Chap16-Part3

Chap16-Part3 -...

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Unformatted text preview: EXAM‐3:
APRIL
15th
2011
(Friday)
 Chapters:
15,
16
and
17
 Weak
Acids
 Most
acids
only
par/ally
dissociate
in
water:
 
HA(aq)

+

H2O(l)

↔

H3O+(aq)

+
A‐(aq)
 
HA(aq)

↔

H+(aq)

+
A‐(aq)
 The
acid
dissocia/on
constant,
Ka,
describes
the
extent
of
 protonated
and
deprotonated
species
at
equilibrium:
 Larger
values
of
Ka
indicate
stronger
acids
(more
H+
generated).
 When
a
compound
is
composed
en/rely
of
C,
H
and
O
atoms,
the
 hydrogen
atoms
which
dissociate
are
almost
always
bonded
 to
O
rather
than
C
atoms.
 CalculaHng
Ka
and
the
Percent
IonizaHon
from
pH
 1)
Make
a
table
with
the
known
informa/on.
 2)
Calculate
[H+]
from
pH.
 3)
Write
the
concentra/on
changes.
 4)
Calculate
the
equilibrium
concentra/ons.
 5)
Find
the
percent
ioniza/on.
 6)
Calculate
Ka.
 Example:

0.085
M
phenylace/c
acid
(HC8H7O2)
with
pH
=
2.68
 
HC8H7O2(aq)

↔

H+(aq)

+
C8H7O2‐(aq)
 1)
Make
a
table
with
the
known
informa/on.
 HC8H7O2 (M) initial change final H+ (M) C8H7O2- (M) 0.085 0 0 2)
Calculate
[H+]
from
pH.

 HC8H7O2 (M) initial H+ (M) C8H7O2- (M) 0.085 0 0 change final 2.1x10-3 3)
Write
the
concentra/on
changes.
 HC8H7O2 (M) H+ (M) C8H7O2- (M) initial 0.085 0 0 change -2.1x10-3 +2.1x10-3 +2.1x10-3 final 2.1x10-3 4)
Calculate
the
equilibrium
concentra/ons.
 HC8H7O2 (M) H+ (M) C8H7O2- (M) initial 0.085 0 0 change -2.1x10-3 +2.1x10-3 +2.1x10-3 final 0.083 2.1x10-3 2.1x10-3 5)
Find
the
percent
ioniza/on.

 HC8H7O2 (M) H+ (M) C8H7O2- (M) initial 0.085 0 0 change -2.1x10-3 +2.1x10-3 +2.1x10-3 final 0.083 2.1x10-3 2.1x10-3 PS:
Generally,
percent
Ioniza/on
for
an
acid
=
[H+]equilib
X
100/
[HA]ini/al
 6)
Calculate
Ka.
 HC8H7O2 (M) H+ (M) C8H7O2- (M) initial 0.085 0 0 change -2.1x10-3 +2.1x10-3 +2.1x10-3 final 0.083 2.1x10-3 2.1x10-3 CalculaHng
pH
from
Ka
 1)
Make
a
table
with
the
known
informa/on.
 2)
Write
the
changes
in
concentra/ons
as
a
func/on
of

 an
unknown
variable
(e.g.,
“x”).
 3)
Calculate
the
equilibrium
concentra/ons
as
a
func/on
of
x.
 4)
Write
an
algebraic
expression
for
Ka
as
a
func/on
of
x.
 5)
Find
the
value
of
x.
 6)
Calculate
the
equilibrium
concentra/ons.
 7)
Calculate
pH
from
[H+].
 Example:

The
ac/ve
ingredient
in
aspirin
is
acetylsalicylicacid
(HC9H7O4).

 Aspirin
is
a
monopro/c
acid
with
Ka
=
3.3x10‐4
at
25oC.

Find
the
pH
of
a
 solu/on
made
by
dissolving
two
500
mg
aspirin
tablets
in
250
mL
water.


 
 
 
HC9H7O4(aq)

↔

H+(aq)

+
C9H7O4‐(aq)
 ionizable
group
 1)
Make
a
table
with
the
known
informa/on.
 HC9H7O4 (M) initial change final H+ (M) C9H7O4- (M) 0.022 0 0 2)
Write
the
changes
in
concentra/ons
as
a
func/on
of

 an
unknown
variable
(e.g.,
“x”).
 HC9H7O4 (M) H+ (M) C9H7O4- (M) initial 0.022 0 0 change -x +x +x final 3)
Calculate
the
equilibrium
concentra/ons
as
a
func/on
of
x.
 HC9H7O4 (M) H+ (M) C9H7O4- (M) initial 0.022 0 0 change -x +x +x final 0.022 - x x x 4)
Write
an
algebraic
expression
for
Ka
as
a
func/on
of
x.
 HC9H7O4 (M) H+ (M) C9H7O4- (M) initial 0.022 0 0 change -x +x +x final 0.022 - x x x 5)
Find
the
value
of
x.
 6)
Calculate
the
equilibrium
concentra/ons.
 HC9H7O4 (M) H+ (M) C9H7O4- (M) initial 0.022 0 0 change -2.5x10-3 +2.5x10-3 +2.5x10-3 final 0.020 +2.5x10-3 +2.5x10-3 7)
Calculate
pH
from
[H+].
 HC9H7O4 (M) H+ (M) C9H7O4- (M) initial 0.022 0 0 change -2.5x10-3 +2.5x10-3 +2.5x10-3 final 0.020 +2.5x10-3 +2.5x10-3 Strong
acids
release
much
more
 H+
into
aqueous
solu/on
than
 weak
acids.
 The
led
flask
contains
1
M

