Acids and Bases

1
Acids and Bases
2
Acids

• Svante Arrhenius, a Swedish chemist, defines an
  acid as a substance that yields hydrogen ions
  (H) when dissolved in water.
• Formulas for acids contain one or more hydrogen
  atoms as well as an anion.

3
Naming Acids

• In some cases two different names seem to be
  assigned to the same chemical formula.
• HCl(g) hydrogen chloride
• HCl(l) hydrogen chloride
• HCl(aq) hydrochloric acid
• The name assigned to the compound depends on its
  physical state. In the gaseous or pure liquid
  state, HCl is a molecular compound called
  hydrogen chloride. When it is dissolved in
  water, the molecules break apart into H and Cl-
  ions in this state, the substance is called
  hydrochloric acid.

4

• Binary acids (formed by hydrogen and one other
  element) are named with a hydro- prefix and an
  -ic ending on the anion root.
• Ex- HCl
• HBr

Hydrochloric acid
Hydrobromic acid
5

• The formulas for oxoacids, (acids that contain
  hydrogen and an anion containing oxygen) are
  usually written with the H first, followed by the
  anion, as illustrated in the following examples
• H2CO3 HNO3 HClO2

Carbonic acid
Nitric acid
Chlorous acid
If the anion ends in -ate then the acid ends in
-ic, if the anion ends in -ite, then the acid
ends in -ous. Remember, ic goes with the
higher oxidation state, N has an oxidation state
of ___ in HNO3 (nitric acid) and ___ in HNO2
(nitrous acid)
5
3
6
Acids

• Acids have a sour taste for example, vinegar
  owes its sourness to acetic acid, and lemons and
  citrus fruits contain citric acid.
• Acids cause color changes in plant dyes for
  example, they change the color of blue litmus
  red.
• Acids react with certain metals to produce
  hydrogen gas.
• Acids react with carbonates and bicarbonates to
  produce carbon dioxide gas.
• Aqueous acid solutions conduct electricity.

7
Brønsted Acid

• Arrheniuss definitions of acids are limited in
  that they apply only to aqueous solutions.
  Broader definitions were proposed by the Danish
  chemist Johannes Brønsted. A Brønsted acid is a
  proton donor.
• HCl(aq) H(aq) Cl-

Remember, the H ion is really just a proton (a
hydrogen atom is one proton and one electron, you
pull off the electron and all you are left with
is
8

• The size of a proton is about 10-15 m, compared
  to the diameter of 10-10 m for an average atom or
  ion. Such an exceedingly small charged particle
  cannot exist as a separate entity in aqueous
  solution owing to its strong attraction for the
  negative region of the polar water molecule.
  Consequently, the proton exists in a hydrated
  form as H3O, and is referred to as the hydronium
  ion
• H H2O H3O

9

• Since the acidic properties of the proton are
  unaffected by hydration, we will generally use
  H(aq) to represent the hydrated proton. This
  notation is for convenience only, because H3O is
  closer to reality. Keep in mind that both
  notations represent the same species in aqueous
  solution.

H(aq) H3O(aq)
10

• Monoprotic acids
• each unit of acid yields one hydrogen upon
  ionization
• Diprotic acids
• each unit of an acid gives up two H ions
• Triprotic acids
• yields three H ions upon ionization

HCl
H2CO3
H3PO4
11

• Diprotic acids give up their two H ions in
  separate steps
• H2SO4(aq) H(aq) HSO4-(aq)
• HSO4-(aq) H(aq) SO4-2(aq)
• Triprotic acids give up their H ions in three
  separate steps.

Is HSO4- a strong or weak acid? Explain
Weak, only partially ionizes
12
Bases

• In another definition formulated by Svante
  Arrhenius, a base can be described as a substance
  that yields hydroxide ions (OH-) when dissolved
  in water. Some examples are
• NaOH
• KOH
• Ba(OH)2

Sodium hydroxide
Potassium hydroxide
Barium hydroxide
13

• Ammonia (NH3) is also classified as a common
  base. At first glance this may seem to be an
  exception to the definition of a base. Note that
  as long as a substance yields hydroxide ions when
  dissolved in water, it need not contain hydroxide
  ions in its structure to be considered a base.
  In fact, when ammonia dissolves in water, the
  following reaction occurs
• NH3 H2O NH4 OH-

Thus it is properly classified as a base.
14
Bases

• Bases have a bitter taste
• Bases feel slippery for example, soaps, which
  contain bases, exhibit this property
• Bases cause color changes in plant dyes for
  example, they change the color of red litmus blue
• Aqueous base solutions conduct electricity

15
Brønsted Base

• A Brønsted base is defined by Johannes Brønsted
  as being a substance capable of accepting a
  proton. NH3 is a Brønsted base because it
  accepts a proton to form NH4

16
Lewis Acids and Bases

• G.N. Lewis formulated a definition for what is
  now called a Lewis base a substance that can
  donate a pair of electrons, and a Lewis acid, a
  substance that can accept a pair of electrons.

