Chemistry II
Chapter 12
Solutions
Thirsty Solutions
Thirsty Solutions
• homogeneous mixtures
 composition may vary from one sample to another
 appears to be one substance, though really contains multiple
materials
• most homogeneous materials we encounter are actually
solutions
 e.g., air, sea water, blood, tea, coffee, ‘alcohol’
• nature has a tendency toward spontaneous mixing
 generally, uniform mixing is more energetically favorable
Thirsty Solutions
• solute is the dissolved substance
 seems to “disappear”
 “takes on the state” of the solvent
• solvent is the substance solute dissolves in
 does not appear to change state
• when both solute and solvent have the same state, the solvent is the
component present in the highest percentage
• solutions in which the solvent is water are called aqueous solutions
Thirsty Solutions
• nature has a tendency toward spontaneous mixing
 generally, uniform mixing is more energetically favorable
Thirsty Solutions
Thirsty Solutions
Seawater is a more concentrated solution than body fluids. As a result
when seawater flows through the digestive tract it draws water out of
the surrounding tissues – leads to dehydration
Types of Solutions and Solubility
Solute
Phase
Solvent
Phase
gas
gas
liquid solutions
gas
liquid
solid
liquid
liquid
liquid
soda (CO2 in H2O)
vodka (C2H5OH in H2O)
seawater (NaCl in H2O)
solid solutions
solid
liquid
Solid
solid
brass (Zn in Cu), steel,
plastics
Hg/Au Amalgam
Solution Phase
gaseous solutions
Example
air (mostly N2 & O2)
Types of Solutions and Solubility
soluble - when one substance (solute) dissolves in another (solvent)
Insoluble - when one substance does not dissolve in another
•
the maximum amount of solute that can be
dissolved in a given amount of solvent is
called the solubility
•
solubility depends on – tendency towards mixing,
and the types of intermolecular attractive forces
•
the solubility of one substance in another varies
with temperature and pressure
Types of Solutions and Solubility
Nature’s Tendency towards Mixing: Entropy
•
Chemical systems tend toward lower PE
 e.g. two opposite charged particles attract
•
Solution formation does not necessarily lower the PE of
the system
 the difference in attractive forces between atoms of
two separate ideal gases vs. two mixed ideal gases
is negligible
 yet the gases mix spontaneously
•
the gases mix because the energy of the system is
lowered through the release of entropy
•
entropy is the measure of energy dispersal throughout
the system
•
energy has a spontaneous drive to spread out over as
large a volume as it is allowed
Types of Solutions and Solubility
The Effect of Intermolecular Forces
• energy changes in the formation of most solutions also involve differences in
intermolecular forces
• The intermolecular forces in the solute and solvent may promote or prevent
solution formation
Types of Solutions and Solubility
The Effect of Intermolecular Forces
Types of Solutions and Solubility
The Effect of Intermolecular Forces
Relative interactions
Solvent-Solute
>
Solute-to-Solute and
Solvent-to-Solvent
Solvent-Solute
=
Solute-to-Solute and
Solvent-to-Solvent
Solution Forms
Solvent-Solute
<
Solute-to-Solute and
Solvent-to-Solvent
Solution May or May
Not Form
Solution Forms
• when the Solvent-Solute attractions are weaker than the sum of the
solute-to-solute and solvent-to-solvent attractions, the solution will only
form if the energy difference is small enough to be overcome by the
entropy
Types of Solutions and Solubility
The Effect of Intermolecular Forces
Examples:
•
e.g. pentane (C5H12) and heptane (C7H16) – both have dispersion forces between
