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12
Subject
Chemistry
Date
Jan 9, 2024
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9.1
The
Acid-Base
Properties
of
Salt
Solutions
9.2
Solubility
Equilibria
9.3
Predicting
the
Formation
of
a
Precipitate
Prerequisite
Concepts
and
Skills
Before
you
begin
this
chapter,
review
the
following
concepts
and
skills:
m
solving
equilibrium
problems
(Chapter
7,
section
7.3)
m
identifying
conjugate
acid-
base
pairs
and
comparing
their
strengths
(Chapter
8,
section
8.1)
m
interpreting
acid-base
titration
curves
(Chapter
8,
section
8.4)
Aqueous
Solutions
and
Solubility
Equilibria
The
kidneys
are
sometimes
called
the
master
chemists
of
the
body.
They
work
to
maintain
the
constant
composition
of
the
blood
by
helping
to
bal-
ance
water
and
the
various
ions
that
are
present
in
the
blood.
A
very
important
equilibrium
in
the
blood,
which
the
kidneys
help
to
control,
involves
calcium
ions
and
phosphate
ions.
3Ca2+(aq)
+
2PO43_(aq)
=
Ca3(PO4)2(S)
This
equilibrium
sometimes
causes
problems,
however.
For
example,
calcium
phosphate
helps
to
give
bones
their
rigidity.
If
the
kidneys
remove
too
many
phosphate
ions
from
the
blood
due
to
disease,
the
position
of
equilibrium
shifts
to
the
left,
as
predicted
by
Le
Chételier’s
principle.
More
calcium
phosphate
dissolves
in
the
blood,
reducing
bone
density.
The
loss
of
too
much
calcium
phosphate
can
lead
to
osteoporosis.
If
the
kidneys
remove
too
many
calcium
ions
from
the
blood,
the
equilibrium
position
in
the
kidneys
shifts
to
the
right.
Solid
calcium
phosphate
can
form
in
the
kidneys,
producing
kidney
stones.
Kidney
stones,
which
are
painful,
can
also
form
as
the
result
of
calcium
oxalate
precipitating
in
the
kidneys.
Precipitates
of
other
compounds
can
affect
different
areas
of
the
body:
gallstones
in
the
gall
bladder
and
gout
in
the
joints
are
two
examples.
In
this
chapter,
you
will
continue
your
study
of
acid-base
reactions.
You
will
find
out
how
ions
in
aqueous
solution
can
act
as
acids
or
bases.
Then,
by
applying
equilibrium
concepts
to
ions
in
solution,
you
will
be
able
to
predict
the
solubility
of
ionic
compounds
in
water
and
the
forma-
tion
of
a
precipitate.
R
F
g
s
This
kidney
stone
formed
when
potassium
oxalate
precipitated
out
of
solution
in
a
kidney.
How
can
you
predict
whether
a
precipitate,
such
as
potassium
oxalate,
will
form
in
an
aqueous
solution?
\i
5
A
2
!
)
L
v
LR
T
.
.
T
e
418
MHR
«
Unit
4
Chemical
Systems
and
Equilibrium
The
Acid-Base
Properties
of
Salt
Solutions
At
the
end
of
Chapter
8,
you learned
that
when
just
enough
acid
and
Section
Preview/
base
have
been
mixed
for
a
complete
reaction,
the
solution
can
be
acidic,
-
Specific
Expectations
neutral,
or
basic.
The
reaction
between
an
acid
and
a
base
forms
an
In
this
section,
you
will
aqueous
solution
of
a
salt.
The
properties
of
the
dissolved
salt
determine
a
predict
qualitatively
whether
the
pH
of
a
titration
solution
at
equivalence.
Figure
9.1
shows
the
acidic,
a
solution
of
a
specific
basic,
or
neutral
properties
of
three
salt
solutions.
salt
will
be
acidic,
basic,
In
this
section,
you
will
learn
how
to
predict
the
pH
of
an
aqueous
or
neutral
solution
of
a
salt.
Predicting
the
pH
is
useful
when
you
are
performing
a
m
solve
problems
that
involve
titration
experiment, because
you
need
to
choose
an
indicator
that
acid-base
titration
data
and
changes
colour
at
a
pH
value
that
is
close
to
the
pH
at
equivalence.
the
pH
at
the
equivalence
point
Acidic
and
Basic
Properties
of
Salts
=
communicate
your
under-
h
di
basi
£
luti
¢
1
lts
£
standing
of
the
following
The
acidic
or
basic
property
of
an
aqueous
solution
ol
a
salt
results
from
terms:
equivalence
point,
reactions
between
water
and
the
dissociated
ions
of
the
salt.
Some
ions
do
end-point
not
react
with
water.
They
are
neutral
in
solution.
Ions
that
do
react
with
water
produce
a
solution
with
an
excess
of
H3O"(aq)
or
OH™(5q).
The
extent
of
the
reaction
determines
the
pH
of
the
solution.
As
you
will
see,
the
reaction
between
an
ion
and
water
is
really
just
another
acid-base
reaction.
Is
there
a
way
to
classify
salts
so
that
you
can
predict
whether
their
solutions
in
water
will
be
basic,
acidic,
or
neutral?
In
the
following
ExpressLab,
you
will
determine
the
pH
values
of
solutions
of
various
salts.
Then
you
will
analyze
your
results
to
decide
if
there
is
a
pattern.
e
fl——
_w
L\
F
_—
7{
|
s
~|
NENOL
.
b
NN
..i
NaOC
|
sodomon
|
LRER
LY
o
03“1\
L‘
W__E
|
N“quouqm
1
|
N‘\No!lq\
sl
L
i
A
B
Photograph
(A)
shows
three
aqueous
salt
solutions.
Photograph
(B)
shows
the
solutions
after
a
some
universal
indicator
solution
has
been
added
to
each
one.
The
orange
colour
indicates
an
acidic
solution,
the
yellow
colour
indicates
a
neutral
solution,
and
the
purple
colour
indicates
a
basic
solution.
Chapter
9
Agueous
Solutions
and
Solubility
Equilibria
«
MHR
419
2RI
)
Testing
the
pH
of
Salt
Solutions
|
In
this
ExpressLab,
you
will
measure
the
pH
LY
[
|
.
.
.
WD
T
TR
[T
values
of
solutions
of
different
salts
with
the
same
concentration
in
mol/L.
The
pH
of
a
solution
gives
NH4CI
|
the
relative
[H30*],
compared
with
the
[OH™].
From
NaCl
this
information,
you
can
infer
which
ion
in
the
salt
NaCH3;COO
reacted
with
water
to
the
greatest
extent.
K,CO4
unknown
Safety
Precaution
\
3.
Pour
a
small
volume
of
one
of
the
solutions
into
h
o
the
well
of
a
spot
plate.
Test
the
solution
using
e
Some
of
the
salts
are
irritants.
Wear
gloves
and
.
.
universal
pH
paper.
By
comparing
the
colour
of
el
)18
L
Wk
the
paper,
estimate
the
pH
of
the
solution.
i
R
Its
i
le.
Materials
ecord
your
results
in
you
table
dropper
bottles
containing
0.1
mol/L
solutions
of
each
of
the
following
salts:
ammonium
chloride,
4,
Repeat
step
3
for
each
of
the
solutions.
NH4Cl(aq);
sodium
chloride,
NaCliaq);
sodium
Analysis
acetate,
NaCH3COO,q);
potassium
carbonate,
1.
An
ion
that
does
not
react
with
water
has
no
|
K2CO3(aq)
effect
on
the
pH
of
water.
If
both
the
cation
and
-
unknown
salt
the
anion
in
a
salt
do
not
react
with
water,
a
V
universal
pH
paper
solution
of
the
salt
will
be
neutral.
Which
salt
-
spot
plate
consists
of
cations
and
anions
that
do
not
react
with
water?
How
did
you
classify
this
salt
in
terms
of
the
strength
of
the
acid
and
base
used
1.
Copy
the
following
table
into
your
notebook
and
to
form
it?
title
it.
Before you
test
each
salt,
write
the
formula
of
the
salt
in
the
appropriate
box.
(You
will
not
be
testing
any
salt
that
results
from
the
reaction
of
a
weak
base
and
a
weak
acid.)
I
Procedure
2.
If
a
salt
consists
of
one
ion
that
does
not
react
with
water,
and
another
ion
that
does,
the
pH
of
a
solution
of
the
salt
will
be
affected
by
the
ion
that
reacts
with
water.
|
Strong
acid
|
Weak
acid
(a)
Which
salt(s)
formed
acidic
solutions?
Which
ion
accounts
for
the
acidic
nature
of
Strong
base
.
Weak
base
the
solution?
(b)
Which
salt(s)
formed
basic
solutions?
2.
Copy
the
following
table
into
your
notebook,
Which
ion
accounts
for
the
basic
nature
of
to
record
your
observations.
Give
your
table
the
solution?
a
title.
3.
Classify
your
unknown
salt
in
terms
of
the
strength
of
the
base
and
the
acid
from
which
the
salt
was
formed.
Salts
That
Dissolve
and
Form
Neutral
Solutions
In
the
following
equation,
HA(,q)
represents
a
strong
acid,
which
dissociates
completely
in
water.
HAgq)
+
H2Ogp
=
HsO'aq)
+
ATiag)
Interpreting
the
equation
from
right
to left,
the
conjugate
base
of
the
acid, A7,
has
no
tendency
to
combine
with
hydronium
ions
in
water.
Therefore,
it
is
a
weak
base.
In
fact,
the
anion
that
is
formed
by
a
strong
acid
is
a
much
weaker
base
than
water,
so
it
does
not
react
with
water.
Consequently,
when
a
salt
that
contains
the
anion
of
a
strong
acid
is
dissolved
in
water,
the
anion
has
no
effect
on
the
pH
of
the
solution.
420
MHR
«
Unit
4
Chemical
Systems
and
Equilibrium
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For
example,
the
conjugate
base
of
hydrochloric
acid,
HCl,
is
Cl".
The
chloride
ion
is
a
very
weak
base,
so
it
does
not react
significantly
with
water.
Therefore,
the
chloride
ion,
and
the
conjugate
bases
of
other
strong
acids,
do
not
affect
the
pH
of
an
aqueous
solution.
Similarly,
the
cations
that
form
strong
bases
(the
alkali
metals
and
the
metals
below
beryllium
in
Group
2
(IIA))
do
not
tend
to
react
with
hydroxide
ions.
These
cations
are
weaker
acids
than
water.
Therefore,
when
a
salt
contains
one
of
these
ions
(for
example,
Na*)
the
cation
has
no
effect
on
the
pH
of
an
aqueous
solution.
If
a
salt
contains
the
cation
of
a
strong
base
and
the
anion
of
a
strong
acid,
neither
ion
reacts
with
water.
Therefore,
the
solution
has
a
pH
of
7.
Sodium
chloride
is
an
example
of
such
a
salt.
It
is
formed
by
the
reaction
of
sodium
hydroxide
(a
strong
base)
and
hydrochloric
acid
(a
strong
acid).
Salts
of
strong
bases
and
strong
acids
dissolve
in
water
and
form
neutral
solutions.
Salts
That
Dissolve
and
Form
Acidic
Solutions
A
weak
base,
such
as
aqueous
ammonia,
dissociates
very
little.
Therefore,
the
equilibrium
lies
to
the
left.
NH3(aq)
+
HyOy)
=
NH4+(aq)
+
OH~
Interpreting
this
equation
from
right
to
left,
the
ammonium
ion,
(which
is
the
conjugate
acid
of
aqueous
ammonia)
is
a
relatively
strong
acid,
compared
to
water.
In
solution,
ammonium
ions
react
with
water,
resulting
in
an
acidic
solution:
NH4+(aq]
+
HyOp
=
NHj(aq)
+
H30+(aq)
If
a
salt
consists
of
the
cation
of
a
weak
base
and
the
anion
of
a
strong
acid,
only
the
cation
reacts
with
water.
The
solution
has
a
pH
that
is
less
than
7.
Salts
of
weak
bases
and
strong
acids
dissolve
in
water
and
form
acidic
solutions.
Salts
That
Dissolve
and
Form
Basic
Solutions
A
weak
acid,
HA,
forms
a
conjugate
base,
A™,
that
is
relatively
strong.
J
CHEM
The
reaction
of
such
an
anion
with
water
results
in
a
solution
that
is
basic.
For
example,
consider
acetic
acid,
CH3COOH,q),
a
weak
acid.
The
conju-
A
reaction
between
an
ion
gate
base
of
acetic
acid,
CH3COQO™(uq),
is
relatively
strong,
compared
to
and
water
is
sometimes
called
water.
It
reacts
with
water
to
form
a
basic
solution.
a
hydrolysis
reaction.
An
CH3C0O0
(5q)
+
H20()
==
CH3COOH(5q)
+
OH
(5q)
ammonium
ion
hydrolyzes,
but
.
.
.
a
sodium
ion
does
not.
An
If
a
salt
consists
of
the
cation
of
a
strong
base
and
the
anion
of
a
weak
acetate
ion
hydrolyzes,
but
a
acid,
such
as
NaCH3COO,
only
the
anion
reacts
significantly
with
water.
chloride
ion
does
not.
The
reaction
produces
hydroxide
ions.
Therefore,
the
solution
will
have
a
N
J
pH
that
is
greater
than
7.
Salts
of
strong
bases
and
weak
acids
dissolve
in
water
and
form
basic
solutions.
Salts
of
Weak
Bases
and
Weak
Acids
If
a
salt
consists
of
the
cation
of
a
weak
base
and
the
anion
of
a
weak
acid,
both
ions
react
with
water.
The
solution
is
weakly
acidic
or
weakly
basic,
depending
on
the
relative
strength
of
the
ions
that
act
as
the
acid
or
the
base.
You
can
determine
which
ion
is
stronger
by
comparing
the
values
of
K,
and
Ky,
associated
with
the
cation
and
the
anion,
respectively.
If
Ky
>
Ky,
the
solution
is
acidic.
If
Ky,
>
K,,
the
solution
is
basic.
Chapter
9
Agueous
Solutions
and
Solubility
Equilibria
«
MHR
421
The
values
of
K,
and
K,
are
shown
for
acid-base
conjugate
pairs.
A
strong
acid,
such
as
HCl5q),
forms
a
conjugate
base,
Cl™(aq).
that
is
too
weak
to
react
significantly
with
water.
A
weaker
acid,
such
as
CH;COOH,,),
forms
a
conjugate
base,
CH;C007,,),
that
is
strong
enough
to
react
significantly
with
water.
Figure
9.2
shows
the
relationship
between
acid-base
conjugate
pairs.
.CI‘
JHCI
HOBr.
.OBr‘
CH;COO0~
.
.
CH3;COOH
HCN
,
JCN-
NH."
NH3
K./
Ky
[P
IR
NN
AU
(U
TN
NN
SNNrUN
ST
A
T
N
AR
R
N
R
B
Y
I
1
|
1
I
1
|
1
I
|
1
|
T
|
1
|
T
I
1
|
1
I
1
|
1
I
T
|
1
|
»~
10720
10716
10712
1078
107
109
10*
108
’—
weaker
stronger
The
following
list
summarizes
the
acidic
and
basic
properties
of
salts.
Table
9.1
shows
the
acid-base
properties
of
salts
in
relation
to
the
acids
and
bases
that
react
to
form
them.
Acidic
and
Basic
Properties
of
Salts
e
The
salt
of
a
strong
acid/weak
base
dissolves
to
form
an
acidic
solution.
e
The
salt
of
a
weak
acid/strong
base
dissolves
to
form
a
basic
solution.
e
The
salt
of
a
weak
acid/weak
base
dissolves
to
form
an
acidic
solution
if
K;
for
the
cation
is
greater
than
Ky
for
the
anion.
The
solution
is
basic
if
Ky,
for
the
anion
is
greater
than
K,
for
the
cation.
