1. ACIDS AND BASES
Chapter 4.4 & 18.1-5 Silberberg

2. 
See the Learning Objectives on page 821.

Understand these Concepts: 18.1-20,
23-25.

Master these Skills: 18.1-5, 6-10, 12-14, 17.


3. Table 18.1
Some Common Acids and Bases and their Household
Uses.

4. 
The Savante Arrhenius Theory (1884)


Acid-ionizes in aqueous solution to produce H+ ions.
Base-ionizes in aqueous solution to produce OHions.

5. Svante August Arrhenius

Svante Arrhenius was born
on February 19, 1859 and
he died October 2, 1927.
He was a Swedish
physical chemist best
known for the
development of his acidbase theory. In 1903 he
was awarded the Nobel
Prize for Chemistry.

6. Strong and Weak Acids
A strong acid dissociates completely into ions in water:
HA(g or l) + H2O(l) → H3O+(aq) + A-(aq)
A dilute solution of a strong acid contains no HA molecules.
A weak acid dissociates slightly to form ions in water:
HA(aq) + H2O(l)
H3O+(aq) + A-(aq)
In a dilute solution of a weak acid, most HA molecules are
undissociated.
Kc =
[H3O+][A-]
has a very small value.
[HA][H2O]

7. Classifying the Relative Strengths of Acids

Strong acids include



the hydrohalic acids (HCl, HBr, and HI) and
oxoacids in which the number of O atoms exceeds the number
of ionizable protons by two or more (eg., HNO3, H2SO4,
HClO4.)
Weak acids include

the hydrohalic acid HF,

acids in which H is not bonded to O or to a halogen (eg.,
HCN),

oxoacids in which the number of O atoms equals or exceeds
the number of ionizable protons by one (eg., HClO, HNO2),
and

carboxylic acids, which have the general formula RCOOH
(eg., CH3COOH and C6H5COOH.)

8. 
HBr + H2O


H3O+ + Br -1
HBr is a strong acid--it completely dissociates.
HF + H2O

------>
=
H3O+ + F
HF is a weak acid--it only partially ionizes.

9. 



H+ ions are bare protons.
In aqueous solutions these protons are
hydrated H(H2O)n+ where n is a small
number.
The hydrated hydrogen ion is normally
represented as H3O+.
In many reactions when it is obvious that
aqueous solutions are involved, H+ will be
used to represent H3O+.

10. 






HCl
HBr
HI
HNO3
HClO3
HClO4
H2SO4

11. Classifying the Relative Strengths of Bases

Strong bases include

water-soluble compounds containing O2- or OH- ions.

The cations are usually those of the most active metals:
 M2O or MOH, where M = Group 1A(1) metal (Li, Na, K, Rb, Cs)
 MO or M(OH)2 where M = group 2A(2) metal (Ca, Sr, Ba).

Weak bases include

ammonia (NH3),

amines, which have the general formula

The common structural feature is an N atom with a lone
electron pair.

12. 
NaOH ------> Na+ + OH-1


NaOH is a strong base--it completely dissociates.
NH3 + H2O

=
NH4+ + OH-1
NH3 is a weak base-- it only partially ionizes to
produce OH- ions.

13. 







LiOH
NaOH
KOH
RbOH
CsOH
Ca(OH)2
Ba(OH)2
Sr(OH)2

14. 


Neutralization is the combination of hydrogen
ions(acid) with hydroxide ions(base) to
produce water(neutral).
The net ionic equation for the reaction of a
strong acid with a strong base would be
H+ + OH= H2O

15. 





Acid- a proton (H+) donor
Base- a proton(H+) acceptor
An acid-base reaction is the transfer of a proton
from an acid to a base.
NH3 + H2O
=
NH4+ + OH-1
base
acid
conjugate conjugate
acid
base

16. Johannes Nicolaus Bronsted

Johannes Brønsted was
born on February 22,
1879 and died on
December 17, 1947. He
was a Danish physical
chemist known for a
widely applicable acidbase concept
introduced in 1923.
His work was
independent of Lowry.

