Acids and
Bases
L. Scheffler
Lincoln High School
1
Acids and Bases




The concepts acids and bases were loosely defined as
substances that change some properties of water.
One of the criteria that was often used was taste.
Substances were classified
 salty-tasting
 sour-tasting
 sweet-tasting
 bitter-tasting
Sour-tasting substances would give rise to the word
'acid', which is derived from the Greek word oxein,
which mutated into the Latin verb acere, which means
'to make sour'
2
Acids
•
React with certain metals to produce
hydrogen gas.
React with carbonates and bicarbonates
to produce carbon
dioxide gas
Vinegar is a solution of acetic acid.
Citrus fruits contain citric acid.
•
•
•
Bases
•
•
•
Have a bitter taste
Feel slippery.
Many soaps contain bases.
3
Properties of Acids
 Produce H+ (as H3O+) ions in water (the hydronium ion
is a hydrogen ion attached to a water molecule)
 Taste sour
 Corrode metals
 Electrolytes
 React with bases to form a salt and water
 pH is less than 7
 Turns blue litmus paper to red “Blue to Red A-CID”
4
Properties of Bases
 Generally produce OH- ions in water
 Taste bitter, chalky
 Are electrolytes
 Feel soapy, slippery
 React with acids to form salts and water
 pH greater than 7
 Turns red litmus paper to blue
“Basic Blue”
5
Arrhenius Definition
Arrhenius
Acid - Substances in water that increase the
concentration of hydrogen ions (H+).
Base - Substances in water that increase
concentration of hydroxide ions (OH-).
Categorical definition – easy to sort substances
into acids and bases
Problem – many bases do not actually contain
hydroxides
6
Bronsted-Lowry Definition
Acid - neutral molecule, anion, or cation
that donates a proton.
Base - neutral molecule, anion, or cation
that accepts a proton.
Ex
HA + :B

HB+
+
:A-
HCl + H2O

H3O+ +
Cl-
Acid

Base
Conj Acid
Conj Base
Operational definition - The classification
depends on how the substance behaves in a
chemical reaction
7
Conjugate Acid Base Pairs
Conjugate Base - The species remaining after an acid
has transferred its proton.
Conjugate Acid - The species produced after base
has accepted a proton.
HA & :A- - conjugate acid/base pair
:A- - conjugate base of acid HA
:B & HB+ - conjugate acid/base pair
HB+ - conjugate acid of base :B
8
Examples of BronstedLowry Acid Base Systems
Note: Water can act as acid or base
Acid
Conjugate Acid
Base
Conjugate Base
+
H2O

H3O+ +
Cl-
H2PO4- +
H2O

H3O+
+
HPO42-
NH4+
+
H2O 
H3O+
+
NH3
Base
:NH3
+
Acid
Conjugate Acid Conjugate Base
H2O 
NH4+ +
OH-
PO43-
+
HCl
H2O 

HPO42- +
OH9
G.N. Lewis Definition
Lewis
Acid - an electron pair acceptor
Base - an electron pair donor
10
The pH Scale
pH
[H3O+ ]
[OH- ]
pOH
11
pH and acidity
1. Acidity or Acid Strength depends on Hydronium Ion
Concentration [H3O+]
2. The pH system is a logarithmic representation of
the Hydrogen Ion concentration (or OH-) as a means
of avoiding using large numbers and powers.
pH
= - log [H3O+] = log(1 / [H3O+])
pOH = - log [OH-]
= log(1 / OH-])
3. In pure water [H3O+] = 1 x 10-7 mol / L (at 20oC)
 pH = - log(1 x 10-7) = - (0 - 7) = 7
4. pH range of solutions:
0 - 14
pH < 7 (Acidic) [H3O+] > 1 x 10-7 m / L
pH > 7 (Basic) [H3O+] < 1 x 10-7 m / L
12
pH and acidity
Kw = [H3O+] [OH-] = 1.0 x10-14
In water
[H3O+] = [OH-] = 1.0 x10-7
pH + pOH = 14
13
Calculating the pH
pH = - log [H3O+]
Example 1: If [H3O+] = 1 X 10-10
pH = - log 1 X 10-10
pH = - (- 10)
pH = 10
Example 2: If [H3O+] = 1.8 X 10-5
pH = - log 1.8 X 10-5
pH = - (- 4.74)
pH = 4.74
14
Indicators
15
pH and acidity
The pH values of several
common substances are
shown at the right.
Many common foods are
weak acids
Some medicines and many
household cleaners are
bases.
16
Neutralization
An acid will neutralize a base,
giving a salt and water as products
Examples
HCl
H2SO4
H3PO4
2 HCl
+ NaOH
+ 2 NaOH
+ 3 KOH
+ Ca(OH) 2




