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Chapter 16 - Acid-Base
Equilibria
Homework:
13, 14, 15, 16, 17, 18, 20, 21, 22, 23,
24, 25, 27, 29, 30, 31, 34, 36, 37, 38,
39, 40, 41, 42, 43, 44, 45, 46, 47, 50,
52, 53, 55, 56, 61, 62, 65, 71, 72, 73,
74, 75, 81, 82, 83, 85, 86, 87, 88, 92,
16.2 - Bronsted-Lowry Acids
and Bases

In chapter 4, introduced the Arrhenius
concept of acids and bases




An Arrhenius acid was something that produced a
H+ ion in an aqueous solution
An Arrhenius base was something that produced a
OH- ion in an aqueous solution
Unfortunately, Arrhenius concept limits acids
and bases to aqueous solutions.
Bronsted and Lowry expanded and made a
more general definition of acids and bases.
The H+ Ion in Water.


HCl ( g ) 
 H (aq)  Cl (aq)
H 2O

Given this equation, we see that hydrogen
chloride ionizes in water to form H+(aq).



An H+ ion is a proton with no valence electrons
This small, positively charged particle interacts
strongly with the nonbonding electrons pairs of the
water molecules
This forms hydrated hydrogen ions
The hydronium ion

When H+ interacts with water, it forms the
hydronium ion


H3O+
The hydronium ion will then form bonds with other
water molecules, making large clusters of
hydrated hydrogen ions


H5O2+ or H9O4+
Chemists will use H+(aq) and H3O+(aq)
interchangeably.

It is the hydrated proton that is responsible for the
properties of aqueous solutions of acids
Proton-Transfer Reactions

When we closely look at the reaction
when HCl dissolves in water, we fid the
HCl molecule actual transfers an H+ ion
(proton) to the water molecule


So we can represent the reaction between
an HCl molecule and a water molecule to
form hydronium and chloride ions
HCl(g) + H2O(l)  H3O+(aq) + Cl-(aq)
Defining Acids and Bases

Bronsted and Lowry defined acids and
bases in terms of their ability to transfer
protons


An acid is a substance (molecule or ion) that
can donate a proton to another substance
A base in a substance that can accept a proton

So when HCl dissolves in water, HCl acts as a
Bronsted-Lowry acid


It donates a proton to H2O
And H2O would act as a Bronsted-Lowry base
because it accepts a proton

Because the emphasis is on proton
transfer the concept also applies to
reactions that do not involve solutions

HCl + NH3  Cl- + NH4+


HCl is the Bronsted-Lowry acid
NH3 is the Bronsted-Lowry base
Another Look

NH3(aq) + H2O(l)



NH4+(aq) + OH-(aq)
Ammonia is an Arrhenius base because
adding it to water produces OHIt is a Bronsted-Lowry base because it
accepts a proton from H2O
The H2O is the Bronsted-Lowry acid
because it donates a proton to the NH3
molecule

Acids and bases always work together to
transfer a proton



A substance can function as an acid ONLY if there
is another substance to simultaneously act like a
base
To be a Bronsted-Lowry acid, a molecule or ion
MUST have a hydrogen atom it can lose as an H+
ion
To be a Bronsted-Lowry base, a molecule or ion
MUST have a nonbinding pair of electrons to use
to bind the H+ ion
Both?

Some substances can act like an acid in one
reaction and a base in another

H2O is a Bronsted-Lowry base when reacting with HCl


H2O is a Bronsted-Lowry acid when reacting with NH3


HCl(g) + H2O(l)  H3O+(aq) + Cl-(aq)
NH3(aq) + H2O(l)
NH4+(aq) + OH-(aq)
A substance that is capable of being both an acid
and a base is called amphiprotic


Acts a base when combined with something more
acidic than itself
Acts as an acid when combined with something more
basic than itself
Conjugate Acid-Base Pairs


In any acid-base equilibrium both the
forward and reverse reaction involve proton
transfers
Example


Consider the reaction of an acid (HX) with
water
HX(aq) + H2O(l) X-(aq) + H3O+(aq)

In the forward reaction HX donates a proton to H2O


So HX is the acid, and H2O is the base
In the reverse reaction, H3O+ donates a proton to the
X- ion

An acid and a base such as HX and Xthat differ only because of the presence
of a proton are called a conjugate acidbase pair

Every acid has a conjugate base

Formed from the removal of a proton from the
acid


OH- is the conjugate base of H2O
X- is the conjugate base of HX
Likewise

Every base has a conjugate acid

Formed by adding a proton to the base


H3O+ is the conjugate acid of H2O
HX is the conjugate acid of X-
Examples with Equations


In any acid-base reaction, we can identify
two sets of conjugate acid-base pairs.
Nitrous acid and water
remove H+

HNO2(aq) + H2O(l)
Acid
Base
NO2-(aq) + H3O+(aq)
Conjugate base
add H+
Conjugate acid
NH3and H2O reaction
add H+

NH3(aq) + H2O(l)
Base
Acid
NH4+(aq) + OH-(aq)
Conjugate acid
remove H+
Conjugate base
Example

What is the conjugate base of each of
the following acids

HClO4


H2S


HS-
PH4+


ClO4-
PH3
HCO3
CO3-2
Example 2

What is the conjugate acid for each of
the following bases?

