Acids and Bases 2
The Brønsted Definitions

Brønsted Acid  proton donor
•
Brønsted Base  proton acceptor
Conjugate acid - base pair  an acid and its
conjugate base or a base and its conjugate
acid

Example Acid-Base Reactions

Look at acetic acid dissociating
CH3COOH(aq)  CH3COO-(aq) + H+(aq)


Brønsted acid
Conjugate base

Look at NH3(aq) in water
NH3(aq) + H2O(l)  NH4+(aq) + OH-(aq)


Brønsted base
conjugate acid
Representing Protons in Aqueous
Solution
CH3COOH(aq)  CH3COO-(aq) + H+(aq)
CH3COOH(aq) + H2O(l)  CH3COO-(aq) +
H3O+(aq)
HCl (aq)  Cl-(aq) + H+(aq)
HCl(aq) + H2O(l)  Cl-(aq) + H3O+(aq)
Representing Protons

Both representations of the proton are
equivalent

H5O2+ (aq), H7O3+ (aq), H9O4+ (aq) have
been observed

We will use either H+(aq) or H3O+(aq)
What is H+ (aq)?
+
H
H
H
O+
H+
H
H2O
H
O
H +
H3O+
H
H5O2+
OH2
H
H
O
H
+
O
H
H2O
H9O4+
H
The Hydroxide Bases

KOH, RbOH, NaOH, are not strictly Brønsted
Bases since none of these substances
accepts a proton
KOH(aq)  K+(aq) + OH-(aq)
NaOH(aq)  Na+(aq) + OH-(aq)
OH-(aq) + H3O+(aq)  2 H2O(l)
The Autoionization of Water


Water autoionizes (self-dissociates) to a
small extent
2H2O(l)  H3O+(aq) + OH-(aq)
H2O(l)  H+(aq) + OH-(aq)
These are both equivalent definitions of the
autoionization reaction. Water is acting as a
base and an acid in the above reaction 
water is amphoteric.
The Autoionization Equilibrium
–
from the preceding

 chapter
[H 3O ][OH ]
Kc =
[H 2O]

[H  ][OH  ] 
 or

[H 2O] 

– but we know [H2O] is constant
The Defination of Kw
Kc [H2O] = Kw = [H+][OH-]
Ion product constant for water, Kw, is the
product of the molar concentrations of H+
and OH- ions in pure water at a temperature
of 298.15 K
Kw = [H+][OH-] = 1.0x10-14 at 298.2 K
Using Kw


In a solution containing [H+] = [OH-] = 1.0 x
10-7 M, this solution is neutral at 298.2 K.
What if we adjust [OH-] in solution by adding
a small amount of a proton acceptor (a
base)?
–
e.g. [OH-] = 1.0 x 10-4 M
From the Kw expression
Kw = [H+][OH-] = 1.0x10-14
Kw / [OH-] = [H+]
 [H+] = 1.0 x 10-14 / 1.0 x 10- 4
[H+] = 1.0 x 10-10 M
The Definition of an Acidic
Solution

We define an acidic solution as one where
the [H+] in the solution is greater than the the
[H+] in pure water
acidic solution  [H+] > 1.0 x 10-7 mole/L at
298.2 K
The Definition of a Basic Solution


Basic solutions are those where the [H+] in
the solution is less than its concentration in
pure water at 298.2 K.
Basic solution  [H+] < 1.0 x 10-7 mole/L
An alternative definition of a a basic solution
is as follows
Basic solution  [OH-] > 1.0 x 10-7 mole/L
The Definition of a Neutral Solution

A neutral solution is defined as one where
the [H+] in the solution is equal to the
hydrogen ion concentration in pure water
Neutral solution  [H+] = [OH-] = 1.0 x 10-7
mole/L at 298.2 K!
The Dependence of Kw on
Temperature

In our definitions of an acidic, basic, and a
neutral solutions, we had explicitly stated the
temperature as 298.2 K. Why?
Kw is temperature dependent

How will that affect our definition of an acidic,
basic, or a neutral solution?
Neutrality at Body Temperature

