Acids and Bases

Acids and Bases - CHEM
50:
Acids
and
Bases


Info iconThis preview shows page 1. Sign up to view the full content.

View Full Document Right Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: CHEM
50:
Acids
and
Bases
 Instructor:
Ram
Subramaniam
 
 Topics
 
 • Solutions,
solute,
and
solvent
 • Concentration
of
solutions:
 o Mass
to
mass
percent
 o Mass
to
volume
percent
 o Molarity
 • Definitions
of
Acids
and
Bases:
 o Arrhennius
 o Bronsted‐Lowry
 o Strong
vs.
weak
 • Conjugate
acid/base
pair
 • Autoionization
of
water,
KW
 • The
pH
scale
 • Concentration
to
pH
(and
vice
versa)
calculations
 • Stoichiometric
calculations
in
acid‐base
reactions
 
 Solutions
 
 A
solution
is
a
combination
of
a
solute
and
a
solvent
(Solution
=
Solute
+
Solvent).
 The
solute
is
the
substance
in
smaller
quantity
and
the
solvent
is
the
substance
in
 larger
quantity.
A
solution
is
a
homogeneous
mixture
and
therefore
has
uniform
 properties.
In
aqueous
solutions,
the
solvent
is
water.

 
 Concentration
of
solutions
 
 The
concentration
of
a
solution
is
a
determinant
of
the
amount
of
solute
that
is
 present
in
a
solution.
This
amount
can
be
expressed
in
many
different
ways
and
 three
of
those
possible
ways
will
be
examined
here.
 
 Mass
to
Mass
percent:
This
is
also
written
as
m/m
%.

 
 GramsSolute m/m
%
=
 × 100% 
 GramsSolution 
 Mass
to
Volume
percent:
This
is
also
written
as
m/v
%.
 
 € GramsSolute m/v%
=
 × 100% 
 mL, Solution 
 Molarity:
This
is
the
moles
of
solute
in
a
liter
of
the
solution.
This
is
also
written
as
 “M”.

 € Molarity = € molesSolute 
 Liters, Solution 
 Definitions
of
Acids
and
Bases
 
 Arrhenius
Definitions
 
 Acid:
Releases
H+
ions
in
solution.
When
an
acid
undergoes
dissociation
in
water,
it
 will
release
H+
ions.
 
 2O 
 Example:
HCl
 H→ 
H+(aq)
+
 Cl‐(aq)
 
 Base:
Releases
OH‐
ions
in
solution.
When
a
base
undergoes
dissociation
in
water,
it
 will
release
OH‐
ions.
 € 
 2O 
 Example:
NaOH
 H→ 
Na+(aq)
 +
 OH‐(aq)
 
 Bronsted­Lowry
Defnitions
 
 € Proton:
The
same
as
H+
ion.

 
 Acid:
Donates
a
proton
 
 
 Example:
HCl
+
H2O
 → 
H3O+(aq)
+
Cl‐(aq)
 
 Base:
Accepts
a
proton
 
 € 
 Example:
NH3
+
H2O
 ⇔
NH4+(aq)
+
OH‐(aq)
 
 Weak
and
Strong
Acids
and
Bases
 
 € Strong
Acid:
Dissociates
completely
in
water.
There
are
only
six
strong
acids,
they
 are:
HCl,
HBr,
HI,
H2SO4,
HNO3,
HClO4.
All
other
acids
are
considered
to
be
weak
 acids.
When
writing
dissociations
of
a
strong
acid,
you
draw
the
arrow
going
in
one
 direction
only
to
indicate
complete
dissociation.

 
 Weak
Acid:
Does
not
dissociate
completely
in
water.
When
writing
dissociations
of
a
 weak
acid,
you
draw
the
arrow
going
in
both
directions
to
indicate
that
some
of
the
 weak
acid
is
going
to
exist
as
molecules
and
will
not
undergo
any
dissociation.


 
 Strong
Base:
Dissociates
completely
in
water.
All
the
hydroxides
of
alkali
metals
and
 alkaline
earth
metals
are
considered
as
strong
bases.
Therefore:
LiOH,
NaOH,
KOH,
 CsOH,
RbOH,
Mg(OH)2,
Ca(OH)2,
Ba(OH)2,
Sr(OH)2.
When
writing
dissociations
of
a
 strong
base,
you
draw
the
arrow
going
in
one
direction
only
to
indicate
complete
 dissociation.
 