CN

SO42

HSO4-
H2O


HCN
H3O+
HCO3
H2CO3
Example 3

The hydrogen sulfite ion (HSO3-) is
amphiprotic

Write an equation for the reaction of HSO3with water, in which the ion acts as an acid


Show the conjugate acid-base pairs
Then write an equation for the reaction of
HSO3- with water, in which the ion acts as
an base

Also identify the conjugate acid-base pairs
HSO3- as acid then base

HSO3-(aq) + H2O(l)
H3O+(aq)

SO32- (aq) +
Conjugate pairs
HSO3-(aq) and SO32- (aq) are acid and conjugate
base, respectively
+
 H2O(l) and H3O (aq) are base and conjugate acid,
respectively
 HSO3-(aq) + H2O(l)
H2SO3(aq) + OH
(aq)

Conjugate pairs


HSO3-(aq) and H2SO3(aq) are base and conjugate
acid, respectively
H O(l) and OH-(aq) are acid and conjugate base,
Relative Strengths of Acids
and Bases



Some acids are better proton donors than
others
Likewise, some bases are better proton
acceptors than others
If we arrange acids in order of their ability to
donate a proton...


We find the easier they give up a proton, the less
easily its conjugate base accepts a proton
Similarly, the more easily a base accepts a proton,
the less easily the conjugate acid gives up a
proton
Generally

The stronger an acid (more easily it
gives up a proton), the weaker it s
conjugate base.



If we know something about the strength of
an acid, we also know something about the
strength of its conjugate base.
Example table shown on pg. 674
We can break up acids and bases into 3
broad categories, based on their behavior
in water
Table
Three Broad Categories
1.
Strong Completely
transfer their
protons to water


No undissociated
molecules in
solution
Conjugate bases
have a negligible
tendency to attract
protons
Three Broad Categories
2.
Weak acids/Weak
Base



Only partially
dissociate in water
Exist in solution as
a mixture of acid
molecules and their
constituent ions
Conjugate bases
show slight ability
to attract protons
Three Broad Categories
3.
Negligible
acids/Strong Base


Do not act like
acids in water
Conjugate bases
are very strong
Equilibrium?

In every acid-base reaction, the position of
the equilibrium favors the transfer of the
proton to the stronger base.

Which means the equilibrium mixture has more of
the weaker acid and weaker base and less of the
stronger acid and stronger base


Because strong acid has a weak conjugate base (and
vice versa)
Which really means the reaction will favor the side
with the stronger base
Example


For the following proton-transfer
reaction (i.e. acid-base reaction), predict
whether the equilibrium lies
predominantly to the left (Kc < 1) or to
the right (Kc > 1)
HSO4-(aq) + CO32-(aq) SO42-(aq) + HCO3(aq)

Use table on pg. 674 to determine the stronger
acid/base
HSO4-(aq) + CO32-(aq)
(aq)

SO42-(aq) + HCO3-
Equilibrium will favor the proton going to the
stronger of the two bases

The two bases in this equation are CO32- (in the
forward reaction) and SO42- (in the reverse
reaction)

Looking at the chart on pg. 674, I see




CO32- is lower on the chart for bases than SO42This means CO32- is a stronger base
Reaction favors side with stronger base, so this reaction
favors the products
Kc < 1
16.3 - The Autoionization of
Water

One of the most important properties of water
is its ability to act as wither an acid or base.



In the presence of an acid, water acts like a base
(proton acceptor)
In the presence of a base, water acts like an acid
(proton donor)
In fact, one water molecule can donate a
proton to another water molecule

This is called the autoionization of water


H2O + H2O H3O+ + OHWhen this happens, no individual
no individual molecule remains ionized
molecule remains ionized for long
ionized for long


Very rapid reaction
At room temperature, about 2 out of every
about 2 out of every 109 molecule are
The Ion Product of Water

Because the ionization of water is an
equilibrium process, we can write the
following equilibrium-constant
expression for it


H2O(l) + H2O(l) H3O+(aq) + OH-(aq)
Kc = [H3O+][OH-]

Where did the H2O go?

