Chap13_part3

Chap13_part3 - Factors
Affec,ng
Solubility:


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Unformatted text preview: Factors
Affec,ng
Solubility:


 1)
Solute‐Solvent
Interac,ons
 Liquids
that
readily
mix
to
form
solu3ons
are
called
miscible.
 Liquids
that
do
not
readily
mix
to
form
solu3ons
are
immiscible.
 a)
Two
polar
liquids
tend
to
be
miscible
because
of
favorable
dipole‐dipole

 interac3ons.
 b)
Polar
liquids
as
well
as
those
capable
of
forming
hydrogen
bonds
tend

 to
be
miscible
in
water.
 c)
Nonpolar
liquids
tend
to
be
immiscible
in
water
because
the
solute‐
 solvent
interac3ons
are
not
strong
enough
to
overcome
the
hydrogen
bonding

 in
water.
 d)
Two
nonpolar
liquids
are
usually
miscible.
 2
 Factors
Affec,ng
Solubility:


 1)
Solute‐Solvent
Interac,ons
 e)
Alcohols
become
increasingly
insoluble
in
the
polar
solvent
water
as
their
 hydrocarbon
components
increase
in
length,
because
the
nonpolar
character
of
the
 alcohol
molecules
begins
to
dominate
the
polar
character.
 f)
Alcohols
become
increasingly
soluble
in
the
nonpolar
solvent
hexane
as
their
 hydrocarbon
components
increase
in
length,
because
the
London
dispersion
 interac3ons
increase
in
magnitude.
 3
 Factors
Affec,ng
Solubility:


 1)
Solute‐Solvent
Interac,ons
 (a)  hydrogen
bonding
between
two
ethanol
molecules
 (b)
hydrogen
bonding
between
ethanol
and
water
molecules
 The
more
similar
the
intermolecular
a0rac1ons,
the
more
likely
one
substance
is
to
 be
soluble
in
another.
 4
 Factors
Affec,ng
Solubility:


 1)
Solute‐Solvent
Interac,ons
 Adding
OH
groups
to
a
nonpolar
molecule
increases
the
solubility
in
water.
 5
 Factors
Affec,ng
Solubility:


 1)
Solute‐Solvent
Interac,ons
 Some
vitamins
like
vitamin
A
(re3nol)
are
primarily
nonpolar.

They
are

 stored
in
faLy
3ssue
and
can
build
to
toxic
levels.

Other
vitamins
like

 vitamin
C
(ascorbic
acid)
are
water
soluble,
readily
excreted,
and
must
be
 con3nually
replenished.
 6
 Factors
Affec,ng
Solubility:


 2)
Pressure
 The
solubili3es
of
solids
and
liquids
are
not
significantly
changed
by
pressure.
 The
solubili3es
of
gases
in
liquids
are
appreciably
affected
by
pressure.
 When
the
pressure
on
a
gas
dissolved
in
a
liquid
is
increased,
the
 concentra3on
of
gas
in
the
liquid
is
increased.
 7
 Henry’s
Law:


 Gas
Solubility
in
Liquids
&
Pressure
 “The
solubility
of
a
gas
in
a
liquid
increases
in
direct
propor3on
to
its
 par3al
pressure
above
the
solu3on”.
 S g = kPg Sg
=
solubility
of
the
dissolved
gas
 Pg
=
par3al
pressure
of
the
gas
over
the
liquid
 k
=
Henry’s
law
constant
 The
constant
k
depends
on
the
iden3ty
of
the
gas,
the
iden3ty
of
the

 liquid,
and
the
temperature.
 Higher
Pg
means
higher
Sg.
 8
 Carbonated
beverages
are
 boLled
under
pressure.
 When
the
boLle
is
opened,
 the
par3al
pressure
of
CO2
 above
the
liquid
is
reduced.
 The
CO2
solubility
decreases.
 The
CO2
leaves
the
solu3on
 by
forming
bubbles
which
rise
 and
join
the
air.
 9
 The
Henry’s
Law
constant
for
CO2
in
water
is
3.1
x
10‐2
M‐atm‐1
at
25oC.


 (a)  Find
the
solubility
of
CO2
(in
M)
dissolved
in
water
at
25oC
when
the
 pressure
of
pure
CO2
gas
above
the
water
is
10
atm.
 (b)
Find
the
solubility
of
CO2
(in
M)
dissolved
in
water
at
25oC
when
the

 par3al
pressure
of
CO2
(in
air)
is
0.05
atm.
 10
 The
Henry’s
Law
constant
for
CO2
in
water
is
3.1
x
10‐2
M‐atm‐1
at
25oC.