their molecules AND there are dispersion forces between pentane and heptane
molecules
•
All 3 interactions are of similar size –
solvent-solute = solute-solute = solvent-solvent
•
they are soluble or miscible liquids
Types of Solutions and Solubility
The Effect of Intermolecular Forces
Examples:
•
e.g. hexane (C6H14) and water (H2O) – water molecules has strong H-bonds to each
other but cannot bond to hexane
•
Energy required to separate H2O molecules is too great, H2O interactions >>>>
H2O/Hexane interactions
•
solvent-solute < solute-solute and solvent-solute < solvent-solvent
•
they are insoluble or immiscible liquids
Types of Solutions and Solubility
The Effect of Intermolecular Forces
• Chemist’s Rule of Thumb –
•
•
Like Dissolves Like
a chemical will dissolve in a solvent if it has a similar structure to the
solvent
when the solvent and solute structures are similar, the solvent molecules
will attract the solute particles at least as well as the solute particles to
each other
Classifying Solvents
Solvent
Class
Structural
Feature
Water, H2O
polar
O-H
Methyl Alcohol, CH3OH
polar
O-H
Ethyl Alcohol, C2H5OH
polar
O-H
Acetone, C3H6O
polar
C=O
Toluene, C7H8
nonpolar
C-C & C-H
Hexane, C6H14
nonpolar
C-C & C-H
Diethyl Ether, C4H10O
nonpolar
C-C, C-H & C-O,
(nonpolar > polar)
Carbon Tetrachloride
nonpolar
C-Cl, but symmetrical
Draw the above molecules
Example 12.1a  predict whether the
following vitamin is soluble in fat or water
OH
OH
H2C
C
H
H
C
O
C
C
C
HO
OH
Vitamin C
O
Example 12.1b  predict whether the
following vitamin is soluble in fat or water
O
H
C
C
CH3
HC
C
C
HC
C
CH
C
H
C
O
Vitamin K3
Energetics of Solution Formation
•
•
•
Separation of Solute
 must overcome IMF or ion-ion attractions in solute
 requires energy, ENDOTHERMIC ( + DHsolute)
Separation of Solvent
 must overcome IMF of solvent particles
 requires energy, ENDOTHERMIC (+ DHsolvent)
Interaction of Solute & Solvent
 attractive bonds form between solute particles and solvent particles
 “Solvation” or “Hydration” (where water = solvent)
 releases energy, EXOTHERMIC (- DHmix)
DHsoln = DHsolute + DHsolvent + DHmix
•
•
•
DHmix depends on what you are mixing.
If molecules can attract each other DHmix is large and negative.
If molecules can’t attract DHmix is small and negative
Energetics of Solution Formation
DHsolute + DHsolvent < DHmix
overall process will be exothermic
DHsolute + DHsolvent > DHmix
overall process will be endothermic
Energetics of Solution Formation
Heats of Hydration
•
for aqueous ionic solutions, the energy added to overcome the attractions between
water molecules and the energy released in forming attractions between the water
molecules and ions is combined into a term called the heat of hydration
 attractive forces in water = H-bonds
 attractive forces between ion and water = ion-dipole
DHhydration = DHsolvent + DHmix = heat released when 1 mole of gaseous ions dissolves in water
DHsoln = DHsolute + [DHsolvent + DHmix]
ΔHsoln = ΔHsolute + ΔHhydration
(+ve)
(-ve)
24
Energetics of Solution Formation
Heats of Hydration
•
|ΔHsolute| < |ΔHhydration|
ΔHsoln = -ve = exothermic reaction, e.g. KOH
KOH(s) → K+(aq) + OH-(aq)
•
|ΔHsolute| > |ΔHhydration|
ΔHsoln = +ve = endothermic reaction, e.g. NH4NO3
NH4NO3(s) → NH4+(aq) + NO3-(aq)
•
ΔHsoln = -57.56 kJ/mol
ΔHsoln = +25.67 kJ/mol
|ΔHsolute| = |ΔHhydration|
ΔHsoln ~ 0, e.g. NaCl
NaCl(s) → Na+(aq) + Cl-(aq)
ΔHsoln = +3.88 kJ/mol
Energetics of Solution Formation
Heats of Hydration
ΔHsol’n = Δ Hsolute (+ve) + Δ Hhydration (-ve)
Energetics of Solution Formation
Heats of Hydration
Which is NH4NO3, which is KOH?