Table
9.1
The
Acid-Base
Properties
of
Various
Salts
T
A
RO
R
LG
LUTGLE
CIATCL
RO
E
R
G
D
ETHT
Reaction
with
water:
only
the
anion
Solution:
basic
Examples:
NaCH;COO,
KF,
Reaction
with
water:
neither
ion
Solution:
neutral
Examples:
NaCl,
K3SOy,
Cation
derived
from
a
strong
base
Ca(NO3),
Mg(HSO4)
Cation
derived
Reaction
with
water:
Reaction
with
water:
from
a
weak
base
|
only
the
cation
both
ions
Solution:
acidic
Solution:
neutral
if
K,
=
K},
Examples:
NH,Cl,
NH4;NOs,
acidic
if
K>
Ky,
NH,CIO,
basic
if
K},
>
Ky
Examples:
NH,CN
(basic),
{(NHa)2S
(basic),
NH/NO;
(acidic)
Sodium
Fluoride:
A
Basic
Salt
That
Protects
Teeth
Sugar
is
a
common
ingredient
in
prepared
foods.
When
sugar
remains
on
your
teeth,
bacteria
in
your
mouth
convert
it
into
an
acid.
The
principal
constituent
of
tooth
enamel
is
a
mineral
called
hydroxyapatite,
Ca;0(PO4)s(OH),.
Hydroxyapatite
reacts
with
acids
to
form
solvated
ions
and
water.
(Solvated
ions
are
ions
surrounded
by
solvent
particles.)
Eventually,
a
cavity
forms
in
the
enamel.
To
help
prevent
acid
from
damaging
tooth
enamel,
many
water
treat-
ment
plants
add
small
concentrations
(about
1
ppm])
of
salts,
such
as
NaF,
422
MHR
«
Unit
4
Chemical
Systems
and
Equilibrium
to
the
water.
You
may
notice
that
a
fluoride
salt
has
also
been
added
to
J
CHEM
your
toothpaste.
Fluoride
ions
displace
OH™
from
hydroxyapatite
to
form
FACT
fluoroapatite,
Ca;9(PO4)sF2,
as
shown
in
the
equation
below.
If
you
cannot
brush
your
teeth
Ca19(PO4)6(OH)z(s)
+
2F
(aq)
=
Ca10(PO4)eF2(s)
+
20H
(5
after
a
sweet
snack,
try
chew-
The
fluoride
ion
is
a
much
weaker
base
than
the
hydroxide
ion,
and
ing
sugar-free
gum.
Chewing
gum
stimulates
the
production
fluoroapatite
is.
therefore
less
reactive
towards
acids.
.
N
of
saliva,
which
is
usually
The
following
Sample
Problem
shows
you
how
to
predict
the
acidity
weakly
alkaline.
The
saliva
or
basicity
of
aqueous
salt
solutions.
helps
to
neutralize
acids
on
your
teeth.
Eating
cheese
for
dessert
also
helps
to
neutralize
Sample
Problem
acid,
because
cheese
is
basic.
N
)
Predicting
the
Acidity
or
Basicity
of
Salts
Problem
Predict
the
acid/base
property
of
an
aqueous
solution
of
each
of
the
following
salts.
If
you
predict
that
the
solution
is
not
neutral,
write
the
equation
for
the
reaction
that
causes
the
solution
to
be
acidic
or
basic.
(a)
sodium
phosphate,
NagPO,
(commonly
called
trisodium
phosphate,
TSP,
used
in
detergents
and
dishwashing
compounds)
(b)
ammonium
nitrate,
NH4NOj
(used
as
a
fertilizer)
(e)
sodium
chloride,
NaCl,
(used
to
de-ice
winter
roads)
(d)
ammonium
hydrogen
carbonate,
NH,HCOj3;
(used
in
baked
foods
as
a
leavening
agent)
What
Is
Required?
Predict
whether
each
aqueous
solution
is
acidic,
basic,
or
neutral.
Write
equations
that
represent
the
solutions
that
are
acidic
or
basic.
What
Is
Given?
The
formula
of
each
salt
is
given.
Plan
Your
Strategy
Determine
whether
the
cation
is
from
a
strong
or
weak
base,
and
whether
the
anion
is
from
a
strong
or
weak
acid.
Ions
derived
from
weak
bases
or
weak
acids
react
with
water
and
affect
the
pH
of
the
solution.
If
both
ions
react
with
water,
compare
the
equilibrium
constants
(K,
and
Kj)
to
determine
which
reaction
goes
farthest
to
completion.
Act
On
Your
Strategy
(a)
Sodium
phosphate,
NazPQy,
is
the
salt
of
a
strong
base
(NaOH)
and
a
weak
acid
(HPO4?%").
Only
the
phosphate
ions
react
with
water.
PO43_(aq)
+
HZO(K)
=
HPO42_(aq)
+
OH_(aq)
The
solution
is
basic.
(b)
Ammonium
nitrate,
NH;NOs,
is
the
salt
of
a
weak
base
(NH3(,q))
and
a
strong
acid
(HNQOj3).
Only
the
ammonium
ions
react
with
water.
NHa'(aq)
+
H2Og)
=
NHa(ag)
+
H30"(aq)
The
solution
is
acidic.
(¢)
Sodium
chloride,
NaCl,
is
the
salt
of
a
strong
base
(NaOH)
and
a
strong
acid
(HClg).
Neither
ion
reacts
with
water,
so
the
solution
is
neutral.
\_
Chapter
9
Agueous
Solutions
and
Solubility
Equilibria
«
MHR
423
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==
-
{d)
Ammonium
hydrogen
carbonate,
NH,HCOs,
is
the
salt
of
a
weak
base
(NH3(aq))
and
a
weak
acid
(H;CO3(aq)).
Both
ions
react
with
water.
NH4+(aq)
+
HZO(g)
=
NH3(aq)
+
H30+[aq]
HCO3_(aq)
+
HOpy)
=
H2C03(aq)
+
OH_(aq)
The
equilibrium
that
lies
farther
to
the
right
has
the
greater
influence
on
the
pH
of
the
solution.
The
equilibrium
constants
you
need
are
K
for
NH,*
and
K;,
for
HCO3™.
Each
ion
is
the
conjugate
of
a
compound
for
which
the
appropriate
constant
is
given
in
tables.
From
Appendix
E,
Kj,
for
NH;z(aq)
is
1.8
x
107°.
Using
KK},
=
Ky,
+_
1.0x104
K
for
NHy"
=
27005
=
5.6
x
10710
Similarly,
from
Appendix
E,
K,
for
H,CO
=
4.5
x
107".
__1.0x101
Ky,
for
HCO5™
=
25
<107
=2.2x1078
Because
K,
for
the
hydrogen
carbonate
reaction
is
larger
than
K;
for
the
ammonium
ion
reaction,
the
solution
is
basic.
Check
Your
Solution
The
equations
that
represent
the
reactions
with
water
support
the
prediction
that
NH4NOj
dissolves
to
form
an
acidic
solution
and
NazPO,
dissolves
to
form
a
basic
solution.
Calcium
chloride
is
the
salt
of
a
strong
base-strong
acid,
so
neither
ion
reacts
with
water
and
the
solution
is
neutral.
Both
ions
in
ammonium
hydrogen
carbonate
react
with
water.
Because
K;
for
HCOj;™
is
greater
than
K,
for
NH4*,
the
salt
dissolves
to
form
a
weakly
basic
solution.
I
Practice
Problems
1.
Predict
whether
an
aqueous
solution
of
each
salt
is
neutral,
acidic,
or
basic.
(a)
NaCN
(e)
Mg(NQO3)
(b)
LiF
(d)
NH,I
2.
Is
the
solution
of
each
salt
acidic,
basic,
or
neutral?
For
solutions
that
are
not
neutral,
write
equations
that
support
your
predictions.
(a)
NH,BrO,
(c)
NaOBr
(b)
NaBrQO,
(d)
NH,Br
3.
K;
for
benzoic
acid,
CgHs;COQOH,
is
6.3
x
107°.
K,
for
phenol,
CsHs0H,
is
1.3
x
1071%,
Which
is
the
stronger
base,
CsHsCOO
(aq)
or
CeH50
(aq)?
Explain
your
answer.
4.
Sodium
hydrogen
sulfite,
NaHSO3,
is
a
preservative
that
is
used
to
prevent
the
discolouration
of
dried
fruit.
In
aqueous
solution,
the
hydrogen
sulfite
ion
can
act
as
either
an
acid
or
a
base.
Predict
whether
NaHSO;3;
dissolves
to
form
an
acidic
solution
or
a
basic
solution.
(Refer
to
Appendix
E
for
ionization
data.)
—
—
424
MHR
«
Unit
4
Chemical
Systems
and
Equilibrium
Calculating
pH
at
Equivalence
In
an
acid-base
titration,
you
carefully
measure
the
volumes
of
acid
and
base
that
react.
Then,
knowing
the
concentration
of
either
the
acid
or
the
base,
and
the
stoichiometric
relationship
between
them,
you
calculate
the
concentration
of
the
other
reactant.
The
equivalence
point
in
the
titration
occurs
when
just
enough
acid
and
base
have
been
mixed
for
a
complete
reaction
to
occur,
with
no
excess
of
either
reactant.
As
you
learned
in
Chapter
8,
you
can
find
the
equivalence
point
from
a
graph
that
shows
pH
versus
volume
of
one
solution
added
to
the
other
solution.
To
determine
the
equivalence
point
experimentally,
you
need
to
measure
the
pH.
Because
pH
meters
are
expensive,
and
the
glass
electrodes
are
fragile,
titrations
are
often
performed
using
an
acid-base
indicator.
An
acid-base
indicator
is
usually
a
weak,
monoprotic
organic
acid.
The
indicator
is
in
equilibrium
between
the
undissociated
acid,
which
is
one
colour,
and
its
conjugate
base,
which
is
a
different
colour.
HIngq)
+
H;0)
=
H30+(aq)
+
InT(aq)
colour
1
colour
2
For
example,
one
indicator,
called
methyl
red,
is
red
when
the
pH
is
below
4.2,
and
yellow
when
the
pH
is
above
6.2.
The
point
in
a
titration
at
which
an
indicator
changes
colour
is
called
the
end-point.
The
colour
change
occurs
over
a
range
of
about
2
pH
units,
but
the
pH
of
a
solution
changes
rapidly
near
the
equivalence
point.
Often
a
single
drop
of
base
causes
the
shift
from
colour
1
to
colour
2.
For
methyl
red,
the
end-point
occurs
over
the
pH
range
4.2
to
6.2.
Therefore,
this
indicator
is
used
when
an
acid-base
titration
results
in
a
moderately
acidic
solution
at
equiva-
lence.
Figure
9.3
shows
the
colours
and
end-points
of
various
indicators.
pH
1
Crystal
violet
B
Thymol
blue
.
2,4-Dinitrophenol
Bromophenol
blue
R
Bromocresol
green
=
Methyl
red
|
Alizarin
B
|
I
Bromothymol
blue
I
Phenol
red
a
Phenolphthalein
n
Alizarin
yellow
R
_n
EMZXE]
A
range
of
indicators
are
available
with
different
end-points
are
available.
Thymol
blue
and
alizarin
are
diprotic.
They
change
colour
over
two
different
ranges.
To
choose
the
appropriate
indicator
for
a
particular
titration,
you
must
know
the
approximate
pH
of
the
solution
at
equivalence.
As
you learned
in
Chapter
8,
an
acid-base
titration
curve
provides
this
information.
What
if
you
do
not
have
a
titration
curve?
You
can
calculate
the
pH
at
equivalence
using
what
you
know
about
salts.
Chapter
9
Aqueous
Solutions
and
Solubility
Equilibria
«
MHR
425
For
example,
suppose
that
you
are
titrating
a
strong
acid
solution
against
a
weak
base.
At
the
equivalence
point,
the
flask
contains
an
aqueous
solution
of
the
salt
that
is
formed
by
the
reaction.
The
solution
of
a
salt
of
a
weak
base
and
a
strong
acid
is
acidic,
so
methyl
red
may
be
an
appropriate
indicator.
The
pH
at
equivalence
in
a
titration
is
the
same
as
the
pH
of
an
aqueous
solution
of
the
salt
formed.
I
Sample
Problem
Calculating
pH
at
Equivalence
Problem
2.0
x
10"
mL
of
0.20
mol/L
NHj(g)
is
titrated
against
0.20
mol/L
HCl(g).
Calculate
the
pH
at
equivalence.
Use
Figure
9.3
to
select
an
appropriate
indicator.
What
Is
Required?
You
need
to
calculate
the
pH
at
equivalence.
As
well,
you
need
to
find
an
indicator
that
changes
colour
at
a
pH
that
is
close
to
the
same
pH
at
equivalence.
What
Is
Given?
You
know
that
20
mL
of
0.20
mol/L
NHj(agq)
base
reacts
with
0.20
mol/L
HClaq).
Tables
of
K,
and
K;,
values
are
in
Appendix
E.
Figure
9.3
shows
the
pH
ranges
and
colour
changes
for
various
indicators.
Plan
Your
Strategy
Step1
Write
the
balanced chemical
equation
for
the
reaction
Step2
Calculate
the
amount
(in
mol)
of
NHj(g),
using
its
volume
and
concentration.
Then,
determine
the
amount
(in
mol)
of
HClq)
that
is
needed
to
react
with
all
the
NHjag).
Step3
Find
the
volume
of
HClq),
based
on
the
amount
and
concentration
of
HClg).
Step
4
Calculate
the
concentration
of
the
salt
formed,
based
on
the
amount
(in
mol)
and
the
total
volume
of
the
solution.
Decide
which
ion
reacts
with
water.
Write
the
equation
that
represents
the
reaction.
Step5
Determine
the
equilibrium
constant
for
the
ion
that
is
involved
in
the
reaction.
Then
divide
the
[NH;"]
by
K;
to
determine
whether
the
change
in
concentration
of
the
ammonium
ion
can
be
ignored.
Step6
Set
up
an
ICE
table
for
the
ion
that
is
involved
in
the
reaction
with
water.
Let
x
represent
the
change
in
the
concentration
of
the
ion
that
reacts.
Step
7
Write
the
equilibrium
expression.
Substitute
the
equilibrium
concentrations
into
the
expression,
and
solve
for
x.
Step
8
Use
the
value
of x
to
determine
the
[H3O"].
Then
calculate
the
pH
of
the
solution.
Using
Figure
9.3,
choose
an
indicator
that
changes
colour
in
a
range
that
includes
the
pH
of
the
solution.
\\
_J/
426
MHR
«
Unit
4
Chemical
Systems
and
Equilibrium
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-
~
Act
on
Your
Strategy
Step
1
The
following
chemical
equation
represents
the
reaction.
NH3(aq]
+
HCl(aq)
—>
NH4C1(aq)
Step2
Amount
(in
mol)
of
NHjz(aq)
=
0.20
mol/L
x
0.020
L
=
4.0
X
10~
mol
NHj(aq)
and
HCl(aq)
react
in
a
1:1
ratio.
Therefore,
4.0
x
10~
mol
of
HCl(aq)
is
needed.
Step3
Amount
(in
mol)
HCl(yq)
=
4.0
x
107°
mol
=
0.20
mol/L
x
Volume
HClq)
(in
L)
4.0
x
1073
mol
Volume
HClyq)
=
0.20
mol/L
=0.020
L
Step4
From
the
chemical
equation,
the
amount
of
NH4Cl
that
is
formed
is
equal
to
the
amount
of
acid
or
base
that
reacts.
Therefore,
4.0
x
10~
mol
of
ammonium
chloride
is
formed.
Total
volume
of
solution
=
0.020
L.
+
0.020
L
=0.040
L
_
4.0x
10
mol
INHLCI]
=
4.0
x
102
L,
=
0.10
mol/L
The
salt
forms
NH4"(aq)
and
Cl7(5q)
in
solution.
NH,"(5q)
is
the
conjugate
acid
of
a
weak
base,
so
it
reacts
with
water.
Cl7(yq)
is
the
conjugate
base
of
a
strong
acid,
so
it
does
not
react
with
water.
The
pH
of
the
solution
is
therefore
determined
by
the
extent
of
the
following
reaction.
