Chapter 14 — Solutions
1
Solutions
Brief Review of Major Topics in Chapter 13, Intermolecular forces
2
IonIon-Ion Forces (Ionic Bonding)
Chapter 14
Na+—Cl- in salt
These are the
strongest forces.
Lead to solids with
high melting
temperatures.
NaCl,
NaCl, mp = 800 oC
MgO,
MgO, mp = 2800 oC
3
4
Attraction Between
Ions and Permanent
Dipoles
Covalent Bonding Forces
••
C=C, 610 kJ/mol
-δ
C–H, 413 kJ/mol
••
C–C, 346 kJ/mol
water
dipole
H
H +δ
O
Water is highly polar
and can interact with
positive ions to give
hydrated ions in
water.
CN, 887 kJ/mol
5
Attraction Between
Ions and Permanent
Dipoles
••
••
water
-δ
dipole
O
H
H +δ
Water is highly polar
and can interact with
positive ions to give
hydrated ions in
water.
6
Attraction Between
Ions and Permanent
Dipoles
Many metal ions are hydrated. This
is the reason metal salts dissolve
in water.
Chapter 14 — Solutions
7
Attraction Between
Ions and Permanent
Dipoles
DipoleDipole-Dipole Forces
Such forces bind molecules having permanent
dipoles to one another.
Attraction between ions and dipole depends on ion
charge and ionion-dipole distance.
distance.
Measured by ∆H for Mn+ + H2O -->
--> [M(H2O)x]n+
δ- H
• • •O
Mg2+
H
δ+
-1922 kJ/mol
δ- H
• • •O
Na +
8
δ- H
• • •O
H
δ+
H
δ+
+
Cs
-405 kJ/mol -263 kJ/mol
9
DipoleDipole-Dipole Forces
Hydrogen Bonding
10
A special form of dipoledipole-dipole attraction, which
enhances dipoledipole-dipole attractions.
Influence of dipoledipole-dipole forces is seen in the
boiling points of simple molecules.
Compd
Mol. Wt.
Boil Point
N2
28
-196 oC
CO
28
-192 oC
Br2
160
59 oC
ICl
162
97 oC
H-bonding is strongest when X and Y are N, O, or F
11
Hydrogen Bonding in H2O
H-bonding is especially
strong in water because
• the O—
O—H bond is very
polar
• there are 2 lone pairs on
the O atom
Accounts for many of
water’
water’s unique
properties.
FORCES INVOLVING
INDUCED DIPOLES
How can nonnon-polar molecules such as O2 and I2
dissolve in water?
The water dipole INDUCES a dipole
in the O2 electric cloud.
DipoleDipole-induced
dipole
12
Chapter 14 — Solutions
FORCES INVOLVING
INDUCED DIPOLES
13
FORCES INVOLVING
INDUCED DIPOLES
14
Solubility increases with mass the gas
• Process of inducing a
dipole is
polarization
• Degree to which
electron cloud of an
atom or molecule can
be distorted in its
polarizability.
15
IM FORCES — INDUCED DIPOLES
Consider I2
dissolving
in ethanol,
CH3CH2OH.
FORCES INVOLVING
INDUCED DIPOLES
16
Formation of a dipole in two nonpolar I2 molecules.
Induced dipoledipoleinduced dipole
-δ
I-I
-δ O
R
H
+δ
I-I
The alcohol
temporarily
creates or
INDUCES a
dipole in I2.
+δ
-δ O
R
H
+δ
17
Intermolecular Forces
Summary
18
Liquids
In a liquid
• molecules are in constant
motion
• there are appreciable
intermolecular forces
• molecules close together
• Liquids are almost
incompressible
• Liquids do not fill the
container
Chapter 14 — Solutions
19
20
Liquids—
Liquids—Evaporation
Liquids
To evaporate, molecules
must have sufficient
energy to break IM
forces.
The two key properties we need to describe are
EVAPORATION and its opposite—
opposite—
CONDENSATION
evaporation-->
evaporation--->
LIQUID
Add energy VAPOR
break IM bonds
Breaking IM forces
requires energy. The
process of
evaporation is
endothermic.
endothermic.
make IM bonds
Remove energy
<---condensation
---condensation
Phase Diagrams
21
Solutions: Definitions
22
A solution is a
One constituent is usually
regarded as the SOLVENT
mixture of 2 or more
and the others as SOLUTES.
substances in a single phase.
