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Solutions

Solution – a homogeneous mixture of
pure substances (occur in all phases, but
we will focus on aqueous solutions)

The SOLVENT is the medium in which
the SOLUTES are dissolved. (The
solvent is usually the most abundant
substance.)
– Example:
• Solution: Salt Water
• Solute: Salt
• Solvent: Water
Concentration of Solution

Concentration refers to the amount of
solute dissolved in a solution.
Ways of measuring concentration…
MOLARITY (M)
mol solute
Molarity (M) 
L sol' n
MOLALITY (m)
mol solute
molality (m) 
kg solvent
Arrr…Don’t be confusing this
with molarity.
MASS PERCENT (%)
mass solute
mass % 
 100
total mass of sol' n
Recall that ppt, ppm and ppb are
similar measurements, but the multiplier
becomes 103, 106 or 109 respectively.
MOLE FRACTION ()
For mol fract. of component A (  A ),
mol A
A 
total moles
NORMALITY (N)
N  # equivalents per L of sol'n, or
N  M  total pos. ox
See page 487
Practice Problems (to turn in):
Concentration Units

Complete the table below for aqueous
solutions of urea, CO(NH2)2:
Solution
molality
(m)
(a)
2.577
(b)
(c)
Mass
percent
solvent
ppm
solute
mol
fraction
solvent
45.0
4968
(d)
Check your answers with online key
0.815
Energy of Making Solutions


Heat of solution ( DHsoln ) is the energy
change for making a solution.
Most easily understood if broken into
steps.
1. Break apart solvent
2. Break apart solute
3. Mixing solvent and solute
1. Break apart Solvent

Have to overcome attractive forces. DH1 >0
2. Break apart Solute.

Have to overcome attractive forces. DH2 >0
3. Mixing solvent and solute




DH3 depends on what you are mixing.
Molecules can attract each other - DH3 is
large and negative.
Molecules can’t attract - DH3 is small and
negative.
This explains the rule “like dissolves
like”

Size of DH3 determines whether a
Solute and
solution will form
Solvent
E
n
e
r
g
y
Solvent
Reactants
DH1
DH2
DH3
Solution
DH3
Solution
Types of Solvent and solutes


If DHsoln is small and positive, a solution
will still form because of entropy.
There are many more ways for them to
become mixed than there is for them to
stay separate.
Solution Formation – Factors
Favoring Spontaneity

Processes in which the energy content of the
system decreases (exothermic) tend to occur
spontaneously. DH < 0

Processes in which the disorder (entropy) of
the system increases tend to occur
spontaneously. DS < +
DG = DH - TDS
“LIKE DISSOLVES LIKE”
DH1
DH2
DH3
DHsoln
Polar Solvent,
Polar Solute
Large
Large
Large
(-)
Small
Solution
Forms
Polar Solvent,
NonPolar Solute
Small
Large
Small
Large
(+)
No Solution
Forms
NonPolar Solvent,
NonPolar Solute
Small
Small
Small
Small
Solution
Forms
NonPolar Solvent,
Polar Solute
Large
Small
Small
Large
(+)
No Solution
Forms
It seems the extent to which
one substance dissolves in
another depends on the nature
of both the solute and the
solvent. (“like dissolves like”).
It also depends on temperature
and at least for gases, on
pressure.
As a rule, network covalent
solids, such as graphite and
quartz (SiO2), do not dissolve
in any solvent. Nor do metals,
technically speaking “dissolve.”
Ionic and molecular compounds,
on the other hand will dissolve
in certain solvents.
Structure and Solubility

Water soluble molecules must have
dipole moments -polar bonds.

To be soluble in non polar solvents the
molecules must be non polar.
Soap
O-
CH2
CH3
CH2
CH2
P
CH2
CH2
CH2
CH2
O-
O-
Soap
CH2
CH3
CH2

O-
CH2
P
CH2
CH2
CH2
CH2
Hydrophobic non-polar end
O-
O-
Soap
O-
CH2
CH3
CH2
CH2
P
CH2
CH2
CH2
CH2

O-
O-
Hydrophilic
polar end
O-
CH2
CH3
CH2
CH2
P
CH2
CH2
CH2
CH2
_
O-
O-




A drop of grease in water
Grease is non-polar
Water is polar
Soap lets you dissolve the non-polar
in the polar.
Hydrophobic ends
dissolve in grease
Hydrophilic ends
dissolve in water


Water molecules can surround and
dissolve grease.
Helps get grease out of your way.
Pressure effects

Changing the pressure doesn’t affect
the amount of solid or liquid that
dissolves
– They are incompressible.