At T = 310.15 K (physiological temperature)  Kw =
2.4 x 10-14
•
A neutral solution has [H+] = [OH-] = (Kw)½
•
At 310.15 K, a neutral solution is one where [H+] =
[OH-] = 1.5 x 10-7 M
(UNLESS OTHERWISE INDICATED, ALL
CALCULATIONS WILL BE AT 298.15 K)
The pH scale
Sørenson - 1909 pH = -log [H+]
Solution Type [H+ ] / M
pH range
neutral
solutions
[H+ ] = [OH-] pH = 7.00
= 1.0x10-7
basic
solutions
[H+ ] <
1.0x10-7
pH > 7.00
acid solutions [H+ ] >
1.0x10-7
pH < 7.00
The Relationship between pH and
pOH
pH = -log [H+]
pOH = -log [OH-]
From the Kw expression
Kw = [H+][OH-] = 1.0x10-14 at 298.2 K
-log (1 x 10-14) = -log [H+] -log [OH-]
14.00 = pH + pOH
Acid Strength and % Dissociation
CH3COOH(aq) ⇄ CH3COO-(aq) + H+(aq)
HCOOH(aq) ⇄ HCOO-(aq) + H+(aq)

both weak acids < 5% ionized
Other examples of weak acids  HF, HNO2,
HCN
Acid Strength

The strength of an acid is directly dependent
on its % dissociation ( value)

For an acid
 = % dissociati on =
–
H 

nacid  o
x 100%
n = the number of groups that donate a proton
Base Strength and % Dissociation
Strong Bases  also 100% ionized in water

NaOH(aq)  Na+(aq) + OH-(aq)
Ba(OH)2(aq)  Ba2+(aq) + 2OH-(aq)
Some bases are weak bases; they don’t
ionize completely.
NH3(aq) + H2O(l)  NH4+ (aq) + OH-(aq)
< 5% ionized in aqueous solution
Base Strength

The strength of a base is also directly
dependent on its % dissociation ( value)

For a base
 = % dissociati on =
OH 

mbase o
x 100%
–
[base]o  the original concentration of base
–
m = the number of basic groups in the molecule
Conjugate Acid-Base Strengths
CH3COOH (aq) ⇄ CH3COO-(aq) + H+(aq)
 Note that the conjugate base of acetic
acid is a reasonable proton acceptor
CH3COO-(aq) + H2O (l)  CH3COOH
(aq) + OH-(aq)
Other Examples
HNO3 (aq)  H+(aq) + NO3-(aq)

conjugate base (very weak)
HCOOH (aq) ⇄ HCOO-(aq) + H+(aq)

conjugate base is relatively strong
NH3(aq) + H2O(l) ⇄ NH4+ (aq) + OH-(aq)

relatively strong conjugate acid
Other Examples
HCl (aq)  Cl-(aq) + H+(aq)
 The conjugate base, Cl- ion, is extremely
weak
Cl-(aq) + H2O (l)  HCl (aq) + OH-(aq)
Other Examples her Examples


S2-(aq) + H2O (l)  HS-(aq) + OH- (aq)
The conjugate acid, the HS- ion, is extremely
weak
HS-(aq) ⇄ H+ (aq) + S2- (aq)
The equilibrium lies very far to the left for this
reaction
Other Examples




The greater the acid strength (large Ka), the
weaker the conjugate base of that acid
The weaker the acid (smaller Ka), the
stronger its conjugate base
If the base strength is high (Kb is large), its
conjugate acid is very weak
The weaker the base (small Kb value), the
stronger the conjugate acid of the base
Calculating the pH of Solution of
Strong Acids
For the dissolution of HCl, HI, or any of the
other seven strong acids in water
HCl (aq)  H+ (aq) + Cl- (aq)
 = 100%
HI (aq)  H+ (aq) + I- (aq)
 The pH of these solutions can be easily
obtained from the molarity of the dissolved
acid
pH = -log [H+] = -log[HCl]

}
Calculating the pH of Solution of
Strong Bases

For the dissolution of NaOH, Ba(OH)2,
or any of the other strong bases in
water
NaOH (aq)  Na+ (aq) + OH- (aq)
Ba(OH)2 (aq)  Ba2+ (aq) + 2OH- (aq)
}  = 100%
Calculating the pH of Solution of
Strong Bases