17

• The significance of the Lewis concept is that it
  is much more general than other definitions. For
  example, the reaction between boron trifluoride
  and ammonia is a Lewis acid-base reaction.

18
Strength of Acids and Bases

• Strong acids are strong electrolytes, which, for
  practical purposes, are assumed to ionize
  completely in water. That means that at
  equilibrium, solutions of strong acids will not
  contain any nonionized acid molecules.
• Like strong acids, strong bases are all strong
  electrolytes that ionize completely in water.

19

• The most common strong acids are HClO4, HCl, HNO3
  and H2SO4. Hydroxides of alkali metals and
  alkaline Earth metals are strong bases (like
  NaOH, KOH and Ba(OH)2).
• Other strong acids and strong bases are listed on
  your Relative Strengths of Acids and Bases
  Reference Sheet.

20

• The strength of an acid is measured by its
  tendency to ionize
• HX ? H X-
• The strength of the H-X bond influences the
  extent to which an acid undergoes ionization.
  The stronger the bond (the higher the bond
  dissociation energy in kJ/mol), the more
  difficult it is for the HX molecule to break up
  and hence the weaker the acid.

21
Bond Dissociation Energies for Hydrogen Halides
and Acid Strengths
Bond Bond Dissociation Energy (kJ/mol) Acid Strength
H-F 568.2
H-Cl 431.9
H-Br 366.1
H-I 298.3
weak
strong
strong
strong
22
The Strength of Oxoacids
Oxoacids contain hydrogen, oxygen, and one other
element Z, which occupies a central position. To
compare oxoacid strength, it is convenient to
separate the oxoacids into two groups.
23
Oxoacids having different central atoms that are
from the same group of the periodic table and
that have the same oxidation number. Within this
group, acid strength increases with increasing
electronegativity of the central atom.
HClO3 gt HBrO3
The Cl pulls more strongly on the electron pair
shared with the O, making the O-H bond more
polar, therefore making it easier to ionize (and
the acid stronger)
24
Oxoacids having the same central atom but
different numbers of attached groups. Within
this group, acid strength increases as the
oxidation number of the central atom increases.
HClO4 gt HClO3 gt HClO2 gt HClO
The greater the number of O atoms pulling on the
Cl, the more that the electrons are pulled away
from the O-H bond, making the O-H bond more
polar, therefore making it easier to ionize (and
the acid stronger)
25

• You are going to remember these trends because
• HCl is a strong acid and HF (with the higher bond
  dissociation energy) isnt.
• HClO4 is a strong acid and HClO3 (where the Cl
  has a lower oxidation number because it has fewer
  oxygen atoms attached to it) isnt.

26

• Note H3O is the strongest acid that can exist
  in aqueous solutions. Acids stronger than H3O
  react with water to produce H3O and their
  conjugate bases.
• Thus, HCl, which is a stronger acid than H3O,
  reacts with water completely to form H3O and
  Cl-.
• HCl(aq) H2O(l) H3O(aq) Cl-(aq)

27

• The OH- ion is the strongest base that can exist
  in aqueous solution. Bases stronger than OH-
  react with water to produce OH- and their
  conjugate acids.
• For example, the oxide ion, (O-2) is a stronger
  base than OH-, so it reacts with water completely
  as follows
• O-2(aq) H2O(l) ? 2OH-(aq)
• For this reason, the oxide ion does not exist in
  aqueous solutions.

28
Amphoteric Compounds
As you could see from the previous two examples,
water will act as either an acid or a base,
depending on the strength of the acid or base
with which it is reacting. Any species that can
react as either an acid or a base is described as
amphoteric.
H2SO4(aq) H2O(l) ? H3O(aq) HSO4-(aq)
Proton acceptor (base)

• NH3(g) H2O(l) ? OH-(aq) NH4(aq)

Proton donor (acid)
29

• An extension of the Brønsted definition of acids
  and bases is the concept of the conjugate
  acid-base pair

CH3COOH(aq) H2O(l) CH3COO-(aq)
H3O(aq)
A conjugate acid-base pair is defined as an acid
and its conjugate base (whats left after the H
was removed from the acid) or a base and its
conjugate acid (substance formed by the addition
of the H to the base).
Because the acid and base are always stronger
than the conjugate acid and conjugate base, the
direction of the reaction proceeds from acid/base
? conjugate acid/conjugate base.
30

• Identify the acid, base, conjugate acid and
  conjugate base in the following reaction
  (Reaction proceeds from stronger to weaker)
• NH3(aq) H2O(l) NH4(aq) OH-(aq)

31
The Acid-Base Properties of Water

• Water is a very weak electrolyte and therefore a
  poor conductor of electricity, but it does
  undergo ionization to a small extent
• H2O(l) H(aq) OH-(aq)

This reaction is sometimes called the
autoionization of water.
32

• In the study of acid-base reactions in aqueous
  solutions, the hydrogen ion concentration is the
  key, because it indicates the acidity or
  alkalinity of the solution. Expressing the
  hydrogen ion as H, we can write the equilibrium
  constant for the autoionization of water as

kw
33

• kw HOH-
• kw is called the ion-product constant, and is the
  product of the molar concentrations of H and OH-
  ions at a particular temperature.