Not in expression, because is a pure liquid
Kc

=
[H3O+][OH-]
Because this
equilibrium-costat
expressio refers
specifically to the
autoioizatio of
water, we use the
symbol Kw to
represet this


Kw is the io-product
costat for water
At 25ºC, Kw = 1.0x10-14
Playing with the equation

Since H+(aq) and H3O+(aq) both mean
the hydrated proton, we can also say




H2O(l) H+(aq) + OH-(aq)
Likewise, Kw can also be expressed as
Kw = [H+][OH-] = 1.0x10-14
This equilibrium-constant expression
and Kw are VERY IMPORTANT

Remember these!
Why is this important?


It is not only applicable to pure water, but to
any aqueous solution
Although equilibrium between H+ and OH(along with other ionic equilibria) are affected
by the presence of other ions in solution, we
tend to ignore those ions except in work
requiring a LOT of accuracy

So the equilibrium-constant expression we just
made can be used to find either [H+] or [OH-] for a
dilute aqueous solution

A solution where [H+] = [OH-] is called
neutral


Most solutions are not neutral.
In most solutions, as one of the above
increases, the other decreases



This keeps the product of their concentrations
equal to 1.0x10-14
When [H+] > [OH-], the solution is acidic
When [H+] < [OH-], the solution is basic
Example

Determine the values of [H+] and [OH-]
in a neutral solution at 25ºC




Remember, that in a neutral solution [H+] = [OH-]
Kw= [H+][OH-]=(x)(x) = 1.0x10-14
x2 = 1.0x10-14
x = 1.0x10-7 M = [H+] = [OH-]

In an acidic solution [H+]
Example 2

Calculate the concentration of H+(aq) in


a solution in which [OH-] is 0.010 M
a solution in which [OH-] is 1.8x10-9 M


Unless told otherwise, assume temperature is
25ºC
Why is this important?
A solution in which [OH-] is
0.010 M





[H+][OH-] = 1.0x10-14
[H+] = 1.0x10-14/ [OH-]
[H+] = 1.0x10-14/ 0.010
[H+] = 1.0x10-12
Is this solution acidic or basic?


Basic
[OH-] > [H+]
A solution in which [OH-] is
1.8x10-9 M





[H+][OH-] = 1.0x10-14
[H+] = 1.0x10-14/ [OH-]
[H+] = 1.0x10-14/ 1.8x10-9
[H+] = 5.6x10-6
Is this solution acidic or basic?


Acidic
[OH-] < [H+]
The pH scale

The concentration of H+(aq) is usually
very small

For convenience, we usually express this
concentration in terms of pH


pH is the negative logarithm (in base 10)
pH = -log[H+]
Solution pH

Neutral solution





[H+] = [OH-] = 1.0x10-7
pH = -log(1.0x10-7)
pH = -(-7.00) = 7.00
So the pH of a neutral solution is 7.00 at 25ºC
Acidic solution






Acidic solution has [H+] > 1.0x10-7
Let’s have a pH of 1.0x10-3
pH = -log(1.0x10-3)
pH = -(-3.00) = 3.00
So the pH for an acidic solution is less than 7.00
The more acidic the solution, the lower the pH

Basic solution



[H+] < [OH-]
Suppose [OH-] = 2x10-3 M
First we find [H+] (because we need it for pH)





[H+] = Kw / [OH-] = 1.0x10-14 / 2.0x10-3 = 5.0x10-12M
pH = -log(5.0x10-12)
pH = 11.30
So the pH of a basic solution is greater than 7.00
The more basic the solution, the higher the pH
Notes About pH

A change in [H+] by a factor of 10
causes the pH to change by 1


So a solution with a pH of 6 has 10x the
concentration of H+ than a solution of pH 7
Also, a solution with a pH of 1 has 1000x
the concentration of [H+] than a solution
with a pH of 4.
pH Table
Other

Scales
The negative log is also useful for expressing
the amounts of other small numbers


p
So really,
A common





p
p
just means –log
number is pOH
pOH = -log[OH-]
Why is this useful?
-log[H+] + (-log[OH-]) = -logKw
or
pH + pOH = 14.00
Measuring pH


The pH of a solution can be measured
quickly and accurately with a pH meter
We can also use chemicals called
indicators

An indicator is a colored substance that itself
can exist in either an acid or a base form


The two forms have different colors
If you know the pH at which the indicator turns from
one color to the other, you can determine whether a
solution has a higher or lower pH value than this
Indicators
16.5 - Strong Acids and Bases