Solution Equilibrium
•
•
•
•
the dissolution of a solute in a solvent is an equilibrium process
initially, when there is no dissolved solute, the only process possible is dissolution
shortly, solute particles can start to recombine to reform solute molecules – but the rate
of dissolution >> rate of deposition and the solute continues to dissolve
eventually, the rate of dissolution = the rate of deposition – the solution is saturated with
solute and no more solute will dissolve
Solution Equilibrium
Solubility Limit
• At dynamic equilibrium a solution has the maximum amount of solute
dissolved in it - said to be saturated
 depends on the amount of solvent
 depends on the temperature
 and pressure of gases
• a solution that has less solute than saturation is said to be unsaturated
• a solution that has more solute than saturation is said to be supersaturated
Solution Equilibrium
How Can You Make a Solvent Hold More Solute Than It Is Able To?
• solutions can be made saturated at non-room conditions – then
allowed to come to room conditions slowly
• for some solutes, instead of coming out of solution when the
conditions change, they get stuck in-between the solvent molecules
and the solution becomes supersaturated
• supersaturated solutions are unstable and lose all the solute above
saturation when disturbed
 e.g., shaking a carbonated beverage
Adding Solute to a Supersaturated
Solution of NaC2H3O2
http://www.youtube.com/watch?v=uy6eKm8IRdI
Solution Equilibrium
Temperature Dependence of Solubility of Solids in Water
• solubility is generally given in grams of solute that will dissolve in 100 g of
water
• for most solids, the solubility of the solid increases as the temperature
increases
 when DHsolution is endothermic
• solubility curves can be used to predict
whether a solution with a particular amount
of solute dissolved in water is saturated
(on the line), unsaturated (below the line),
or supersaturated (above the line)
Solution Equilibrium
Factors Affecting the Solubility of Gases in Water
• solubility is generally given in moles of solute that will dissolve in 1
Liter of solution
• Gases have generally lower solubility than ionic or polar covalent
solids because most are nonpolar molecules
• for all gases, the solubility of the gas decreases as the temperature
increases
 the DHsolution is exothermic because you do not need to overcome
solute-solute attractions
Solubility of Gases in Water at Various
Temperatures
0.4
Solubility, g
in 100 g water
0.35
0.3
0.25
CO2
acetylene
0.2
0.15
0.1
0.05
0
0
10
20
30
40
50
60
70
80
90
100 110
Temperature, °C
Solubility of Gases in Water at Various Temperatures
Solubility, g
in 100 g water
0.008
0.007
0.006
0.005
N2
O2
0.004
0.003
0.002
0.001
0
0
10
20
30
40
50
60
70
Temperature, °C
80
90
100
110
Solution Equilibrium
Factors Affecting the Solubility of Gases in Water
Pressure
• the larger the partial pressure of a gas in contact with a liquid, the
more soluble the gas is in the liquid
35
persrst
At lower pressure CO2 is less soluble and
bubbles out of solution
Solubility of Gases in Water at Various
Pressures (temp = 20°C)
Solubility, g in 100 g water
0.25
0.2
0.15
O2
CO2
0.1
0.05
0
0
200
400
600
Partial Pressure
800
1000
1200
Solution Equilibrium
Factors Affecting the Solubility of Gases in Water
Henry’s Law
• the solubility of a gas (Sgas) is directly
proportional to its partial pressure, (Pgas)
Sgas = kHPgas
Often written KH = [X] / P
• kH is called Henry’s Law Constant
Ex 12.2 – What pressure of CO2 is required to
keep the [CO2] = 0.12 M at 25°C?