 (a)  Find
the
solubility
of
CO2
(in
M)
dissolved
in
water
at
25oC
when
the
 pressure
of
pure
CO2
gas
above
the
water
is
10
atm.
 S g = kPg = (3.1x10 − 2 M )(10 atm) = 0.31 M atm (b)
Find
the
solubility
of
CO2
(in
M)
dissolved
in
water
at
25oC
when
the

 par3al
pressure
of
CO2
(in
air)
is
0.05
atm.
 11
 The
Henry’s
Law
constant
for
CO2
in
water
is
3.1
x
10‐2
M‐atm‐1
at
25oC.


 (a)  Find
the
solubility
of
CO2
(in
M)
dissolved
in
water
at
25oC
when
the
 pressure
of
pure
CO2
gas
above
the
water
is
10
atm.
 S g = kPg = (3.1x10 − 2 M )(10 atm) = 0.31 M atm (b)
Find
the
solubility
of
CO2
(in
M)
dissolved
in
water
at
25oC
when
the

 par3al
pressure
of
CO2
(in
air)
is
0.05
atm.
 S g = kPg = (3.1x10 −2 M )(0.05 atm) = 0.002 M atm 12
 William
Henry
(English
Chemist,
1775‐1836)
 ‐ 
English
Chemist
 ‐ 
Henry’s
Law
 ‐ 
Worked
on
Illumina3ng
gasses
and
Fire‐
damps
 ‐ 
Composi3on
of
Hydrochloric
acid
and
Ammonia
 13
 The
par3al
pressure
of
O2
in
air
at
sea
level
is
0.21
atm.

The
solubility
of
pure
O2
in
water,
at

 1
atm
and
20oC,
is
1.38
x
10‐3
M.

Find
the
molar
concentra3on
of
O2
at
the
surface
water
of
a

 mountain
lake
saturated
with
air
at
20oC
and
at
an
atmospheric
pressure
of
665
torr.


 (a)
Find
the
Henry’s
Law
constant.
 (b)
Find
the
pressure
at
the
lake
in
units
of
atm.
 (c)
Find
the
par3al
pressure
of
the
O2
at
the
lake
in
units
of
atm.
 (d)
Find
the
molar
concentra3on
of
O2
in
the
lake.
 14
 The
par3al
pressure
of
O2
in
air
at
sea
level
is
0.21
atm.

The
solubility
of
pure
O2
in
water,
at

 1
atm
and
20oC,
is
1.38
x
10‐3
M.

Find
the
molar
concentra3on
of
O2
at
the
surface
water
of
a

 mountain
lake
saturated
with
air
at
20oC
and
at
an
atmospheric
pressure
of
665
torr.


 (a)
Find
the
Henry’s
Law
constant.
 S g = kPg Sg 1.38 x 10 −3 M M ⇒ k= = = 1.38 x 10 −3 Pg 1 atm atm (b)
Find
the
pressure
at
the
lake
in
units
of
atm.
 (c)
Find
the
par3al
pressure
of
the
O2
at
the
lake
in
units
of
atm.
 (d)
Find
the
molar
concentra3on
of
O2
in
the
lake.
 15
 The
par3al
pressure
of
O2
in
air
at
sea
level
is
0.21
atm.

The
solubility
of
pure
O2
in
water,
at

 1
atm
and
20oC,
is
1.38
x
10‐3
M.

Find
the
molar
concentra3on
of
O2
at
the
surface
water
of
a

 mountain
lake
saturated
with
air
at
20oC
and
at
an
atmospheric
pressure
of
665
torr.


 (a)
Find
the
Henry’s
Law
constant.
 S g = kPg Sg 1.38 x 10 −3 M M ⇒ k= = = 1.38 x 10 −3 Pg 1 atm atm (b)
Find
the
pressure
at
the
lake
in
units
of
atm.
 P = 665 torr ( 1 atm ) = 0.875 atm 760 torr (c)
Find
the
par3al
pressure
of
the
O2
at
the
lake
in
units
of
atm.
 (d)
Find
the
molar
concentra3on
of
O2
in
the
lake.
 16
 The
par3al
pressure
of
O2
in
air
at
sea
level
is
0.21
atm.

The
solubility
of
pure
O2
in
water,
at

 1
atm
and
20oC,
is
1.38
x
10‐3
M.

Find
the
molarconcentra3on
of
O2
at
the
surface
water
of
a

 mountain
lake
saturated
with
air
at
20oC
and
at
an
atmospheric
pressure
of
665
torr.


