New chm 152 unit 3 power points sp13

ACIDS AND BASES
Chapter 4.4 & 18.1-5 Silberberg



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.



Table 18.1

Some Common Acids and Bases and their Household
Uses.



The Savante Arrhenius Theory (1884)



Acid-ionizes in aqueous solution to produce H+ ions.
Base-ionizes in aqueous solution to produce OHions.

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.

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]

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.)



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.








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+.









HCl
HBr
HI
HNO3
HClO3
HClO4
H2SO4

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.



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.










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







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









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

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.

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.

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.

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.

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

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-

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-.

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-.

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).







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

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)








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

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.

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.

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.

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.

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.

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.

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.




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





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.

New chm 152 unit 3 power points sp13






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.

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.











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

Figure 18.7

Methods for measuring the pH of an aqueous solution.

pH (indicator) paper

pH meter

Figure 18.5

The pH values of
some familiar
aqueous solutions.

pH = -log [H3O+]

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.

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).

Figure 18.5

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

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

Sample Problem 18.3
[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

[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







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.

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.








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.





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.








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



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

New chm 152 unit 3 power points sp13

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