34

• In pure water at 25 oC, the concentrations of H
  and OH- ions are equal and found to be H 1.0
  x 10-7 M and OH- 1.0 x 10-7 M. Thus,
• kw HOH-
• kw (1.0 x 10-7)(1.0 x 10-7)
• kw 1.0 x 10-14

35

• Whether we have pure water or a solution of
  dissolved species, the following relation ALWAYS
  holds at 25 oC
• kw HOH- 1.0 x 10-14

36
Calculate the concentration of OH- ions in an HCl
solution whose hydrogen ion concentration is 1.3
M.
Because HCl is a strong acid HCl ? H Cl-








HCl H Cl-
1.3 0.0 0.0
-1.3 1.3 1.3
0.0 1.3 1.3
kw HOH- 1.0 x 10-14 (1.3)OH- OH-
7.7 x 10-15 M
37

• pH A Measure of Acidity

Because the concentrations of H and OH- ions in
aqueous solutions are frequently very small
numbers and therefore inconvenient to work with,
Soren Sorensen in 1909 proposed a more practical
measure called pH. The pH of a solution is
defined as the negative logarithm of the hydrogen
ion concentration (in mol/L) pH -log H
pH is a dimensionless quantity (it will not have
a label)
38
Since pH is simply a way to express hydrogen ion
concentration, acidic and basic solutions at 25
oC can be distinguished by their pH values, as
follows Acidic solutions H gt 1.0 x 10-7 M,
pH lt 7.00 In an acidic solution there is an
excess of H ions H gt OH- Basic
solutions H lt 1.0 x 10-7 M, pH gt 7.00 In a
basic solution there is an excess of OH- ions
OH- gt H Neutral solutions H 1.0 x 10-7
M, pH 7.00 Whenever H OH-, the aqueous
solution is said to be neutral.
Note when concentration has two significant
figures, pH will have two numbers TO THE RIGHT OF
THE DECIMAL!
39

• Calculate the pH of a 1.0 x 10-3 M HCl solution.

Since HCl is a strong acid, it is completely
ionized in solution HCl(aq) ? H(aq)
Cl-(aq)
HCl(aq) H(aq) Cl-(aq)
I 1.0 x 10-3 0.0 0.0
C -1.0 x 10-3 1.0 x 10-3 1.0 x 10-3
E 0.0 1.0 x 10-3 1.0 x 10-3
Thus, H 1.0 x 10-3 M pH -log(1.0 x
10-3) pH 3.00
40

• The concentration of H ions in a bottle of table
  wine was 3.2 x 10-4 M right after the cork was
  removed. Only half of the wine was consumed.
  The other half, after it had been standing open
  to the air for a month, was found to have a
  hydrogen ion concentration equal to 1.0 x 10-3 M.
  Calculate the pH of the wine on these two
  occasions.

When the wine was first opened pH -log
H pH -log (3.2 x 10-4) 3.49 After the
wine sat for a month pH -log H pH -log
(1.0 x 10-3) 3.00
Why did the acidity increase?
Some of the ethanol converted to acetic acid, a
reaction that takes place in the presence of O2.
41

• Given the pH of a solution, you can figure out
  the H concentration by using the simple
  formula
• H 10-pH

What is the hydrogen ion concentration of an acid
with a pH of 3.00?
H 10-pH H 10-3.00 H 1.0 x 10-3
42
A pH meter is commonly used in the laboratory to
determine the pH of a solution. Although many pH
meters have scales marked with values from 1 to
14, pH values can, in fact, be less than 1 and
greater than 14.
43
A pOH scale analogous to the pH scale can be
devised using the negative logarithm of the
hydroxide ion concentration of a solution. Thus
we define pOH as pOH -logOH-
44
Now consider again the ion-product constant for
water kw HOH- 1.0 x 10-14 Taking the
negative logarithm of both sides we
obtain -logH -logOH- -log(1.0 x 10-14)
Logs make adders multiply
-logH -logOH- 14.00
pH
pOH
45

• In a NaOH solution OH- is 2.9 x 10-4 M.
  Calculate the pH of the solution.

First, figure out the pOH
pOH -log OH- pOH -log (2.9 x 10-4) pOH
3.54
Then use the pOH to figure out the pH
pH pOH 14.00 pH 14.00 pOH pH 14.00
3.54 10.46
46

• Calculate the pH of a 0.020 M Ba(OH)2 solution.

