SOLUTIONS AND THEIR
BEHAVIOR
CHAPTER 14
1
CHAPTER OVERVIEW
• This chapter examines homogeneous
mixtures called solutions, which are
made up of a solute and a solvent.
• Concentrations of solutions can be
expressed in a variety of units.
• Properties of solutions that depend
only on the number of solute
particles and not their type are called
colligative properties.
2
14.1 UNITS OF CONCENTRATION
• Molarity, M, moles solute per liter
solution
• Molality, m, moles of solute per
kilogram solvent
• Mole fraction, XA, moles A divided
by moles total
• Weight percent (mass percent),
wt.% A,
(mass A divided by mass total) x 100%
3
Molarity -vs- Molality
Each flask contains
19.4 g of K2CrO4
Water to the 1.00 L mark
Exactly 1.00 kg of water added
4
UNITS OF CONCENTRATION
• parts per million, ppm, is calculated
like percent, but multiply by 106
• Remember that the mass of the
solution equals the mass of the
solute plus solvent.
• Conversions between molarity and
the other concentration units
requires the density of the solution.
5
Solutions
Why does a raw egg swell or shrink when
placed in different solutions?
6
Some Definitions
A solution is a
HOMOGENEOUS
mixture of 2 or
more substances
in a single phase.
One constituent is
usually regarded
as the SOLVENT
and the others as
SOLUTES.
7
14.2 THE SOLUTION PROCESS
• The key to understanding the solution
process is intermolecular forces: solvent solvent; solute - solute; solute - solvent.
8
THE SOLUTION PROCESS
• What prevents solubility is an energy
barrier when the latter interaction is
significantly weaker than the former
interactions.
• Like dissolves like is a general rule, but
not and explanation of the solubility
phenomenon.
9
• Liquids Dissolving in Liquids
–Miscible liquids are soluble in all
proportions.
–Immiscible liquids do not mix, but form
separate layers.
• Isopropanol is miscible with water but gasoline is not. Explain why.
10
SOLUTIONS
• A saturated solution is one which has
reached its equilibrium solubility at
that temperature.
• An unsaturated solution is one that has
not reached its equilibrium solubility.
• A supersaturated solution is one in
which the equilibrium solubility has
been temporarily exceeded.
11
Definitions
Solutions can be classified as
unsaturated or
saturated.
A saturated solution
contains the
maximum quantity of
solute that dissolves
at that temperature.
SUPERSATURATED
SOLUTIONS contain
more than is possible
and are unstable.
12
Energetics of the Solution
Process
13
If the enthalpy of
formation of the
solution is more
negative than that of
the solvent and solute,
the enthalpy of
solution is negative.
The solution process is
exothermic!
Solids Dissolving in Liquids
The same rules apply.
• Compare the intermolecular forces.
• I2 is quite soluble in CCl4, but not very soluble in
water. Explain why?
14
Supersaturated
Sodium Acetate
• One application of a
supersaturated
solution is the sodium
acetate “heat pack.”
• Sodium acetate has an
ENDOthermic heat of
solution.
15
Ionic Solutions
• The heat of solution for ionic compounds
is the sum of the lattice energy (+), bonds
breaking, and the hydration energy (-),
bonds forming.
• It may be positive (endo) or negative (exo)
depending on the relative magnitudes of
these energies.
16
DHsol’n can be calc’d using Hess' Law.
821 kJ/mol – 819 kJ/mol = +2 kJ/mol (endo)
17
DHsol’n can be calc’d using Hess' Law.
18
Ion Size also
determines
Solubility
Remember Coulombs Law
+(charge n)(charge
n)
Force of Attraction =k
d2
19
Ionic Solutions
• Temperature has a significant effect on
solubility for salts and is consistent with
Le Chatelier's principle.
20
The heat of solution for
many salts is positive,
endothermic, as seen by
the positive slope of the
graph.
21
Supersaturated
Sodium Acetate
Sodium acetate has an ENDOthermic heat of
solution.
NaCH3CO2 (s) + heat ---->
Na+(aq) + CH3CO2-(aq)
Therefore, formation of solid sodium acetate
from its ions is EXOTHERMIC.