NH4+(aq)
+
HZO(@
=
NH3(aq)
+
H30+(aq)
Step5
From
Appendix
E,
Kp,
for
NHj(aq)
is
1.8
x
107°.
K,
for
the
conjugate
acid,
NH;"(aq),
can
be
calculated
using
the
relationship
KK},
=
Ky
K,
1.0x10™4
K=%
~18x107
=5.6
x
10710
[NH,
]
_
_
0.10
K,
5.6
x
10710
=
1.8
x
108
This
is
well
above
500,
so
the
change
in
[NH,],
x,
can
be
ignored.
Step
6
Concentration
(mol/L)
NHi'aq
+
HOp
=
NHiag
+
HiO'g
Initial
0.10
—
0
~0
Change
—X
—
+x
+x
|
Equilibrium
0.10—x
—
X
X
|
(~0.10)
|
\
Chapter
9
Agueous
Solutions
and
Solubility
Equilibria
«
MHR
427
PROBLEM
TIP
When
you
are
determining
the
concentration
of
ions
in
solution,
remember
to
use
the
final
volume
of
the
solution.
In
Practice
Problem
6,
for
example,
the
final
volume
of
the
solution
is
50
mL
because
20
mL
of
base
is
added
to
30
mL
of
acid.
You
can
assume
that
the
total
volume
is
the
sum
of
the
reactant
volumes,
because
the
reactant
solutions
are
dilute.
Coninod
.
_
[INH;3][H30%]
Step
7
K,
=
NFL']
_10
_
(x)(x)
5.6
X
107"
=
010
X
=4/5.6
x
1011
(only
the
positive
value
=17.5
X
10°®
mol/I.
makes
sense)
Step8
x =
[H30]
=
7.5
X
107°
mol/L
pH
=
—log[H;0%]
=-log
7.5
x
10°°
=5.13
Methyl
red,
which
changes
colour
over
pH
4.2
t0
6.2,
is
a
good
choice
for
an
indicator.
If
methyl
red
was
not
available,
bromocresol
green,
which
changes
colour
in
the
pH
range
3.8-5.2,
could
be
used.
Check
Your
Solution
The
titration
forms
an
aqueous
solution
of
a
salt
derived
from
a
weak
base
and
a
strong
acid.
The
solution
should
be
acidic,
which
is
supported
by
the
calculation
of
the
pH.
\.
/
I
Practice
Problems
—
5.
After
titrating
sodium
hydroxide
with
hydrofluoric
acid,
a
chemist
determined
that
the
reaction
had
formed
an
aqueous
solution
of
0.020
mol/L
sodium
fluoride.
Determine
the
pH
of
the
solution.
6.
Part
way
through
a
titration,
2.0
x
10*
mL
of
0.10
mol/L
sodium
hydroxide
has
been
added
to
3.0
x
10!
mL
of
0.10
mol/L
hydrochloric
acid.
What
is
the
pH
of
the
solution?
7.
0.025
mol/L
benzoic
acid,
CgHsCOOQOH,
is
titrated
with
0.025
mol/L
sodium
hydroxide
solution.
Calculate
the
pH
at
equivalence.
8.
50.0
mL
of
0.10
mol/L
hydrobromic
acid
is
titrated
with
0.10
mol/L
aqueous
ammonia.
Determine
the
pH
at
equivalence.
\—
J/
Section
Summary
In
this
section,
you learned
why
solutions
of
different
salts
have
different
pH
values.
You
learned
how
to
analyze
the
composition
of
a
salt
to
predict
whether
the
salt
forms
an
acidic,
basic,
or
neutral
solution.
Finally,
you learned
how
to
apply
your
understanding
of
the
properties
of
salts
to
calculate
the
pH
at
the
equivalence
point
of
a
titration.
You
used
the
pH
to
determine
a
suitable
indicator
for
the
titration.
In
section
9.2,
you
will
further
investigate
the
equilibria
of
solutions
and
learn
how
to
predict
the
solubility
of
ionic
compounds
in
solution.
428
MHR
«
Unit
4
Chemical
Systems
and
Equilibrium
Section
Review
(1]
Sodium
carbonate
and
sodium
hydrogen
carbonate
both
dissolve
to
form
basic
solutions.
Comparing
solutions
with
the
same
concentra-
tion,
which
of
these
salts
forms
the
more
basic
solution?
Explain.
(2]
Determine
whether
or
not
each
ion
reacts
with
water.
If
the
ion
does
react,
write
the
chemical
equation
for
the
reaction.
Then
predict
whether
the
ion
forms
an
acidic
solution
or
a
basic
solution.
(a)
Br~
(b)
CH3NH;"
(c)
ClO4~
(d)
OCI~
(3]
Predict
whether
sodium
dihydrogen
citrate,
NaH,CgH507,
dissolves
to
form
an
acidic
or
a
basic
solution.
(Refer
to
Appendix
E)
O
©
A
chemist
measures
the
pH
of
aqueous
solutions
of
Ca(OH),,
CaF>,
NH4NO3;,
KNOj3,
and
HNO3;.
Each
solution
has
the
same
concentration.
Arrange
the
solutions
from
most
basic
to
most
acidic.
©
©
A
student
uses
a
transfer
pipette
to
put
25.00
mL
of
0.100
mol/L
acetic
acid
into
an
Erlenmeyer
flask.
Then
the
student
adds
sodium
hydroxide
from
a
burette
to
the
flask,
and records
the
following
read-
ings
of
volume
and
pH.
VM)
0.00
6.00
10.00
12.00
14.00
14.40
14.60
14.80
15.20
15.40
16.00
1]
2.8
4.2
5.1
5.5
6.2
6.5
6.8
7.6
9.8
10.5
11.4
(a)
Draw
a
graph
that
shows
the
results.
Plot
pH
on
the
vertical
axis
and
the
volume
of
base
added
on
the
horizontal
axis.
Use
speadsheet
software
if
available.
(b)
From
your
graph,
what
is
the
pH
at
the
equivalence
point?
(c)
Determine
the
initial
concentration
of
the
sodium
hydroxide.
(d)
Use
Figure
9.3
to
suggest
a
suitable
indicator
for
this
titration.
©
O
Congo
red
is
an
indicator
that
changes
colour
in
the
pH
range
3.0
to
5.0.
Suppose
that
you
are
titrating
an
aqueous
sodium
hydroxide
solution
with
nitric
acid
added
from
a
burette.
Is
congo
red
a
suitable
indicator?
If
so,
explain
why.
If
not,
will
the
end-point
occur
before
or
after
the
equivalence
point?
Explain
your
answer.
(7]
If
you
were
titrating
a
strong
base
with
a
weak
acid,
which
of
these
indicators
might
be
suitable:
bromphenol
blue,
bromthymol
blue,
or
phenolphthalein?
(Refer
to
Figure
9.3.)
©
©
0.10
mol/L
hydrochloric
acid
is
titrated
with
0.10
mol/L
methylamine,
CH3NH,.
Calculate
the
pH
at
the
equivalence
point.
©
D
Sodium
phosphate,
NagPQy,
is
sold
at
hardware
stores
as
TSP
(trisodium
phosphate).
Crystals
of
NagPO,
are
dissolved
in
water
to
make
an
effective
cleaning
solution.
TSP
can
be
used,
for
example,
to
prepare
a
surface
before
painting.
(a)
Explain
why
TSP
is
an
effective
cleaning
solution.
(b)
Suggest
safety
precautions
you
should
take
when
using
TSP.
(c)
Explain
why
you
should
never
mix
a
solution
of
TSP
with
other
household
cleaning
products.
Chapter
9
Aqueous
Solutions
and
Solubility
Equilibria
«
MHR
429
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Section
Preview/
Specific
Expectations
In
this
section,
you
will
m
perform
an
experiment
to
determine
the
solubility
product
constant
(Ksp)
for
calcium
hydroxide
solve
equilibrium
problems
that
involve
concentrations
of
reactants
and
products,
and
Kp
calculate
the
molar
solubility
of
a
pure
substance
in
water
or
in
a
solution
of
a
common
ion,
given
K,
communicate
your
under-
standing
of
the
following
term:
solubility
product
constant
(Ksp)
GETNER-28
Barium
sulfate,
BaSO0,,
is
a
sparingly
soluble
ionic
compound
that
is
used
to
enhance
X-ray
imaging.
Solubility
Equilibria
In
Chapter
7,
you learned
that
a
saturated
solution
containing
undissolved
solute
is
an
example
of
a
system
at
equilibrium.
Recall
from
your
previous
chemistry
course
that
the
solubility
of
a
solute
is
the
amount
of
solute
that
will
dissolve
in
a
given
volume
of
solvent
at
a
certain
temperature.
In
other
words,
the
solubility
of
a
solute
indicates
how
much
of
that
solute
is
present
in
a
solution
at
equilibrium.
Data
tables
often
express
solubility
as
(g
solute)/(100
mL
solvent).
Molar
solubility,
on
the
other
hand,
is
always
expressed
in
terms
of
(mol
solute/L
solvent).
The
solubility
of
ionic
solids
in
water
covers
a
wide
range
of
values.
Knowing
the
concentration
of
ions
in
aqueous
solution
is
important
in
medicine
and
in
chemical
analysis.
In
this
section,
you
will
continue
to
study
equilibrium.
You
will
examine
the
solubility
equilibria
of
ionic
compounds
in
water.
Solubility
as
an
Equilibrium
Process
In
Chapter
7,
you learned
that
three
factors—change
in
enthalpy
(AH),
change
in
entropy
(AS),
and
temperature
(7)—determine
whether
or
not
a
change
is
favoured.
The
same
three
factors
are
important
for
determining
how
much
of
a
salt
will
dissolve
in
a
certain
volume
of
water.
These
factors
are
combined
in
the
following
equation,
where
AG
is
the
change
in
free
energy
of
the
system.
AG
=AH
-
TAS
As
you
know
from
Chapter
7,
a
change
is
favoured
when
AG
is
negative.
When
a
salt
dissolves,
the
entropy
of
the
system
always
increases,
because
ions
in
solution
are
more
disordered
than
ions
in
a
solid
crystal.
An
increase
in
entropy
favours
the
formation
of
a
solution
because
the
term
—TAS
is
negative.
Most
solids
dissolve
to
a
greater
extent
at
higher
solution
temperatures,
because
the
term
—TAS
becomes
more
negative.
Explaining
the
overall
enthalpy
change
for
a
dissolving
salt
is
more
complicated
because
it
involves
a
number
of
energy
changes.
Cations
must
be
separated
from
anions,
which
requires
energy.
Water
molecules
then
surround
each
ion
in
solution,
which
releases
energy.
If
the
overall
enthalpy
change
is
negative,
the
formation
of
a
solution
is
favoured.
If
the
enthalpy
increases,
the
formation
of
a
solution
is
not
favoured
because
AG
is
less
negative.
Heterogeneous
Equilibrium:
A
Solubility
System
Consider
barium
sulfate,
a
sparingly
soluble
salt.
X-ray
technicians
give
patients
a
barium
sulfate
suspension
to
drink
before taking
an
X-ray
of
the
large
intestine.
A
suspension
of
barium
sulfate
is
opaque
to
X-rays,
which
helps
to
define
this
part
of
the
body.
(See
Figure
9.4.)
When
you
add
barium
sulfate
crystals
to
water,
barium
ions
and
sulfate
ions
leave
the
surface
of
the
solid
and
enter
the
solution.
Initially,
the
concentration
of
these
ions
is
very
low.
Thus,
the
forward
change,
dissolving,
occurs
at
a
greater
rate
than
the
reverse
change:
BaSOyu)
=
Ba*(ag)
+
SO
(aq)
430
MHR
«
Unit
4
Chemical
Systems
and
Equilibrium
As
more
ions
enter
the
solution,
the
rate
of
the
reverse
change,
recrystallisation,
increases.
Eventually,
the
rate
of
recrystallisation
becomes
equal
to
the
rate
of
dissolving.
As
you
know,
when
the
forward
rate
and
the
backward
rate
of
a
process
are
equal,
the
system
is
at
equilibrium.
Because
the
reactants
and
the
products
are
in
different
phases,
the
reaction
is
said
to
have
reached
heterogeneous
equilibrium.
For
solubility
systems
of
sparingly
soluble
ionic
compounds,
equilibrium
exists
between
the
solid
ionic
compound
and
its
dissociated
ions
in
solution.
The
Solubility
Product Constant
When
excess
solid
is
present
in
a
saturated
solution,
you
can
write
the
equilibrium
constant
expression
for
the
dissolution
of
the
solid
in
the
same
way
that
you
wrote
the
equilibrium
constant
expression
for
a
homogeneous
equilibrium
in
Chapter
7.
For
example,
the
equilibrium
constant
expression
for
barium
sulfate
is
written
as
follows:
_
[Baz+(aq)]
[8042_(aq)]
[BaSOqs)]
The
concentration
of
a
solid,
however,
is
itself
a
constant
at
a
constant
temperature.
Therefore,
you
can
combine
the
term
for
the
concentration
of
the
solid
with
the
equilibrium
constant
to
arrive
at
a
new
constant:
K[BaSO4s)]
=
[Ba®*(aq)][SO4*T(ag)]
Ksp
=
[Baz+(aq)]
[8042_(aq)]
The
new
constant
is
called
the
solubility
product
constant,
K.
The
following
Sample
Problem
shows
how
to
write
expressions
for
Ksp.
K
I
Sample
Problem
Writing
Solubility
Product
Expressions
Problem
Write
the
solubility
product
expression
for
each
compound.
(a)
barium
carbonate
(b)
calcium
iodate
(c)
copper(Il)
phosphate
Solution
First
write
a
balanced
equation
for
the
equilibrium
between
excess
solid
and
dissolved
ions
in
a
saturated
aqueous
solution.
Then
use
the
balanced
equation
to
write
the
expression
for
K.
(a)
BaCOj3)
=
Baz+(aq)
+
C032_(aq]
Kp
=
[Ba*'][CO3*]
(b)
Ca(I03)y5)
=
Ca2+(aq)
+
2103
(ag)
Ksp
=
[Ca?*][1057]?
(¢)
Cus(POg)as)
=
3Cu*'(aq)
+
2POL*
(o)
Kep
=
[Cu*P[POST?
Check
Your
Solution
The
K,
expressions
are
based
on
balanced
equations
for
saturated
solutions
of
slightly
soluble
ionic
compounds.
The
exponents
in
the
K
expressions
match
the
corresponding
coefficients
in
the
chemical
equa-
tion.
The
coefficient
1
is
not
written,
following
chemical
convention.
\
J
PROBLEM
TIP
lon
concentration
terms
in
solubility
product
expressions
are
usually
written
without
their
phases,
as
you see
here.
.
J
Chapter
9
Agqueous
Solutions
and
Solubility
Equilibria
«
MHR
431
I
Practice
Problems
—
9.
Write
the
balanced
chemical
equation
that
represents
the
dissociation
of
each
compound
in
water.
Then
write
the
corresponding
solubility
product
expression.
(a)
copper(I)
chloride
(e)
silver
sulfate
(b)
barium
fluoride
(d)
calcium
phosphate
10.
Write
a
balanced
dissolution
equation
and
solubility
product
expression
for
silver
carbonate,
Ag,COs;.
11.
Write
a
balanced
dissolution
equation
and
solubility
product
expression
for
magnesium
ammonium
phosphate,
MgNH,PO,.
12.
Iron(IIl)
nitrate
has
a
very
low
solubility.
(a)
Write
the
solubility
product
expression
for
iron(III)
nitrate.
(b)
Do
you
expect
the
value
of
K,
of
iron(Ill)
nitrate
to
be
larger
or
smaller
than
the
K,
for
aluminum
hydroxide,
which
has
a
slightly
higher
solubility?
N—
_/
You
know
intuitively
that
solubility
depends
on
temperature.
For
most
ionic
compounds,
more
solute
can
dissolve
in
a
solvent
at
higher
temperatures.
Chemists
determine
the
solubility
of
an
ionic
compound
by
\J
N
FACT
One
compound
that
becomes
less
soluble
at
higher
tempera-
experiment,
and
then
use
the
solubility
data
to
determine
K;p.