The chemistry of aqueous solutions often
critically depends on the pH of the solution
Therefore, it is important to look at how the
pH of the solution relates to strong acids and
bases
Strong acids as bases are called strong
electrolytes

This means they exist in aqueous solutions
entirely as ions
Strong Acids

The seven most common strong acids
include six monoprotic acids and one
diprotic acid



HCl, HBr, Hi, HNO3, HClO3 and HClO4
H2SO4
For these strong acids, when in solution
with these, the solutions exists entirely
of H3O+ and its conjugate base
Example


Nitric acid example
HNO3(aq) + H2O(l)  H3O+(aq) + NO3(aq)



No equilibrium arrows
Reaction goes to completion to products
This reaction is also seen as

HNO3(aq)  H+(aq) + NO3-(aq)

In an aqueous solution of a strong acid, the
acid is normally the only significant source of
H+ ions


If pH is more than 6, then we also need to
consider the H+ ions from the autoionization of
H2O
Finding the pH of a strong monoprotic acid is
simple

The [H+] equals the original concentration of the
acid
Example

If I have a 0.20 M solution of HNO3(aq)


Then the [H+] also equals 0.20 M
Diprotic acids (like H2SO4) are a bit
more complex

We will deal with these later
Example 2

What is the pH of a 0.040 M solution of
HClO4?


If HClO4 has a concentration of 0.040, then
[H+] also has a concentration of 0.040
pH = -log[H+] = -log(0.040) = 1.40
Strong Bases


There are relatively few common strong
bases
The most common soluble strong bases are

ionic hydroxides of the alkali metals and the
alkaline earth metals


NaOH, LiOH, Ca(OH)2, Mg(OH2), etc.
These compounds completely dissociate into ions in an
aqueous solution


So 0.30 M NaOH is really 0.30 M Na+(aq) and 0.30 M OH(aq)
In other words, we do not find undissociated NaOH in a
solution

Because strong bases dissociate
entirely (in an aqueous solution), finding
their pH is also relatively
straightforward.
Example
What is the pH of a 0.028 M solution
of NaOH?


Note: Two ways to do this
1.
2.

Find the [H+]
Find pOH, then convert to pH
I will do both
Find the [H+]



[OH-][H+] = 1.0x10-14
[H+] = 1.0x10-14 / 0.028 = 3.57x10-13M
pH = -log(3.57x10-13) = 12.45
Find pOH, then convert to pH




pOH = -log(0.028) = 1.55
pH + pOH = 14.00
14.00 - pOH = pH
14.00-1.55 = 12.45
Example 2

Ca(OH)2 is a strong base that
dissociates in water to give TWO OHions per formula unit


This means that the concentration of OHwill be twice that of the concentration of
Ca(OH)2
What is the pH of a 0.0011 M solution of
Ca(OH)2?



[OH-] = 2x0.0011 M = 0.0022M
pOH = -log(0.0022) = 2.66
pH = 14.00 - pOH = 14.00 - 2.66 =
11.34
Other Strong Bases

If a substance reacts with water to form OH(aq), it will be strongly basic


Most commonly these will be compounds with the
oxide (O2-) ion
Ionic metal oxides (Na2O and CaO) are often used
in these cases


With the metal oxides, each mole of the O2- reacts to
form 2 moles of OH-
Ionic hydrides (H-) and nitrides (N3-) also will act
as bases in water
16.6 - Weak Acids

Most acidic substances are weak acids



Will only partially ionize in aqueous
solution
We use the equilibrium constant for the
ionization reaction to find out how much a
weak acid ionizes
For example purposes, we will
represent a weak acid as HA
Therefore

We can write a reaction with a weak
acid as such
HA (aq) + H2O(l) H3O+(aq) + A-(aq)
or
HA H+(aq) + A-(aq)

Each of these means the same thing
Equilibrium Constant
Expression


Kc
Because [H2O] is a liquid (and solvent), we
leave it out of the equilibrium-constant
expression
So, depending on which reaction equation
we used





H O A 

3
HA
Kc

H A 

HA
Slight changes

As we did for the ion-product constant for the
autoionization of water, we change the
subscript constant to tell the type of equation
to which it belongs


Weak acids, so change the Kc to Ka
Ka is called the acid-dissociation constant


Many Ka values given in Appendix D of book
The magnitude of Ka tells us the tendency of
the acid to ionize in water


The larger the value of Ka, the stronger the acid
Ka will usually be less than 10-3
Finding Ka from pH

To find one or the other of these things,
we treat these problems like an
equilibrium problem (from the last
chapter)
Example

We have a 0.10 M solution of formic
acid (HCHO2) and measured the pH
using a pH meter.