Ba(OH)2 is a strong base each Ba(OH)2 unit
produces two OH-ions Ba(OH)2(aq) ? Ba2(aq)
2OH-(aq)
Ba(OH)2(aq) Ba2(aq) OH-(aq)
I (M) 0.020 0.00 0.00
C (M) -0.020 0.020 2(0.020)
E (M) 0.00 0.020 0.040
Thus, OH- 0.040 M pOH -log 0.040 1.40 pH
14.00 pOH pH 14.00 1.40 12.60
47

• To determine the hydroxide ion when given the
  pOH, you need to use the formula
• 10-pOH OH-

What is the molarity of a NaOH solution that has
a pH of 11.30?
14.00 pH pOH 14.00 11.30 2.70 10-2.70
2.0 x 10-3 OH-
Because NaOH is a strong base, the OH- at the
end is equal to the initial concentration of NaOH.
The molarity of the NaOH 2.0 x 10-3 M
48
Weak Acids and Acid Ionization Constants

• Most acids are weak acids, which ionize only to a
  limited extent in water. At equilibrium, aqueous
  solutions of weak acids contain a mixture of
  nonionized acid molecules, H3O ions, and the
  conjugate base.
• The limited ionization of weak acids is related
  to the equilibrium constant for ionization, which
  is represented as ka.

49

• Consider a weak monoprotic acid, HA. Its
  ionization in water is represented by
• HA(aq) H2O(l) H3O(aq) A-(aq)
• Weak acid
  conjugate base
• or simply
• HA(aq) H(aq) A-(aq)

50

• Write the equilibrium expression for the
  ionization of HA.

ka, the acid ionization constant, is the
equilibrium constant for the ionization of an
acid.
51

• At a given temperature, the strength of the acid
  HA is measured quantitatively by the magnitude of
  ka. The larger ka, the stronger the acid that
  is, the greater the concentration of H ions at
  equilibrium due to its ionization. Keep in mind,
  however, that only weak acids have ka values
  associated with them.

52
Percent Ionization

• Another measure of the strength of an acid is its
  percent ionization, which is defined as

H concentration at equilibrium Initial
concentration of acid
Percent ionization
X 100
53

• The stronger the acid, the greater the percent
  ionization, for example, a strong acid like HCl
  ionizes 100.
• The extent to which a weak acid ionizes depends
  on the initial concentration of the acid. The
  more dilute the solution, the greater the percent
  ionization.

54

• You have a reference sheet that lists a number of
  weak acids and their ka values at 25 oC.
  Although all of the acids on that sheet are weak,
  within the group there is great variation in
  their strengths.
• For example, ka for HF (6.8 x 10-4) is about 1.5
  million times greater than that for HCN (6.2 x
  10-10).

55

• Generally, we can calculate the hydrogen ion
  concentration or pH of an acid solution at
  equilibrium, given the initial concentration of
  the acid and its ka value.
• Alternatively, if we know the pH of a weak acid
  solution and its initial concentration, we can
  determine its ka.

56

• Suppose we are asked to calculate the pH of a
  0.50 M HF solution at 25 oC. The ionization of
  HF is given by

57

• The first step is to identify all the species
  present in solution that may affect its pH.
  Because weak acids ionize to a small extent, at
  equilibrium the major species present are
  nonionized HF and some H and F- ions.
• Another major species is H2O, but its very small
  kw (1.0 x 10-14) means that water is not a
  significant contributor to the H ion
  concentration. Therefore, unless otherwise
  stated, we will always ignore the H or OH- ions
  produced by the autoionization of water.

58
HF(aq) H(aq) F-(aq) We can
summarize the changes in the concentrations of
HF, H, and F- in the table below
HF(aq) H(aq) F-(aq)
I 0.50 0.00 0.00
C -x x x
E 0.50 x x x
59

• The equilibrium concentrations of HF, H and F-,
  expressed in terms of the unknown x, are
  substituted into the ionization constant
  expression to give

(x)(x) 0.50 - x
ka
6.8 x 10-4
Rearranging this expression, we write
x2 6.8 x 10-4x 3.4 x 10-4 0
60

• This is a quadratic equation which can be solved
  using the quadratic formula. Or, we can try
  using a shortcut to solve for x.
• Because HF is a weak acid and weak acids ionize
  only to a slight extent, we reason that x must be
  small compared to 0.50 Therefore we can make the
  approximation
• 0.50 x 0.50

61

• Now the ionization constant expression becomes

x2 0.50 - x
x2 0.50
6.8 x 10-4

62

• Thus we have solved for x without having to use
  the quadratic equation. At equilibrium, we have

HF H F-
(0.50 0.018) M 0.48 M
0.018 M
0.018 M
And the pH of the solution is
pH -log(0.018) 1.74
63

• How good is this approximation? The
  approximation is valid if the following
  expression is equal to or less than 5

Molarity of H at equilibrium
Initial concentration of weak acid
If this percent ionization is greater than 5,
you must use the quadratic formula
64
The Quadratic Equation

• -b vb2 4ac
• 2a

x
If the percent ionization had been greater than
5, we would have had to use the quadratic
equation.
x2 6.8 x 10-4x 3.4 x 10-4 0
a 1 b 6.8 x 10-4 c -3.4 x 10-4
65