Like
the
tures
is
calcium
carbonate,
equilibrium
constant,
K,
is
temperature-dependent.
Therefore,
different
CaCOs.
If
you
have
an
electric
experiments
must
be
carried
out
to
determine
K,
at
different
temperatures.
kettle,
you
have
probably
Chemical
reference
books
list
the
solubility
of
a
wide
variety
of
noticed
a
build-up
of
scale
on
the
heating
element.
The
scale
is
mostly
calcium
carbonate,
which
precipitates
out
of
the
ter
as
it
heats
up.
|
warerastheatstp
)
Sample
Problem
compounds.
In
the
next
Sample
Problem,
you
will
learn
how
to
use
measured
solubilities
to
determine
Kp.
Determing
Ksp
from
Measured
Solubilities
Problem
A
chemist
finds
that
the
solubility
of
silver
carbonate,
Ag,COs,
is
1.3
X
107
mol/L
at
25°C.
Calculate
Ksp
for
silver
carbonate.
What
Is
Required?
You
need
to
find
the
value
of
K,
for
AgzCO;3
at
25°C.
What
Is
Given?
You
know
the
solubility
of
Ag,CO;
at
25°C.
Plan
Your
Strategy
Step1
Write
an
equation
for
the
dissolution
of
Ag,COj.
Step
2
Use
the
equation
to
write
the
solubility
product
expression.
Step3
Find
the
concentration
(in
mol/L)
of
each
ion.
Step
4
Substitute
the
concentrations
into
the
solubility
product
expression,
and
calculate
Kgp.
432
MHR
«
Unit
4
Chemical
Systems
and
Equilibrium
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-
~
Act
on
Your
Strategy
Step1
AgoCOs35)
=
2Ag8
ag)
+
CO32_(aq)
Step2
K,
=
[Ag']2[CO52]
Step3
[Ag']
=
2
x
[Ag,CO3]
=2
X
1.3
X
10~*
mol/L
=
2.6
X
10~*
mol/L
[COgZ_]
=
[A82CO3]
=3.2X
10_2
mol/L
Step4
K,
=
[Ag'1?[CO;]
=
(2.6
x
107%*(1.3
x
107%)
=
8.8
x
10712
Based
on
the
given
solubility
data,
Ky
for
silver
carbonate
is
8.8
x
1072
at
25°C.
Check
Your
Solution
The
value
of
Ky
is
less
than
the
concentration
of
the
salt,
as
expected.
It
has
the
correct
number
of
significant
digits.
Check
that
you
wrote
the
balanced chemical
equation
and
the
corresponding
Ky,
equation
correctly.
Pay
attention
to
molar
relationships
and
to
the
exponent
of
each
term
in
the
K,
equation.
\
-7
i
Practice
Problems
13.
The
maximum
solubility
of
silver
cyanide,
AgCN,
is
1.5
X
10~®
mol/L
at
25°C.
Calculate
K,
for
silver
cyanide.
14.
A
saturated
solution
of
copper(Il)
phosphate,
Cu3(PO,4),,
has
a
concentration
of
6.1
X
1077
g
Cu3(PO,);
per
1.00
x
10?>
mL
of
solution
at
25°C.
What
is
Kgp
for
Cuz(PO,);
at
25°C?
15.
A
saturated
solution
of
CaF,
contains
1.2
x
10%°
formula
units
of
calcium
fluoride
per
litre of
solution.
Calculate
Ky,
for
CaF,.
16.
The
concentration
of
mercury(I)
iodide,
Hg,l,,
in
a
saturated
solution
at
25°C
is
1.5
X
107
ppm.
(a)
Calculate
Kgp
for
HgyI,.
The
solubility
equilibrium
is
written
as
follows:
ngIz
=
ngz+
+
2T
(b)
State
any
assumptions
that
you
made
when
you
converted
ppm
to
mol/L.
\
J/
Determining
a
Solubility
Product Constant
In
Investigation
9-A,
you
will
collect
solubility
data
and
use
these
data
to
determine
a
Ky,
for
calcium
hydroxide,
Ca(OH),.
When
you
calculate
K,
you
assume
that
the
dissolved
ionic
compound
exists
as
independent
hydrated
ions
that
do
not
affect
one
another.
This
assumption
simplifies
the
investigation,
but
it
is
not
entirely
accurate.
Ions
do
interfere
with
one
another.
As
a
result, the
value
of
K,
that
you
calculate
will
be
just
an
approximation.
Ky,
values
that
are
calculated
from
data
obtained
from
experiments
such
as
Investgation
9-A
are
generally
higher
than
the
actual
values.
Chapter
9
Agueous
Solutions
and
Solubility
Equilibria
«
MHR
433
gnvestigatio
-
‘Analyzing
and
interpreting
Determining
K,
for
Calcium
Hydroxide
The
value
of
K,
for
a
basic
compound,
such
as
3.
Ca(OH),,
can
be
determined
by
performing
an
acid-base
titration.
Question
What
is
the
value
of
K,
for
Ca(OH),?
Predictions
Look
up
the
solubility
of
Ca(OH),
in
a
solubility
table.
Do
you
expect
a
large
or
a
small
value
of
K;p
for
Ca(OH),?
Materials
6
50
mL
beaker
white
sheet
of
paper
2
identical
microscale
pipettes
Place
the
beaker
on
the
white
sheet
of
paper.
.
To
transfer
a
small
amount
of
indicator
to
the
acid
solution,
place
one
drop
of
phenolph-
thalein
solution
on
a
watch
glass.
Touch
the
drop
with
the
glass
rod.
Transfer
this
small
amount
of
indicator
to
the
HCI
in
the
beaker.
.
Add
saturated
Ca(OH),
solution,
drop
by
drop,
to
the
HCI
solution
until
the
solution
[
turns
a
permanent
pale pink.
Swirl
the
flask,
!
particularly
as
the
pink
begins
to
appear.
Record
the
number
of
drops
of
Ca(OH),
that
you
added.
Conduct
as
many
trials
as
you
can.
.
Dispose
of
all
solutions
as
directed
by
your
teacher.
Do
not
dispose
of
any
chemicals
down
the
drain.
wash
bottle
that
contains
distilled
water
Analysis
phenolphthalein
in
dropping
bottle
glass
stirring
rod
Petri
dish
or
watch
glass
10
mL
of
recently
filtered
saturated solution
of
C&(OH)Z
2.
10
mL
of
0.050
mol/L
HCI
Safety
Precautions
If
you
get
acid
or
base
on
your
skin,
flush
with
4.
plenty
of
cold
water.
Procedure
1.
Read
the
Procedure,
and
prepare
an
appropri-
5.
ate
data
table.
Remember
to
give
your
table
Determine
the
average
number
of
drops
of
Ca(OH);
solution
that
you
needed
to
neutralize
,
10
drops
of
0.050
mol/L
HCI.
-
Write
the
balanced
chemical
equation
for
the
dissociation
of
calcium
hydroxide
in
water.
.
Write
the
balanced
chemical
equation
that
represents
the
neutralization
of
aqueous
Ca(OH),
with
HCl(aq).
Use
the
balanced chemical
equation
for
the
neutralization
of
HCI
with
Ca(OH),
to
determine
[OH™]
in
mol/L.
Hint:
Why
do
you
not
need
to
know
the
volume
of
one
drop?
From
[OH~],
determine
[Ca%?*]
in
mol/L.
a
title.
Conclusion
'
2.
Add
exactly
10
drops
of
0.050
mol/L
HCl
to
a
6.
50
mL
beaker.
To
ensure
uniform
drop
size,
use
the
same
type
of
pipette
to
dispense
the
acid
and
the
base.
Also,
hold
the
pipette
vertically
when
dispensing
the
solution.
Always
record
the
number
of
drops
added.
434
MHR
«
Unit
4
Chemical
Systems
and
Equilibrium
Calculate
K;p
for
Ca(OH),.
If
possible, look
up
the
accepted
value
of
K,
for
Ca(OH),.
Calculate
your
percent
error.
Using
the
Solubility
Product Constant
You
can
use
the
value
of
Ky,
for
a
compound
to
determine
the
concentra-
tion
of
its
ions
in
a
saturated
solution.
The
following
Sample
Problem
shows
you
how
to
do
this.
You
will
use
an
approach
that
is
similar
to
the
approach
you
used
in
section
7.2
to
find
equilibrium
amounts
using
K.
for
homogeneous
equilibria.
Table
9.2
shows
the
values
of
K,
for
selected
ionic
compounds
at
25°C.
The
compounds
are
organized
from
the
largest
K,
to
the
smallest.
You
will
find
a
more
comprehensive
table
of
K,
values
in
Appendix
E.
Table
9.2
Values
of
K,
for
Some
lonic
Compounds
at
25°C
Compound
|
Ksp
magnesium
sulfate,
MgSQO,
5.9x
1073
lead(II)
chloride,
PbCI,
1.7
x
107
barium
fluoride,
BaF,
1.5
x
1078
cadmium
carbonate,
CdCOg3
1.8
x
10714
copper(II)
hydroxide,
Cu(OH),
2.2
x
10720
silver
sulfide,
Ag,S
8
x
10748
I
Sample
Problem
Calculating
Molar
Solubility
From
Kjsp
Problem
Lead(II)
iodide,
Pbl,,
films
are
being
investigated
for
their
usefulness
in
X-ray
imaging.
Pbl,
is
also
used
for
decorative
work,
such
as
mosaics,
because
of
its
attractive
golden yellow
colour.
At
25°C,
K,
for
Pbl,
is
9.8
x
107%,
What
is
the
molar
solubility
of
Pbl,
in
water
at
25°C?
What
Is
Required?
You
need
to
determine
the
solubility
(in
mol/L)
of
PbI,
at
25°C.
What
Is
Given?
At
25°C,
Ky
for
PbI,
is
9.8
x
107,
Plan
Your
Strategy
Step
1
Write
the
dissociation
equilibrium
equation.
Step
2
Use
the
equilibrium
equation
to
write
an
expression
for
K.
Step3
Set
up
an
ICE
table.
Let
x
represent
molar
solubility.
Use
the
stoichiometry
of
the
equilibrium
equation
to
write
expressions
for
the
equilibrium
concentrations
of
the
ions.
Step
4
Substitute
your
expressions
into
the
expression
for
Ky,
and
solve
for
x.
Act
on
Your
Strategy
Step
1
PbIz(s)
=
Pb2+(aq)
+
ZI_(aq)
Step2
K,
=
[Pb**][I7]
\
CONCEPT
CHECK
Suppose
that
two
different
salts,
AX
and
BY,,
have
the
same
K.
Are
the
salts
equally
soluble
at
the
same
temperature?
Explain
your
answer.
Chapter
9
Agueous
Solutions
and
Solubility
Equilibria
«
MHR
435
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Coninod
.
Step
3
Concentration
(mol/L)
Pbly
=
Pb**
g
+
217
()
Initial
—
0
0
Change
—
+x
+2x
Equilibrium
—
X
2x
Step4
Ky,
=
[Pb*][I']*=9.8
x
107°
=
x
X
(2x)?
=X
X
4X2
=
4x3
~4x°=9.8
x
107°
_
39.8x107°
X
=
-
4
=
1.3
x
107
mol/L
The
molar
solubility
of
Pbl,
in
water
is
1.3
x
1073
mol/L.
Check
Your
Solution
Recall
that
x
=
[Pb2+]eq
and
2x
=
[I]eq.
Substitute
these
values
into
the
K,
equation.
You
should
get
the
given
K.
\.
J
I
Practice
Problems
17.
Ky,
for
silver
chloride,
AgCl,
is
1.8
x
107?
at
25°C.
(a)
Calculate
the
molar
solubility
of
AgCl
in
a
saturated
solution
at
25°C.
(b)
How
many
formula
units
of
AgCl
are
dissolved
in
1.0
L
of
saturated
silver
chloride
solution?
(c)
What
is
the
percent
(m/v)
of
AgCl
in
a
saturated
solution
at
25°C?
18.
Iron(IIl)
hydroxide,
Fe{OH);,
is
an
extremely
insoluble
compound.
Ky
for
Fe(OH);
is
2.8
x
10739
at
25°C.
Calculate
the
molar
solubility
of
Fe(OH);
at
25°C.
19.
Ky,
for
zinc
iodate,
Zn(IO;),
is
3.9
x
107°
at
25°C.
Calculate
the
solubility
(in
mol/L
and
in
g/L)
of
Zn(IO;),
in
a
saturated
solution.
20.
What
is
the
maximum
number
of
formula
units
of
zinc
sulfide,
ZnS,
that
can
dissolve
in
1.0
L
of
solution
at
25°C?
Ky,
for
ZnS
is
2.0
x
10722,
\
J/
The
Common
lon
Effect
So
far,
you
have
considered
solubility
equilibria
for
pure
substances
dissolved
in
water.
What
happens
to
the
solubility
of
an
ionic
compound
when
it
is
added
to
a
solution
that
already
contains
one
of
its
ions?
Consider
a
saturated
solution
of
lead(II)
chromate.
See
Figure
9.5(A)
on
the
next
page.
The
following
equation
represents
this
equilibrium.
PbCrOy)
=
Pb2+(aq)
+
CTO42_(aq)
436
MHR
«
Unit
4
Chemical
Systems
and
Equilibrium
A
solution
of
a
salt
that
contains
chromate
ions,
such
as
sodium
chromate,
Na;CrOy(,q),
is
added
to
the
mixture.
More
yellow
lead(II)
chromate
precipitates
out
of
solution,
as
shown
in
Figure
9.5(B).
As
you learned
in
Chapter
8,
this
phenomenon
is
called
the
common
ion
effect.
The
observed
result
is
predicted
by
Le
Chatelier’s
principle.
Adding
a
common
ion
to
a
solution
increases
the
concentration
of
that
ion
in
solution.
As
a
result,
equilibrium
shifts
away
from
the
ion.
In
this
example,
adding
chromate
ions
causes
the
equilibrium
to
shift
to
the
left,
and
lead(II)
chromate
precipitates.
GITMZXX]
Adding
chromate
ions
to
an
equilibrium
system
of
lead(ll)
chromate
(A)
causes
the
equilibrium
position
to
shift
to
the
left.
As
a
result,
more
solid
lead(ll)
chromate
precipitates
(B).
PbCrOy
—
Pb2+(aq)+CrO42‘(aq)
PbCrQy
—
Pb2+(aq)+Cr042‘(aq)(added)
Analyzing
the
expression
for
the
solubility
product
constant
gives
the
same
result.
Ksp
=
[Pb2+]
[CI'O42_]
=2.3x
107"
at
25°C
Since
Ky,
is
a
constant
at
a
given
temperature,
an
increase
in
the
concentration
of
one
ion
must
be
accompanied
by
a
decrease
in
the
concentration
of
the
other
ion,
achieved
by
the
formation
of
a
precipitate.
This
explains
why
solid
lead(II)
chromate
precipitates
out
of
solution.
The
next
Sample
Problem
shows
how
to
predict
the
solubility
of
an
ionic
compound
when
a
common
ion
is
present
in
solution.
I
Sample
Problem
The
Effect
of
a
Common
Ion
on
Solubility
Problem
The
solubility
of
pure
PbCrO,)
in
water
is
4.8
x
1077
mol/L.
(a)
Qualitatively
predict
how
the
solubility
will
change
if
PbCrOy
is
added
to
a
0.10
mol/L
solution
of
sodium
chromate,
Na,CrO,.
(b)
Ksp
for
PbCrOy)
is
2.3
x
10713,
Determine
the
solubility
of
PbCrOy
in
a
0.10
mol/L
solution
of
Na,CrQO,.
What
Is
Required?
You
need
to
predict,
and
then
determine,
the
solubility
of
PbCrO,4
(in
mol/L)
in
a
solution
of
Nay,CrO;,.
What
Is
Given?
You
know
K,
for
PbCrO,
and
the
concentration
of
the
salt
with
the
cominon
ion.
—
Chapter
9
Agueous
Solutions
and
Solubility
Equilibria
«
MHR
437
Coninued
.