NaCl
Na2SO4
K3PO4
CaCl2
+ H2O
+ 2 H2O
+ 3 H2O
+ 2 H 2O
17
Neutralization Calculations
If the concentration of acid or base is
expressed in Molarity or mol dm-3
then:
--The volume in dm3 multiplied by the
concentration yields moles.
-- If the volume is expressed in cm3
the same product yields millimoles
18
Neutralization Problems
If an acid and a base combine in a 1 to 1 ratio,
then the volume of the acid multiplied by the
concentration of the acid is equal to the volume of
the base multiplied by the concentration of the
base
Vacid C acid = V base C base
If any three of the variables are known it is
possible to determine the fourth
19
Neutralization Problems
Example 1: Hydrochloric acid reacts with potassium
hydroxide according to the following reaction:
HCl + KOH  KCl + H2O
If 15.00 cm3 of 0.500 M HCl exactly neutralizes 24.00 cm3 of
KOH solution, what is the concentration of the KOH
solution?
Solution:
Vacid Cacid = Vbase Cbase
(15.00 cm3 )(0.500 M) = (24.00 cm3 ) Cbase
Cbase = (15.00 cm3 )(0.500 M)
(24.00 cm3 )
Cbase = 0.313 M
20
Neutralization Problems
Whenever an acid and a base do not combine in a 1 to 1
ratio, a mole factor must be added to the neutralization
equation
n Vacid C acid = V base C base
The mole factor (n) is the number of times the moles the
acid side of the above equation must be multiplied so as to
equal the base side. (or vice versa)
Example
H2SO4 + 2 NaOH  Na2SO4 + 2 H2O
The mole factor is 2 and goes on the acid side of the
equation. The number of moles of H2SO4 is one half that of
NaOH. Therefore the moles of H2SO4 are multiplied by 2 to
equal the moles of NaOH.
21
Neutralization Problems
Example 3: Phosphoric acid reacts with potassium
hydroxide according to the following reaction:
H3PO4 + 3 KOH  K3PO4 + 3 H2O
If 30.00 cm3 of 0.300 M KOH exactly neutralizes 15.00 cm3 of
H3PO4 solution, what is the concentration of the H3PO4
solution?
Solution:
In this case the mole factor is 3 and it goes on the acid side,
since the mole ratio of acid to base is 1 to 2. Therefore
3 Vacid Cacid = Vbase Cbase
3 (15.00 cm3 )(Cacid) = (30.00 cm3 ) (0.300 M)
Cacid = (30.00 cm3 )(0.300 M)
(3) (15.00 cm3 )
Cacid = 0.200 M
22
Neutralization Problems
Example 2: Sulfuric acid reacts with sodium hydroxide
according to the following reaction:
H2SO4 + 2 NaOH  Na2SO4 + 2 H2O
If 20.00 cm3 of 0.400 M H2SO4 exactly neutralizes 32.00 cm3 of
NaOH solution, what is the concentration of the NaOH
solution?
Solution:
In this case the mole factor is 2 and it goes on the acid side,
since the mole ratio of acid to base is 1 to 2. Therefore
2 Vacid Cacid = Vbase Cbase
2 (20.00 cm3 )(0.400 M) = (32.00 cm3 ) Cbase
Cbase = (2) (20.00 cm3 )(0.400 M)
(32.00 cm3 )
Cbase = 0.500 M
23
Neutralization Problems
Example 4: Hydrochloric acid reacts with calcium
hydroxide according to the following reaction:
2 HCl +
Ca(OH)2  CaCl2 + 2 H2O
If 25.00 cm3 of 0.400 M HCl exactly neutralizes 20.00 cm3 of
Ca(OH)2 solution, what is the concentration of the Ca(OH)2
solution?
Solution:
In this case the mole factor is 2 and it goes on the base side,
since the mole ratio of acid to base is 2 to 1. Therefore
Vacid Cacid = 2 Vbase Cbase
(25.00 cm3) (0.400) = (2) (20.00 cm3) (Cbase)
Cbase = (25.00 cm3 ) (0.400 M)
(2) (20.00 cm3 )
Cbase = 0.250 M
24
Acid Base Dissociation
Acid-base reactions are equilibrium processes.
The relationship between the relative concentrations
of the reactants and products is a constant for a
given temperature. It is known as the Acid or Base
Dissociation Constant.
The stronger the acid or base, the larger the value of
the dissociation constant.
For an acid in water
K eq
[: A - ][H3 O  ]