• -b vb2 4ac
• 2a

x
-6.8 x 10-4 v(6.8 x 10-4)2 4(1)(-3.4 x 10-4)
2(1)
x
-6.8 x 10-4 ?.0014 2(1)
x
Note this is the same value as we estimated
earlier!
x .018 M or -.018 M
The second solution is physically impossible
because the concentration of ions produced as a
result of ionization cannot be negative.
pH -log(0.018) 1.74
66
If the percent ionization for .50 M HF is 3.6,
what is the percent ionization of HF if you
decrease the molarity to 0.050M?
HF(aq) H(aq) F-(aq)
HF(aq) H(aq) F-(aq)
I 0.050 0.00 0.00
C -x x x
E 0.050 x x x
67
(x)(x) 0.050 - x
ka 6.8 x 10-4
3.4 x 10-5 6.8 x 10-4x x2 0 x2 6.8 x
10-4x - 3.4 x 10-5 using the quadratic
equation, x.0055,
68
Weak Bases and Base Ionization Constants

• Weak bases, like weak acids, are weak
  electrolytes.
• B H2O HB OH-
• weak base conjugate
  acid
• Ammonia is a weak base that ionizes only to a
  limited extent in water
• NH3(aq) H2O(l) NH4 OH-(aq)

69

• The equilibrium constant is given by

NH4OH- NH3
kb
Where kb is called the base ionization constant.
70

• Follow the same procedures you used with weak
  acids when solving problems involving weak bases.
  
• The main difference is that you will calculate
  OH- first, rather than H.

71
The Relationship Between the Ionization Constants
of Acids and Their Conjugate Bases

• For any conjugate acid-base pair it is always
  true that
• kakb kw

72

• Calculate the kb of the conjugate base of acetic
  acid

kakb kw
(1.8 x 10-5)(kb) 1.0 x 10-14
kb 5.6x 10-10
73
Diprotic and Polyprotic Acids

• The treatment of diprotic and polyprotic acids is
  more involved than that of monoprotic acids
  because these substances yield more than one
  hydrogen atom per molecule.
• These acids ionize in a stepwise manner, that is,
  they lose one proton at a time.
• An ionization constant expression should be
  written for each ionization step.

74

• Oxalic acid (H2C2O4) is a poisonous substance
  used chiefly as a bleaching and cleansing agent
  (for example, to remove bathtub rings).
  Calculate the concentrations of all the species
  present at equilibrium in a 0.10 M solution.

75

• H2C2O4 H HC2O4-

H2C2O4(aq) H(aq) HC2O4-(aq)
I (M) 0.10 0.00 0.00
C (M) -x x x
E (M) 0.10 x x x
76
HHC2O4- H2C2O4

• Ka

5.6 x 10-2
(x)(x) 0.10 - x
Ka
5.6 x 10-2
Let me save you some work, you need to use the
quadratic formula for this one
x2 5.6 x 10-2x - 5.6 x 10-3 0
x 0.052
77

• When the equilibrium for the first stage of
  ionization is reached, the concentrations are

H HC2O4- H2C2O4
0.052 M
0.052 M
(0.10 - 0.052) M 0.048 M
78

• Next we consider the second stage of ionization.
• At this stage, the major species will be HC2O4-,
  (this serves as the acid in the second stage),
  H, and C2O4-2 (the conjugate base).

79
HC2O4- H C2O4-2
HC2O4-(aq) H(aq) C2O4-2(aq)
I (M) 0.052 0.052 0.00
C (M) -y y y
E (M) 0.052 y 0.052 y y
80
HC2O4-2 HC2O4-
Ka
5.4 x 10-5
(0.052 y)(y) 0.052 - y
Ka
5.4 x 10-5
Let me save you some work, you DONT need to use
the quadratic formula for this one, 0.052 y and
0.052 y 0.052
(0.052)(y) 0.052
Ka
5.4 x 10-5
y 5.4 x 10-5
81

• Testing the approximation - checking the 5 rule

5.4 x 10-5 M 0.052 M
X 100 .10
The approximation is valid.
82

• At equilibrium

H2C2O4 HC2O4- H C2O4-2
0.048 M
(0.052 5.4 x 10-5) M 0.052 M
(0.052 5.4 x 10-5) M 0.052 M
5.4 x 10-5 M
83

• This example shows that for diprotic acids, if
  Ka1 Ka2, then we can assume that the
  concentration of H ions is the product of only
  the first stage of ionization.

84
Acid Base Properties of Salts

• Salts (which are one of the products of an
  acid-base neutralization reaction) are strong
  electrolytes that completely dissociate into ions
  in water. The term salt hydrolysis describes the
  reaction of an anion or a cation of a salt, or
  both, with water. Salt hydrolysis usually
  affects the pH of a solution.