pH was measured to be 2.38
Find the Ka for formic acid at this
temperature
Step 1 - Write the Equation


HCHO2(aq) H+(aq) + CHO2-(aq)
Once we have the equation, we can
write the equilibrium-constant
expression
Step 2 - Determine
concentration of H+



pH = -log[H+] = 2.38
log[H+] = -2.38
[H+] = 10-2.38 = 4.2x10-3M
Step 3 - Ice

HCHO2(aq)
Initial
l
Change
e
H+(aq) + CHO2-(aq)
HCHO2
H+
CHO2-
0.10 M
M
0
0
-4.2x10-3 M
(0.10-4.2x10-
+4.2x10-3 M +4.2x10-3 M
-3
-3
Step 4 - Neat Assumption


Remember, in a weak acid, we will end
up with only a small amount of H+
Equilibrium concentration of formic acid



(0.10 - 4.2x10-3)
.10 - .0042 = .096
Here s the deal: 0.096 is so close to .10
(only off by 4 thousands of M), that we just
assume the concentration of formic acid to
be 0.10
What does this assumption
mean for us?


It means generally we can assume the
change in the concentration of the weak
acid is so small, its equilibrium
concentration will be the same as its
initial!
We can now plug our values into our
equilibrium-constant expression
Step 5 - Plug Everything Into
Equilibrium-Constant Expression

HCHO2(aq)

H CHO  4.2 10 4.2 10 


 1.8 10


Ka
H+(aq) + CHO2-(aq)
3
3
2
HCHO2 

0.10
This is a reasonable answer

Most weak acids have a Ka between 10-3 and 1010
4
Using Ka to Calculate pH

Knowing the Ka AND the initial
concentration of the weak acid, we can
find [H+]

Which then lets us find the pH of the weak
acid solution
Example



0.30 M solution of acetic acid, HC2H3O2
HC2H3O2(aq) H+(aq) + C2H3O2+(aq)
Ka at 25ºC is 1.8x10-5


According to the formula for acetic acid,
the hydrogen that ionizes is the one
separate from the rest (it is attached to the
oxygen)
We write the formula in this way to
emphasize that only that one hydrogen is
ionized
Step 1 - Write the EquilibriumConstant Expression


HC2H3O2(aq) H+(aq) + C2H3O2-(aq)
Ka at 25ºC is 1.8x10-5
Step

2 - ICE
HC2H3O2(aq)  H+(aq) +
C2H3OHC
2 2(aq)
H3O2( H+(aq) C2H3O2aq)
(aq)
Iitia
l
0.30
M
Chag
e
-x
M
Equili (0.30
brium
-x) M
0
+x

0
M
M
+x
x
M
M
EquilibriumCostat
Expressio

H C H O 

 1.8 x10


Ka
2
3
2
HC 2 H 3O2 
( x)( x)
5
Ka 
 1.8 x10
0.30  x
5
( x)( x)
Ka 
 1.8 x10 5
Simplify 0.30  x


This will tur ito the
quadratic
But we ca simplify
this, by remember
that Ka is VERY SMALL


Because Ka is so small,
we recogize equilibrium
will be FAR to the left,
ad that x will be VERY
small compared to the
iitial cocetratio of
acetic acid
Thus, we ca say that x
New(better)
equatio!




Ka =
x2/0.30 =
1.8x10-5
x2 =
(0.30)(1.8x10-5)
= 5.4x10-6
x = (5.4x10-6).5 =
2.3x10-3
[H+] = x = 2.3x10-3
M
Was Our
Assumptio
Valid?


We ca check to see
if our assumptio that
the iitial
cocetratio - x was
approximately equal
to the iitial
cocetratio
Fid the percet of
the acid that ioized

To do this, divide the
[H+] by the iitial
cocetratio of the
Assumptio Check




[H+]/[HC2H3O2] x 100%
0.0023 M/ 0.30 M
x 100%
=0.77%
Our assumptio was
valid
Weak Acid
Properties

We fid the
cocetratio of H+
is oly a very small
percetage of the
cocetratio of the
acid


Which meas oly a
small amout of the
acid actually doates
a proto
Because there is

Properties iclude


Electrical
coductivity
Rate of reactio
a active metal
with
Polyprotic
Acids

May acids have
more tha oe
ioizable H atom



These acids are
called polyprotic
acids
These ted to ioize
i steps
Sulfurous Acid
example (H SO )
Ka1

?
The umbers o the
costats refer to the
particular proto that
ioizig



ad Ka2
So ka1 refers
reactio that
first proto
Ad ka2 refers
reactio that
secod proto
Ka1

>
Ka2
to the
ioizes
is
the
to the
ioizes the
After the first proto left,
we have a egatively
charged compoud.