HOMOGENEOUS
Definitions
Solute(s):
Solute(s): That which is dissolved, generally in lower
concentration
Solvent: The medium doing the dissolving of the solute
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24
Dissolving An Ionic Solid
Solutions can be classified as unsaturated or
saturated.
saturated.
A saturated solution contains the maximum
quantity of solute that dissovles at that
temperature.
SUPERSATURATED SOLUTIONS contain
more than is possible and are unstable.
Writing Equations: NaCl(s) Na+(aq) + Cl-(aq)
You do CaCl2 and Al2 (SO4) 3
Chapter 14 — Solutions
25
Energy
of the Solution Process
26
Energy
of the Solution Process
27
Lattice Energy – E to form lattice (note negative sign)
Hydration Energy – E evolved when gaseous ions are
dissolved
28
Energy - KF
If the enthalpy of formation of the solution is
more negative that that of the solvent and
solute, the enthalpy of solution is negative.
The solution process is exothermic!
29
Supersaturated
Sodium Acetate
• One application of a
supersaturated
solution is the sodium
acetate “heat pack.”
pack.”
• Sodium acetate has an
ENDOthermic heat
of solution.
30
Supersaturated
Sodium Acetate
Sodium acetate has an ENDOthermic heat of
solution.
NaCH3CO2 (s) + heat ---->
---->
Na+(aq)
(aq) + CH3CO2-(aq)
Therefore, formation of solid sodium acetate from
its ions is EXOTHERMIC.
EXOTHERMIC.
Na+(aq)
(aq) + CH3CO2-(aq) --->
--->
NaCH3CO2 (s) + heat
Chapter 14 — Solutions
31
Concentration Units
32
Concentration Units
An IDEAL SOLUTION is one
where the properties depend
only on the concentration of
solute.
MOLE FRACTION, X
For a mixture of A, B, and C
X A = mol fraction A =
Need conc. units to tell us the
number of solute particles per
solvent particle.
mol A
mol A + mol B + mol C
MOLALITY, m
m of solute =
The unit “molarity”
molarity” does not do
this!
mol solute
kilograms solvent
WEIGHT % = grams solute per 100 g solution
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Calculating Concentrations
Dissolve 62.1 g (1.00 mol) of ethylene glycol in 250. g of
H2O. Calculate mol fraction, molality,
molality, and weight % of
glycol.
Calculating Concentrations
34
Consider a solution that is 20.00 g of NaOH dissolved in 250 mL of total
aqueous solution (the final density of solution is 1.10 g/mL)
Calculate M, m, mole fraction NaOH and w%
Calculating Concentrations
35
Calculating Concentrations
Consider a solution that is 20.00 g of NaOH dissolved in 250 mL of total
aqueous solution (the final density of solution is 1.10 g/mL)
Consider a solution that is 20.00 g of NaOH dissolved in 250 mL of total
aqueous solution (the final density of solution is 1.10 g/mL)
Calculate M, m, mole fraction NaOH and w%
Calculate M, m, mole fraction NaOH and W%
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Chapter 14 — Solutions
37
What Affects Solubility??
Dissolving Gases & Henry’
Henry’s
Law
38
• Pressure – Inc. P over gas, increase solubility
• Gas + Solvent Solution
• Temperature – Reactions are exothermic (∆H =
negative) or endothermic (∆H = positive).
Gas solubility (mol/L) = kH • Pgas
• Endo: NaNO3 (s) + Heat Na+(aq) + NO3-(aq)
• Increase temperature increase solubility (most salts)
kH for H2 = 1.07 x 10-6 M/mmHg
When Pgas drops, solubility drops.
• Exo: NH4Cl (s) NH4+(aq) + Cl-(aq) + Heat
• Increase temperature decrease solubility
Colligative Properties
What is solubility of H2 at 1 atm?
atm? And 0.33 atm?
atm?
39
On adding a solute to a solvent, the props. of the
solvent are modified.
• Vapor pressure
decreases Will
• Melting point
decreases
• Boiling point
increases
• Osmosis is possible (osmotic pressure)
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.