The pH of these solutions is obtained by first
calculating the pOH from the molarity of the
dissolved base
pOH = -log [OH-] = -log[NaOH]
pOH = -log [OH-] = -log{2 [Ba(OH)2]}
pH = 14.00 - pOH
The Seven Strong Acids
chloric acid
HClO3
hydrobromic acidHBr
hydrochloric acid HCl
hydroiodic acid
HI
nitric acid
HNO3
perchloric acid
HClO4
sulphuric acid
H2SO4
 What about the relative strength of the strong
acids?
The Leveling Effect

H+ (aq) (or H3O+(aq)) is the strongest acid
that can exist in aqueous solution. Any acid
stronger than H+(aq) reacts with water
completely to produce H+(aq) and the weak
conjugate base.
The Leveling Effect


HNO3 (aq) is a stronger acid than H+(aq)
(H3O+) \ reacts with water completely to
form H+(aq)
HNO3 (aq)  H+ (aq) + NO3- (aq)
Acids weaker than H+(aq) have the
equilibrium lying primarily to the left.
HNO2(aq)  H+(aq) + NO2- (aq)
The Leveling Effect

The OH- ion is the strongest base that can
exist in aqueous solution. Bases stronger
than OH-(aq) react with water to produce the
hydroxide ion (OH-).
The Leveling Effect

NH2- (the amide ion) is an extremely strong base
(much stronger than OH-). Therefore,
NaNH2 (aq) + H2O (l)  NH3 (aq) + NaOH(aq)

NH2- cannot exist in aqueous solution.

NH3 is a much weaker base than OH-. Therefore,
when it reacts with water, the equilibrium favours the
reactants
NH3 (aq) + H2O (l)  NH4+ (aq) + OH-(aq)
The Leveling Effect Defined

Any acid that is stronger than H+(aq) means that we
have 100% ionisation of the acid. For acids like
HCl(aq), HClO4(aq), HNO3(aq), the appearance is
one of equal acid strength.

Water is said to have a leveling effect on the acid
strength, in that the solvent (in this case water)
cannot differentiate among the relative strength of all
acids stronger than H+(aq), which is water’s
conjugate acid.
Equilibria in Aqueous Solutions of
Weak Acids/ Weak Bases

By definition, a weak acid or a weak base
does not ionize completely in water (
<<100%). How would we calculate the pH of
a solution of a weak acid or a weak base in
water?
Weak Acid/Bases and pH


For the dissolution of HF(aq) in water.
HF (aq)  H+ (aq) + F- (aq)
Ka = 7.1 x 10-4
The small value of Ka indicates that this acid
is only ionized to a small extent at
equilibrium.
The Ka Value



To obtain the pH of a weak acid solution, we must
apply the principles of chemical equilibrium
Define the acid dissociation constant Ka
For a general weak acid reaction
HA (aq)  H+ (aq) + A- (aq)

H A 
=

Ka

HA 
The Ka Value

For a solution of hydrofluoric acid in water
HF (aq)  H+ (aq) + F- (aq)

H F 
=
= 7.1x10

Ka

HF
4
Equilibria of Weak Bases in Water
To calculate the percentage dissociation of a
weak base in water (and the pH of the
solutions)
CH3NH2 (aq) + H2O  CH3NH3+(aq) + OH- (aq)
 We approach the problem as in the case of
the weak acid above, i.e., from the chemical
equilibrium viewpoint.

The Kb Value


Define the base dissociation constant Kb
For a general weak base reaction with water
B (aq) + H2O (l)  B+ (aq) + OH- (aq)

B OH 
=

Kb

B
• For the above system
Kb

CH NH OH 
=
3

3
CH3 NH2 

Diprotic/Polyprotic Acids

Look at the following system.
H2C2O4 (aq)  HC2O4- (aq) + H+ (aq)
HC2O4- (aq)  C2O42- (aq) + H+ (aq)

Ka1
Ka2
In general, we find that for the dissociation of
diprotic and polyprotic acids, the magnitudes
of the dissociation constants decrease in the
direction
Ka1 > Ka2 > Ka3 etc.
Example


For oxalic acid in water,
Ka1= 6.5 x 10-2
Ka2= 6.1 x 10-5
Since Ka1>> Ka2, the [H+] (and the pH) in the
solution is due primarily to the first
dissociation ONLY.
Obtaining the Relationship
between Ka and Kb