Expressing Solution Concentration
• solutions have variable composition
• to describe a solution, need to describe components and relative
amounts
• the terms dilute and concentrated can be used as qualitative
descriptions of the amount of solute in solution
concentration = amount of solute in a given amount of solution
Expressing Solution Concentration
Molarity, M
• moles of solute per liter of solution
• used because it describes how many molecules of solute in
each liter of solution
• if a sugar solution concentration is 2.0 M, 1 liter of solution
contains 2.0 moles of sugar, 2 liters = 4.0 moles sugar, 0.5
liters = 1.0 mole sugar
Molarity, C (M) =
moles of solute, n (mol)
liters of solution, V (L)
n = CV
Preparing 1 L of a 1.00 M NaCl Solution
Expressing Solution Concentration
Molarity, M
Dilution
• often, solutions are stored as concentrated stock solutions
• to make solutions of lower concentrations from these stock solutions,
more solvent is added
 the amount of solute doesn’t change, just the volume of solution
moles solute in solution 1 = moles solute in solution 2
• the concentrations and volumes of the stock and new solutions are
inversely proportional
C1∙V1 = C2∙V2
Expressing Solution Concentration
Molarity, M
e.g. find the volume of 12 M HCl required to make 1.0 L of 1.0 M HCl
C1∙V1 = C2∙V2
Expressing Solution Concentration
Molarity and Dissociation
• the molarity of the ionic compound allows you to determine the
molarity of the dissolved ions
CaCl2(aq) → Ca+2(aq) + 2 Cl-(aq)
• A 1.0 M CaCl2(aq) solution contains 1.0 moles of CaCl2 in each
liter of solution
 1 L = 1.0 moles CaCl2, 2 L = 2.0 moles CaCl2
• Because each CaCl2 dissociates to give one Ca+2 = 1.0 M Ca+2
 1 L = 1.0 moles Ca+2, 2 L = 2.0 moles Ca+2
• Because each CaCl2 dissociates to give 2 Cl- = 2.0 M Cl 1 L = 2.0 moles Cl-, 2 L = 4.0 moles Cl-
Expressing Solution Concentration
Molality, m
• moles of solute per 1 kilogram of solvent
 defined in terms of amount of solvent, not solution
 like the others
• does not vary with temperature
 because based on masses, not volumes
Molality, (m) =
moles of solute (mol)
mass of solvent (kg)
Expressing Solution Concentration
Parts by Mass (percent, pph)
• parts of solute in every 100 parts solution
• mass percent = mass of solute in 100 parts solution by mass
 if a solution is 0.9% by mass, then there are 0.9 grams of solute in
every 100 grams of solution
 or 0.9 kg solute in every 100 kg solution
Mass of Solute, g
Mass Percent 
100%
Mass of Solution, g
Mass of Solute  Mass of Solvent  Mass of Solution
Expressing Solution Concentration
Using Concentrations as Conversion Factors
• concentrations show the relationship between the amount of solute
and the amount of solvent
 12%(m/m) sugar(aq) means 12 g sugar  100 g solution
 or 12 kg sugar  100 kg solution; or 12 lbs.  100 lbs. solution
 5.5%(m/v) Ag in Hg means 5.5 g Ag  100 mL solution
 22%(v/v) alcohol(aq) means 22 mL EtOH  100 mL solution
• The concentration can then be used to convert the amount of solute
into the amount of solution, or vice versa
Preparing a Solution
• need to know amount of solution and concentration of solution
• calculate the mass of solute needed
 start with amount of solution
 use concentration as a conversion factor
 5% by mass 5 g solute  100 g solution
 “Dissolve the grams of solute in enough solvent to total the total
amount of solution”
Example –
How would you prepare 250.0 g of 5.00% by
mass glucose solution (normal glucose)?