Pressure does affect solubility of gases.
Dissolving Gases


Pressure affects the
amount of gas that can
dissolve in a liquid.
The dissolved gas is at
equilibrium with the gas
above the liquid.


The gas is at
equilibrium with the
dissolved gas in this
solution.
The equilibrium is
dynamic.


If you increase the
pressure the gas
molecules dissolve
faster.
The equilibrium is
disturbed.


The system reaches a
new equilibrium with
more gas dissolved.
Henry’s Law:
P= kC
Pressure=(constant)(gas concentration)
Temperature Effects



Increased temperature increases the
rate at which a solid dissolves.
We can’t predict whether it will increase
the amount of solid that dissolves.
We must read it from a graph of
experimental data.
Solubilities of Various Solids in Water
Solubility
(g solute/
100 g H2O)
20
40
60
Temp (C)
80
100
Gases are predictable



As temperature
increases, solubility
decreases.
Gas molecules can
move fast enough to
escape.
Thermal pollution.
Vapor Pressure of Solutions


A nonvolatile solvent lowers the vapor
pressure of the solution.
The molecules of the solvent
must overcome the force of
both the other solvent
molecules and the
solute molecules.
Raoult’s Law:
Psoln = solvent Psolvent
Sol’n vap. press =
mole fraction of solvent x vapor pressure of pure solvent

Applies only to an ideal solution where the
solute doesn’t contribute to the vapor
pressure.
To determine whether a sol’n is IDEAL…

Liquid-liquid solutions where both are volatile.

Modify Raoult’s Law to
Ptotal = PA + PB = APA0 + BPB0
• Ptotal = vapor pressure of mixture
• PA0= vapor pressure of pure A


If this equation works then the solution is
ideal.
Solvent and solute are alike.
Deviations

If Solvent has a strong affinity for solute
(H bonding)…
– Lowers solvents ability to escape.
– Lower vapor pressure than expected.
– Negative deviation from Raoult’s law.
DHsoln is large and negative exothermic.

Endothermic DHsoln indicates positive
deviation.
Colligative Properties of Solutions
Colligative properties = physical
properties of solutions that depend
on the # of particles dissolved, not
the kind of particle.
Colligative Properties




Lowering vapor pressure
Raising boiling point
Lowering freezing point
Generating an osmotic pressure
Boiling Point Elevation

a solution that contains a nonvolatile
solute has a higher boiling pt than the
pure solvent; the boiling pt elevation is
proportional to the # of moles of solute
dissolved in a given mass of solvent.
 Ok…
but WHY???
Consider evaporation….only the solvent
moleculesBut
evaporate.
you see some of the solute molecules
block the surface so that fewer solvent
molecules can reach the surface and
evaporate.
This lowers vapor pressure because fewer
molecules evaporate.
Lowering Vapor Pressure
The thing to remember is that this is
a physical effect. The solute
molecules get in the way, and prevent
some of the solution molecules from
reaching the surface where they can
evaporate.
Lowering Vapor Pressure
It doesn’t matter much what the
solute molecules are, just how many
of them there are. Thus, this is a
colligative property.
Measuring Vapor Pressure
Pure
Water
Aqueous
Solution
Pure
Water
Aqueous
Solution
Pure
Water
Aqueous
Solution
Pure
Water
Aqueous
Solution
Pure
Water
Aqueous
Solution
Lowering vapor pressure,
affects boiling point. Remember
that boiling occurs when the vapor
pressure of a liquid equals the
ambient pressure.
Also, recall that vapor pressure of
water depends on temperature
Vapor Pressure of Water
Room
Temp
110oC
100oC
1074 Torr
760
75oC
300
20oC
17
0o C
4.6
Water Temp
Room
Pressure
Vapor Pressure
Temp oC
Room Temp
Vapor Pressure
(Torr)
110o
1074
100o
760
75o
300
20o
17
0o
4.6
Room Pressure
Now add some solute to the liquid…
Temp oC
110o
Room Temp
Vapor Pressure
(Torr)
1074
100o
760
75o
300
20o
17
0o
4.6
Room Pressure
Temp oC
110o
100o
75o
Room Temp
20o
0o
Vapor Pressure
(Torr)
1074
760
300
17
4.6
Room Pressure
Temp oC
Vapor Pressure
Water now boils at a higher
(Torr)
temperature.
1074
110o
760
Room Pressure
100o
300
75o
Room Temp
17
20o
Lower Vapor Pressure
4.6
0o
Vapor Pressure Changes with Temperature
Boiling Point Elevation
 DTb
where:
= kbm
DTb = elevation of boiling pt
m = molality of solute
kb = the molal boiling pt elevation constant
for a particular solvent

kb for water = 0.52 °C/m
Ex: What is the normal boiling pt of a 2.50
m glucose, C6H12O6, solution?