Na+(aq) + CH3CO2-(aq) --->
NaCH3CO2 (s) + heat
22
Dissolving Gases
• Gas solubility decreases with increasing
temperature which means ΔHsolution < 0 ,
exothermic.
• Gas solubilities increase with increasing
pressure.
• Write the general equation for the solubility
of a gas showing that the process is
exothermic and show how increasing the
temperature decreases the solubility.
Gas + Solvent ⇄ Solution + Heat
23
24
25
Colligative Properties
 On adding a solute to a solvent, the properties
of the solvent are modified.
• Vapor pressure
decreases
• Melting point
decreases
• Boiling point
increases
• Osmosis is possible (osmotic pressure)
26
Colligative Properties
 These changes are called
COLLIGATIVE PROPERTIES.
 They depend only on the
NUMBER of solute particles
relative to solvent particles, not
on the KIND of solute particles.
Concentration Units
An IDEAL SOLUTION is one
where the properties depend only
on the concentration of solute.
Need concentration units to tell us
the number of solute particles per
solvent particle.
The unit “molarity” does not do
this!
27
Concentration Units
28
MOLE FRACTION, X
For a mixture of A, B, and C
mol A
X A = mol fraction A =
mol A + mol B + mol C
MOLALITY, m
mol solute
m of solute =
kilograms solvent
WEIGHT % = grams solute per 100 g solution
Calculating Concentrations
Dissolve 62.1 g (1.00 mol) of ethylene
glycol in 250. g of H2O.
Calculate mole fraction, molality, and
weight % of glycol.
30
Calculating Concentrations
Dissolve 62.1 g (1.00 mole) of ethylene glycol in 250. g
of H2O. Calculate X, m, and % of glycol.
250. g H2O = 13.9 mole
1.00 mol glycol
X glycol =
1.00 mol glycol + 13.9 mol H2O
X glycol = 0.0672
Calculating Concentrations
31
Dissolve 62.1 g (1.00 mol) of ethylene glycol in 250. g
of H2O. Calculate X, m, and % of glycol.
Calculate molality
1.00 mol glycol
conc (molality) =
= 4.00 molal
0.250 kg H2 O
Calculate weight %
62.1 g
% glycol =
x 100% = 19.9%
62.1 g + 250. g
Dissolving Gases &
Henry’s Law
Gas solubility (M) = kH • Pgas
kH for O2 = 1.66 x 10-6 M/mmHg
When Pgas drops, solubility drops.
32
33
Lake Nyos, Cameroon
COLLIGATIVE PROPERTIES
• Changes in Vapor Pressure: Raoult's Law
• The presence of a solute in the solvent
lowers the vapor pressure of the solvent.
Psolvent = Xsolvent Posolvent
• If the solute is also volatile, a similar
equation applies to the solute.
Psolute = Xsolute Posolute
• The total pressure for the solution is given
by:
Ptotal = Psolvent + Psolute
34
COLLIGATIVE PROPERTIES
• If the solute is nonvolatile, the total
pressure is just the pressure of the solvent
and is lower than that of the pure
solvent.
• Study examples and exercises.
35
Understanding
Colligative Properties
To understand colligative properties, study
the LIQUID-VAPOR EQUILIBRIUM for a
solution.
surface
H—O
H—O
H
H—O
H
H
H—O
H
36
Understanding
Colligative Properties
To understand
colligative
properties,
study the
LIQUID-VAPOR
EQUILIBRIUM
for a solution.
37
Understanding
Colligative Properties:
Raoults’s Law
VP of H2O over a solution depends on the
number of H2O molecules per solute molecule.
Psolvent proportional to Xsolvent
OR
Psolvent = Xsolvent • Posolvent
VP of solvent over solution =
(Mol frac solvent)•(VP pure solvent)
RAOULT’S LAW
38
Raoult’s Law
39
An ideal solution is one that obeys Raoult’s law.
PA = XA • PoA
Because mole fraction of solvent, XA, is always
less than 1, then PA is always less than PoA.