Expressing Solution Concentration
Parts by Mass (parts per million, ppm)
• grams of solute per 1,000,000 g of solution
• mg of solute per 1 kg of solution
• 1 liter of water = 1 kg of water
ppm =
mass solute (g)
mass solution (g)
ppm =
mass solute (mg)
mass solution (kg)
ppm =
mass solute (mg)
volume solution (L)
x 106
Solution Concentrations
Mole Fraction, XA
• the mole fraction is the fraction of the moles of one component in
the total moles of all the components of the solution
• total of all the mole fractions in a solution = 1
• unitless
• the mole percentage is the percentage of the moles of one
component in the total moles of all the components of the solution
 = mole fraction x 100%
mole fraction of A = XA = moles of components A
total moles in the solution
Ex 12.4a – What is the molarity of a solution
prepared by mixing 17.2 g of C2H6O2 with 0.500 kg
of H2O to make 515 mL of solution?
Given: 17.2 g C2H6O2, 0.500 kg H2O, 515 mL soln.
Find: M
Concept Plan: g C2H6O2
mol C2H6O2
mol
M
M
L
mL soln.
L soln.
Relationships: M = mol/L, 1 mol C2H6O2 = 62.07 g, 1 mL = 0.001 L
C2H
Solve:
0.2717mol
1 mol
C6OH2 O
17.2 g C 2 H 6O 2 
M
2
6  20.2771 mol
62.07 g C 2 H 6O 2
0.515 L
0.001 L
mL  M
 0.515 L
M 5150.538
1 mL
Check: the unit is correct, the magnitude is reasonable
Ex 12.4d – What is the mole percent of a solution
prepared by mixing 17.2 g of C2H6O2 with 0.500 kg
of H2O to make 515 mL of solution?
Given: 17.2 g C2H6O2, 0.500 kg H2O, 515 mL sol’n
Find: C
Concept Plan: g C2H6O2
mol C2H6O2  %  mol A 100%
mol total
g H2O
mol H2O
C%
Relationships: C = molA/moltot, 1 mol C2H6O2 = 62.07 g, 1 mol H2O = 18.02 g
1 mol
O 2H O
Solve:
2 H 6C
0.27
71Cmol
17
.
2
g
C
H
O

 %  2 6 2 62.07 g C H 2O 6 02.2771 mol 100%
0.2771 mol C2 H 6O2 2 6 227.75 mol H 2O 
.500
kg H 2%
O
0%
 0.989
1000 g 1 mol H 2O

 27.75 mol H 2O
1 kg 18.02 g H 2O
Check: the unit is correct, the magnitude is reasonable
Converting Concentration Units
• assume a convenient amount of solution






given %(m/m), assume 100 g solution
given %(m/v), assume 100 mL solution
given ppm, assume 1,000,000 g solution
given M, assume 1 liter of solution
given m, assume 1 kg of solvent
given X, assume you have a total of 1 mole of solutes in the solution
• determine amount of solution in non-given unit(s)
 if assume amount of solution in grams, use density to convert to mL and
then to L
 if assume amount of solution in L or mL, use density to convert to grams
• determine the amount of solute in this amount of solution, in grams and moles
• determine the amount of solvent in this amount of solution, in grams and moles
• use definitions to calculate other units
Colligative Properties
• Colligative properties – property that depends on the number or
particles in solution not the type of particle
• e.g.




Vapor pressure lowering
Freezing point depression
Boiling point elevation
Osmotic pressure
Colligative Properties
Vapor Pressure Lowering
• What is the effect of a nonvolatile solute on the VP of a pure
liquid?