 HC2H3O2
(a
weak
acid).
 The
right
flask
contains
1
M
 HCl
(a
strong
acid).
 The
balloons
both
contain
 the
same
amount
of
magnesium.
 When
equilibrium
is
reached,
 the
strong
acid
on
the
right
 releases
much
more
H+
than
 the
weak
acid
on
the
led.
 The
released
H+
converts
 to
H2
gas,
and
the
balloon
 on
the
right
is
more
highly
 inflated.
 The
percent
ioniza/on
and
the
pH
depend
on
the
acid
concentra/on
 ace/c
acid
 CalculaHng
pH
from
Ka
 1)
Make
a
table
with
the
known
informa/on.
 2)
Write
the
changes
in
concentra/ons
as
a
func/on
of

 an
unknown
variable
(e.g.,
“x”).
 3)
Calculate
the
equilibrium
concentra/ons
as
a
func/on
of
x.
 4)
Write
an
algebraic
expression
for
Ka
as
a
func/on
of
x.
 5)
Find
the
value
of
x.
 6)
Calculate
the
equilibrium
concentra/ons.
 7)
Calculate
pH
from
[H+].
 Calculate
the
percent
ioniza/on
and
the
pH
for

 (a)  0.100
M
HF
and
(b)
0.00100
M
HF.


 The
value
of

Ka
is
6.8
x
10‐4.
 1a)
Make
a
table
with
the
known
informa/on.
 initial change final HF (M) 0.100 H+ (M) 0 F- (M) 0 2a)
Write
the
changes
in
concentra/ons
as
a
func/on
of
an
unknown
 variable
(e.g.,
“x”).
 HF (M) H+ (M) F- (M) initial 0.100 0 0 change -x +x +x final 3a)
Calculate
the
equilibrium
concentra/ons
as
a
func/on
of
x.
 HF (M) H+ (M) F- (M) initial 0.100 0 0 change -x +x +x final 0.100 - x x x 4a)
Write
an
algebraic
expression
for
Ka
as
a
func/on
of
x.
 HF (M) H+ (M) F- (M) initial 0.100 0 0 change -x +x +x final 0.100 - x x x 5a)
Find
the
value
of
x.
 6a)
Calculate
the
equilibrium
concentra/ons.
 HF (M) H+ (M) F- (M) initial 0.100 0 0 change -7.9x10-3 +7.9x10-3 +7.9x10-3 final 0.092 +7.9x10-3 +7.9x10-3 7a)
Calculate
pH
from
[H+].
 HF(M) H+ (M) F- (M) initial 0.100 0 0 change -7.9x10-3 +7.9x10-3 +7.9x10-3 final 0.092 +7.9x10-3 +7.9x10-3 1b)
Make
a
table
with
the
known
informa/on.
 HF (M) initial change final H+ (M) F- (M) 0.00100 0 0 2b)
Write
the
changes
in
concentra/ons
as
a
func/on
of

 an
unknown
variable
(e.g.,
“x”).
 HF (M) H+ (M) F- (M) initial 0.00100 0 0 change -x +x +x final 3b)
Calculate
the
equilibrium
concentra/ons
as
a
func/on
of
x.
 HF (M) H+ (M) F- (M) initial 0.00100 0 0 change -x +x +x final 0.00100 - x x x 4b)
Write
an
algebraic
expression
for
Ka
as
a
func/on
of
x.
 HF (M) H+ (M) F- (M) initial 0.00100 0 0 change -x +x +x final 0.00100 - x x x 5b)
Find
the
value
of
x.
 6b)
Calculate
the
equilibrium
concentra/ons.
 HF (M) H+ (M) F- (M) initial 0.00100 0 0 change -5.5x10-4 +5.5x10-4 +5.5x10-4 final 0.00045 +5.5x10-4 +5.5x10-4 7b)
Calculate
pH
from
[H+].
 HF (M) H+ (M) F- (M) initial 0.00100 0 0 change -5.5x10-4 +5.5x10-4 +5.5x10-4 final 0.00045 +5.5x10-4 +5.5x10-4 The
percent
ioniza/on
and
the
pH
depend
on
the
acid
concentra/on
 For
higher
acid
concentra/ons,
the
percent
ioniza/on
decreases,
 but
the
concentra/on
of
[H+]
increases
and
the
pH
decreases.
 (a)  Which
is
the
strongest
acid?


 (a)  Which
is
the
strongest
acid?


 HY
isthe
strongest
acid
because
it
is
completely
dissociated.
 (b)
Which
has
the
smallest
Ka? 

 (b)
Which
has
the
smallest
Ka? 






HX
 (c)
Which
solu/on
has
the
highest
pH?
 (c)
Which
solu/on
has
the
highest
pH?
 The
solu/on
of
HX
has
the
highest
pH
because
it
has
the
 lowest
concentra/on
of
H+.

 PolyproHc
Acids
 Polypro/c
acids
have
more
than
one
ionizable
proton.
 PolyproHc
Acids
 It
is
always
easier
to
remove
the
first
proton
as
compared
to
the
second.
 Note
that
Ka1
>
Ka2
>
Ka3.
 ...
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