Usually the Ka
values for the first
ad secod
disassociatio
costat differs by
at least a factor
of 103
Note: Sulfuric acid
(H2SO4) has a Ka1 as
large
Makig Life
Easy

Because Ka1 is so
much larger tha
ay of the other K
values for a give
acid, almost all of
the [H+] comes from
the first ioizatio

As log as the other
Ka values differ by
a factor of 103 or
Example

The solubility of
CO2 i pure water at
25ºC at 0.1 atm is 0.0037 M. The common practice is to
assume that all of the dissolved CO2 is in the form of the
carbonic acid (H2CO3), which is produced by the reaction
between the CO2 and H2O.
CO2(aq) + H2O(l)  H2CO3(aq)




What is the pH of a
0.0037 M solutio
of H2CO3?
Ka1 = 4.3x10-7
Ka2 = 5.6x10-11
Note: Sice Ka2
differs from Ka1 by
more tha 103, we
treat the acid
H2CO3(aq) 
+
H (aq) + HCO3
(aq)
H2CO3
Iitia 0.003
l
7 M
Chag
e
-x
M
Equili (0.00
brium
37 x)M
H+
HCO3-
0
0
+x

M
M
+x
x
M
M
equilibrium
costatexpressio

H HCO 




K a1
3
H 2CO3 
x x 
.0037  x
 4.3x10
7
We make same
assumptio
x x 
.0037  x
 4.3 x10
7
x x   4.3x107
.0037
5

x  4.0 x10  [ H ]

5
pH   log[ H ]   log( 4.0 x10 )  4.40
16.7 - Weak
Bases

May substaces
behave as weak
bases i water


They react with
water, ad pull
protos from H2O
Formig the
cojugate acid of
the base ad OH- ios

B(aq) +
H2O  HB+ +
OH-

The most commoly
ecoutered weak base
is ammoia



NH3(aq) + H2O(l) 
NH4+(aq) + OH-(aq)
So it would  have
the

NH 4 OH
equilibrium-costat
Kb 
expressio
of
NH



3


Kb is the special
equilibrium costat used
Usig Kb to
calculate [OH-]

Frakly, use the
same procedure you
would have used to
fid [H+] usig Ka


Use ICE
Treat the equilibrium
cocetratio of the
weak base as the
same as the iitial
cocetratio of the
weak base
Types of
Bases


Weak
It ca be a little
tricky to determie
a weak base from a
chemical formula
However, weak bases
fall ito two
geeral categories
Category 1

The first category
icludes eutral
substaces with a
atom with a
obodig pair of
electros

These obodig pairs
of electros are
where the base
accepts the proto
Category 2

This category
icludes the aios
of weak acids

I other words: The
cojugate base of a
weak acid
Determie the
Cocetratio
of a Salt

A solutio is made
by addig solid
sodium hypochlorite
(NaClO) to eough
water to make 2.00
L of solutio has a
pH of 10.50. How
may moles of NaClO
were added to the
water?
Basic Necessary
Data


pH of solutio is
10.50
Kb for NaClO is
3.33x10-7 (foud
data table)
i
Step 1: Fid
[OH-]



pOH = 14.00 - pH =
14.00 - 10.50 =
3.50
-log[OH-] = 3.50
[OH-] = 10-3.50 =
3.16x10-4 M

So this is the
equilibrium
cocetratio of
OH-
Step
2 - ICE
ClO-
HClO
OH-
M
0
0
Chag
e
3.16x
10-4M
+3.16
x104M
+3.16
x104M
Equili
(x-
3.16x
3.16x
Iitia
l

Step 3 - Fid
[ClO-]
Kb

HClO OH


ClO 


4
4
(3.16 x10 )(3.16 x10 )
7
Kb 

3
.
3
x
10
( x  3.16 x104 )
Mathmathmathmathmath
4 2
(3.16 x10 )
4
x
 3.16 x10  0.30M
7
3.3x10
Step 4:
moles
Fid
Molarity
0.30M
NaClO

of
is
0.30 moles 2.00 L

 0.60 mol
1L
Relatioship
Betwee Kaad
Kb




Ka x Kb = Kw
or
pKa + pKb = pKw=
14.00
What does this mea?

If we kow the K
value for a
cojugate acid/base
of what we are
Example

Fid the basedissociatio costat
for the fluoride
io.