We have already seen that there is a relationship
between the strength of an acid and the ability of its
conjugate base to hydrolyse.
HCOOH (aq)  HCOO- (aq) + H+ (aq)
Ka (HCOOH) = 1.8 x 10-4
Examine the reverse reaction, the hydrolysis
(reaction of the substance with water) of HCOOHCOO-(aq) + H2O (l)  HCOOH (aq) + OH-(aq)
Obtaining the Kb of the Conjugate
Base
HCOOH (aq)  HCOO- (aq) + H+(aq)
HCOO- (aq) + H2O (l)  HCOOH (aq) + OH- (aq)
K (1) = Ka (HCOOH)
K (2) = Kb (HCOO-)

Add the two reactions together
HCOOH (aq)  HCOO- (aq) + H+(aq)
HCOO- (aq) + H2O (l)  HCOOH (aq) + OH- (aq)
The Ka and Kb Value


We are left with the overall reaction
H2O (l)  H+ (aq) + OH- (aq)
Kw = [H+][ OH-]
From our rules for the equilibria of multiple
reactions.
Kw = K (1) K (2)
K w = Kb K a
Variation of Conjugate base
Strength with Ka
HCOOH (aq)  HCOO- (aq) + H+ (aq)
Ka (HCOOH) = 1.8 x 10-4
Kb (HCOO-) = 5.6 x 10-11
CH3COOH (aq)  CH3COO- (aq) + H+ (aq)
Ka (CH3COOH) = 1.8 x 10-5
Kb (CH3COO-) = 5.6 x 10-10
Salts of Conjugate Bases

Look at the dissolution of CH3COONa in
water.
CH3COONa (aq)  Na+ (aq) + CH3COO- (aq)
But we know that the acetate ion, CH3COO(aq), will hydrolyze in aqueous solution
according to the following reaction.
CH3COO- (aq) + H2O (l)  CH3COOH (aq) +
OH-(aq)

Salts of Conjugate Bases

CH COOHOH 
K CH COO  =
CH COO 

b

3

3
3

Since the hydrolysis reaction produces OH-,
dissolving a salt similar to CH3COONa (the
salt of the conjugate base of CH3COOH) in
water produces a basic solution.
Salts of Conjugate Acids

Look at the dissolution of NH4Cl in water
NH4Cl (aq)  NH4+ (aq) + Cl- (aq)

But we know that the ammonium ion, NH4+
(aq), will donate a proton aqueous solution
NH4+ (aq)  NH3 (aq) + H+(aq)
Salts of Conjugate Acids

K a NH4


NH H 
 = NH 

3

4
Since the hydrolysis reaction produces H+,
dissolving a salt similar to NH4Cl (the salt of
the conjugate acid of NH3) in water produces
an acidic solution.
Both the cation and anion
Hydrolyse


What about salts in which both the cation and
anion hydrolyze?
The pH of the solution will depend on the
magnitude of the Ka and the Kb values of the
respective acidic and basic ions.
Ka vs. Kb
Ka > Kb
Type of
Solution
Acidic Solution
Ka < Kb
Basic Solution
Ka = Kb
Neutral
Solution
Salts of Weak Acids/Strong Bases


How do we prepare a solution of HCOONa?
Titration of HCOOH with NaOH according to
the following reaction
HCOOH (aq) + NaOH (aq)  HCOONa (aq) + H2O (l)



Weak Acid Strong Base Basic Salt
Dissolution of the salt of a weak acid/strong base
produces a basic solution (pH > 7.00).
The Strong Acid/Weak Base Case
How do we prepare a solution of NH4Cl?
 Titration of HCl with NH3 according to the
following reaction
HCl (aq) + NH3 (aq)  NH4Cl (aq)



Strong
Weak
Acidic
Acid
Base
Salt

Dissolution of the salt of a strong acid/weak
base produces a acidic solution (pH < 7.00).
The Weak Acid/Weak Base Case


What would be the pH of a solution of
CH3COONH4?
Look at the following reactions
NH4+ (aq)  NH3 (aq) + H+ (aq)
K = Ka (NH4+)
CH3COO- (aq) + H2O (l)  CH3COOH (aq) +
OH- (aq)
K = Kb (CH3COO-)
The pH of a solution the salt of a weak
acid/weak base depends on the magnitudes of the
equilibrium constants.
Ka vs. Kb
Ka > Kb
Type of
Solution
Acidic Solution
Ka < Kb
Basic Solution
Ka = Kb
Neutral
Solution