“normal” implies 1 atm of pressure
DTb = kbm
DTb = (0.52 C/m)(2.50 m)
DTb = 1.3 C

Tb = 100.0 C + 1.3 C = 101.3 C



Ex: How many grams of glucose, C6H12O6,
would need to be dissolved in 535.5 g of water
to produce a solution that boils at 101.5°C?



DTb = kbm
1.5 C= (0.52 C/m)(m)
m = 2.885
x
2.885 m 
0.5355 kg
x  1.5449 mol  180 g  278.1 g  280g
1 mol
Freezing/Melting Point Depression

The freezing point of a solution is
always lower than that of the pure
solvent.
Freezing Point Depression
When a solution freezes, crystals of
pure solvent usually separate out; the solute
molecules are not normally soluble in the
solid phase of the solvent.
As a result, the part of the
phase diagram that represents
the vapor pressure of the solid
is the same as that for the pure
liquid. The vapor pressure
curves for the liquid and solid
phases meet at the triple point.
Pressure
liquid
solid
gas
Temp
The part of the
phase diagram
that represents
the vapor
pressure of the
solid is the same
as that for the
pure liquid.
Pressure
liquid
solid
gas
Temp
Freezing point
is lowered at
all pressures
Pressure
liquid
solid
gas
Temp
Boiling
temperature
is raised at
all pressures
Pressure
liquid
solid
gas
Temp
As a result, the
liquid range
temperature is
increased at
both ends.
Freezing/Melting Point
Depression
 DTf
where:
= kfm
DTf = lowering of freezing point
m = molality of solute
kf = the freezing pt depression constant

kf for water = 1.86 °C/m
Ex: Calculate the freezing pt of a
2.50 m glucose solution.

DTf = kfm
DTf = (1.86 C/m)(2.50 m)
DTf = 4.65 C

Tf = 0.00C - 4.65 C = -4.65C


Ex: When 15.0 g of ethyl alcohol, C2H5OH, is
dissolved in 750 grams of formic acid, the freezing
pt of the solution is 7.20°C. The freezing pt of pure
formic acid is 8.40°C. Determine Kf for formic acid.
15.0 g C2H5OH 1mol

 0.3261mol
46 g
0.3261 mol
 0.4348 m
0.75 kg
DTf = kfm
1.20 C= (kf)( 0.4348 m)
kf = 2.8 C/m
Ex: An antifreeze solution is prepared containing 50.0 cm3
of ethylene glycol, C2H6O2, (d = 1.12 g/cm3), in 50.0 g
water. Calculate the freezing point of this 50-50 mixture.
Would this antifreeze protect a car in Chicago on a day
when the temperature gets as low as –10° F?
(-10 °F = -23.3° C)
50.0 cm3 C2H6O2
1.12 g
1mol


 0.90323 mol
3
62.0 g
cm
0.90323 mol
 18.1m
0.050 kg
DTf = kfm
DTf = (1.86C/m)(18.1 m)
DTf = 33.7 C
Tf = 0 C – 33.7 C = -33.7 C
YES!
Electrolytes and Colligative
Properties
• Colligative properties depend on the # of
particles present in solution.
• Because ionic solutes dissociate into ions,
they have a greater effect on freezing pt and
boiling pt than molecular solids of the same
molal conc.
Electrolytes and Colligative
Properties

For example, the freezing pt of water is
lowered by 1.86°C with the addition of any
molecular solute at a concentration of 1 m.
– Such as C6H12O6, or any other covalent
compound

However, a 1 m NaCl solution contains 2
molal conc. of IONS. Thus, the freezing pt
depression for NaCl is 3.72°C…double that of
a molecular solute.
– NaCl  Na+ + Cl- (2 particles)
Electrolytes - Boiling Point Elevation and
Freezing Point Depression
The relationships are given by the following equations:

DTf = imkf or
DTb = imkb
DTf/b = f.p. depression/elevation of b.p.
m = molality of solute
kf/b = b.p. elevation/f.p depression constant
i = # particles formed from the dissociation of
each formula unit of the solute
(van’t Hoff factor)
JacobusVan’t Hoff
(1852 – 1911)

i = The number of particles that the solute
molecule breaks into the Van’t Hoff factor

Assumes that when a salt dissolves, it
completely dissociates into its component
ions, which then move around independently.
This assumption is not always true because
of ion pairing.