85
Salts that Produce Neutral Solutions

• NaNO3 Na NO3-

86
Salts that Produce Neutral Solutions

• NaNO3 Na NO3-

87
Salts that Produce Neutral Solutions

• NaNO3 Na NO3-

Consequently, NaNO3 and other salts formed from a
strong acid and a strong base do not affect the
pH of a solution.
88
Salts that Produce Basic Solutions

• NaCH3COO Na CH3COO-

89
Salts that Produce Basic Solutions

• NaCH3COO Na CH3COO-

90
Salts that Produce Basic Solutions

• CH3COO- OH- CH3COOH

Because the CH3COO- ions would bond with the H
ions, the OH- ions (which are left behind when
the H ions come off of the water molecules)
affect the pH of the solution. In other words,
solutions produced by salts made from strong
bases and weak acids will be basic in nature.
91
Salts that Produce Acidic Solutions

• NH4Cl NH4 Cl-

92
Salts that Produce Acidic Solutions

• NH4Cl NH4 Cl-

NH3 H
NH4 is a conjugate acid, and it will dissociate
producing excess H ions. In other words,
solutions produced by salts made from strong
acids and weak bases will be acidic in nature.
93

• When the cation of a salt comes from a weak base
  and the anion comes from a weak acid, you need to
  compare the Ka and Kb values to determine if the
  solution will be acidic or basic. For example,
• NH4NO2 NH4 NO2-
• Ka 5.7 x 10-10 NH4 NH3 H
• Kb 1.4 x 10-11 NO2- HNO2 OH-

If Kb gt Ka, the solution will be basic, if Ka gt
Kb, the solution will be acidic..
therefore an aqueous solution of NH4NO2 will be
acidic.
94

• Oxides can be classified as acidic, basic, or
  amphoteric.
• Most alkali metal oxides and all alkaline earth
  metal oxides form bases when they react with
  water. These oxides are sometimes called basic
  anhydrides.
• Beryllium oxide and several metallic oxides in
  the boron family (Group 3A) and carbon family
  (Group 4A) are amphoteric.

Na2O H2O ?
2NaOH
2Ba(OH)2
BaO H2O ?
95

• Nonmetalic oxides that react with water to form
  acids are sometimes referred to as acidic
  anhydrides.

CO2 H2O ?
H2CO3
SO3 H2O ?
H2SO4
N2O5 H2O ?
2HNO3
P4O10 H2O ?
4H3PO4
Cl2O7 H2O ?
2HClO4
96
Buffer Solutions

• A buffer solution is a solution of (1) a weak
  acid or a weak base and (2) its salt both
  components must be present. The solution has the
  ability to resist changes in pH upon the addition
  of small amounts of either acid or base.
• Buffers are very important to chemical and
  biological systems. The pH in the human body
  varies greatly from one fluid to another for
  example, the pH of blood is about 7.4, whereas
  the gastric juice in our stomach has a pH of
  about 1.5. These pH values, which are crucial
  for proper enzyme function and the balance of
  osmotic pressure, are maintained by buffers in
  most cases.

97
CH3COOH CH3COO- H
Weak acid

• NaCH3COO CH3COO- Na

Conjugate base
If you add a base to this solution, the OH- will
be neutralized by the acetic acid in the buffer,
therefore you will not notice a significant
difference in the pH of the solution.
98
CH3COOH CH3COO- H

• NaCH3COO CH3COO- Na

If you add an acid to this solution, the acetate
ion will bond to most of the added H, so no
appreciable change in pH will be observed.
99

• The buffering capacity, that is, the
  effectiveness of the buffer solution, depends on
  the amount of acid and conjugate base from which
  the buffer is made. The larger the amount, the
  greater the buffering capacity.

100

• In general, a buffer system can be represented as
  weak acid/salt or weak acid/conjugate base. Thus
  the sodium acetate/acetic acid buffer system we
  discussed can be written as
• CH3COOH/CH3COONa
• or simply
• CH3COOH/CH3COO-

101

• Which of the following are buffer systems?

NaClO4/HClO4
KF/HF
102

1. Calculate the pH of a buffer system containing
   1.0 M CH3COOH and 1.0 M CH3COONa.
2. What is the pH of the buffer system after the
   addition of 0.10 mole of gaseous HCl to 1 L of
   the solution. Assume that the volume of the
   solution does not change when the HCl is added.

103
CH3COOH CH3COO- H
I 1.0 1.0 0.0
C -x x x
E 1.0 - x 1.0 x 0.0 x
Assume ionization is negligible
104
CH3COOH CH3COO- H
I 1.0 1.0 0.0
C -x x x
E 1.0 1.0 x
105
CH3COOH CH3COO- H

• ka

1.8 x 10-5
H
1.0 x 10-5
106

• H

1.0 x 10-5
-logH
pH
4.74
When the concentration of the acid and the
conjugate base are the same, the pH of the buffer
is equal to the pKa, which is determined by
taking the log of the ka, of the acid
pH pka -log ka
107

• After the addition of HCl, complete ionization of
  the 1.0 M HCl acid occurs
• HCl ? H Cl-
• 0.10 mol 0.10 mol

This 0.10 mol of H ions will bond with 0.10 mol
of the CH3COO- ions, DECREASING the amount of
CH3COO- by .10 mol and INCREASING the amount of
CH3COOH by .10 mol
108

• So now instead of 1.0 mol of CH3COO- ions, there
  will only be 0.90 mol (1.0 - .10)

And since there were originally 1.0 mol of
CH3COOH, you now have 1.10 mol of CH3COOH with
the added .10 mol
CH3COO-
.90 M
CH3COOH
1.10 M
109

• ka

1.8 x 10-5
H
2.2 x 10-5
110
2.2 x 10-5

• H

pH -log(2.2 x 10-5)
pH
4.66
This is a very slight change in pH, before the
HCl was added, the H 1.8 x 10-5, after the
addition of HCl, the H 2.2 x 10-5, this is
an increase by a factor of 1.2 with a pH change
from 4.74 to 4.66
111

• To appreciate the effectiveness of the
  CH3COONa/CH3COOH buffer, let us find out what
  would happen if 0.10 mol HCl were added to 1 L of
  water, and compare the increase in H ion
  concentration.