S
Plan
Your
Strategy
{a)
Use
Le
Chatelier’s
principle
to
make
a
prediction.
(b)
Step
1
Write
the
equilibrium
equation.
Step2
Use
the
equilibrium
equation
to
write
an
expression
for
K.
Step3
Set
up
an
ICE
table.
Let
x
represent
the
concentration
of
chromate
(the
common
ion) that
is
contributed
by
PbCrQ,.
Initial
conditions
are
based
on
the
solution
of
Na,CrQ,.
Step4
Solve
for
x,
and
check
your
prediction.
Act
on
Your
Strategy
(a)
Based
on
Le
Chatelier’s
principle,
the
solubility
of
PbCrO,
in
a
solution
that
contains
a
common
ion
(chromate}
will
be
less
than
the
solubility
of
PbCrQO,
in
water.
{b)
Step1
PbCrOy,)
=
Pb2+(aq)
+
CI‘O42_[aq)
Step2
K;p
=
[Pb*'][CrO,%]
=
2.3
x
10713
Step
3
Concentration
(mol/L)
PbCrOy
=
Pb?*(5q)
+
CrO4*
(ag)
Initial
—_
0
0.10
Change
—
+x
+x
Equilibrium
—
X
0.10
+
x
N
Step4
Since
Kjp
is
very
small,
you
can
assume
that
x
is
much
smaller
than
0.10. To
check
the
validity
of
this
assumption,
determine
whether
or
not
0.10
is
more
than
500
times
greater
than
Ky:
0.10
_
0.10
Kp
23x10718
=4.3
x
10"
>
500
Therefore,
in
the
(0.10
+
x)
term,
x
can
be
ignored.
In
other
words,
(0.10
+
x)
is
approximately
equal
to
0.10.
Therefore,
you
can
simplify
the
equation
as
follows:
Ksp
=
[Pb*][CrO,*]
=
2.3
x
107"
=
(x)(0.10
+
x)
=~
(x)(0.10)
~(x)(0.10)
=
2.3
x
107"
x
=
2.3
X
107?
mol/L
The
molar
solubility
of
PbCrQO,
in
a
solution
of
0.10
mol/L
Na,CrQ,
is
2.3
x
10712
mol/L.
Your
prediction
was
correct.
The
solubility
of
PbCrO,4
decreases
in
a
solution
of
common
ions.
Check
Your
Solution
The
approximation
in
step
4
is
reasonable.
The
solubility
of x
is
much
smaller
than
0.10.
To
prove
that
the
approximation
works,
substitute
the
calculated
value
of x
into
the
initial
expression
for
K.
438
MHR
«
Unit
4
Chemical
Systems
and
Equilibrium
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I
Practice
Problems
Note: Refer
to
Appendix
E,
as
necessary.
21.
Determine
the
molar
solubility
of
AgCl
(a)
in
pure
water
(b)
in
0.15
mol/L
NaCl
0.050
mol/L
Nal.
(a)
in
pure
water
(b)
in
0.25
mol/L
Na;SO,
solubility
of
PbCl,
(a)
in
pure
water
(b)
in
0.10
mol/L
CaCl,
\
22.
Determine
the
molar
solubility
of
lead(Il)
iodide,
Pbl,,
in
23.
Calculate
the
molar
solubility
of
calcium
sulfate,
CaSO,,
24.
Ky,
for
lead(Il)
chloride,
PbCl,,
is
1.6
x
107°.
Calculate
the
molar
R
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Partner
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more
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equilibria,
go
to
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12
Electronic
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Partner.
]
J
Chemist
P
canacians
163
-
Dr.
Joseph
Maclnnis
has
a
medical
degree,
but
he
does
not
work
in
a
clinic,
a
hospital,
or
a
typical
doctor’s
office.
His
office
is
the
ocean,
and
he
is
an
expert
in
deep-sea
diving.
He
combines
his
medical
expertise
with
his
knowledge
of
the
properties
of
gases
to
study
the
effects
of
deep-sea
diving
on
humans.
Maclnnis
was
born
in
Barrie,
Ontario,
in
1937.
In
1963,
after
graduating
from
the
University
of
Toronto’s
medical
school,
he
received
a
fellowship
to
study
diving
medicine
at
the
University
of
Pennsylvania.
Diving
medicine
is
a
specialized
field,
dealing
with
the
challenges
that
humans
face
when
exposed
to
undersea
environments.
For
every
10
m
below
sea
level
that
a
diver
descends
under
water,
ambient
pressure
increases
by
one
unit
of
atmospheric
pressure.
This
increase
in
pressure,
and
its
effect
on
heterogeneous
equilibria
in
the
body,
is
one
of
the
most
profound
challenges
that
a
diver
encounters.
As
a
diver
descends,
and
is
subjected
to
increased
pressure,
gases
such
as
nitrogen
dissolve
in
the
blood
and
tissues.
In
the
equilibrium
between
dissolved
nitrogen
and
gaseous
nitrogen,
increased
pressure
favours
dissolved
nitrogen,
according
to
Le
Chatelier’s
principle.
As
the
diver
ascends
to
the
ocean
surface
and
pressure
decreases,
equilibrium
favours
the
nitrogen
coming
out
of
solution.
If
a
diver
ascends
too
quickly,
the
lungs
cannot
remove
the
nitrogen
fast
enough.
Gas
bubbles
form
in
the
blood,
causing
decompression
sickness.
Decompression
sickness
is
also
known
as
“the
bends,”
because
one
of
its
symptoms
is
an
inability
to
bend
the
joints.
Maclnnis
was
a
consultant
to
the
team
that
dis-
covered
the
Titanic,
and
he
was
the
first
Canadian
to
explore
the
wreck.
In
1991,
Maclnnis
and
his
team
used
small
submarines
with
pressurized
hulls
to
film
the
wreck
in
IMAX
format.
The
wreck
is
about
4000
m
below
sea
level,
but
the
pressure
inside
the
sub-
marines was
kept
similar
to
the
pressure
above
the
surface
of
the
sea.
Maclnnis
is
passionate
about
the
environment,
both
below
and
above
the
sea.
Since
1996,
Maclnnis
has
been
Chair
of
the
Friends
of
the
Environment
Foundation,
a
non-profit
organization
that
funds
projects
that
help
to
protect
Canada’s
environment.
In
recognition
of
his
accomplishments,
Maclnnis
was
made
a
member
of
the
Order
of
Canada
in
1976.
Chapter
9
Agueous
Solutions
and
Solubility
Equilibria
«
MHR
439
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COURSE
EHAiLENG
How
do
buffers
in
human
blood
maintain
the
blood’s
pH?
In
the
Chemistry
Course
Challenge,
you
will
learn
about
the
buffer
system
in
blood.
How
would
you
design
an
investigation
to
demonstrate
how
the
system
resists
changes
in
pH?
Buffers
and
the
Common
lon
Effect
In
Chapter
8,
you
learned
that
buffers
resist
changes
in
pH
when
acids
or
bases
are
added.
Buffers
have
many
uses
in
a
chemistry
lab.
For
example,
commercial
buffers
with
certified
pH
values
are
available.
Chemists
in
analytical
laboratories
use
these
acidic,
basic,
and
neutral
buffers
to
ensure
the
accuracy
of
their
pH
meters
across
the
pH
range.
Recall,
from
Chapter
8,
that
a
buffer
consists
of
a
weak
acid/conjugate
base
mixture
or
a
weak
base/conjugate
acid
mixture.
One
buffer
that
you
examined
previously
contains
acetic
acid
and
sodium
acetate.
The
common-ion
effect
applies
to
this
buffer.
The
equilibrium
of
the
acetic
acid
is
affected
by
the
common
acetate
ion
from
sodium
acetate.
Since
buffers
are
an
example
of
the
common-ion
effect,
the
concentra-
tion
of
ions
(hence
the
pH)
of
a
buffer
solution
can
be
calculated
using
a
method
that
is
similar
to
the
method
you
used
to
determine
solubility
in
a
common
ion
effect
problem.
You
will
use
K,,
however,
instead
of
Kgp,.
Also,
buffers
are
in
homogeneous
equilibria,
unlike
saturated
solubility
systems,
which
are
in
heterogeneous
equilibria.
Therefore,
you
need
to
consider
the
initial
concentration
of
the
reactants.
(The
reactants
are
not
solids,
so
their
concentration
is
not
constant.)
I
Sample
Problem
pH
of
a
Buffer
and
the
Common-Ion
Effect
Problem
A
buffer
solution
was
made
by
mixing
equal
volumes
of
0.20
mol/L
solutions
of
acetic
acid
and
sodium
acetate
at
25°C.
What
is
the
pH
of
the
buffer
solution?
What
Is
Required?
You
need
to
determine
the
pH
of
the
buffer
solution.
What
Is
Given?
0.20
mol/L
CH3COOH
and
0.20
mol/L.
CH;COONa
are
mixed
in
equal
volumes.
K;
for
CH;COOH
is
1.8
x
107°.
Plan
Your
Strategy
Step1
Determine
the
initial
concentrations
of
CH;COOH
and
CH3;COO~
in
the
buffer
solution.
Step2
Write
the
reaction
for
the
dissociation
of
CH3COOH.
Set
up
an
ICE
table,
including
the
initial
concentration
of
CH;COO~
in
the
buffer.
Step3
Write
the
equation
for
K,,
and
substitute
equilibrium
terms
into
the
equation.
Step
4
Solve
the
equation
for
x.
Assume
that
x
is
small
compared
with
the
initial
concentrations.
Check
the
validity
of
this
assumption
when
you
find
the
value
of
x.
Step5
pH
=
-log[H;0%]
\—
_/
440
MHR
«
Unit
4
Chemical
Systems
and
Equilibrium
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Act
on
Your
Strategy
Step1
When
the
solutions
are
mixed,
the
total
volume
is
twice
the
volume
of
each
component
volume.
Therefore,
the
concentration
of
each
component
is
halved.
[CH3COO0H]
=
[CH3COO|
=
0.10
mol/L
Step
2
Concentration
(mol/L)
CH3COOHq
+
H;0p
==
CH3C00
(g
+
H3O%g)
Initial
0.10
—
0.10
~0
Change
-X
—
+x
+x
Equilibrium
0.10—x
—
0.10
+
x
x
_
[CH3CO0O1[H307]
Step3
K
=
"cH,coon]
_
(0.10
+
x)(x)
(0.10
—
x)
Step
4
Assume
that
(0.10
+
x)
=~
0.10
and
(0.10
—
x)
=~
0.10.
PROBLEM
TIP
K,
=
(0.10)(x)
_
4
g
%
10-5
You can
check
whether
it
rea-
(0.10)
sonable
to
ignore
x
by
dividing
~.x
=
1.8
x
107
mol/L
0.10
by
K;.
The
value
is
greater
Step5
pH
=
—log(1.8
x
107
than
500,
so
the
approximation
is
reasonable.
=
4.74
L
)
The
pH
of
the
buffer
solution
is
4.74.
Check
Your
Solution
The
value
of x
(1.8
x
107°)
is
negligible
compared
with
the
initial
concentration
of
each
component
{(0.10).
A
buffer
that
is
made
using
a
weak
acid
and
its
conjugate
base
should
have
a
pH
that
is
less
than
7.
I
Practice
Problems
_/
25.
A
buffer
solution
is
made
by
mixing
250
mL
of
0.200
mol/L
aqueous
ammonia
and
400
mL
of
0.150
mol/L.
ammonium
chloride.
Calculate
the
pH
of
the
buffer
solution.
26.
A
buffer
solution
is
made
by
mixing
200
mL
of
0.200
mol/L
aqueous
ammonia
and
450
mL
of
0.150
mol/L.
ammonium
chloride.
Calculate
the
pH
of
the
buffer
solution.
21.
A
buffer
solution
contains
0.200
mol/L
nitrous
acid,
HNO;(q),
and
0.140
mol/L
potassium
nitrite,
KNOj(q).
What
is
the
pH
of
the
buffer
solution?
28.
A
buffer
solution
is
prepared
by
dissolving
1.80
g
of
benzoic
acid,
CeHsCOOH,
and
1.95
g
of
sodium
benzoate,
NaCgHsCOO),
in
800
mL
of
water.
Calculate
the
pH
of
the
buffer
solution.
Chapter
9
Agqueous
Solutions
and
Solubility
Equilibria
«
MHR
441
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Section
Summary
In
this
section,
you
determined
the
solubility
product
constant,
K,
based
on
solubility
data.
You
obtained
your
own
solubility
data
and
used
these
data
to
calculate
a
value
for
K,.
You
determined
the
molar
solubility
of
ionic
solutions
in
pure
water
and
in
solutions
of
common
ions,
based
on
their
K,
values.
In
section
9.3,
you
will
further
explore
the
implications
of
Le
Chételier’s
principle.
You
will
use
a
reaction
quotient,
Qsp,
to
predict
whether
a
precipitate
forms.
As
well,
you
will
learn
how
selective
precipitation
can
be
used
to
identify
ions
in
solution.
Section
Review
O
The
values
of
K,
for
binary
ionic
compounds
that
contain
either
sodium,
potassium,
or
ammonium
cations
are
not
listed
in
a
K,
table.
State
why.
(2]
A
saturated
solution
of
a
slightly
soluble
ionic
compound
does
not
contain
any
of
the
solute
in
solid
form.
Is
this
system
at
equilibrium?
Explain
your
answer
briefly.
©
O
Does
a
larger
value
of
K,
necessarily
mean
a
greater
solubility?
To
illustrate
your
answer,
calculate
and
compare
the
molar
solubility
of
silver
chloride,
AgCl
(Ksp
=
1.8
X
10719),
with
silver
chromate,
Ag,CrO,
(Ksp
=
2.6
x
10712),
O
O
Explain
why
there
is
no
simple
relationship
between
K,
for
a
given
compound
and
the
solubility
of
the
compound
in
g/L.
©
©
The
maximum
solubility
of
barium
fluoride,
BaF,,
at
25°C,
is
1.3
g/L.
(a)
Calculate
K,
for
BaF,
at
25°C.
(b)
Calculate
the
solubility
of
BaF,
in
formula
units
of
BaF,/L.
O
O
A
chemist
adds
20.0
g
of
sodium
hydroxide,
NaOH,
to
a
1.0
L.
solution
of
1.0
mol/L
acetic
acid.
(a)
Explain
why
the
resulting
solution
resists
pH
changes
when
an
acid
or
a
base
is
added.
(b)
Calculate
the
pH
of
the
solution.
Assume
that
there
is
no
change
in
the
volume
of
the
solution
when
NaOH
is
added.
©
©
A
buffer
solution
is
prepared
by
adding
2.1
g
of
NaF
to
5.00
x
10>
mL
of
0.10
mol/L
HF,q.
Calculate
the
pH
of
the
buffer
solu-
tion.
Assume
that
no
change
in
volume
results
from
the
addition
of
NaF
to
the
HF
solution.
€)
©
A
solution
of
BaCl,
is
added
to
a
solution
of
Na,SO,.
(a)
What
do
you
expect
to
observe?
Include
a
balanced chemical
equation
in
your
answer.
(b)
Calculate
the
solubility
(in
mol/L
and
in
g/L)
of
BaSO,
in
pure
water.
(e)
Calculate
the
solubility
(in
mol/L
and
in
g/L)
of
BaSO,
in
0.085
mol/L
Na,SQO,.
©
O
Explain
why
the
solubility
of
nickel(II)
carbonate,
NiCOs,
is
unaf-
fected
by
the
addition
of
NaCl
to
a
saturated solution
of
NiCOj.
442
MHR
«
Unit
4
Chemical
Systems
and
Equilibrium
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Predicting
the
Formation
of
a
Precipitate
In
section
9.2,
you
worked
with
concentrations
of
ions
in
solutions
at
equilibrium.
You
used
K,
to
determine
the
solubility
of
ionic
substances
in
pure
water
and
in
solutions
of
common
ions.
The
systems
you
worked
with
were
saturated
solutions
that
contained
excess,
undissolved
solute.
How
do
you
predict
whether
a
given
concentration
of
ions
will
result
in
the
precipitation
of
an
ionic
compound? How
can
you
tell
if
a
solution
is
saturated?