17. Thomas Martin Lowry


Thomas Lowry was born
on October 26, 1874 and
died on September 2,
1936.
He was an English chemist
widely - known for an
acid-base concept identical
to that of Johannes N.
Bronsted.

18. Brønsted-Lowry Acid-Base Definition
An acid is a proton donor, any species that donates an H+
ion.
• An acid must contain H in its formula.
A base is a proton acceptor, any species that accepts an H+
ion.
• A base must contain a lone pair of electrons to bond to H+.
An acid-base reaction is a proton-transfer process.

19. Conjugate Acid-Base Pairs
In the forward reaction:
NH3 accepts a H+ to form NH4+.
H2S + NH3
HS- + NH4+
H2S donates a H+ to form HS-.
In the reverse reaction:
NH4+ donates a H+ to form NH3.
H2S + NH3
HS- + NH4+
HS- accepts a H+ to form H2S.

20. Conjugate Acid-Base Pairs
H2S + NH3
HS- + NH4+
H2S and HS- are a conjugate acid-base pair:
HS- is the conjugate base of the acid H2S.
NH3 and NH4+ are a conjugate acid-base pair:
NH4+ is the conjugate acid of the base NH3.
A Brønsted-Lowry acid-base reaction occurs when an acid
and a base react to form their conjugate base and conjugate
acid, respectively.
acid1 + base2
base1 + acid2

21. Table 18.4
The Conjugate Pairs in some Acid-Base Reactions
Conjugate Pair
Acid
+
Base
Base
+
Acid
Conjugate Pair
Reaction 1
HF
+
H2O
F-
+
H3O+
Reaction 2
HCOOH +
CN-
HCOO- +
HCN
Reaction 3
NH4+
+
CO32-
NH3
+
HCO3-
Reaction 4
H2PO4-
+
OH-
HPO42-
+
H2O
Reaction 5
H2SO4
+
N2H5+
HSO4-
+
N2H62+
Reaction 6
HPO42-
+
SO32-
PO43-
+
HSO3-

22. Sample Problem 18.4
Identifying Conjugate Acid-Base Pairs
PROBLEM: The following reactions are important environmental processes.
Identify the conjugate acid-base pairs.
(a) H2PO4-(aq) + CO32-(aq)
(b) H2O(l) + SO32-(aq)
HPO42-(aq) + HCO3-(aq)
OH-(aq) + HSO3-(aq)
PLAN: To find the conjugate pairs, we find the species that donated an H+
(acid) and the species that accepted it (base). The acid donates an H+
to becomes its conjugate base, and the base accepts an H+ to
becomes it conjugate acid.
SOLUTION
:
(a) H2PO4-(aq) + CO32-(aq)
acid1
base2
HPO42-(aq) + HCO3-(aq)
base1
acid2
The conjugate acid-base pairs are H2PO4-/HPO42- and CO32-/HCO3-.

23. Sample Problem 18.4
(b) H2O(l) + SO32-(aq)
acid1
base2
OH-(aq) + HSO3-(aq)
base1
acid2
The conjugate acid-base pairs are H2O/OH- and SO32-/HSO3-.

24. Figure 18.8
Strengths of conjugate acid-base pairs.
The stronger the acid is, the
weaker its conjugate base.
When an acid reacts with a base
that is farther down the list, the
reaction proceeds to the right
(Kc > 1).

25. 



Water can act as both an acid and a base.
Water is said to be amphoteric (amphiprotic).
H2O + H2O = H3O+ + OH -1
base
acid
conjugate conjugate
acid
base

26. Autoionization of Water
Water dissociates very slightly into ions in an equilibrium
process known as autoionization or self-ionization.
2H2O (l)
H3O+ (aq) + OH- (aq)

27. 



Acid--species that accepts a share in a pair of
electrons--an electron pair acceptor
Base--species that donates a share in a pair of
electrons--an electron pair donor
:NH3 + H2O
base
acid
=
H:NH3+ + OH-1
conj.acid conj.base

28. Gilbert Newton Lewis

G. N. Lewis was born on
October 23, 1875 and died on
March 23, 1946. He was a
famous American physical
chemist, who in 1923,
developed the electron-pair
theory of acid-base chemical
reactions. He is also known for
the creation of Lewis structures
for drawing chemical
molecules.