 (a)
Find
the
Henry’s
Law
constant.
 S g = kPg Sg 1.38 x 10 −3 M M ⇒ k= = = 1.38 x 10 −3 Pg 1 atm atm (b)
Find
the
pressure
at
the
lake
in
units
of
atm.
 P = 665 torr ( 1 atm ) = 0.875 atm 760 torr (c)
Find
the
par3al
pressure
of
the
O2
at
the
lake
in
units
of
atm.
 P(0 2 ) = 0.21(0.875 atm) = 0.184 atm (d)
Find
the
molar
concentra3on
of
O2
in
the
lake.
 17
 The
par3al
pressure
of
O2
in
air
at
sea
level
is
0.21
atm.

The
solubility
of
pure
O2
in
water,
at

 1
atm
and
20oC,
is
1.38
x
10‐3
M.

Find
the
molar
concentra3on
of
O2
at
the
surface
water
of
a

 mountain
lake
saturated
with
air
at
20oC
and
at
an
atmospheric
pressure
of
665
torr.


 (a)
Find
the
Henry’s
Law
constant.
 S g = kPg Sg 1.38 x 10 −3 M M ⇒ k= = = 1.38 x 10 −3 Pg 1 atm atm (b)
Find
the
pressure
at
the
lake
in
units
of
atm.
 P = 665 torr ( 1 atm ) = 0.875 atm 760 torr (c)
Find
the
par3al
pressure
of
the
O2
at
the
lake
in
units
of
atm.
 P(0 2 ) = 0.21(0.875 atm) = 0.184 atm (d)
Find
the
molar
concentra3on
of
O2
in
the
lake.
 S (0 2 ) = kP(O2 ) = (1.38 x 10 −3 M / atm)(0.184 atm) = 2.54 x10 −4 M 18
 Deep‐Sea
Divers
and
N2
Solubility
in
Water
 Workers
in
caissons
(underwater
construc3on
sites)
breathe
air
at
more
 than
normal
pressure.

If
they
are
returned
to
the
surface
too
rapidly,
N2,
 dissolved
in
their
blood
at
high
pressure,
comes
out
of
solu3on
and
may
 cause
emboli
(gas
bubbles
in
the
bloodstream).

Blood
at
37oC
and
1
atm
 dissolves
1.3
mL
of
pure
N2
in
100
mL
of
blood.

 Calculate
the
volume
of
N2
liberated
from
the
blood
of
a
caisson
worker
 returned
to
1
atm
pressure
aaer
prolonged
exposure
to
pressure
300
m
 below
sea
level
where
the
pressure
is
30
atm.