You
substitute
the
concentrations
of
the
ions
into
an
expres-
sion
that
is
identical
to
the
solubility
product
expression.
Because
these
concentrations
may
not
be
the
same
as
the
concentrations
that
the
equilibrium
system
would
have,
however,
the
expression
has
a
different
name:
the
ion
product.
Comparing
the
lon
Product
With
the
Solubility
Product
The
ion
product,
Qp,
is
an
expression
that
is
identical
to
the
solubility
product
constant,
but
its
value
is
calculated
using
concentrations
that
are
not
necessarily
those
at
equilibrium.
(The
relationship
between
the
expression
for
solubility
product,
K;p,
and
the
expression
for
the
ion
product,
Qsp,
is
analogous
to
the
relationship
between
the
equilibrium
constant,
K;,
and
the
reaction
quotient,
(Q..)
For
example,
suppose
that
you
slowly
add
an
ionic
compound,
such
as
magnesium
sulfate,
MgSQ,,
to
water
in
a
beaker.
The
following
equation
represents
the
dissociation.
MgSO4(S)
=
Mg2+(aq)
+
8042_(aq)
KSP
=5.9x
1073
Initially,
all
the
magnesium
sulfate
dissolves.
The
solution
is
not
at
equilibrium
as
long
as
more
solid
can
dissolve.
As
you
add
more
solid,
however,
the
solution
reaches
equilibrium.
The
solution
is
saturated,
and
no
more
salt
will
dissolve.
If
you
continue
to
add
magnesium
sulfate
to
the
water,
you
will
see
an
increasing
amount
of
solid
in
the
beaker.
The
ion
product
expression
for
magnesium
sulfate
is
written
as
follows:
Qsp
=
[M82+]
[8042_]
You
calculate
Q;,
by
substituting
the
concentration
of
each
ion
into
the
expression.
If
Q;,
is
larger
than
K,
the
product
of
the
concentrations
of
the
ions
is
greater
than
it
would
be
at
equilibrium.
For
the
system
to
attain
equilibrium,
some
of
the
ions
must
leave
the
solution
by
precipitation.
Conversely,
if
Q,,
is
less
than
K,
the
product
of
the
concentration
of
the
ions
is
smaller
than
it
is
at
equilibrium.
Therefore,
the
solution
is
not
yet
saturated
and
more
ions
can
be
added
to
the
solution
without
any
precipitation.
The
relationship
between
Q, and
K;;
for
the
dissociation
of
a
slightly
soluble
ionic
compound
is
summarized
on
the
next
page.
Use
the
following
general
equation
as
a
reference.
Solid
ionic
compound
=
Dissociated
ions
in
solution
Section
Preview/
Specific
Expectations
In
this
section,
you
will
m
predict
the
formation
of
precipitates
by
using
the
solubility
product
constant
solve
equilibrium
problems
involving
concentrations
of
reactants
and
products
and
Kp
communicate
your
under-
standing
of
the
following
terms:
ion
product
(Qgp),
fractional
precipitation,
qualitative
analysis,
quantitative
analysis
Chapter
9
Agueous
Solutions
and
Solubility
Equilibria
«
MHR
443
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A
chemist
can
test
for
the
presence
of
chloride
ions
in
a
solution
by
adding
silver
nitrate
solution.
If
a
white
precipitate
of
silver
chloride
forms,
the
test
is
positive
for
chloride.
Qp
<
K,
*
The
system
attains
equilibrium
by
moving
to
the
right,
favouring
dissociation.
More
solid
can
dissolve.
Qp
=
Ky
The
system
is
at
equilibrium.
e
No
more
solid
can
dissolve.
e
No
precipitate
forms.
Qp
>
K,
*
The
system
attains
equilibrium
by
moving
to
the
left,
favouring
precipitation.
e
A
precipitate
forms
until
equilibrium
is
reached.
Using
the
lon
Product
Expression
You
can
use
the
relationship
between
the
ion
product
expression
and
the
solubility
product
expression
to
predict
whether
a
precipitate
will
form
in
a
given
system.
One
common
system
involves
mixing
solutions
of
two
soluble
ionic
compounds,
which
react
to
form
an
ionic
compound
with
a
very
low
solubility.
If
Q,,
>
K,;,
based
on
the
initial
concentrations
of
the
ions
in
solution,
the
sparingly
soluble
compound
will
form
a
precipitate.
How
do
you
know
which
ionic
compounds
are
soluble
and
which
are
not?
In
your
previous
chemistry
course,
you
learned
a
set
of
solubility
guidelines.
Table
9.3
summarizes
these
guidelines.
Remember:
the
higher
guideline
takes
precedence.
(For
instance,
guideline
3
says
that
carbonates
have
very
low
solubility.
Sodium
carbonate
is
soluble,
however,
because
guideline
1
says
that
ionic
compounds
containing
sodium
are
soluble.)
Chemists
do
not
usually
work
with
solubility
products
for
soluble
compounds.
Thus,
you
will
not
find
soluble
ionic
compounds
listed
in
K,
tables.
If
you
see
a
compound
in
a
Kj,
table,
you
know
that
it
has
a
low
solubility
relative
to
compounds
such
as
sodium
chloride.
Table
9.3
General
Solubility
Guidelines
Guideline
Cations
OIS
Result
3
(T
1
Li*,
Na*,
K*
NO;~,
CH;COO-,
|
soluble
|
Ca(ClOs),
is
Rb*,
Cs*,
Nh,"
ClO5~
insoluble.
2
Ag*,
Pb*,
Hg*
CO4%,
PO,*,0%,
|
very
low
|
BaO
and
Ba(OH),
S*,
OH-
solubility
|
are
soluble.
Group
2
(ITA)
sulfides
tend
to
decompose.
3
Cl.,
Br,I”
soluble
4
Ba?*,
Ca®*,
Sr?*
very
low
solubility
5
Mg?*,
Cu?*,Zn?%",
|
SO,*
soluble
FeZ+
Fe3+
A13+
For
example,
silver
nitrate
and
sodium
chloride
are
both
soluble
ionic
compounds.
When
they
are
mixed
in
solution,
however,
they
react
to
form
silver
chloride,
as
shown
in
Figure
9.7.
The
equation
for
the
reaction
is:
AgNO3(aq)
+
NaCl(aq)
<=
AgCl)
+
NaNOj(yq)
Silver
chloride
has
a
very
low
solubility.
Therefore,
a
precipitate
will
form
even
if
the
initial
concentration
of
chloride
and
silver
ions
is
very
small.
The
following
Sample
Problems
show
you
how
to
decide
whether
a
precipitate
will
form
under
a
given
set
of
conditions.
444
MHR
«
Unit
4
Chemical
Systems
and
Equilibrium
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I
Sample
Problem
Predicting
Precipitation
I
Problem
A
common
test
for
chloride
ions
in
a
solution
involves
adding
AgNO3(4g
to
the
solution
being
tested.
If
chloride
ions
are
present
in
sufficient
quantity,
the
chemist
will
observe
a
white
cloudiness
that
indicates
the
formation
of
a
precipitate.
Ksp
for
silver
chloride
is
1.8
x
107°.
A
drop
(0.050
mL)
of
6.0
mol/L
silver
nitrate
is
added
to
1.0
L
of
0.10
mol/L
sodium
chloride.
Does
a
precipitate
of
silver
chloride
form?
What
Is
Required?
Will
silver
chloride
precipitate
under
the
given
conditions?
What
Is
Given?
You
know
the
concentration
and
volume
of
the
silver
nitrate
and
sodium
chloride
solutions.
For
AgNOj;,
¢
=
6.0
mol/L
and
V
=
0.050
mL.
For
NaCl,
¢
=
0.10
mol/L
and
V
=
1.0
L.
You
also
know
Kj,
for
silver
chloride.
Plan
Your
Strategy
Step
1
Determine
the
concentrations
of
silver
ions
and chloride
ions
in
the
reaction
mixture.
Step
2
Substitute
the
ion
concentrations
into
the
ion
product
expression
for
silver
chloride
to
determine
Qp.
Step3
Compare
Qs
with
Ky,
and
predict
whether
or
not
a
precipitate
will
form.
Act
on
Your
Strategy
Step
1
Because
the
volume
of
the
AgNO;
solution
is
much
smaller
than
the
volume
of
the
NaCl
solution,
you
can
ignore
it.
In
other
words,
assume
that
the
final
volume
of
the
reaction
mixture
is
1.0
L.
NaCl(aq)
=
Na+(aq)
+
Cl_(aq)
».[Cl7]
=
[NaCl]
=
0.10
mol/L
AgNO;()
=
Ag'ag)
+
NO3
(ag)
[Ag+]
_
[AgNO3]
_C
X
Vinitial
Vfinal
_
(6.0
mol/L)(0.050
L.
x
1.0
x
10~%
/mL
1.0k
=
3.0
x
10~
mol/L
Step2
Qs
=
[Ag'][C]]
=
(3.0
x
107%)(0.10)
=3.0x107°
Step3
Since
Qs
>
Kip,
AgCl
will
precipitate
until
Qsp
=
1.8
x
10719,
Check
Your
Solution
The
units
are
correct
in
the
calculation
of
[Ag'].
It
seems
reasonable
that
a
precipitate
formed,
since
Ky,
for
silver
chloride
is
very
small
compared
with
the
concentration
of
chloride
ions
and
silver
ions.
|
\
-7
Chapter
9
Agueous
Solutions
and
Solubility
Equilibria
«
MHR
445
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PROBLEM
TIP
For
Practice
Problems
31
and
32,
you
will
need
to
look
up
values
of
Ksp
in
Appendix
E.
I
Practice
Problems
—
29.
A
solution
contains
0.15
mol/L
of
NaCl
and
0.0034
mol/L
Pb(NOs)s.
Does
a
precipitate
form?
Include
a
balanced
chemical
equation
for
the
formation
of
the
possible
precipitate.
Ky,
for
PbCl,
is
1.7
x
107,
30.
One
drop
(0.050
mL)
of
1.5
mol/L
potassium
chromate,
K,CrQOy,
is
added
to
250
mL
of
0.10
mol/L
AgNQOs.
Does
a
precipitate
form?
Include
a
balanced
chemical
equation
for
the
formation
of
the
possible
precipitate.
Kg,
for
Ag,CrO,
is
2.6
x
1072,
31.
A
chemist
adds
0.010
g
of
CaCl;
to
5.0
x
10>
mL
of
0.0015
mol/L
sodium
carbonate,
Na;COj3.
Does
a
precipitate
of
calcium
carbonate
form?
Include
a
balanced chemical
equation
for
the
formation
of
the
possible
precipitate.
32.
0.10
mg
of
magnesium
chloride,
MgCl,,
is
added
to
2.5
x
10?
mL
of
0.0010
mol/L.
NaOH.
Does
a
precipitate
of
magnesium
hydroxide
form?
Include
a
balanced chemical
equation
for
the
formation
of
the
possible
precipitate.
The
next
Sample
Problem
shows
how
to
predict
whether
or
not
a
precipi-
tate
will
form
when
two
significant
volumes
of
reactant
are
mixed
together.
You
will
need
to
use
the
solubility
guidelines
to
decide
whether
or
not
an
insoluble
compound
forms.
I
Sample
Problem
Predicting
Precipitation
II
Problem
A
chemist
mixes
100.0
mL
of
0.25
mol/L
Ca(NOs3),
with
200.0
mL
of
0.070
mol/L
NaF.
Does
a
precipitate
form?
What
Is
Given?
You
know
the
initial
concentration
and
volume
of
the
calcium
nitrate
and
sodium
fluoride
solutions.
For
Ca(NOs),,
¢
=
0.25
mol/L
and
V
=100.0
mL.
For
NaF,
¢
=
0.070
mol/L
and
V
=
200.0
mL.
As
well,
you
have
the
solubility
guidelines
in
Table
9.3.
Plan
Your
Strategy
Step1
Decide
whether
a
compound
with
low
solubility
forms
when
calcium
nitrate
and
sodium
fluoride
are
mixed,
using
the
solubility
guidelines.
Step
2
If
an
insoluble
compound
forms,
write
an
equation
that
represents
the
reaction.
Look
up
K,
for
the
compound
in
Appendix
E.
Step
3
Determine
the
concentrations
of
the
ions
that
make
up
the
compound.
Step
4
Substitute
the
concentrations
of
the
ions
into
the
ion
product
expression
for
the
compound
to
determine
Q.
Step5
Compare
(s,
with
Ky,
and
predict
whether
or
not
a
precipitate
forms.
\—
J/
446
MHR
«
Unit
4
Chemical
Systems
and
Equilibrium
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4
")
Act
on
Your
Strategy
Step1
According
to
the
solubility
guidelines,
NaF
and
Ca(NOs);
are
both
soluble,
because
ionic
compounds
that
contain
sodium
or
nitrate
ions
are
soluble.
When
you
mix
the
solutions,
NaNOj,
and
CaF,
can
also
form.
NaNOQOj
is
soluble,
but
CaF;
is
not,
according
to
guideline
4.
Step2
NaF(yq)
+
Ca(NO;;)z(aq)
=
NaNOQOj(yq)
+
Galy
From
Appendix
E,
Ky,
for
CaF;
is
3.2
x
1071,
Step3
When
calculating
the
volume
of
the
reaction
mixture,
assume
that
it
is
equal
to
the
sum
of
the
volumes
of
the
solutions.
This
is
a
reasonable
assumption,
because
both
solutions
are
dilute.
[Ca?']
=
[Ca(NO3),]
=
%
final
_
(0.25
mol/L)(100.0
=)
(100.0
mE.
+
200.0
mE
=
8.3
X
102
mol/L
[F]
=
[NaF]
=
¢
X
Vinitial
Viinal
_
(0.070
mol/1)(200.0
mt)
(100.0
mL.
+
200.0
mkL.)
=
4.7
x
107
mol/L
Step4
Qs
=
[Ca*'][F]?
=
(8.3
x
107%)(4.7
x
107%)?
=1.8x
10™*
Step5
Since
Qs
>
Ksp,
CaF,
precipitates
until
Qsp
=
3.2
x
10711,
Check
Your
Solution
The
units
in
the
calculation
of
the
concentrations
of
the
ions
are
correct.
It
seems
reasonable
that
a
precipitate
formed,
since
Ky
for
calcium
fluoride
is
very
small
compared
with
the
concentrations
of
the
chloride
ions
and
silver
ions.
\—
_J
Practice
Problems
J/
33.
1.0
X
102
mL
of
1.0
X
102
mol/L
Pb(NQs;);
is
added
to
40
mL
of
0.040
mol/L
NaCl.
Does
a
precipitate
form?
Include
a
balanced
chemical
equation
for
the
formation
of
the
possible
precipitate.
34.
2.3
X
10?2
mL
of
0.0015
mol/L
AgNQ;
is
added
to
1.3
x
10?2
mL
of
0.010
mol/L
calcium
acetate,
Ca(CH3COQ);.
Does
a
precipitate
form?
Include
a
balanced
chemical
equation
for
the
formation
of
the
possible
precipitate.
Ky,
for
AgCH3;COO
is
2.0
x
1073,
35.
25
mL
of
0.10
mol/L
NaOH
is
added
to
5.0
x
102
mL
of
0.00010
mol/L
cobalt(Il)
chloride,
CoCl,.
Does
a
precipitate
form?
Include
a
balanced
chemical
equation
for
the
formation
of
the
possible
precipitate.
36.
250
mL
of
0.0011
mol/L
Al,(SQO,);
is
added
to
50
mL
of
0.022
mol/L
BaCl;.
Does
a
precipitate
form?
Include
a
balanced chemical
equation
for
the
formation
of
the
possible
precipitate.
Chapter
9
Agueous
Solutions
and
Solubility
Equilibria
«
MHR
447
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Chemists
use
a
centrifuge
to
separate
a
precipitate
from
a
solution.
The
centrifuge
spins
the
samples
quickly,
and
the
resulting
force
pushes
the
precipi-
tate
to
the
bottom
of
the
test
tube.
The
solution
can
then
be
decanted.
