
1.3x106
6.6x10-4
Why is HF a weak acid and all the rest strong?
Binary Acid Strength
HI
HBr
Bond Length (pm)
160.9 141.2
Bond Dissociation energy (Kj/molr) 297 368
Acid Ionization Constant K
109
108
HCl
127.4
431
HF
91.7
569
1.3x106
6.6x10-4
Why is HF a weak acid and all the rest strong?
Probably due to the fact that the H3O+ and F- are
held together by hydrogen bonds thus keeping the
hydronium ion concentration low
Binary Acid Strength
Polarity of the HA bond is important within a period
• For example period number 2 acids
• Thus HA strength increases as the
electronegativity of A increases from left to right
on the periodic chart.
• HCH3<HNH3<HOH<HF With in a period since Arelatively the same size.
For the binary acids we see that within a group bond length
is the deciding factor, but with in a period bond polarity is a
factor, since size is not changing as much as it does down
a group.
Acid-Base Properties of Salts
Cation
neutral
neutral
conj acid
of weak
base
conj acid
of weak
base
Anion
neutral
conj base
of weak
acid
neutral
conj base
of weak
acid
Acidic
or Basic
neutral
Example
NaCl
basic
NaF
acidic
NH4Cl
depends
on Ka & Kb (NH4)2SO4
values
Ka/Kb Relationship
HA(aq) + H2O(l )
A-(aq) + H2O (l)
+][A-]
[H
O
H3O+(aq) + A-(aq) Ka= 3
[HA]
[OH-][HA]
OH-(aq) + HA (aq) Kb=
[A-]
HA(aq)+2H2O(l)+A-(aq)
2 H2O (l)
H3O+(aq)+A-(aq)+OH-(aq)+HA (aq)
H3O+ (aq) + OH- (aq)
[H3O+][A-][OH-][HA]
K=
[HA] [A-]
Kw=[H3O+][OH-]
Ka x Kb=
[H3O+][A-][OH-][HA]
[HA] [A-]
Kw = Ka x Kb= [H+][OH-]
Practice
Calculate the pH of a 0.575 M of sodium
acetate.
Practice
Calculate the pH of a 0.880 M of ammonium
nitrate solution.
Common-Ion Effect
• The common-ion effect is the shift in the
position of an equilibrium caused by the
addition of an ion taking part in the reaction.
• An example would be combining a solution of
Na2CO3 with a solution of H2CO3.
• The presence of Na2CO3 would cause stress
of the H2CO3 equilibrium so that less would
ionize.
HENDERSON HESSELBAUCH EQUATION
[H3O+][A-]
Ka =
[HA]
Now take the log of both sides of the equation
log Ka = log [H3
O +]
-log [H3O+] = - log Ka
[A-]
pH = pKa + log
[HA]
+ log
+ log
[A-]
[HA]
[A-]
[HA]
Henderson-Hasselbalch Equation
pH = pKa
[base]
+ log
[acid]
• This equation is used to calculate an
approximate pH of a common-ion effect
or buffer solution.
• The pH of this solution does not change
when it is diluted with deionized water.
pH Buffers
• Buffer: a solution of a weak acid or base
and its conjugate partner (HA and A-).
Buffers resist changes in pH.
• The weak acid reacts with any added OH• HA + OH- <--> H2O + A• The weak base reacts with any added H3O+
• A- + H3O+ <--> HA + H2O
pH Change in a Buffer as a
Function of Buffer Concentration
Addition of strong base
Addition of strong acid
Buffer Capacity
• Buffer capacity is the quantity of acid or
base that can be added to a pH buffer
without significantly changing the pH of
the buffer.
• Buffer capacity is a function of:
 concentrations of buffer components
 pH of the buffer
Practice
Calculate the pH of a solution containing 0.22 M
acetic acid and 0.13 M sodium acetate.
Practice
Determine the number of moles of sodium
acetate that must be added to 250.0 mL of 0.16
M acetic acid in order to prepare a pH 4.70
buffer (Ka = 1.76 x 10-5).