[HA] [H 2 O]
For a base in water
K eq
[HB][ OH - ]

[: B - ] [H 2 O]
Note :
H3 O   [H  ]

[H 2 O] in dilute solutions is constant.
K eq [H2 O]  K a
[: A - ] [H  ]

[HA]
K eq [H 2 O]  K b
[HB] [OH - ]

[: B - ]
25
Acid Strength
Strong Acid
- Transfers all of its protons to water;
- Completely ionized;
- Strong electrolyte;
- The conjugate base is weaker and has a
negligible tendency to be protonated.
Weak Acid
- Transfers only a fraction of its protons to
water;
- Partly ionized;
- Weak electrolyte;
- The conjugate base is stronger, readily
accepting protons from water
 As acid strength decreases, base strength increases.
 The stronger the acid, the weaker its conjugate base
 The weaker the acid, the stronger its conjugate base
26
Acid Dissociation Constants
Dissociation constants for some weak acids
27
Base Strength
Strong Base - all molecules accept a proton;
- completely ionizes;
- strong electrolyte;
- conjugate acid is very weak, negligible
tendency to donate protons.
Weak Base
- fraction of molecules accept proton;
- partly ionized;
- weak electrolyte;
- the conjugate acid is stronger. It more
readily donates protons.
 As base strength decreases, acid strength increases.
 The stronger the base, the weaker its conjugate acid.
 The weaker the base the stronger its conjugate acid.
28
Common Strong Acids/Bases
Strong Acids
Strong Bases
Hydrochloric Acid
Sodium Hydroxide
Nitric Acid
Potassium Hydroxide
Sulfuric Acid
*Barium Hydroxide
Perchloric Acid
*Calcium Hydroxide
*While strong bases they are
not very soluble
29
Water as an Equilibrium System
Water has the ability to act as either a BronstedLowry acid or base.
Autoionization – spontaneous formation of low
concentrations of [H+] and OH-] ions by proton
transfer from one molecule to another.
Equilibrium Constant for Water
Kc
[H 3 O  ] [OH - ]