Kb
for F- is ot
listed i ay tables
However, Ka for HF is
give
Ka for HF is 6.8x10-4
K =
K
/
K =
Base Properties
of Salt
Solutios

We ca assume that
whe salts dissolve i
water, they are
completely
dissociated


Most salts are strog
electrolytes
Cosequetly, the
acid-base properties
of salt solutios are
Ability to
React with
Water

I geeral, a
aio (-) i
solutio ca be
cosidered the
cojugate base of
a acid


Cl- is the cojugate
base of HCl ad C2H3O2is the cojugate base
of HC2H3O2
Whether a aio
Origial Acid
from Cojugate
Base

To idetify the acid
ad assess its stregth,
we simply add a proto
to the aios formula


- plus a
gives H
proto
(H+)
If the acid is a
strog acid, the the
aio i questio will
be a egligible base

Therefore, the aio
will ot affect the pH of

However, if the H
is NOT oe of the
seve strog acids,
the it is a weak
acid


I this case, the
ao is a weak base
Which meas the aio
will react (to a
small extet) with

Aios that have
ioizable protos
(such as HSO3-) are
amphiprotic


They will act as
either a acid or a
base
Behavior depeds o
the relative sizes of
Ka ad Kb for the io
Ability to
React with
Water

Polyatomic catios
whose formulas cotai
oe or more protos
ca be cosidered the
cojugate acid of
weak basis



NH4+ is the cojugate
acid of the weak base
NH3
So NH4+ will react with
water to lower the pH
However,
ios
of
of Catio ad
Aio i
Solutio

If a aqueous salt
solutio cotais




a aio that does
NOT react with water
a catio that does
NOT react with water
we expect the pH to
be eutral
If the solutio
cotais

a
aio
that
reacts

If





solutio
cotais
A catio that reacts with
water to form a
hydroium io
A aio that does ot
react with water
We expect the pH to be
acidic
If

the
the
solutio
cotais
A catio that reacts with
water to form a
hydroium io
A aio that reacts with
water to produce a
I
1.
2.
3.
Summary
A aio that is the
cojugate base of a
strog acid will ot
affect the pH of a
solutio
A aio that is the
cojugate base of a
weak acid will cause
a icrease i pH
A catio that is the
cojugate acid of a
weak base will cause
a decrease i pH
Other metal ios
will cause a
decrease i pH
6. Whe
a solutio
cotais both the
cojugate base of
a weak acid ad
the cojugate acid
of a weak base,
5.
Example

List the followig
i order of
icreasig pH
0.1 M Ba(C2H3O2)2
ii.0.1
M NH4Cl
iii.0.1
M NH3CH3Br
iv.0.1
M KNO3
i.
Aalysis
i.


ii.


0.1
M
Ba(C2H3O2)2
Ba2+ is the io of oe
of the heavy group 2
elemets, will ot
affect pH
C2H3O2- is the cojugate
base of a weak acid.
Will make solutio basic
0.1
M
NH4Cl
Cl- is the cojugate
base of a strog acid.
Will ot affect pH
NH4+ is the cojugate
iii.
0.1


NH3CH3Br
NH3CH3+ is the cojugate
acid of a weak base.
Will make solutio acid
Br- is the cojugate
base of a strog acid.
Will ot affect pH


M
Kb of NH3 is 1.8x10-5
Kb of NH2CH3 is 4.4x104

Sice
NH3 has smaller
Kb, ad is weaker
0.1 M
iv.




KNO3
K+ io is the of the
strog base KOH
NO3- is the cojugate
base of the strog
acid HNO3
Neither will react
with water, makig
solutio eutro
Overall icreasig
pH
Example 2

Predict whether the
salt Na2HPO4 will
form a acidic or
basic solutio o
dissolvig with water


Na2HPO4 is
amphiprotic
2 Possible reactios


HPO42-(aq)  H+ (aq) +
PO43-(aq)
HPO42-(aq) + H2O 
H PO -(aq) + OH-(aq)

HPO42-(aq)
PO43-(aq)






H+ (aq)
+
This is oe of the give
equatios withi the
chapter.
Ka = 4.2x10-13
HPO42-(aq)
H2PO4-(aq)


+
+
H2O 
OH-(aq)
We ca fid the Kb for
this by its cojugate
acid, H2PO4Ka = 6.2x10-8
Ka x Kb = Kw
Therefore, Kb= 1.6x10-
Base Behavior
ad Chemical
Structure


Whe a substace is
dissolved i water,
it may behave as
a acid, base, or
be either.
We will look at how
the chemical
structure of a
substace determies
Factors that
Affect Acid
Stregth

A molecule cotaiig H
will trasfer a proto
ONLY if the H- bod is
polarized



Meaig that the  has a
relatively large
electroegativity
Which is why HCl is so
acidic (Cl has a large
electroegativity)
Why NaH is basic

Because the H atom possesses
a egative charge ad will
But that
all!

is
ot
A secod factor that
helps determie
whether a molecule
that has a H-
bod will doate a
proto is the
stregth of the
bod

Very strog bods are
less easily dissociated
Ad still ot
all!