112

• Before the addition of HCl
• After the addition of HCl

H 1.0 x 10-7
H 0.10 M
1.0 x 106
This would be a millionfold increase as the pH
changed from 7.00 to 1.00!
113
A "very convenient" equation for dealing with
buffer solutions is the Henderson-Hasselback
equation.
A- HA
pH pKa log
114
Previous question - Calculate the pH of a buffer
system containing 1.0 M CH3COOH and 1.0 M
CH3COONa.
pH pka
pH -log(ka)
pH -log(1.8 x 10-5)
pH 4.74
115
Just as we did previously, lets calculate the pH
of this buffer system after the addition of 0.10
mole of gaseous HCl to 1 L of the solution.
Assume that the volume of the solution does not
change when the HCl is added.
pH -log(1.8 x 10-5) -.087
pH 4.65
116
Calculate the pH of a buffer system containing
.50 M C6H5COOH and 1.0 M C6H5COOK.
pH -log(6.5 x 10-5) .30
pH 4.49
117
What is the pH of the previous buffer system
after the addition of 0.10 mole of gaseous NaOH
to 1 L of the solution. Assume that the volume
of the solution does not change when the base is
added.
Adding a .10 mol base will neutralize .10 mol of
the acid, C6H5COOH, decreasing the molarity of
the acid to .40M, and, for every .10 mol of
C6H5COOH you neutralize, you will form .10 mol
more of C6H5COO-
C6H5COO-
1.10 M
CH3COOH
.40 M
118
C6H5COO-
1.10 M
CH3COOH
.40 M
pH -log(6.5 x 10-5) .44
pH 4.63
119
Acid Base Titrations

• Quantitative studies of acid-base neutralization
  reactions are most conveniently carried out using
  a technique known as titration. In titration, a
  solution of accurately known concentration,
  called a standard solution, is added gradually to
  another solution of unknown concentration, until
  the chemical reaction between the two solutions
  is complete.

120

• If we know the volumes of the standard solution,
  we can calculate the concentration of the unknown
  solution.
• Sodium hydroxide is one of the bases commonly
  used in the laboratory. However, it is difficult
  to obtain solid sodium hydroxide in a pure form
  because it is hygroscopic, (it has a tendency to
  absorb water from air), and its solution reacts
  with carbon dioxide. For these reasons, a
  solution of sodium hydroxide must be standardized
  before it can be used in accurate analytical
  work.

121

• We can standardize the sodium hydroxide solution
  by titrating it against an acid solution of
  accurately known concentration. The acid often
  chosen for this task is a monoprotic acid called
  potassium hydrogen phthalate (abbreviated as
  KHP), for which the molecular formula is KHC8H4O4.

122

• The procedure for the titration of KHP and NaOH
  is as follows
• Add a known amount of KHP to an Erlenmeyer flask.
  Add some distilled water to make up a solution.
• Next, carefully add NaOH solution from a buret
  until the equivalence point is reached. The
  equivalence point is the point at which the acid
  has completely reacted with or been neutralized
  by the base.

123

• The equivalence point is usually signaled by a
  sharp change in the color of an indicator. In an
  acid-base titration, indicators are substances
  that have distinctly different colors in acidic
  and basic solutions. One common indicator is
  phenolphthalein.

Phenolphthalein indicates the presence of a/n
_________. Phenolphthalein is ________ in acidic
solutions, _______ in neutral solutions and
________ in basic solutions.
base
colorless
colorless
pink
124

• At the equivalence point, all the KHP present has
  been neutralized by the added NaOH and the
  solution is still colorless. However, if we add
  just one more drop of NaOH solution from the
  buret, the solution will immediately turn pink
  because the solution is now basic.