The
total
blood
volume
of
 an
average
adult
is
3.2
L;
air
contains
78%
N2.
 19
 (a)
Find
the
number
of
mols
of
N2
in
1.3
mL
at
1
atm
and
37oC.
 (b)
Find
the
molarity
of
N2
in
blood
at
1
atm
and
37oC.
 (c)
Find
the
Henry’s
Law
constant
for
N2
in
blood.
 (d)
Find
the
molarity
of
N2
in
blood
at
30
atm
and
37oC.
 20
 (a)
Find
the
number
of
mols
of
N2
in
1.3
mL
at
1
atm
and
37oC.
 PV n= = RT (1 atm)(1.3 x10 −3 L) = 5.1x10 −5 mol N 2 L − atm (0.08206 )(310.1 K ) mol − K (b)
Find
the
molarity
of
N2
in
blood
at
1
atm
and
37oC.
 (c)
Find
the
Henry’s
Law
constant
for
N2
in
blood.
 (d)
Find
the
molarity
of
N2
in
blood
at
30
atm
and
37oC.
 21
 (a)
Find
the
number
of
mols
of
N2
in
1.3
mL
at
1
atm
and
37oC.
 PV n= = RT (1 atm)(1.3 x10 −3 L) = 5.1x10 −5 mol N 2 L − atm (0.08206 )(310.1 K ) mol − K (b)
Find
the
molarity
of
N2
in
blood
at
1
atm
and
37oC.
 5.1x10 −5 mol S1 = = 5.1x10 − 4 M N 2 0.1 L (c)
Find
the
Henry’s
Law
constant
for
N2
in
blood.
 (d)
Find
the
molarity
of
N2
in
blood
at
30
atm
and
37oC.
 22
 (a)
Find
the
number
of
mols
of
N2
in
1.3
mL
at
1
atm
and
37oC.
 PV n= = RT (1 atm)(1.3 x10 −3 L) = 5.1x10 −5 mol N 2 L − atm (0.08206 )(310.1 K ) mol − K (b)
Find
the
molarity
of
N2
in
blood
at
1
atm
and
37oC.
 5.1x10 −5 mol S1 = = 5.1x10 − 4 M N 2 0.1 L (c)
Find
the
Henry’s
Law
constant
for
N2
in
blood.
 S1 5.1x10 −4 M M k= = = 6.5 x10 − 4 P1 0.78(1 atm) atm (d)
Find
the
molarity
of
N2
in
blood
at
30
atm
and
37oC.
 23
 (a)
Find
the
number
of
mols
of
N2
in
1.3
mL
at
1
atm
and
37oC.
 PV n= = RT (1 atm)(1.3 x10 −3 L) = 5.1x10 −5 mol N 2 L − atm (0.08206 )(310.1 K ) mol − K (b)
Find
the
molarity
of
N2
in
blood
at
1
atm
and
37oC.
 5.1x10 −5 mol S1 = = 5.1x10 − 4 M N 2 0.1 L (c)
Find
the
Henry’s
Law
constant
for
N2
in
blood.
 S1 5.1x10 −4 M M k= = = 6.5 x10 − 4 P1 0.78(1 atm) atm (d)
Find
the
molarity
of
N2
in
blood
at
30
atm
and
37oC.
 S 30 = kP30 = (6.5 x10 − 4 M )(0.78)(30 atm) = 0.015 M atm 24
 (e)
Find
the
number
of
mols
of
N2
in
the
body
at
30
atm
and
37oC.
 (f)
Find
the
number
of
mols
of
N2
in
the
body
at
1
atm
and
37oC.
 (g)
Find
the
number
of
mols
of
N2
released
upon
returning
to
1
atm.
 (h)
Find
the
volume
of
air
(via
N2)
released
upon
returning
to
1
atm.
 25
 (e)
Find
the
number
of
mols
of
N2
in
the
body
at
30
atm
and
37oC.
 n30 = (0.015 mol )(3.2 L) = 0.048 mol L (f)
Find
the
number
of
mols
of
N2
in
the
body
at
1
atm
and
37oC.
 (g)
Find
the
number
of
mols
of
N2
released
upon
returning
to
1
atm.
 (h)
Find
the
volume
of
air
(via
N2)
released
upon
returning
to
1
atm.
 26
 (e)
Find
the
number
of
mols
of
N2
in
the
body
at
30
atm
and
37oC.
 n30 = (0.015 mol )(3.2 L) = 0.048 mol L (f)
Find
the
number
of
mols
of
N2
in
the
body
at
1
atm
and
37oC.
 n1 = (5.1x10 − 4 mol )(3.2 L) = 0.0016 mol L (g)
Find
the
number
of
mols
of
N2
released
upon
returning
to
1
atm.
 (h)
Find
the
volume
of
air
(via
N2)
released
upon
returning
to
1
atm.
 27
 (e)
Find
the
number
of
mols
of
N2
in
the
body
at
30
atm
and
37oC.
 n30 = (0.015 mol )(3.2 L) = 0.048 mol L (f)
Find
the
number
of
mols
of
N2
in
the
body
at
1
atm
and
37oC.
 n1 = (5.1x10 − 4 mol )(3.2 L) = 0.0016 mol L (g)
Find
the
number
of
mols
of
N2
released
upon
returning
to
1
atm.
 n = (0.048 − 0.0016) mol = 0.046 mol (h)
Find
the
volume
of
air
(via
N2)
released
upon
returning
to
1
atm.
 28
 (e)
Find
the
number
of
mols
of
N2
in
the
body
at
30
atm
and
37oC.
 n30 = (0.015 mol )(3.2 L) = 0.048 mol L (f)
Find
the
number
of
mols
of
N2
in
the
body
at
1
atm
and
37oC.
 n1 = (5.1x10 − 4 mol )(3.2 L) = 0.0016 mol L (g)
Find
the
number
of
mols
of
N2
released
upon
returning
to
1
atm.
 n = (0.048 − 0.0016) mol = 0.046 mol (h)
Find
the
volume
of
air
(via
N2)
released
upon
returning
to
1
atm.
 V= nRT = P L − atm )(310.1 K ) mol − K = 1.5 L 0.78(1 atm) (0.046 mol )(0.08206 29
 Factors
Affec,ng
Solubility:


 3)
Temperature
 The
solubility
of
most
solids
in
water
increases
with
temperature.
 30
 Factors
Affec,ng
Solubility:


 3)
Temperature
 The
solubility
of
most
gases
in
water
decreases
with
temperature.
 Warming
up
cooler
water

bubbles.
 31
 Ways
of
Expressing
Concentra,on
 Qualita3ve: 
“dilute”
or
“concentrated”
 Quan3ta3ve: 

 
 
1)
mass
percentage
 
 
2)
mole
frac3on
 
 
3)
molarity
 
 
4)
molality
 

 1)
Mass
Percentage
(wt%)
 mass of component mass % of component = x 100 total mass of solution The
mass
%
is
some3mes
referred
to
as
the
weight
%.
 Example:

2
gm
NaCl
in
150
gm
water
 2 mass % = x 100 = 1.3% 150 + 2 Note
that
the
denominator
is
152
not
150.
 1)
Mass
Percentage
(wt%)
 For
dilute
solu3ons,
some3mes
the
mass
%
is
referred
to

 as
“parts
per
million”:
 mass of component ppm = x 10 6 total mass of solution For
very
dilute
solu3ons,
some3mes
the
mass
%
is
referred
to

 as
“parts
per
billion”:
 mass of component ppb = x 10 9 total mass of solution For
dilute
and
very
dilute
solu3ons,
the
total
mass
 of
the
solu3ons
is
≈
the
total
mass
of
the
solvent.
 2)
Mol
Frac,on
 mols of component mol fraction of component = X = total mols in solution 3)
Molarity
 mols of solute molarity = M = liters of solution 4)
Molality
 mols of solute molality = m = kg solvent When
the
solvent
is
water,
1
L
≈
1
kg.

Thus,
when
the
solu3on
is
 dilute,
the
denominators
of
3)
and
4)
are
equivalent.
 Calcula,ng
Molarity
and
Molality
 A
solu3on
containing
80.5
gm
of
vitamin
C
(C6H8O6)
in
210
gm
water
has
a
 density
of
1.22
gm/mL.
Calculate
the
(a)
mass
percentage;
(b)
mol
frac3on;

 (c)
molality;
and
(d)
molarity.


 (a)
 A
solu3on
containing
80.5
gm
of
vitamin
C
(C6H8O6)
in
210
gm
water
has
a
 density
of
1.22
gm/mL.
Calculate
the
(a)
mass
percentage;
(b)
mol
frac3on;

 (c)
molality;
and
(d)
molarity.


 (a)
 (b)
 mass solute 80.5 x100 = = 27.7% mass solution 80.5 + 210 A
solu3on
containing
80.5
gm
of
vitamin
C
(C6H8O6)
in
210
gm
water
has
a
 density
of
1.22
gm/mL.
Calculate
the
(a)
mass
percentage;
(b)
mol
frac3on;

 (c)
molality;
and
(d)
molarity.


 (a)
 (b)
 mass solute 80.5 x100 = = 27.7% mass solution 80.5 + 210 molecular weight solute = 6(12) + 8(1) + 6(16) = 176 80.5 gm = 0.457 mol 176 gm / mol gm molecular weight solvent = 18 mol 210 gm mol solvent = = 11.7 mol 18 gm / mol mol solute = mol fraction = 0.457 = 0.038 11.7 + 0.457 gm mol A
solu3on
containing
80.5
gm
of
vitamin
C
(C6H8O6)
in
210
gm
water
has
a
 density
of
1.22
gm/mL.
Calculate
the
(a)
mass
percentage;
(b)
mol
frac3on;

 (c)
molality;
and
(d)
molarity.


 (c)
 A
solu3on
containing
80.5
gm
of
vitamin
C
(C6H8O6)
in
210
gm
water
has
a
 density
of
1.22
gm/mL.
Calculate
the
(a)
mass
percentage;
(b)
mol
frac3on;

 (c)
molality;
and
(d)
molarity.


 (c)
 mol solute = 0.457 mol mass solvent = 210 gm = 0.210 kg 0.457 mol mol molality = = 2.18 0.210 kg kg (d)
 A
solu3on
containing
80.5
gm
of
vitamin
C
(C6H8O6)
in
210
gm
water
has
a
 density
of
1.22
gm/mL.
Calculate
the
(a)
mass
percentage;
(b)
mol
frac3on;

 (c)
molality;
and
(d)
molarity.


 (c)
 mol solute = 0.457 mol mass solvent = 210 gm = 0.210 kg 0.457 mol mol molality = = 2.18 0.210 kg kg (d)
 mol solute = 0.457 mol (210 + 80.5) gm volume solution = = 238 mL = 0.238 L 1.22 gm / mL 0.457 mol mol molarity = = 1.92 0.238 L L ...
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