29. The Lewis Acid-Base Definition
A Lewis base is any species that donates an electron pair to form
a bond.
A Lewis acid is any species that accepts an electron pair to form
a bond.
The Lewis definition views an acid-base reaction as the
donation and acceptance of an electron pair to form a
covalent bond.

30. Lewis Acids and Bases
A Lewis base must have a lone pair of electrons to donate.
Any substance that is a Brønsted-Lowry base is also a Lewis base.
A Lewis acid must have a vacant orbital (or be able to
rearrange its bonds to form one) to accept a lone pair and form
a new bond.
Many substances that are not Brønsted-Lowry acids are Lewis acids.
The Lewis definition expands the classes of acids.

31. Electron-Deficient Molecules as Lewis Acids
B and Al often form electron-deficient molecules, and these
atoms have an unoccupied p orbital that can accept a pair of
electrons:
BF3 accepts an electron pair from ammonia to form a covalent bond.

32. Lewis Acids with Polar Multiple Bonds
Molecules that contain a polar multiple bond often function as
Lewis acids:
The O atom of an H2O molecule donates a lone pair to the S of SO2, forming
a new S‒O σ bond and breaking one of the S‒O p bonds.

33. Metal Cations as Lewis Acids
A metal cation acts as a Lewis acid when it dissolves in water to
form a hydrated ion:
The O atom of an H2O molecule donates a lone pair to an available orbital on
the metal cation.

34. Sample Problem 18.15
Identifying Lewis Acids and Bases
PROBLEM: Identify the Lewis acids and Lewis bases in the following
reactions:
(a) H+ + OHH2O (b) Cl- + BCl3
BCl4- (c) K+ + 6H2O
K(H2O6)+
PLAN: We examine the formulas to see which species accepts the electron
pair (Lewis acid) and which donates it (Lewis base) in forming the
adduct.
SOLUTION
: The H+ ion accepts the electron pair from OH-. H+ is the Lewis acid and
(a)
OH- is the Lewis base.
(b) BCl3 accepts an electron pair from Cl-. Cl- is the Lewis base and BCl3 is
the Lewis acid.
(c) An O atom from each H2O molecule donates an electron pair to K+. H2O
is therefore the Lewis base, and K+ is the Lewis acid.

35. 

ionize or dissociate completely
Strong acids


100%
H+ + Cl-1
Strong bases


HCl
NaOH
100%
Na+ + OH-1
Soluble salts

NaCl
100%
Na+
+ Cl-1

36. 

Determine the ion concentrations in 0.050M
nitric acid solution.
Determine the concentration of all ions in a
0.020M solution of calcium hydroxide,
Ca(OH)2.

38. 



H2O
= H+ + OH-1
KW = [H+][OH-1]
At 25OC KW = 1.0x10-14
Calculate the [H+] and [OH-1 ] in 0.050M HCl
solution.

40. The pH Scale
pH = -log[H3O+]
The pH of a solution indicates its relative acidity:
In an acidic solution,
In a neutral solution,
In basic solution,
pH < 7.00
pH = 7.00
pH > 7.00
The higher the pH, the lower the [H3O+] and the less acidic the
solution.

41. 





a convenient way of expressing the acidity
and basicity of dilute aqueous solutions.
pH = -log[H+]
This applies to other ion concentrations as
well
pOH = -log[OH -1 ]
Another useful relationship
pH + pOH = 14

42. Figure 18.7
Methods for measuring the pH of an aqueous solution.
pH (indicator) paper
pH meter

43. Figure 18.5
The pH values of
some familiar
aqueous solutions.
pH = -log [H3O+]

44. Table 18.3
The Relationship between Ka and pKa
Acid Name (Formula)
Ka at 25°C
pKa
Hydrogen sulfate ion (HSO4-)
1.0x10-2
1.99
Nitrous acid (HNO2)
7.1x10-4
3.15
Acetic acid (CH3COOH)
1.8x10-5
4.75
Hypobromous acid (HBrO)
2.3x10-9
8.64
Phenol (C6H5OH)
1.0x10-10
10.00
pKa = -logKa
A low pKa corresponds to a high Ka.