125
Apparatus for acid-base titration.
A NaOH solution is added from the buret to a KHP
solution in an Erlenmeyer flask.
A faint pink color appears when the equivalence
point is reached. If your solution turns
fuchsia, you have gone past the equivalence point
126

• The neutralization reaction between NaOH and KHP
  is one of the simplest types of acid-base
  neutralization known. A neutralization reaction
  is a reaction between an acid and a base.
• Aqueous strong acid-strong base reactions produce
  water and a salt, (an ionic compound made up of a
  cation other than H and an anion other than OH-
  or O-2)

127

• The reaction between KHP and sodium hydroxide is
• KHC8H4O4(aq) NaOH(aq) ? KNaC8H4O4(aq) H2O(l)

acid
base
salt
water
128

• You can use the following format to solve
  acid-base neutralization problems

129
What is the molarity of the acid if 16.1 mL of
0.610 M NaOH was required to neutralize 20.0 mL
of H2SO4?
.0200
.0161
.610
130
H2SO4 2NaOH ? Na2SO4 2H2O
Acid
Base
M
M
mol
mol
L
L
.0161
.610
.0200
.00982
.00491
.00491 mol H2SO4
.0161 L NaOH x
x
131
Acid
Base
M
M
mol
mol
L
L
.0200
.0161
.610
.00982
.00491
.0246
.0246 M
132
Titration Curve
The reaction between HCl, a strong acid, and
NaOH, a strong base, can be represented by
HCl(aq) NaOH(aq) ? NaCl(aq) H2O(l)
The pH profile of a titration of this
neutralization reaction is known as a titration
curve.
133

• Consider the addition of 0.10 M NaOH solution
  (from a buret) to an Erlenmeyer flask containing
  25 mL of 0.10 M HCl.

134

• It is possible to calculate the pH of a solution
  at every stage of titration.
• What is the pH of the solution after the addition
  of 10.0 mL of 0.10 M NaOH to 25.0 mL of 0.10 M
  HCl?

135
Because both NaOH and HCl completely dissociate

• 0.10 mol OH-
• 1 L

x
0.0100 L OH-
0.00100 mol OH-
0.10 mol H 1 L
x
0.0250 L H
0.00250 mol H
0.00250 mol H - 0.00100 mol OH-
0.00150 mol H
136

• To determine the pH, you need to calculate H

.043M H
.0350 L
pH -log (.043)
pH 1.37
137

• At the equivalence point of a titration between a
  weak acid and a strong base, the pH will be
  greater than 7 because

138
at neutralization

• CH3COOH NaOH ? CH3COONa H2O

Na CH3COO-
This acetate ion has an affinity for the H ion
in the water, (thus leaving the OH- behind,
making the solution basic.)
139
The pka of a weak acid can be determined
experimentally
140
12.5 mL
25.0 mL
pH pka
141
At half-way to the equivalence point, pH
pka Taking the negative log of both sides
yields 10-pH ka This gives us an experimental
way to determine the ka of a weak acid, and using
a ka table, the identity of an unknown weak acid.
12.5 mL
25.0 mL
4.74 pka 10-4.74 ka ka 1.8 x 10-5
Using a data table, you could match the ka of 1.8
x 10-5 to CH3COOH
142

• A slightly different curve results when you
  titrate a strong acid vs a weak base. At the
  equivalence point of a titration between a strong
  acid and a weak base, the pH will be less than 7,
  because

143

• A weak base, like NH3 would form an acidic salt
  when titrated vs HCl
• HCl NH3 ? NH4Cl

and, at neutralization the NH4Cl would dissociate
NH3 H
resulting in a pH of less than 7 is due to the
presence of H ions formed by the hydrolysis of
NH4
144
Acid-Base Indicators

• An indicator is usually a weak organic acid or
  base that has distinctly different colors in its
  nonionized (molecular) form and ionized form.
  The end point of a titration occurs when the
  indicator changes color.
• However, not all indicators change color at the
  same pH, so the choice of indicator for a
  particular titration depends on the nature of the
  acid and base used in the titration (that is,
  whether they are strong or weak).
• By choosing the proper indicator for a titration,
  we can use the end point to determine the
  equivalence point.

145

• Let us consider a weak monoprotic acid that we
  will call HIn. To be an effective indicator, HIn
  and its conjugate base, ___, must have distinctly
  different colors.

In-
One color
A different color
146

• If the indicator is in a sufficiently acidic
  environment, the equilibrium, according to Le
  Chateliers principle, shifts to the __________
  and the predominant color will be that of ______,

left
HIn
147

• In a basic environment, the equilibrium shifts to
  the right because

The H and the OH- will form water, thus removing
the H from the system and therefore shifting the
equilibrium to the right. The predominant color
will then be that of In-.
148

• The end point of an indicator does not occur at a
  specific pH rather, there is a range of pH
  within which the end point will occur. In
  practice, we choose an indicator whose end point
  lies on the steep part of the titration curve.
  Because the equivalence point also lies on the
  steep part of the curve, this choice ensures that
  the pH at the equivalence point will fall within
  the range over which the indicator changes color.

149
Some Common Acid-Base Indicators
Indicator In acid In base pH range
Thymol blue Red Yellow 1.2 - 2.8
Bromophenol blue Yellow Bluish purple 3.0 4.6
Methyl orange Orange Yellow 3.1 4.4
Methyl red Red Yellow 4.2 6.3
Chlorophenol blue Yellow Red 4.8 6.4
Bromothymol blue Yellow Blue 6.0 7.6
Cresol red Yellow Red 7.2 8.8
Phenolphthalein Colorless Reddish pink 8.3 10.0
The pH range is defined as the range over which
the indicator changes from the acid color to the
base color.