[H2 O] 2
K c [H2 O] 2  [H 3 O  ] [OH - ]
K w  [H 3 O  ] [OH - ]  1.0 x 10 -14 (at 25 o C)
K w  [H  ] [OH - ]
 1.0 x 10 -14 (at 25 o C)
In Pure Water :
[H  ]  [OH - ]  1.0 x 10 -7
30
Weak Acid Equilibria
A weak acid is only partially ionized.
Both the ion form and the unionized form exist at
equilibrium
HA + H2O  H3O+ + AThe acid equilibrium constant is
Ka = [H3O+ ] [A-]
[HA]
Ka values are relatively small for most weak acids.
The greatest part of the weak acid is in the
unionized form
31
Weak Acid Equilibrium
Constants
Sample problem . A certain weak acid dissociates in water as
follows:
HA + H2O  H3O+ + AIf the initial concentration of HA is 1.5 M and the equilibrium
concentration of H3O+ is 0.0014 M. Calculate Ka for this acid
Solution
Ka = [H3O+ ] [A-]
[HA]
I
C
E
Substituting
[HA] 1.5 -x 1.5-x
Ka =
(0.0014)2 = 1.31 x 10-6
[A-]
0 +x
x
1.4986
[H3O+ ] 0 +x
x
x = 0.0014
1.5-x = 1.4986
32
Weak Base Equilibria
Weak bases, like weak acids, are partially
ionized. The degree to which ionization
occurs depends on the value of the base
dissociation constant
General form: B + H2O  BH+ + OHKb
= [BH+][OH-]
[B]
Example
NH3 + H2O  NH4+ + OHKb
= [NH4+][ OH-]
[NH3]
33
Weak Base Equilibrium
Constants
Sample problem . A certain weak base dissociates in water as
follows:
B + H2O  BH+ + OHIf the initial concentration of B is 1.2 M and the equilibrium
concentration of OH- is 0.0011 M. Calculate Kb for this base
Solution
Kb = [BH+ ] [OH-]
[B]
I
C
E
[B]
1.2 -x 1.2-x
[OH-]
0 +x
x
[BH+ ] 0 +x
x
x = 0.0011
1.2-x = 1.1989
Substituting
Kb =
(0.0011)2 = 1.01 x 10-6
1.1989
34
Weak Acid Equilibria
Concentration Problems
Problem 1. A certain weak acid dissociates in water as follows:
HA + H2O  H3O+ + AThe Ka for this acid is 2.0 x 10-6. Calculate the [HA] [A-], [H3O+ ]
and pH of a 2.0 M solution
Solution
Ka = [H3O+ ] [A-] = 2.0 x 10-6
[HA]
I
C
E
Substituting
[HA] 2.0 -x 2.0-x
Ka =
x2
= 2.0 x 10-6
[A-]
0 +x
x
2.0-x
[H3O+ ] 0 +x
x
If x <<< 2.0 it can be dropped
from the denominator
The x2 = (2.0 x10-6)(2.0) = 4.0 x10-6 x = 2.0 x 10-3
[A-] = [H3O+ ] = 2.0 x10-3 [HA] = 2.0 - 0.002= 1.998
pH = - log [H3O+ ] =-log (2.0 x 10-3) = 2.7
35
Weak Acid Equilibria
Concentration Problems
Problem 2. Acetic acid is a weak acid that dissociates in water as
follows: CH3COOH + H2O  H3O+ + CH3COOThe Ka for this acid is 1.8 x 10-5. Calculate the [CH3COOH],[CH3COO-]
[H3O+ ] and pH of a 0.100 M solution
Solution
Ka = [H3O+ ] [CH3COO- ] = 1.8 x 10-5
[CH3COOH]
I
C
E
Substituting
[CH3COOH] 0.100 -x 0.100-x
Ka =
x2
= 1.8 x 10-5
[CH3COO- ]
0
+x
x
0.100-x
+
[H3O ]
0
+x
x
If x <<< 0.100 it can be dropped
from the denominator
The x2 = (1.8 x10-5)(0.100) = 1.8 x10-6 x = 1.3 x 10-3
[CH3COO--] = [H3O+ ] = 1.3 x10-3 [CH3COOH ] = 0.100 - 0.0013 = 0.0987
pH = - log [H3O+ ] =-log (1.3 x10-3) = 2.88
36
Weak Base Equilibria
Example1. Ammonia dissociates in water according to the
following equilibrium
NH3 + H2O  NH4+ + OHKb = [NH4+][ OH-] = 1.8 x 10-5
[NH3]
Calculate the concentration of [NH4+][ OH-] [NH3 ]and the pH of
a 2.0M solution.
I
C
E
Substituting
[NH3] 2.0 -x 2.0-x
Kb =
x2
= 1.8x 10-5
[OH-]
0 +x
x
2.0-x
[NH4+]
0 +x
x
If x <<< 2.0 it can be dropped
from the denominator
The x2 = (1.8 x10-5)(2.0) = 3.6 x10-5 x = 6.0 x 10-3
[OH-] = [NH4+] = 6.0 x10-3 [NH3] = 2.0- 0.006= 1.994
pOH = - log [OH-] =-log (6.0 x10-3) = 2.22
pH = 14-pOH = 14-2.22 = 11.78
37
Amphoteric Solutions