Titration
A method for determining concentrations of
unknown solutions
Titrant- Substance in burette
Analate-Unknown solution
Standard solution-A stable precisely know solution
End point-When the indicator changes color
Equivalence point-No excess reactants present
Indicator-Weak organic acid used to indicate end point
Inflection point-When slope of titration curve changes sign
Titration curve-Graph pH vs. volume titrant added
End point vs Equivalence point-hopefully the same
Titration
Titrating a SA with a SB
 Equivalence point should be around pH=7
 Choose an indicator that changes around 7,
phenolphthalein is typically the indicator of
choice, but it changes at 8.5 is this OK?
 Go thru the pH changes vs ml NaOH added when
titrating 10.0 ml of 0.10M HCl with 0.10 M NaOH
Titration
NaOH
mL
0
1
2
5
8
9
10
11
12
HCl
moles
0.001
0.0009
0.0008
0.0005
0.0002
0.0001
0.000
0.000
0.000
NaOH
moles
0.00
0.00
0.00
0.00
0.00
0.00
0.00
0.0001
0.0002
[HCl]
[NaOH]
pH
0.10
0.082
0.067
0.033
0.011
0.0053
0.000
0.000
0.000
0.000
0.000
0.000
0.000
0.000
0.000
0.000
0.0048
0.0091
1
1.09
1.12
1.48
1.95
2.26
7.00
11.7
12.0
Total
Volume
10 ml
11 ml
12 ml
15 ml
18 ml
19 ml
20 ml
21 ml
22 ml
Titration
Titration WA with a SB (NaOH and acetic acid)
 The titration produces a salt of a conjugate
base, which is relatively strong since it came
from a WA
 The initial pH is higher than in the titration of
a strong acid (Weak only partially ionizes.
 There is an initial rather sharp increase in pH
at the start of the titration, because a common
ion (acetate) shifts the equilibrium back to the
left, thus raising the pH
 After the sharp increase and before the
equivalence point is a relatively flat region
called the buffer region
Titration
 [H+] = [A-] at the half point of the equivalence
point and the pH = pKa according to the
Henderson/Heselbalch equation
 At the equivalence point the pH>7 due to the
hydrolysis of acetate ion producing more OH The steep portion of the curve is between pH710, less than a drop of titrant to get there
 More limitation in indicators since steep part
starts at pH 7, i.e. cannot use a indicator that
changes color less than pH7
 The salt will react with water (hydrolysis) to
produce OH- thus a basic equivalence point
Acid/Base Indicators
Figure 16.13
Choosing an Indicator
Titration Curves for NH3 and NaOH
Titration Curves for Acetic and
Hydrochloric Acids
Titration Curve for Carbonate Ion
A titration curve is a graph of pH of a solution as
titrant is added.
Practice
A 40.0 mL solution of 0.100 M sodium
hypochlorite is titrated with a 0.100 M
hydrochloric acid solution. Calculate the pH
after 20.0 mL and 30.0 mL additions of the acid.
THE END
ChemTour: Acid Rain
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PC | Mac
This ChemTour explores the effects of fossil fuel burning on
the pH of rainwater, as well as the resulting environmental
and industrial consequences.
ChemTour: Acid-Base Ionization
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This ChemTour explores the differences among Brønsted–
Lowry acids, Brønsted–Lowry bases, Lewis acids, and
Lewis bases.
ChemTour: Self-Ionization of
Water
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This ChemTour illustrates the process by which water
molecules act as both a proton acceptor (base) and a
proton donor (acid), and explores the equilibrium constant
(Kw) for the self-ionization of water.
ChemTour: pH Scale
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This ChemTour introduces the pH scale and uses
interactive graphs to explain the relationship between pH,
pOH [H3O+], and [OH–].
ChemTour: Acid Strength and
Molecular Structure
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Students learn to determine relative acid strength based on
the molecular and electronic structure of the acid.
ChemTour: Buffers
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Students learn to use the Henderson–Hasselbalch equation
to predict the pH of a buffer.