 Weak
Base:
Does
not
dissociate
completely
in
water.
When
writing
dissociations
of
a
 weak
base,
you
draw
the
arrow
going
in
both
directions
to
indicate
that
some
of
the
 weak
acid
is
going
to
exist
as
molecules
and
will
not
undergo
any
dissociation.
 
 Conjugate
Acid/Base
Pair
 
 When
an
acid
dissociates
and
loses
a
proton,
the
resulting
species
is
called
the
 conjugate
base.

 
 Examples:
 HCl
(acid),
Cl‐
(conjugate
base)
 
 
 
 H2CO3
(acid),
HCO3‐
(conjugate
base)
 
 
 
 
 
 HCO3‐
(acid),
CO32‐
(conjugate
base)
 
 When
a
base
accepts
a
proton,
the
resulting
species
is
called
the
conjugate
acid.

 
 Examples:
 NH3
(base),
NH4+
(conjugate
acid)
 
 
 
 PO43‐
(base),
HPO42‐
(conjugate
acid)
 
 
 
 HPO42‐
(base),
H2PO4‐
(conjugate
acid)
 
 Autoionization
of
Water
 
 Water
can
act
as
both
an
acid
and
a
base
as
per
the
Bronsted‐Lowry
definition.
That
 means
that
a
molecule
of
water
can
donate
a
proton
while
another
molecule
of
 water
can
accept
a
proton.
In
pure
water
this
is
what
happens.
 
 
 H2O
 +
 H2O
 ⇔
 H3O+
 
 
 +
 OH‐
 
 Acid
 
 Base
 
 Conjugate
Acid
 
 Conjugate
Base
 
 In
pure
water
the
concentrations
of
H3O+
(hydronium
ion)
and
OH‐
(hydroxide
ion)
 € are
equal.
The
molarity
of
the
hydronium
ion
and
hydroxide
ion
are
both
equal
to
 1×10‐7M.
This
solution
would
therefore
be
considered
to
be
a
neutral
solution.

 
 The
ionic
product
of
water
is
the
product
of
the
concentrations
of
the
hydronium
ion
 and
the
hydroxide
ion
and
is
given
the
symbol
KW.
 
 KW
=
[H3O+]×[OH‐]
=
(1×10‐7)×(
1×10‐7)
=
1×10‐14
 
 NOTE:
The
square
brackets,
[X]
means:
“concentration
of”
X
and
specifically,
 molarity
of
X.

 
 
 The
pH
scale
 
 Solutions
in
which
concentrations
of
H3O+
(hydronium
ion)
and
OH‐
(hydroxide
ion)
 are
equal
(to
1×10‐7M
each)
are
considered
to
be
neutral.

 
 Since
acids
increase
the
concentration
of
protons,
that
would
imply
that
acids
 increase
the
concentration
of
the
hydronium
ion.
Such
solutions
in
which
the
[H3O+]
 >
1×10‐7M
are
considered
to
be
acidic.

 
 Since
bases
increase
the
concentration
of
hydroxide
ions,
solutions
in
which
the
 concentration
of
hydroxide
is
more
than
the
hydronium
ion
are
considered
to
be
 basic.
Therefore
in
basic
solutions
[OH‐]
>
1×10‐7M
 
 Definition
of
pH:
pH
=
‐log[H3O+]
 
 Example:

For
a
solution
with
[H3O+]

=
1
x
10−4

 
 
 
 pH
=
 −log
[1
x
10−4
]
 pH
=
‐
[‐4.0]
 pH
=

4.0
 
 
 
 
 
 
 Note:

The
number
of
decimal
places
in
the
pH
equals
the
significant
figures
in
the
 coefficient
of
[H3O+].

 
 As
per
the
pH
scale:
pH
<
7
is
acidic,
pH
=
7
is
neutral,
and
pH
>
7
is
basic.

 
 Concentration
to
pH
calculations
 
 For
all
practical
purposes,
H+
and
H3O+
are
considered
to
be
identical.
If
the
 concentration
of
the
H+
or
H3O+
is
known,
then
pH
can
be
calculated
by
applying
the
 formula
pH
=
‐
log
[H+].