A chemical compound able to react with both an
acid or a base is amphoteric.
Water is amphoteric. The two acid-base couples of
water are H3O+/H2O and H2O/OHIt behaves sometimes like an acid, for example

And sometimes like a base :

Hydrogen carbonate ion HCO3- is also amphoteric,
it belongs to the two acid-base couples
H2CO3/HCO3- and HCO3-/CO32-

38
Common Ion Effect
The common ion effect is a consequence of Le
Chatelier’s Principle
When the salt with the anion (i.e. the conjugate
base) of a weak acid is added to that acid,
It reverses the dissociation of the acid.
 Lowers the percent dissociation of the
acid.

A similar process happens when the salt with the
cation (i.e, conjugate acid) is added to a weak
base.
These solutions are known as Buffer Solutions.
39
Buffer Solutions - Characteristics




A solution that resists a change in pH.
It is pH stable.
A weak acid and its conjugate base
form an acid buffer.
A weak base and its conjugate acid
form a base buffer.
We can make a buffer of any pH by
varying the concentrations of the
acid/base and its conjugate.
40
Buffer Solution Calculations
Calculate the pH of a solution that is 0.50 M CH3COOH and 0.25
M NaCH3COO.
CH3COOH + H2O  H3O+ + CH3COO- (Ka = 1.8 x 10-5)
Solution
Ka = [H3O+ ] [CH3COO- ] = 1.8 x 10-5
[CH3COOH]
I
C
E
.
Substituting
[CH3COOH] 0.50 -x 0.50-x
Ka = x (0.25+x) = 1.8 x 10-5
[CH3COO-]
0.25 +x 0.25+x
(0.50-x)
+
[H3O ]
0
+x
x
If x <<< 0.25 it can be dropped from both
expressions in ( ) since adding or
subtracting a small amount will not
significantly change the value of the ratio
Then the expression becomes x(0.25)/(0.50) = 1.8 x 10-5
x = 3.6 x 10-5 = [H3O+]
pH = - log [H3O+] =-log(3.6 x 10-5 ) = 4.44
41
Buffer Solutions - Equations
=
[H3O+] [A-]
[HA]
1.
Ka
2.
[H3O+] = Ka [HA]
[A-]
The [H3O+] depends on the ratio [HA]/[A-]
Taking the negative log of both sides of equation
2 above

pH = -log(Ka [HA]/[A-])

pH = -log(Ka) - log([HA]/[A-])

pH = pKa + log([A-]/[HA])
42
Henderson Hasselbach
Equation
pH = pKa + log([A-]/[HA])
 pH = pKa + log(Base/Acid)


This expression is known as the HendersonHasselbach equation. It provides a shortcut
from using the I.C.E. model for buffer
solutions where the concentration of both
[A-] and [HA] are significantly greater than
zero.
43
Using the Henderson Hasselbach Equation
pH = pKa + log([A-]/[HA])
Example
Calculate the pH of the following of a mixture that contains
0.75 M lactic acid (HC3H5O3) and 0.25 M sodium lactate (Ka =
1.4 x 10-4)
HC3H5O3 + H2O  H3O+ + C3H5O3Solution
Using the Henderson-Hasselbach equation
pH = - log (1.4 x 10-4) + log ( 0.25/0.75 )
= 3.85 + (-0.477) = 3.37
44
Henderson-Hasselbach
Equation and Base Buffers
For a base a similar expression can be written
pOH = pKb + log ([BH+] / [B])
pOH = pKb + log ([Acid] / [Base])
Example: Calculate the pH of a solution that
contains 0.25 M NH3 and 0.40 M NH4Cl
(Kb = 1.8 x 10-5)
Solution
pOH = - log(1.8 x 10-5) + log (0.40/0.25)
= 4.74 + 0.204 = 4.94
pH = 14 - pOH = 14 - 4.94 = 9.06
45
Henderson-Hasselbach
Equation & Base Buffers
Methyl amine is a weak base with a Kb or 4.38 x 10-4
CH3NH2 + H2O  CH3NH3+ + OHCalculate the pH of a solution that is 0.10 M in methyl
amine and 0.20 M in methylamine hydrochloride.
pOH = pKb + log ([BH+] / [B])
Solution
pOH = -log (4.38 x 10-4) + log (0.20 / 0.10)
= 3.36 + 0.30 = 3.66
pH = 14- 3.66 = 10.34
46
Additional Buffer Problems
How many grams of sodium formate, NaCHOO, would have to
be dissolved in 1.0 dm3 of 0.12 M formic acid, CHOOH, to
make the solution a buffer of pH 3.80? Ka= 1.78 x 10-4
pH = pKa + Log ([A-]/[HA])
Solution
3.80 = -log (1.78 x 10-4) + Log [A-] - Log [0.12]
3.80 = 3.75 + Log [A-] - (-0.92)
Log [A-] = 3.80 - 3.75 - 0.92 = - 0.87
[A-] = 10-0.87 = 0.135 mol dm-3
The molar mass of NaCHOO = 23+12+1+2(16) = 58.0 gmol-1
So (0.135 mol dm-3)(58.0 gmol-1 ) = 7.8 grams per dm-3
47
Relationship of Ka, Kb & Kw