Analytical
Applications
of
Precipitation
Reactions
In
the
previous
Sample
Problems,
you
saw
that
combining
solutions
of
soluble
ionic
compounds
can
cause
the
precipitation
of
a
slightly
soluble
ionic
compound.
Analytical
chemists
use
precipitation
reactions
to
remove
ions
from
solution
or
to
identify
ions
in
an
unknown
solution.
Fractional
Precipitation
Suppose
a
chemist
has
a
solution
that
contains
several
similar
ions.
The
chemist
wants
to
remove
some
of
these
ions,
while
leaving
other
ions
in
solution.
One
way
to
do
this
is
by
fractional
precipitation:
a
process
in
which
ions
are
selectively
precipitated
from
solution,
leaving
other
ions.
For
example,
consider
a
solution
that
contains
three
halide
ions:
CI-,
Br~,
and
I".
Since
these
halides
all
come
from
the
same
group
on
the
periodic
table,
they
share
many
properties.
When
they
are
the
anions
in
slightly
soluble
ionic
compounds,
however,
they
have
different
solubilities.
(See
Table
9.4.)
Therefore,
chemists
can
use
fractional
precipitation
to
separate
them
from
solution.
Table
9.4
Values
of
K,
for
Silver
Halides
Silver
halide
|
Ke,
AgCl
1.8
x
10710
AgBr
3.3x
10713
Agl
1.5
x
10716
Out
of
the
three
compounds,
silver
chloride
is
the
most
soluble
and
silver
iodide
is
the
least
soluble.
(You
can
compare
the
solubilities
of
the
compounds
based
on
their
solubility
products
because
they
are
all
the
same
type.
Each
formula
unit
contains
two
ions.)
To
separate
the
halide
ions,
a
chemist
adds
silver
ions
to
the
solution
until
most
of
the
iodide
ions
precipitate
out
as
Agl.
The
chemist
stops
adding
silver
ions
before
AgBr
begins
to
precipitate.
The
chemist
then
removes
the
solid
Agl
by
filtration
or
by
using
a
centrifuge,
as
shown
in
Figure
9.8.
To
remove
the
bromide
ions,
the
chemist
repeats
the
process
with
the
remaining
solution.
The
chemist
adds
silver
ions
until
most
of
the
bromide
ions
precipitate
out
as
AgBr,
but
stops
adding
ions
before
AgCl
begins
to
precipitate.
By
filtering
or
centrifuging
again,
the
chemist
obtains
a
solution
that
contains
only
chloride
ions.
448
MHR
«
Unit
4
Chemical
Systems
and
Equilibrium
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Qualitative
Analysis
As
well
as
separating
similar
ions
out
of
solution,
chemists
can
also
use
their
understanding
of
solubility
and
precipitation
reactions
to
identify
unknown
ions
in
solution.
Qualitative
analysis
is
the
branch
of
analytical
chemistry
that
involves
identifying
elements,
compounds,
and
ions
in
samples
of
unknown
or
uncertain
composition.
The
other
branch
of
analytical
chemistry
is
quantitative
analysis.
In
quantitative
analysis,
analytical
chemists
determine
how
much
of
a
compound,
element,
or
ion
is
in
a
sample.
One
way
to
identify
cations
in
solution
is
by
selectively
precipitating
them
out
of
solution.
As
you
know,
cations
may
form
soluble
or
insoluble
ionic
compounds,
depending
on
the
anions
that
are
present.
For
example,
copper(Il)
chloride,
CuCl;
is
soluble
in
water.
Copper(lI)
sulfide,
CusS,
is
insoluble
in
an
acidic
solution.
Knowing
about
the
relative
solubility
of
cations
when
combined
with
various
anions
helps
chemists
identify
them.
Using
precipitation
reactions
to
identify cations
was
once
a
common
part
of
qualitative
analyses.
Today,
chemists
usually
identify
and
quantify
unknowns
using
instruments
such
as
spectrophotometers.
However,
examining
the
identification
of
cations
through
precipitation
reactions
allows
you
to
see
how
solubility
equilibria
can
be
manipulated.
It
also
provides
an
opportunity
to
solve
a
chemical
“mystery”—what
is
in
the
unknown
solution?
To
find out
what
cations
are
in
a
solution
using
precipitation
reactions,
chemists
carry
out
reactions
according
to
a
scheme
that
incorporates
tests
for
each
component
that
may
be
present.
One
example
of
such
a
scheme
is
shown
in
Figure
9.9.
This
scheme
shows
how
a
chemist
would
analyze
an
unknown
solution
that
may
contain
all,
some,
or
none
of
the
following
ions:
silver,
Ag';
cadmium,
Cd?*;
and
aluminium,
Al3*.
The
analysis
uses
selective,
step
by
step
precipitation.
www.mcgrawhill.ca/links/
chemistry12
Bauxite
is
the
most
abundant
ore
of
aluminum.
The
first
step
in
extracting
aluminum
from
bauxite
is
called
the
Bayer
process.
The
Bayer
process
involves
a
fractional precipita-
tion
of
impurities,
including
iron(l11)
oxide
and
titanium
dioxide.
Search
the
Internet
to
find
the
history
of
the
Bayer
process
and
learn
how
it
works.
Present
your
findings
as
a
poster.
To
start
your
search,
go
to
the
web
site
above
and
click
on
Web
Links.
add
CI”
add
2
add
OH™
(HCI)
(H,S)
(NH3)
Ag*
Cd2+
@
Cd2+
@
@
o
o
o
2
Z
2
3
=
3
=
3
=
AR
=
A"
£
Alg
g
S
S
S
U
AgClI
CdS
AI(OH)3
1
2
3
4
Initially,
all
If
silverionis
If
cadmiumionis
If
aluminum
is
ions
are
present,
a
white
present,
a
yellow
present,
a
white
dissolved.
precipitate
will
form.
precipitate
will
form.
precipitate
will
form.
This
diagram
shows
a
scheme
for
identifying
silver,
Ag*,
cadmium,
Cd?*,
and
aluminium,
APF*,
ions.
After
each
step,
if
a
precipitate
forms,
it
is
separated
by
centrifugation.
Then
the
solution
is
carefully
poured
into
a
new
test
tube
and
used
in
the
next
step.
Chapter
9
Agueous
Solutions
and
Solubility
Equilibria
«
MHR
449
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The
scheme
that
is
shown
in
Figure
9.9
is
very
simple.
More
complex
qualitative
analyses
involve
many
more
steps
of
isolation
and
identifica-
tion,
including
some
steps
that
are
not
precipitation
reactions.
For
example,
some
ions,
such
as
sodium,
Na*,
and
potassium,
K*,
cannot
be
precipitated
out
of
an
aqueous
solution,
because
the
ionic
compounds
that
contain
them
are
always
soluble.
Instead,
chemists
identify
these
ions
using
a
flame
test.
In
the
following
ThoughtLab,
you
will
simulate
a
qualitative
analysis
that
includes
a
flame
test.
Suppose
that
you
are
an
analytical
chemist.
You
chromium,
Cr¥*;
and
sodium,
Na*.
Your
laboratory
assistant
has
already
performed
some
tests on
known
solutions
of
these
ions
and
has
made
the
following
observations:
e
As®*
forms
a
precipitate
when
H,S,q
is
added.
Cr3
and
Na*
do
not
form
a
precipitate
when
HzS(aq)
is
added.
other
ions
do
not
form
a
precipitate
when
HClq)
is
added.
added.
(Making
the
solution
basic
has
the
same
effect
as
adding
OH™.)
Ag"
also
forms
a
precipitate
in
a
basic
solution.
e
Na*
produces
a
bright
yellow
flame
in
a
flame
test.
]
Procedure
1.
Based
on
these
observations,
devise
a
step-by-
step
scheme
for
testing
an
unknown
solution.
The
solution
may
contain
all,
some,
or
just
one
of
the
possible
ions.
Use
Figure
9.9
to
help
you.
2.
Decide
which
ions
are
in
the
unknown
solution.
(Remember,
the
solution
may
contain
one,
two,
three,
or
all
of
the
possible
ions
listed
above.)
Based
on
your
scheme,
write
down
the
observa-
tions
that
you
expect
to
make
for
this
solution.
Section
Summary
want
to
test
an
unknown
solution
that
contains
some
or
all
of
the
following
ions:
silver,
Ag*;
arsenic,
As®;
e
Ag®
forms
a
precipitate
when
HCl,q)
is
added.
The
e
Cr®*
forms
a
precipitate
when
a
solution
of
NH;
is
A
Qualitative
Analysis
3.
Exchange
schemes
and
observations
with
a
classmate.
Do
not
tell
your
classmate
which
ions
are
in
your
unknown
solution.
Based
on
your
classmate’s
scheme
and
observations,
determine
which
ions
are
in
your
classmate’s
unknown
solution.
4.
Discuss
your
results
with
your
classmate.
Did
you
both
succeed
in
identifying
the
ions
in
the
unknown
solution?
If
not,
you
or
your
classmate
may
need
to
modify
your
schemes
or
observations.
Analysis
1.
Explain
how
qualitative
analysis
depends
on
the
different
solubilities
of
ionic
compounds.
2.
Hg,?*
forms
an
insoluble
chloride,
Hg,Cl,.
If
your
unknown
solution
contained
a
mixture
of
all
or
some
of
the
Hg,?*,
Ag*,
Cr¥*,
and
As*
ions,
would
the
scheme
you
developed
be
sufficient
to
identify
all
the
ions
in
the
solution?
Explain
your
answer.
3.
If
you
were
performing
this
analysis
in
a
laboratory,
what
safety
precautions
would
you
need
to
take?
Explain
your
answer.
In
Chapter
9,
as
in
most
of
Unit
4,
you
learned
about
equilibrium
reactions.
In
this
section,
you
analyzed
precipitation
reactions.
You
mainly
examined
double-displacement
reactions—reactions
in
which
two
soluble
ionic
compounds
react
to
form
a
precipitate.
You
used
the
solubility
product
constant,
Kp,
to
predict
whether
or
not
a
precipitate
would
form
for
given
concentrations
of
ions.
In
Unit
5,
you
will
learn
about
a
class
of
reactions
that
will
probably
be
new
to
you.
You
will
see
how
these
reactions
inter-
convert
chemical
and
electrical
energy.
450
MHR
«
Unit
4
Chemical
Systems
and
Equilibrium
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Section
Review
(1]
Answer
the
following
questions
about
the
ion
product.
(a}
Under
what
circumstances
would
a
chemist
want
to
determine
Qs
for
a
system?
(b)
How
does
Qs
differ
from
K,
?
(¢}
Why
are
there
no
tabulated
(s,
values?
©
©
A
chemist
adds
1.0
mg
of
Nal
to
50
mL
of
a
0.010
mol/L
solution
of
Pb(NO;),.
Does
a
precipitate
form?
K,
for
Pbl,
is
9.8
x
107°.
©
©
How
many
milligrams
of
Na,SO,
will
just
begin
to
precipitate
CaS0O,
from
5.0
X
102
mL
of
a
0.10
mol/L
solution
of
GaCl,?
Ky
for
CaS0,
is
2.4
x
107°.
Hint:
How
does
Qs
compare
with
K,
at
the
point
when
precipitation
just
begins?
O
©
How
many
drops
of
0.0010
mol/L
silver
nitrate
solution
will
just
begin
to
precipitate
AgCl
from
5.0
x
10*
mL
of
a
0.90%
(m/v)
solution
of
NaCl?
(Assume
that
one
drop
equals
0.050
mL.)
K,
for
AgCl
is
1.8
x
10719,
O
o
K;p
for
CaSO,
is
2.4
x
107°,
and
K,
for
SrSO,
is
3.2
x
1077,
Suppose
that
you
have
a
1.0
L
solution
that
is
0.20
mol/L
in
Ca?*"
ions
and
0.20
mol/L
in
Sr?"
ions.
You
slowly
begin
to
add
solid
Na,SQOs,.
(a)
Explain
why
SrSO,
precipitates
first.
(b)
How
many
milligrams
of
Al,(SO,4);
will
just
begin
to
precipitate
SrSQO,
from
the
solution?
O
O
Consider
a
solution
that
contains
Pb%*,
Gu?*,
and
Mg?*
cations,
present
as
their
(soluble)
nitrate
salts.
How
could you
selectively
precipitate
these
cations,
given
solutions
of
NaCl, Na,S,
and
NazPO,?
Present
your
answer
as
a
flowchart,
with
a
chemical
equation
accompanying
each
step.
©
@
To
purify
sodium
chloride,
NaCl,
for
use
as
table
salt,
HCI
is
added
to
a
saturated
solution
of
NaCl.
Explain
how
this
process
works,
using
the
concept
of
the
ion
product.
What
do
you
think
is
the
next
step
in
the
process?
Explain
your
answer.
Chapter
9
Agueous
Solutions
and
Solubility
Equilibria
«
MHR
451
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Review
Reflecting
on
Chapter
9
Summarize
this
chapter
in
the
format
of
your
choice.
Here
are
a
few
ideas
to
use
as
guidelines:
e
Explain
how
to
predict
whether
a
solution
of
a
specific
salt
will
be
acidic,
basic,
or
neutral.
e
Explain
how
to
determine
a
substance’s
solubility
product
constant,
K,
based
on
the
molar
solubility
of
the
substance.
e
Explain
how
to
calculate
the
molar
solubility
of
a
pure
substance
in
water
or
in
a
solution
of
a
common
ion,
given
Kjp.
e
Describe
how
to
use
K,
to
predict
the
formation
of
a
precipitate.
o
Identify
the
effects
of
solubility
equilibria
in
the
human
body.
e
Describe
how
chemists
use
solubility
equilibria
to
remove
specific
ions
from
solution
and
to
identify
ions
in
solution.
Reviewing
Key
Terms
For
each
of
the
following
terms,
write
a
sentence
that
shows
your
understanding
of
its
meaning.
equivalence
point
end-point
solubility
product
constant
(K,)
ion
product
(Qsp)
fractional
precipitation
qualitative analysis
quantitative
analysis
Knowledge/Understanding
1.
Assume
that
the
aqueous
solutions
in
each
pair
have
the
same
concentration.
For
each
pair,
identify
which
of
the
two
solutions
has
the
lower
pH.
(a)
NH4C1(aq)
and
KHSO4(aq)
(b)
H2S(aq)
and
KHSOy(aq)
(c)
NaHPOy(aq)
and
CaBra(yqg)
(d)
AI(NOS)S(aq)
and
AI(NOZ)Z(aq)
2.
Suppose
that
you
have
the
following
indicators
available:
bromphenol
blue,
methyl
red,
bromthymol
blue,
and
phenolphthalein.
Use
Figure
9.3
to
suggest
an
appropriate
indicator
for
each
of
the
following
titrations.
Do
not
do
any
calculations.
(a)
sulfuric
acid
with
sodium
hydroxide
(b)
formic
acid
with
sodium
hydroxide
(c)
pyridine
with
nitric
acid
452
VHR
¢
Unit
4
Chemical
Systems
and
Equilibrium
10.
12.
13.
14.
15.
.
There
are
many
different
acid-base
indicators.
State
two
ways
in
which
they
are
all
similar.
.
What
substance
can
be
added
to
an
aqueous
solution
of
propanoic
acid
to
prepare
a
buffer
solution?
.
Hydrochloric
acid
is
slowly
added
to
a
solution
of
sodium
hydroxide.
How
does
the
pH
of
the
solution
change?
.
A
solution
of
a
certain
salt
is
tested
with
litmus
paper,
and
the
litmus
turns
red.
What
does
this
tell
you
about
the
relative
strengths
of
the
acid
and
base
from
which
the
salt
is
derived?
.
List
the
names
and
formulas
of
three
salts
that
dissolve
in
water
to
form
(a)
an
acidic
solution
(b)
a
basic
solution
.
When
performing
an
acid-base
titration,
you
should
add
only
a
few
drops
of
indicator
rather
than
a
few
millilitres.
Explain
why.
.
Distinguish
between
the
terms
“solubility”
and
“solubility
product
constant.”