 
 For
strong
acids,
the
concentration
of
the
H+
can
be
easily
found
from
the
 concentration
of
the
acid.
For
example:
consider
a
HCl
solution
whose
concentration
 is
0.05
M.
Since
HCl
is
a
strong
acid,
it
dissociates
completely.

 
 HCl

H+
+
Cl‐
 
 This
implies
that
the
concentration
of
H+
is
the
same
as
that
of
the
HCl.
So,
[H+]
=
 0.05
M
as
well.
The
pH
can
then
be
calculated
as
pH
=
‐log(0.05)
=
1.3
 
 For
weak
acids,
the
concentration
of
H+
cannot
be
determined
as
described
above.
 This
topic
will
be
explored
in
a
future
course.

 
 For
strong
bases,
the
concentration
of
the
OH‐
can
be
easily
determined
due
to
the
 fact
that
strong
bases
dissociate
completely.
Consider
a
0.05
M
NaOH
solution.
 
 NaOH

Na+
+
OH‐
 
 This
implies
that
the
concentration
of
OH‐
is
the
same
as
that
of
the
NaOH.
So,
[OH‐]
 =
0.05
M
as
well.

 
 Now,
using
the
[OH‐]
and
KW,
you
will
need
to
calculate
the
[H+].
 
 [H+]
=
KW/[OH‐]
=
(1×10‐14)/0.05
=
2×10‐13
 
 Now
you
can
calculate
the
pH
using
the
formula:
pH
=
‐
log
[H+]
=
‐
log(2×10‐13)
=
 12.7
 
 For
weak
bases,
the
concentration
of
H+
cannot
be
determined
as
described
above.
 This
topic
will
be
explored
in
a
future
course.
 
 Stoichiometric
calculations
in
acid­base
reactions
 
 The
following
examples
illustrate
the
calculations
involved.
 
 EXAMPLE:
What
is
the
molarity
of
an
HCl
solution
if
18.5
mL
of
0.225
M
NaOH
are
 required
to
neutralize
10.0
mL
of
HCl?
 
 





HCl(aq)

+

NaOH(aq)












NaCl(aq)

+

H2O(l)
 
 STEP
1:

Given:
18.5
mL
of
0.225
M
NaOH;
10.0
mL
of
HCl
 
















Need:

Molarity
of
HCl

 
 STEP
2:
18.5
mL




L





moles
NaOH

moles
HCl




M
HCl
 


































































L
HCl

 STEP
3:


 1
L
=
1000
mL






 0.225
mole
NaOH/1
L
NaOH
 














 1
mole
HCl/1
mole
NaOH
 
 STEP
4:
Calculate
the
molarity
of
HCl.
 18.5
mL
NaOH
x


1
L
NaOH







x

0.225
mole
NaOH
 
 
 
 




1000
mL
NaOH
 









1
L
NaOH
 
 









x


1
mole
HCl






=



0.00416
mole
HCl
 













1
mole
NaOH
 














 MHCl


=

0.00416
mole
HCl


=

0.416
M
HCl
 
 









0.0100
L
HCl
 
 
 
 EXAMPLE:
A
25.0
mL
sample
of
phosphoric
acid
is
neutralized
by
42.6
mL
of
1.45
M
 NaOH.

What
is
the
molarity
of
the
phosphoric
acid
solution?
 
 
 3NaOH(aq)

+

H3PO4
(aq)









Na3PO4(aq)
+

3H2O(l)
 0.0426
L
x

1.45
mole
NaOH

x


1
mole
H3PO4

 




























1


L
 
 



3
mole
NaOH
 
 =
0.0206
mole
H3PO4
 
 
 


0.0206
mole
H3PO4

=
0.824
mole/L

=
0.824
M
 
 
 

0.0250
L
 
 Practice
Problems
 
 Page
411,
Chapter
14:
43,
44,
45,
53
to
62,
67,
68
 
 Page
44,
Chapter
15:
9,
10,
13,
14,
15,
17,
27
to
32,
47,
49,
51
to
54,
55,
57,
59,
61

 ...
View Full Document

Ask a homework question - tutors are online