HA weak acid. Its acid ionization is

A- is the conjugate base Its base ionization is

Multiplying Ka and Kb and canceling like terms
48
Titration Curves
A graph showing pH vs volume of acid
or base added
 The pH shows a sudden change near
the equivalence point
 The Equivalence point is the point at
which the moles of OH- are equal to
the moles of H3O+

49
Strong acid-strong base
Titration CurvepH
15_327

pH

At equivalence point, Veq:
13.0
Moles of H3O+ = Moles of
OHThere is a sharp rise in the
pH as one approaches the
7.0
equivalence point
With a strong acid and a
strong base, the equivalence
point is at pH =7
1.0
Neither the conjugate base
0
or conjugate acid is strong
enough to affect the pH


Equivalence
point
50.0
100.0
Vol
3 NaOH added (mL)
cm base added
50
Weak acid-strong base
Titration Curve



The increase in pH is more
gradual as one approaches
the equivalence point
With a weak acid and a
strong base, the
equivalence point is higher
than pH = 7
The strength of the
conjugate base of the weak
acid is strong enough to
affect the pH of the
equivalence point
51
Buffered Weak Acid-Strong
Base Titration Curve



The initial pH is higher than
the unbuffered acid
As with a weak acid and a
strong base, the equivalence
point for a buffered weak
acid is higher than pH =7
The conjugate base is strong
enough to affect the pH
52
Polyprotic Weak Acids


Polyprotic acids have more than one hydrogen
that can be neutralized
Phosphoric Acid has three hydrogen ions.
H3PO4 + H2O  H3O+ + H2PO4H2PO4- +H2O  H3O+ + HPO42HPO42- +H2O  H3O+ + PO43-

At given pH only one acid form and one
conjugate base predominate
pH 0-4.7:
pH 4.7-9.7:
pH 9.7-14:
H3PO4 and H2PO4H2PO4- and HPO42HPO42- and PO4353
Polyprotic Weak Acid-Strong
Base Titration Curve


Phosphoric Acid has three
hydrogen ions.
There are three equivalence
points
H3P04 + H2O  H3O+ + H2PO4H2PO4- +H2O  H3O+ + HPO42HPO42- +H2O  H3O+ + PO43-
54
Salts



A salt is the neutralization product of an acid and
a base.
The anion comes from the acid and the cation
from the base.
Examples
HCl
+ NaOH  NaCl + H2O.
H2SO4 + 2 KOH  K2SO4 + H2O.
55
Salts

If a salt is the result of a
-- Strong acid and a strong base, the pH is near
neutral.
HCl + NaOH  NaCl + H2O.
-- Weak acid and a strong base, the pH will be
greater than 7.
CH3COOH + NaOH  NaCH3COO + H2O
-- Strong acid and a weak base, the pH will be
lower than 7.
NH4OH + HCl  NH4Cl +H2O
-- Weak acid and a weak base, the pH depends on
whether the acid or the base is stronger.
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