The vapor pressure of solvent over a solution is
always LOWERED!
Raoult’s Law
40
Assume the solution containing 62.1 g of glycol in 250. g
of water is ideal.
What is the vapor pressure of water over the solution at
30 oC?
(The VP of pure H2O is 31.8 mm Hg; see App.)
Solution
Xglycol = 0.0672 and so Xwater = ?
Because Xglycol + Xwater = 1
Xwater = 1.000 - 0.0672 = 0.9328
Pwater = Xwater • Powater = (0.9382)(31.8 mm Hg)
Pwater = 29.7 mm Hg
Raoult’s Law
Or (see next slide):
ΔPA = VP lowering = XBPoA
VP lowering is proportional to mole fraction of
the solute!
For very dilute solutions,
ΔPA = K•molalityB
where K is a proportionality constant.
This helps explain changes in melting and
boiling points.
See Exercise 14.6, p. 575
DPsolv = Psolv  Psolv
DPsolv = (  sol Psolv )  Psolv = (1   sol ) Psolv
 solv   solute = 1
so 1   solv =  solute
DPsolv =   solute Psolv
42
Changes in Freezing and
Boiling Points of Solvent
VP Pure solvent
1 atm
VP solvent
after adding
solute
P
BP solution
BP pure
solvent
T
See Figure 14.13
43
Boiling Point Elevation
• If a solute is added to the pure
solvent at its normal boiling point,
the equilibrium vapor pressure will
decrease and the liquid will no longer
boil.
• To reach the new boiling point the
temperature must be increased, thus
boiling point elevation.
Δtbp = Kbp msolute
44
Figure 14.13
45
The boiling point of a
solution is higher than that
of the pure solvent.
46
Elevation of Boiling Point
47
Elevation in BP = DtBP = KBP • m
(where KBP is characteristic of solvent)
VP Pure solvent
1 atm
VP solvent
after adding
solute
P
BP solution
BP pure
solvent
T
47
Change in Boiling Point
48
Dissolve 62.1 g of glycol (1.00 mol) in 250. g of
water. What is the BP of the solution?
KBP = +0.512 oC/molal for water (Table 14.4).
Solution
1.
Calculate solution molality = 4.00 m
2.
DtBP = KBP • m
DtBP = +0.512 oC/molal (4.00 molal)
DtBP = +2.05 oC
BP = 102.05 oC
mol solute
1
=
= 4molal
kg solv
.250
Freezing Point Depression
• The freezing point is lowered by the
presence of a solute since these
particles cannot form the solid and
some of them are occupying the low
energy slots needed to form the solid
solvent.
Δtfp = Kfp msolute
• Note the K is a negative value.
49
Change in Freezing Point
Pure water
50
Ethylene glycol/water solution
The freezing point of a solution is LOWER
than that of the pure solvent.
FP depression = ΔtFP = KFP•m
Freezing Point Depression
Consider equilibrium at melting point
Liquid solvent <------> Solid solvent
• Rate at which molecules go from S to L
depends only on the nature of the solid.
• BUT — rate for L ---> S depends on how
much is dissolved. This rate is SLOWED for
the same reason VP is lowered.
• Therefore, to bring S ---> L and L ---> S rates
into equilibrium for a solution, T must be
lowered.
Thus, FP for solution < FP for solvent
FP depression = ΔtFP = KFP•m
51
Freezing Point Depression
Calculate the FP of a 4.00 molal
glycol/water solution.
KFP = -1.86 oC/molal (Table 14.4, p. 577)
Solution
ΔtFP = KFP • m
= (-1.86 oC/molal)(4.00 m)
ΔtFP =
-7.44 oC
52
Colligative Properties Of
Ionic Solutions
• Ionic compounds dissociate completely into
ions in water.
• All calculations involving water and an ionic
solute must account for the total number of
particles present.
• This factor is called the van't Hoff factor, i.
53
Freezing Point Depression
How much NaCl must be dissolved
in 4.00 kg of water to lower FP to
-10.00 oC?.
Solution
Calculate the required molality.