45. pH, pOH, and pKw
Kw = [H3O+][OH-] = 1.0x10-14 at 25°C
pH = -log[H3O+]
pOH = -log[OH-]
pKw = pH + pOH = 14.00 at 25°C
pH + pOH = pKw for any aqueous solution at any temperature.
Since Kw is a constant, the values of pH, pOH, [H3O+], and
[OH-] are interrelated:
• If [H3O+] increases, [OH-] decreases (and vice versa).
• If pH increases, pOH decreases (and vice versa).

46. Figure 18.5
The relations among [H3O+], pH, [OH-], and pOH.

47. Sample Problem 18.3
Calculating [H3O+], pH, [OH-], and pOH
PROBLEM: In an art restoration project, a conservator prepares copper-plate
etching solutions by diluting concentrated HNO3 to 2.0 M, 0.30
M, and 0.0063 M HNO3. Calculate [H3O+], pH, [OH-], and
pOH of the three solutions at 25°C.
PLAN: HNO3 is a strong acid so it dissociates completely, and [H3O+] =
[HNO3]init. We use the given concentrations and the value of Kw at
25°C to find [H3O+] and [OH-]. We can then calculate pH and pOH.
SOLUTION
:
Calculating the values for 2.0 M HNO3:
[H3O+] = 2.0 M
[OH-]
pH = -log[H3O+] = -log(2.0) = -0.30
Kw
1.0x10-14
=
=
= 5.0x10-15 M
[H3O+]
2.0
pOH = -log[OH-] = -log(5.0x10-15) = 14.30

48. Sample Problem 18.3
Calculating the values for 0.30 M HNO3:
[H3O+] = 0.30 M
[OH-]
pH = -log[H3O+] = -log(0.30) = 0.52
Kw
1.0x10-14
=
=
= 3.3x10-14 M
[H3O+]
0.30
pOH = -log[OH-] = -log(3.3x10-14) = 13.48
Calculating the values for 0.0063 M HNO3:
[H3O+] = 0.0063 M
[OH-]
pH = -log[H3O+] = -log(0.30) = 2.20
Kw
1.0x10-14
=
=
= 1.6x10-12 M
[H3O+]
0.0063
pOH = -log[OH-] = -log(1.6x10-12) = 11.80

49. 


Calculate the pH of a solution in which [H+] =
0.030M
The pH of a solution is 4.597. Determine the
[H+] of this solution.
Determine the [H+], [OH-1 ], pH and pOH for a
0.020M HNO3 solution.

52. Solving Problems Involving
Weak-Acid Equilibria
The notation system
• Molar concentrations are indicated by [ ].
• A bracketed formula with no subscript
indicates an equilibrium concentration.
The assumptions
• [H3O+] from the autoionization of H2O is
negligible.
• A weak acid has a small Ka and its
dissociation is negligible. [HA] ≈ [HA]init.

53. 




Consider the reaction when the weak acid
acetic acid is added to water.
CH3COOH + H2O = H3O+ + CH3COO-1
Ka = [H+][CH3COO-1 ]
[CH3COOH]
Write the equation for the ionization of HCN in
aqueous solution.

55. 

In 0.12M solution, a weak acid HY is 5.0%
ionized. Determine the value for the ionization
constant for this weak acid.
The pH of a 0.10M solution of a weak
monoprotic acid HA is 2.97. Calculate the
value for the ionization constant of this weak
acid.

58. 



Determine the concentrations of all species in
0.15M acetic acid , CH3COOH, solution.
Ka = 1.8x10-5
Determine the concentrations of all species in
0.15M HCN solution. Ka = 4.0x10-10
Determine the concentrations of all species in
0.15M NH3. Kb=1.8x10-5

65. 
The pH of an aqueous NH3 solution is 11.37.
Determine the molarity of this aqueous
ammonia solution. Kb=1.8x10-5