ChemTour: Strong Acid and
Strong Base Titration
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PC | Mac
In this interactive virtual titration lab, students are introduced
to the titration apparatus and learn to determine the
concentration of an unknown acid from the volume of basic
solution added.
ChemTour: Titrations of Weak
Acids
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PC | Mac
Students learn to read and understand the different stages
of a titration curve for a weak acid or polyprotic acid, and
understand what is happening at a molecular level.
If a single drop of water containing 0.02
mmol of HCl is added to 1.0 L of a 10-5 M
solution of AgNO3, will a precipitate form?
AgCl(s)
Ag+(aq) + Cl-(aq)
A) Yes
B) No
Adding a Drop of HCl to a AgNO Solution
Ksp = 1.8×10-10
C) Can’t tell
If a single drop of water containing 0.02
mmol of HCl is added to 1.0 L of a 10-5 M
solution of AgNO3, will a precipitate form?
AgCl(s)
Ag+(aq) + Cl-(aq)
0.02 mmole Cl1.0 L
A) Yes
Ksp = 1.8×10-10
mole
= 2.0 x 10-5 M Cl 103 mmole
B) No
Adding a Drop of HCl to a AgNO Solution
C) Can’t tell
If a single drop of water containing 0.02
mmol of HCl is added to 1.0 L of a 10-5 M
solution of AgNO3, will a precipitate form?
AgCl(s)
Ag+(aq) + Cl-(aq)
0.02 mmole Cl1.0 L
A) Yes
Ksp = 1.8×10-10
mole
-5
=
2.0
x
10
103 mmole
[Cl-] = 10-5
B) No
Adding a Drop of HCl to a AgNO Solution
C) Can’t tell
If a single drop of water containing 0.02
mmol of HCl is added to 1.0 L of a 10-5 M
solution of AgNO3, will a precipitate form?
AgCl(s)
Ag+(aq) + Cl-(aq)
Ksp = 1.8×10-10
mole
-5
=
2.0
x
10
103 mmole
[Ag+] = 10-5
0.02 mmole Cl1.0 L
Q = [Ag+][Cl-] = (2.0X10-5)(10-5) = 2.0 X 10 -10
A) Yes
B) No
Adding a Drop of HCl to a AgNO Solution
C) Can’t tell
If a single drop of water containing 0.02
mmol of HCl is added to 1.0 L of a 10-5 M
solution of AgNO3, will a precipitate form?
AgCl(s)
Ag+(aq) + Cl-(aq)
Ksp = 1.8×10-10
mole
-5
=
2.0
x
10
103 mmole
[Cl-] = 10-5
0.02 mmole Cl1.0 L
Q = [Ag+][Cl-] = (2.0X10-5)(10-5) = 2.0 X 10 -10
Q > Ksp Yes
A) Yes
B) No
Adding a Drop of HCl to a AgNO Solution
C) Can’t tell
Given that Ksp(AgCl) > Ksp(AgBr),
which of the following salts, when
added in excess to an aqueous 0.1 M
AgNO3 solution, will result in the
lowest concentration of Ag+(aq)?
A) AgNO3
B) NaCl
Concentration of Ag+ In Ionic Solutions
C) AgBr
Consider the following arguments for each answer
and vote again:
A. Because of the common-ion effect, the addition of
AgNO3(s) will cause a net decrease in the
concentration of Ag+(aq).
B. Adding NaCl will induce the precipitation of AgCl(s)
from the solution, thus lowering the Ag+(aq)
concentration.
C. AgBr(s) is less soluble than AgCl(s), and so its
addition will cause the greatest decrease in the
Ag+(aq) concentration.
Concentration of Ag+ In Ionic Solutions
Suppose water is slowly added to a
vessel containing a speck of the
sparingly soluble salt BaSO4(s).
Which of the following plots shows
the equilibrium concentration of
Ba2+(aq) in the resulting solution
versus the amount of water added?