ΔtFP = KFP • m
-10.00 oC = (-1.86 oC/molal) • Molality
Concentration = 5.38 molal
54
Freezing Point Depression
How much NaCl must be dissolved in 4.00 kg of water
to lower FP to -10.00 oC?.
Solution
Concentration required = 5.38 molal
This means we need 5.38 mol of dissolved
particles per kg of solvent.
Recognize that m represents the total conc. of all
dissolved particles.
Recall that
1 mol NaCl(aq)
1 mol Na+(aq) + 1 mol Cl-(aq)
55
Freezing Point Depression
How much NaCl must be dissolved in 4.00 kg of
water to lower FP to -10.00 oC?.
Solution
Concentration required = 5.38 molal
We need 5.38 mol of dissolved particles per kg
of solvent.
NaCl(aq) --> Na+(aq) + Cl-(aq)
To get 5.38 mol/kg of particles we need
5.38 mol / 2 = 2.69 mol NaCl / kg
2.69 mol NaCl / kg ---> 157 g NaCl / kg
(157 g NaCl / kg)•(4.00 kg) = 629 g NaCl
56
Boiling Point Elevation and
Freezing Point Depression
Δt = K • m • i
A generally useful equation
i = van’t Hoff factor = number of particles
produced per formula unit.
Compound
Theoretical Value of i
glycol
1
NaCl
2
CaCl2
3
57
58
Osmosis
Salt water
Pure water
• Osmosis occurs when a molecule moves
from a region of high concentration to lower
concentration through a semipermeable
membrane.
59
Osmotic pressure is defined by:
p = cRT
where c is the molarity of the solute
and R is the gas constant.
.08206 atm×L
R=
mol×K
60
61
Osmosis
Solvent
Solution
Semipermeable membrane
The semipermeable membrane should
allow only the movement of solvent
molecules.
Therefore, solvent molecules move
from pure solvent to solution.
62
63
Osmosis
Osmotic
Pressure
Solvent
Solution
The semipermeable membrane
should allow only the movement of
solvent molecules.
Therefore, solvent molecules move
from pure solvent to solution.
Osmosis
Osmotic
Pressure
Solvent
Solution
Equilibrium is reached when pressure
produced by extra solution —
the OSMOTIC PRESSURE, p
p = cRT (where c is conc. in mol/L)
counterbalances pressure of solvent
molecules moving thru the membrane.
64
65
Osmosis
66
Osmosis
• Osmosis of solvent
from one solution
to another can
continue until the
solutions are
ISOTONIC —
they have the same
concentration.
Osmosis
Calculating a Molar Mass
67
Dissolve 35.0 g of hemoglobin in enough water
to make 1.00 L of solution. p measured to be
10.0 mm Hg at 25 °C. Calculate molar mass
of hemoglobin.
Solution
(a) Calculate p in atmospheres
p = 10.0 mmHg • (1 atm / 760 mmHg)
(b)
= 0.0132 atm
Calculate the concentration
Calculating a Molar Mass
Dissolve 35.0 g of hemoglobin in enough water to
make 1.00 L of solution. p measured to be 10.0 mm
Hg at 25 C. Calculate molar mass of hemoglobin.
Solution
Calculate concentration from p = cRT
0.0132 atm
Conc =
(0.0821 L • atm/K • mol)(298K)
(b)
(c)
Concentration = 5.39 x 10-4 mol/L
Calculate the molar mass
Molar mass = 35.0 g / 5.39 x 10-4 mol/L
Molar mass = 65,100 g/mol
68
Reverse Osmosis
69
COLLOIDS
• Colloids are a suspension of very small
particles that do not settle out.
– (milk, jello, ..)
70
71
Hydrophilic
and
hydrophobic
colloids
exist and
emulsions
make use of
molecules
that contain
both.
72
Soap and Surfactants
O
Nonpolar tail
C
Polar head
H
O-
H2O
O
H
H2O
H2O
H2O
Dirt
H2O
H2O
H2O
H2O
H2O
H2O
73
Soap and Surfactants
74
Detergent
Fabric Softener
75