A third factor is the
stability of the
cojugate base, 

I geeral, the greater
the stability of the
cojugate base, the
stroger the acid
So the overall
stregth of a acid is
depedet upo these
three factors
Biary Acids

I geeral, the H-
bod stregth is the
most importat factor
i fidig the acid
stregth amog biary
acids i which  is i
the same group i the
periodic table


A biary acid is oe
cotaiig just hydroge
ad oe other elemet
The stregth of a acid
teds to decrease as the
Across the
Periodic Table

Bod stregth chage is
less movig across a row
i the periodic table
tha dow the table


Bod polarity becomes
the major factor
determiig the acidity
of a biary acids i the
same row.
So acidity geerally
icreases as the
electroegativity of 
Oxyacids


May commo acids,
such as H2SO4 cotai
oe or more O-H bods
Acids i which OH groups
ad possibly additioal
oxyge atoms are
boud to a cetral
atom are called
oxyacids.

We will look at what
factors determie

Cosider a OH group
boud to some atom, Y,
which might i tur have
other groups attached to
it


- Y - O - H
At oe extreme, Y might be
a metal, such as Na, K or
Mg


Because of low
electroegativity, electros
are pulled to the oxyge, ad
ioic compoud cotaiig OH- is
formed
These would be sources of OH-,
ad would be a base
Y as a ometal
cotiued
Agai, as Y becomes
more
electroegative,
compoud becomes
more acidic


Two
1.
2.
reasos
for
this
As electro desity is
draw toward Y, the OH bod becomes weaker
ad more polar,
favorig the loss of H+
Because the cojugate
Eve More
Oxyge!

May oxyacids
cotai more oxyge
atoms boded to the
cetral atom Y


The additioal
oxyges (all
electroegative) pull
electro desity from
the O-H bod
This further icreases
its polarity
Summary

For oxyacids that
have the same
umber of OH groups
ad the same umber
of O atoms


Acid stregth
icreases with
icreasig
electroegativity
the cetral atom
of
For oxyacids that
Additioally

Because oxidatio
umber of the
cetral atom
icreases as the
umber of attached
O atoms icreases

I a series of
oxyacids, the
acidity icrease as
the oxidatio umber
Example

Arrage the compouds
i the followig series
i order of icreasig
acidity

AsH3,


All biary acids
So we look at
electroegativity


HI, NaH, H2O
More electroegative
acidic
NaH has lowest
electroegativity, so
basic
= more
most
Example 2

Arrage the compouds
i the followig series
i order of icreasig
acidity


H2SeO3, H2SeO4, H2O
Number of oxyges boded
to cetral atom is the
biggest determier here

Acidity of oxyacids icreases
with # of oxyge atoms
boded

So H2SeO4
is
a stroger
acid
16.11 - Lewis
Acids ad Bases



For a substace to be
a proto acceptor (a
Bosted-Lowry base) it
must have a ushared
pair of electros to
bid the proto.
G.N. Lewis revised the
defiitio of acidbase reactios to focus
o these pairs of
electros
A Lewis acid is a
So

Every base weve
discussed so far is a
electro pair door


far...
Which meas that
everythig that is a
Brosted-Lowry base is
also a Lewis base
However, i the Lewis
theory, a base ca
doate its electro
pair to somethig

NH3 has
us


bee
a
base
for
But cosider reactio
betwee NH3 ad BF3
BF3 acts as a electro
pair acceptor

Makig it a Lewis acid
Solvets

So far, weve
emphasized water as
our solvet ad the
proto as our source
of acidic properties


Brosted-Lowry acid-base
defiitios most useful
whe doig this
Beefit of the Lewis
acid-base defiitios
is it allows us to treat
Commoalities of
Lewis Acids

May Lewis acids have
a icomplete octet of
electros


Like BF3
Also, may simple
catios ca act as
Lewis acids

Fe3+ will take electros
from cyaide ios to form
ferricyaide io,
Fe(CN)63-
Hydrolysis of
Metal Ios

As we have see,
most metal ios act
like acids i a
aqueous solutio.


Lewis acid-base
theory explais
this
Because metal ios
are + charged,
they attract the
Patters

Acid-dissociatio
costats for
hydrolysis reactios
usually icrease
with icreasig
charge ad
decreasig radius
of the atom

Cu2+ io has smaller
charge ad larger
radius tha Fe3+ is