Why
do
tables
of
K,
values
for
compounds
not
include
values
for
soluble
compounds?
.
Values
of
K,
are
given
for
a
certain
tempera-
ture.
How
do
the
K,
values
for
most
salts
change
if
the
temperature
of
the
solution
is
increased?
Write
an
equation
for
the
dissociation
of
each
compound
in
water.
Then
write
the
corresponding
K,
expression.
(a)
CuBr
(d)
Mgs(PO.),
(b)
CaCI‘O4
(e)
MgNH4PO4
(e)
Ni(OH),
For
a
solubility
system
to
reach
equilibrium,
some
undissolved
solid
must
be
present.
Explain
why.
Compare
the
solubility
of
magnesium
hydroxide
in
a
saturated
solution
of
ammonium
chloride
with
the
solubility
of
magnesium
hydroxide
in
pure
water.
Would
you
expect
magnesium
hydroxide
to
be
more
or
less
soluble,
or
to
have
the
same
solubility,
in
the
ammonium
chloride
solution?
Explain
your
answer.
In
aqueous
solution,
the
dihydrogen
phosphate,
H3POj(aq),
ion
can
act
as
either
an
acid
or
a
base.
When
tested,
a
solution
containing
H,PO,~
was
found
to
be
acidic.
Write
the
equation
for
the
reaction
that
took
place.
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16.
Why
must
s,
usually
be
much
larger
than
K,
before
you
can
observe
a
precipitate
forming?
17.
Silver
halide
salts
are
used
in
black
and
white
photography.
During
the
developing
process,
excess
silver
halide
is
removed
from
the
film
by
using
a
solution
of
sodium
thiosulfate
(commonly
called
“hypo”).
AgBr()
+
252032_(aq)
=
Ag(Szog)z?;l)
+
Braq)
What
is
the
effect
on
this
equilibrium
if
a
more
concentrated
solution
of
hypo
is
used?
Inquiry
18.
Suppose
that
you
have
an
aqueous
solution
of
copper
iodide,
Cul,,
which
you
must
precipitate
and
filter
before
pouring
the
remaining
liquid
into
a
disposal
beaker.
You
decide
to
use
the
common
ion
effect
to
precipitate
the
Cul;.
In
the
laboratory,
you
find
stock
solutions
of
Cu{NQO3),
and
Nal
with
the
same
concentrations.
You
want
to
use
the
smallest
volume
possible.
Which
stock
solution
will
you
choose
to
precipitate
the
Cul,?
Explain
your answer.
19.
Which
has
the
greater
effect
on
the
solubility
of
Pbl;
in
1.0
L
of
a
saturated
solution:
the
addition
of
0.10
mol
of
solid Pb(NQO3;),,
or
the
addition
of
0.10
mol
of
solid
Nal?
Justify
your
answer
with
calculations.
Assume
that
you
can
ignore
the
volume
change
when
solid
Pb(NOs),
is
added.
20.
Lead(II)
iodide and
barium
sulfate
have
almost
the
same
value
for
Ksp.
(Ksp
for
Pbl;
is
9.8
x
107?,
and
Ky,
for
BaSO,
is
1.1
x
10719.)
What
is
the
ratio
of
Pb**(,q)
to
Ba*'(yq
in
satura-
ted
solutions
of
the
salts?
21.
Silver
cyanide
is
used
in
aqueous
solutions
to
plate
a
coating
of
silver
on
ornaments,
jewellery,
and
silver
cutlery.
What
is
the
concentration
of
Ag'(aq)
in
a
saturated
solution
of
AgCN?
22.
Water
with
a
high
chloride
ion
content
is
not
safe
to
drink.
The
following
process
is
used
to
test
for
the
presence
of
Cl7(yq):
Dissolve
5.0
g
of
AgNOQOj;
in
500
mL
of
distilled
water.
Add
10
drops
of
the
solution
to
10
mL
of
the
water
being
tested,
and
look
for
a
cloudy
white
precipitate.
23.
24,
25.
e
G
A,
N
N
What
is
the
smallest
concentration
of
Cl
7
in
the
test
water
that
will
cause
the
formation
of
a
precipitate?
Design
an
experiment
to
determine
K,
for
lead(IT)
chloride,
PbCl;,
using
the
following
reaction.
Pb2+(aq)
+
Zn(s)
=
Pb(s)
+
Zn2+(aq)
Assume
that
you
have
solid
PbCl;,
a
strip
of
zinc
metal,
filter
paper,
an
electronic
balance,
and
any
other
standard
laboratory
equipment.
(a)
Write
the
equation
for
the
dissolution
of
PbCl;
and
the
corresponding
Ky,
equation.
(b)
Provide
a
step-by-step
procedure,
clearly
indicating
what
needs
to
be
measured
and/or
recorded.
(¢)
Explain
how
you
would
use
your
empirical
data
to
determine
K,
for
PbCl,.
Design
an
experiment
to
determine
K,
for
lead(Il)
iodide,
Pbl,,
using
the
reaction
below.
Pb2+(aq)
+
CI‘O42_(aq)
==
PbCrOy
Assume
that
you
have
solid
PbCl;,
a
0.50
mol/L
solution
of
K,CrQy,
filter
paper,
an
electronic
balance,
and
any
other
standard
laboratory
equipment.
(a)
Write
the
equation
for
the
dissolution
of
Pbl,
and
the
corresponding
K;,
equation.
(b)
Provide
a
step-by-step
procedure,
clearly
indicating
what
needs
to
be
measured
and/or
recorded.
(c)
Explain
how
you
would
use
your
empirical
data
to
determine
K,
for
Pbl,.
(d)
If
Kp
for
Pbl,
is
7.9
x
1079,
explain
why
the
solubility
of
PbCrO,
can
be
ignored.
(K,
for
PbCrO,
is
1.8
x
1074,
A
student
titrated
25.0
mL
of
0.280
mol/L
hydrochloric
acid
with
0.360
mol/L
sodium
hydroxide
solution.
The
end-point
was
observed
after
adding
18.80
mL
of
NaOH
).
Some
of
the
steps
that
were
followed
by
the
student
are
listed
below:
1.
A
transfer
pipette
was
rinsed
with
water.
The
pipette
was
then
used
to
obtain
25.0
mL
of
hydrochloric
acid,
which
was
placed
in
an
Erlenmeyer
flask.
2.
About
50
mL
of
water
was
added
to
the
Erlenmeyer
flask.
3.
A
few
drops
of
phenolphthalein
indicator
were
added
to
the
Erlenmeyer
flask.
Chapter
9
Agueous
Solutions
and
Solubility
Equilibria
«
MHR
453
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4.
A
burette
was
rinsed
with
water,
and
then
filled
with
the
sodium
hydroxide
solution.
(a)
Calculate
the
volume
of
NaOH(,q)
that
the
student
should
have
used
to
reach
equivalence.
(b)
What
was
the
percent
error
in
the
titration?
(¢}
Which
step(s)
in
the
titiration
could
account
for
the
error?
26.
Assume
that
you
have
a
solution
that
is
0.0010
mol/L
in
Cu(zq)
and
0.0010
mol/L
in
Ag'(ag)-
If
you
begin
to
add
solid
NaCl
to
the
solution,
which
compound
will
precipitate
first?
Explain.
27.
A
sample
of
well
water
contains
1.0
x
1073
mol/L
calcium
sulfate
and
1.0
x
10~*
mol/L
calcium
carbonate
ion.
A
chemist
adds
barium
chloride
to
the
sample.
Which
precipitate
forms
first,
barium
sulfate
or
barium
carbonate?
28.
Solutions
that
contain
dissolved
ions
have
been
shown
to
conduct
electricity.
(a)
Design
an
experiment
in
which
you
use
a
sensitive
conductivity
meter
to
determine
the
Kp
of
a
slightly
soluble
compound.
(b)
How
do
you
propose
to
calibrate
the
conductivity
meter?
(c)
If
your
teacher
approves
of
your
experimental
design,
and
if
you
have
access
to
a
conduc-
tivity
meter,
carry
out
your
experiment
to
determine
K,
for
Ca(OH)s.
29.
A
student
performs
an
experiment
to
determine
Kp
for
silver
acetate,
AgCH3CO,,
using
the
following
reaction.
2A8
g
+
Cug)
=
2Ag)
+
Cu(z;q)_
The
student
adds
a
strip
of
copper
metal,
with
a
known
mass,
to
a
saturated
solution
of
AgCH3CO,.
After
the
reaction,
the
copper
strip
is
washed,
dried,
and
re-weighed.
(a)
Write
the
equilibrium
equation
that
represents
the
equilibrium
of
AgCH3CO,.
Also
write
the
corresponding
K,
equation.
(b)
K,
for
AgCH;3CO,
is
2.0
x
1072,
A
strip
of
copper
metal,
with
an
initial
mass
of
23.4
g,
was
placed
in
1.00
x
10?
mL
of
saturated
AgCH3C0O;
solution
with
no
solid
AgCH3CO,
present.
Calculate
the
expected
mass
of
the
454
VIHR
-
Unit
4
Chemical
Systems
and
Equilibrium
I
T
I
D
M
R
copper
strip after
the
reaction.
(¢)
Why
must
the
saturated
AgCH3CO;
solution
be
free
of
any
solid
AgCH3;CO,?
30.
A
0.150
mol/L
solution
of
acetic
acid
was
titrated
with
25.0
mL
of
0.250
mol/L
sodium
hydroxide.
(a)
Calculate
the
volume
of
acetic
acid
that
was
needed
at
the
equivalence
point.
(b)
What
is
the
pH
of
the
solution
at
equivalence?
(c)
Suggest
a
suitable
indicator
for
this
titration.
31.
A
buffer
solution
is
prepared
by
adding
30.0
g
of
pure
acetic
acid
to
41.0
g
of
sodium
acetate
in
water,
and
then
diluting
the
solution
to
1.00
L.
What
is
the
pH
of
the
buffer
solution?
Communication
32.
Are
“insoluble”
compounds
really
insoluble?
Explain.
33.
Why
is
the
concentration
of
a
solid,
undissolved
compound
not
included
in
a
K,
expression?
34.
Is
a
supersaturated
solution
an
example
of
an
equilibrium
system?
Explain.
35.
When
a
solution
of
potassium
iodide,
KI,
is
added
to
a
solution
of
lead
(II)
nitrate,
Pb(NOs;).,
a
precipitate
forms.
(a)
Write
a
chemical
equation
for
the
reaction.
Indicate
the
physical
state
of
each
reactant
and
product.
(b)
How
does
the
product
of
the
concentration
of
the
ions
that
make
up
the
precipitate
compare
with
the
value
of
K,
for
the
precipi-
tate
immediately
before
precipitation
begins?
(c)
After
the
precipitate
forms,
how
would
you
describe
the
solution
with
respect
to
the
precipitated
compound:
unsaturated,
saturated,
or
supersaturated?
36.
Explain
“fractional
precipitation”
in
your
own
words.
On
what
principle
is
it
based?
31.
How
is
the
common
ion
effect
an
application
of
Le
Chatelier’s
principle?
Illustrate
your
answer
with
an
example,
including
diagrams
and
chemical
equations.
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N
I
e
e
Making
Connections
8.(b)
1.0
x
10~°
mol/L;
2.4
x
1072
g/L
(¢)
1.3
x
10~°
mol/L;
3.0x
107
g/L
9.3:
2.
Qsp
=
1.7
x
10729
>
K,
therefore
no
38.
The
concentration
of
calcium,
Ca%*,
ions
in
a
S
precipitate
forms.
3.17
mg
4.
one
drop
5.b)
0.18
mg
city’s
water
supply
is
too
high.
(a)
Explain,
with
the
help
of
chemical
equa-
tion(s),
how
to
use
solid
sodium
carbonate
(washing
soda
or
soda
ash),
Nay;COs,
to
reduce
[Ca®']
in
the
water.
(b)
What
challenges
would
engineers
need
to
overcome
when
implementing
the
system
for
reducing
[Ca?*]?
Answers
to
Practice
Problems
and
Short
Answers
to
Section
Review
Questions
Practice
Problems:
1.(a)
basic
(b)
basic
(e)
neutral
(d)
acidic
2(a)
acidic;
NHy"(aq)
+
HoO)
=
NHa(eg)
+
H3O"(aq)
(b)
neutral
{e)
basic;
OBI‘_(aq)
+
HyOy)
=
HOBI‘(aq)
+
OH_(aq)
(d)
acidic;
NH4+(aq)
+
H;0py
=
NH3(aq)
+
H3O+(aq)
3.
CgH50"
is
the
stronger
base
because
K,
for
its
conjugate
acid
is
smaller.
4.
acidic
5.7.74
6.1.70
7.8.15
8.5.28
9.(a)
CuCl(s)
Cu’
(aq)
T
Cl-
(aq),
sp
=
=
[Cu'][CIT]
(b)
BaFy)
=
Ba2+(aq)
+
2F-
(aq)’
sp
=
=
[Ba%*][F]?
()
Ag2SO4(s5)
=
2Ag"(aq)
+
SO*
(ag);
Ksp
=
[Ag'1*[SO.™]
(d)
Ca3(POy)y(s)
=
3Ca’(aq)
+
2POL>
(aq)s
sp
=
[Ca"
3
[PO,3]2
10.
Ag2C035)
=
Ag'(aq)
+
CO3%
(aq);
Ksp
=
[AgT1?[CO3*7]
11.
MgNH,
POy
=
Mg
(ag
T
NH,*
(aq)
+
PO4(a
)’
Ksp
=
[Mg?][NH,'1[PO4*]
12.a)
Kp
=
[Fe*][NO57J?
(b)
smaller
13.2.2x
107
14.1.1
x
107%
15.3.2
x
107
164a)
4.8
X
10729
(b)
1
g
solution
=
1
mL
solution
174a) 1.3
x
1075
mol/L
(b}
7.8
x
108
formula
units
of
AgCl/L
(¢)
1.9
X
107*%
(m/v)
18.
1.0
x
107*°
mol/L
19.
9.9
X
1073
mol/L;
4.1
g/L
20.8.5
x
10'?
formula
units
of
ZnS/L.
21.(a)
1.33
x
10~
mol/L
(b)
1.18
x
107°
mol/L
22.
3.9
x
107°
mol/L
23(a)
7.02
x
1072
mol/L
(b)
1.97
x
10~
mol/L
24(a)
1.6
x
1072
mol/L
(b)
4.0
x
10~*
mol/L
25.9.18
26.
9.03
21.
310
28.4.16
29.
Qsp
=
7.5
X
10~
>
Kyp,
therefore
a
precipitate
forms.
30.
Qsp
=
3.0
x
107%
>
K,
therefore
a
precipitate
forms.
31.
Qsp
=
2.7
X
1077
>
Kyp,
therefore
a
precipitate
forms.
32.
Qsp
=
4.2
x
107"?
<
Kp,
therefore
no
precipitate
forms.
33.
Qsp
=
8.6
x
107®
<
K,
therefore
no
precipitate
forms.
3.
Qsp
=
6.9
X
107°
<
K,
therefore
no
precipitate
forms.
35.
Qsp
=
2.2
X
107
>
Ky,
therefore
a
precipitate
forms.
36.
Qsp
=
1.0
X
107
>
Kyp,
therefore
a
precipitate
forms.
Section
Review:
9.1:
1.
sodium
carbonate
2.(a)
no
reaction
{b)
CH3NH3+(aq)
+
HyO)
=
CH3NHyqg)
+
H3O+(aq)
;
acidic
(¢)
no
reaction
(d)
OCl
(5q)
+
H2O()
=
HOClzg)
+
OH
(aq);
basic
3.
acidic
4.
Ca{OH),
,
CaF,,
KNO3;,
NH4;NO3;,
HNO;
5(b)
8.7
(e)
0.167
mol/L
(d)
phenolphthalein
7.
phenolphthalein
8.5.97
9.2:
5.(a)
1.6
x
1075
(b)
9.6
x
107
formula
units
of
BaF,/L
6.b)
4.74
17.3.17
Chapter
9
Agueous
Solutions
and
Solubility
Equilibria
«
MHR
455
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