Van’t Hoff a Dutch Chemist received the first
Nobel Prize in Chemistry in 1901
FeCl3  Fe+3 + 3Cl
At a given instant a
small percentage of the
+ and – ions are paired,
and thus count as a
single particle. In
general, this becomes
more important in
concentrated solutions.
Cmpd.
i=4
i
i
expected observed
NaCl
2.0
1.9
MgCl2
3.0
2.7
MgSO4
2.0
1.3
FeCl3
4.0
3.4
HCl
2.0
1.9
Ex: What is the freezing pt of:
a) a 1.15 m sodium chloride solution?

NaCl  Na+ + Cl-
n=2

DTf = kf·m·n
DTf = (1.86 C/m)(1.15 m)(2)
DTf = 4.28 C

Tf = 0.00C - 4.28 C = -4.28C


Ex: What is the freezing pt of:
b) a 1.15 m calcium chloride solution?

CaCl2  Ca2+ + 2Cl-
n=3

DTf = kf·m·n
DTf = (1.86 C/m)(1.15 m)(3)
DTf = 6.42 C

Tf = 0.00C – 6.42 C = -6.42C


Ex: What is the freezing pt of:
c) a 1.15 m calcium phosphate solution?


Ca3(PO4)2  3Ca2+ + 2PO43n=5

DTf = kf·m·n
DTf = (1.86 C/m)(1.15 m)(5)
DTf = 10.7 C

Tf = 0.0C – 10.7 C = -10.7C


Determining Molecular Weights
by Freezing Point Depression
Ex: A 1.20 g sample of an unknown compound is dissolved in 50.0 g of
benzene. The solution freezes at 4.92°C. Determine the molecular
weight of the compound. The freezing pt of pure benzene is 5.48°C
and the Kf for benzene is 5.12°C/m.




DTf = 0.56°C
DTf = kf·m
0.56°C = (5.12°C/m)(m)
m = 0.1094
0.1094m 
x mol
0.050 kg
x  0.00547mol
1.20g
Molar Mass 
 219 g mol
0.00547mol
Ex: A 37.0 g sample of a new covalent compound was dissolved in 200.0
g of water. The resulting solution froze at –5.58°C. What is the
molecular weight of the compound?




DTf = 5.58°C
DTf = kf·m
5.58°C = (1.86°C/m)(m)
m = 3.00 m
3.00m 
x mol
0.200 kg
x  0.60mol
37.0g
Molar Mass 
 61.7 g mol
0.60mol
Osmotic Pressure

Experiments show that dependence of the
osmotic pressure on solution concentration is
expressed by the eqn:
 = MRT
• Where,
 = osmotic pressure (atm)
M = molarity (mol/L)
R = gas law constant = 0.08206
T = temp (K)
L atm
mol K
Osmosis
Solvent
Molecule
Solute
Molecule
Semi permeable
Membrane
Water Flow
Osmotic
pressure
Water Flow
A simple rule to remember is:
Salt is a solute, when it is concentrated
inside or outside the cell, it will draw the water in
its direction. This is also why you get thirsty after
eating something salty.
Type of Solutions
1. Isotonic: Solution has the same solute
concentration as the cell.
2. Hypotonic: Solution has less solute
concentration than the cell.
3. Hypertonic: Solution has more solute
concentration the cell.
Isotonic Solution
Water molecules
10% Salt
10% Salt
If the concentration of solute (salt) is equal on
both sides, the water will move back in forth but it
won't have any result on the overall amount of water
on either side. "ISO" means the same
Hypotonic Solution
Water molecules
20% Salt
10% Salt
The word "HYPO" means less, in this case there are less solute (salt)
molecules outside the cell, since salt “sucks”, water will move into the cell.
The cell will gain water and grow larger. In plant cells, the central
vacuoles will fill and the plant becomes stiff and rigid, the cell wall keeps the
plant from bursting
In animal cells, the cell may be in danger of bursting, organelles called
CONTRACTILE VACUOLES will pump water out of the cell to prevent this.
Hypertonic Solution
Water molecules
10% Salt
20% Salt
The word "HYPER" means more, in this case there are more solute (salt)
molecules outside the cell, which causes the water to be sucked in that
direction.
In plant cells, the central vacuole loses water and the cells shrink,
causing wilting. In animal cells, the cells also shrink. In both cases, the cell may
die.
This is why it is dangerous to drink sea water its a myth that drinking sea water will cause you
to go insane, but people marooned at sea will
speed up dehydration (and death) by drinking sea
water.
This is also why "salting fields" was a common
tactic during war, it would kill the crops in the
field, thus causing food shortages.
Animation of cells placed in various
solutions (wait for it to play…)