Understanding
Colligative Properties
VP of H2O over a solution depends on the number
of H2O molecules per solute molecule.
Psolvent proportional to Xsolvent
Psolvent = Xsolvent • Posolvent
VP of solvent over solution
= (Mol frac solvent)•
solvent)•(VP pure solvent)
RAOULT’
RAOULT’S LAW
40
Understanding
Colligative Properties
To understand
colligative
properties,
study the
LIQUIDLIQUID-VAPOR
EQUILIBRIUM
for a solution.
41
Raoult’s Law
An ideal solution is one that obeys Raoult’
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!
42
Chapter 14 — Solutions
43
Raoult’
Raoult’s Law
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
Changes in Freezing and Boiling
Points of Solvent
VP Pure solvent
App. E.)
1 atm
44
VP solvent
after adding
solute
P
BP solution
BP pure
solvent
T
45
46
Elevation of Boiling Point
Vapor
Pressure
Lowering
Elevation in BP = ∆TBP = KBP•m
(where KBP is characteristic of solvent)
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 (see
Table 14.3).
Figure 14.14
47
Elevation of Boiling Point
Elevation in BP = ∆TBP = KBP•m
(where KBP is characteristic of solvent)
VP
VP Pure
Pure solvent
solvent
11 atm
atm
VP
VP solvent
solvent
after
after adding
adding
solute
solute
48
Change in Freezing Point
Pure waterEthylene glycol/water
solution
The freezing point of a
solution is
LOWER than that
of the pure solvent.
FP depression =
∆TFP = KFP•m
P
P
BP
BP solution
solution
BP
BP pure
pure
solvent
solvent
T
T
Calculate the FP of a 4.00 molal
glycol/water solution.
KFP = -1.86 oC/molal (Table 14.4)
Chapter 14 — Solutions
49
Freezing Point Depression
Calculate the FP of a 4.00 molal glycol/water
solution.
KFP = -1.86 oC/molal (Table 14.4)
Freezing Point Depression
KFP = ((-1.86 oC/molal)
C/molal)
51
Freezing Point Depression
52
How much NaCl must be dissolved in 4.00 kg of water to lower FP to 10.00 oC?.
Solution
Solution
53
A general equation
i = van’
van’t Hoff factor = number of particles
produced per formula unit.
Compound
Theoretical Value of i
glycol
1
NaCl
2
CaCl2
3
50
How much NaCl must be dissolved in 4.00 kg of
water to lower FP to -10.00 oC?.
How much NaCl must be dissolved in 4.00 kg of
water to lower FP to -10.00 oC?.
Boiling Point Elevation and
Freezing Point Depression
∆T = K•m•i
Freezing Point Depression
Osmosis
Solvent
Solvent
Solution
Solution
Semipermeable
Semipermeable membrane
membrane
The semipermeable membrane allows only the
movement of solvent molecules.
Solvent molecules move from pure
solvent to solution in an attempt to make
both have the same concentration of
solute.
OSMOTIC PRESSURE, ∏ = cRT
(c is conc. in mol/L)
54
Chapter 14 — Solutions
55
Osmosis
at the Particulate Level
56
Process of Osmosis
Figure 14.17
Osmosic Pressure, ∏
57
Equilibrium is reached when
pressure — the OSMOTIC
PRESSURE,
PRESSURE, ∏ —
produced by extra solution
Osmotic
counterbalances pressure of
pressure solvent molecules moving
thru the membrane.
Osmosis
Calculating a Molar Mass
58
Dissolve 35.0 g of hemoglobin in enough water to
make 1.00 L of solution. Π measured to be 10.0
mm Hg at 25 ˚C. Calc. molar mass of hemoglobin.
∏ = cRT
(c is conc. in mol/L)
Osmosis
Calculating a Molar Mass
Dissolve 35.0 g of hemoglobin in enough water to make 1.00 L of
solution. Π measured to be 10.0 mm Hg at 25 ˚C. Calc. molar mass
of hemoglobin.
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60
Osmosis
• Osmosis of solvent from
one solution to another
can continue until the
solutions are
ISOTONIC — they
have the same
concentration.
Chapter 14 — Solutions
61
Osmosis and Living Cells
Reverse Osmosis
Water Desalination
Water desalination plant in Tampa
62