A)
B)
Dissolution of a Speck of BaSO in H O
C)
Consider the following arguments for each answer
and vote again:
A. As water is added and more BaSO4(s) is dissolved,
the concentration of Ba2+(aq) will increase until the
solution becomes saturated.
B. The concentration of Ba2+(aq) will increase until all
the BaSO4(s) has dissolved, after which additional
water will decrease the Ba2+(aq) concentration.
C. Until the BaSO4(s) has completely dissolved, the
concentration of Ba2+(aq) will remain constant.
Dissolution of a Speck of BaSO in H O
The conductivity of an aqueous solution
is directly proportional to the
concentration of the ions present.
Given this fact, which of the following
plots shows the conductivity of a NaCl
solution as a function of the amount of
AgNO3(s) added?
A)
B)
Conductivity of a NaCl + AgNO Solution
C)
Consider the following arguments for each answer
and vote again:
A. Adding AgNO3(s) will increase the total ion
concentration, so the conductivity will increase until
the solution is saturated.
B. As AgNO3(s) is added, the conductivity of the
solution will decrease because of the precipitation of
AgCl(s) until all of the Cl-(aq) is consumed.
C. Although AgCl(s) will precipitate as AgNO3(s) is
added, the total concentration of ions will remain
constant until the Cl-(aq) is depleted.
Conductivity of a NaCl + AgNO Solution
To the left is a plot of the
autoionization constant, Kw,
versus temperature. What is
the pH of hot water?
A) < 7
Dependence of pH on Temperature
B) 7
C) > 7
Consider the following arguments for each answer
and vote again:
A. At higher temperatures, the concentrations of H3O+
and OH- increase. Therefore, the pH of hot water is
less than 7.
B. Regardless of temperature, the concentrations of H3O+
and OH- remain equal, so the pH remains 7, which is
neutral.
C. At higher temperatures, H+ ions acquire enough
kinetic energy to escape the solution, leaving a
predominance of OH- ions.
Dependence of pH on Temperature
Which of the following, when added to
an NH3(aq) solution, will form a basic
buffer?
A) NaOH
NH Buffer Solution
B) HCl
C) NaCl
Consider the following arguments for each answer
and vote again:
A. NH3, a weak base, is normally an acidic buffer, so to
create a basic buffer, one must add NaOH.
B. By adding HCl to the NH3 solution to form some
NH4+, the solution will become a basic buffer.
C. NH3 is already a weak base, so to create a basic buffer
solution, one need only add a neutral buffering salt
like NaCl.
NH Buffer Solution
To the left is a plot that shows
the pH of a weak acid as it is
titrated with 0.01 M NaOH.
Which of the following plots
would correspond to the same
titration if the same weak acid
were diluted with water and
then titrated with 0.01 M
NaOH?
A)
B)
Titration of a Diluted Weak Base
C)
Consider the following arguments for each answer
and vote again:
A. Diluting a weak acid with water will increase the
initial pH of the solution and decrease the final pH of
the solution.
B. The dilution would have little effect on the initial pH
of the weak acid, especially in the buffer region.
However, the pH after the equivalence point will be
lower.
C. If the weak acid is diluted, the titration will reach the
equivalence point sooner, since the concentration of
the acid will be lower.
Titration of a Diluted Weak Base
Given that the conductivity of an aqueous
solution depends on the concentration of
the ions present, which of the following
graphs shows conductivity (y-axis) plotted
against the acid added (x-axis) for the
titration of the strong base Ba(OH)2 with
the strong acid H2SO4?
A)
B)
Conductivity of a H SO /Ba(OH) Solution
C)
Consider the following arguments for each answer
and vote again:
A. This is a titration of a strong base with a strong acid,
so the conductivity will track the pH of the solution.
B. Although BaSO4(s) will precipitate as H2SO4 is
added, the total concentration of ions will remain
constant until the Ba2+(aq) is depleted.
C. The conductivity will decrease as BaSO4(s) and
H2O(λ) are formed, after which excess H2SO4 will
increase the conductivity.
Conductivity of a H SO /Ba(OH) Solution