liquid  vapor equilibrium
of solvent particles
Colligative Properties
Vapor Pressure Lowering
• the VP of a solution is lower than the VP of the pure solvent
 the solute particles replace some of the solvent molecules
at the surface
Addition of a nonvolatile solute ( ) reduces the rate of
vaporization, decreasing the amount of vapor
Eventually, equilibrium is re-established, but a smaller
number of vapor molecules – therefore the vapor pressure
will be lower
Colligative Properties
Vapor Pressure Lowering
• a concentrated solution will draw solvent molecules toward it due to
the natural drive for materials in nature to mix
• similarly, a concentrated solution will draw pure solvent vapor into it
due to this tendency to mix
• the result is reduction in the conc. of the solution
VP solution <<< VP pure solvent
Colligative Properties
Vapor Pressure Lowering
Raoult’s Law
• the VP of a volatile solvent above a solution is equal to its mole
fraction of its normal vapor pressure, P°
Psolvent in solution = solvent∙P°
 since the mole fraction of the solvent is always less than 1, the
vapor pressure of the solvent in solution will always be less than
the vapor pressure of the pure solvent
Colligative Properties
Vapor Pressure Lowering
e.g. water at 25 °C contains 0.90 mol water and 0.10 mol sucrose
Pure water has VP 23.8 torr,
P soln = H2O∙P°H2O = 0.90 x 23.8 torr = 21.4 torr
VP of solution is lowered in proportion to mole composition of the
solvent, VP is 90% of the VP of pure solvent
Colligative Properties
Vapor Pressure Lowering
Ionic Solutes and Vapor Pressure
• according to Raoult’s Law, the effect of solute on the vapor
pressure simply depends on the number of solute particles
• when ionic compounds dissolve in water, they dissociate – so
the number of solute particles is a multiple of the number of
moles of formula units
• the effect of ionic compounds on the vapor pressure of water is
magnified by the dissociation
 since NaCl dissociates into 2 ions, Na+ and Cl, one mole of
NaCl lowers the vapor pressure of water twice as much as 1
mole of C12H22O11 molecules would
• the van’t Hoff factor, i, is the ratio of moles of solute particles to
moles of formula units dissolved, e.g. NaCl, i = 2
Effect of Dissociation
Dissociation is not ideally complete due to ion pair
formation
Colligative Properties
Vapor Pressures of Solutions Containing a Volatile Solute
•
•
when both the solvent and the solute can evaporate, both molecules will be found in the
vapor phase
the total VP above the solution will be the sum of the VPs s of the solute and solvent
 for an ideal solution
Ptotal = Psolute + Psolvent
•
the solvent decreases the solute VP in the same way the solute decreased the solvent’s
Psolute = solute∙P°solute and Psolvent = solvent∙P°solvent
Colligative Properties
Vapor Pressures of Solutions Containing a Volatile Solute
Ideal vs. Nonideal Solution
• in ideal solutions, the made solute-solvent interactions are equal to the
sum of the broken solute-solute and solvent-solvent interactions
 ideal solutions follow Raoult’s Law
• effectively, the solute is diluting the solvent
• if the solute-solvent interactions are stronger or weaker than the
broken interactions the solution is nonideal
Vapor Pressure of a
Nonideal Solution
•
when the solute-solvent interactions are stronger than the solute-solute + solventsolvent, the total vapor pressure of the solution will be less than predicted by Raoult’s
Law
 because the vapor pressures of the solute and solvent are lower than ideal
•
when the solute-solvent interactions are weaker than the solute-solute + solventsolvent, the total vapor pressure of the solution will be larger than predicted by Raoult’s
Law
Colligative Properties
Freezing Point Depression and Boiling Point Elevation
• the melting/freezing point of a solution is lower than the
melting/freezing point of the pure solvent
Allows salt to melt ice
even if ambient temp is
below freezing
Colligative Properties
Freezing Point Depression and Boiling Point Elevation
VP of soln. is shifted
downward
Net effect:
lower mpt, higher bpt
Depends on amount of
solute not type = colligative
property
Colligative Properties
Freezing Point Depression
• the difference between the freezing point of the solution and freezing
point of the pure solvent is directly proportional to the molal
concentration of solute particles
FPsolvent – FPsolution) = DTf = m∙Kf
• m = molality of solution (mol solute/kg solvent)
• the proportionality constant is called the Freezing Point Depression
Constant, Kf
 the value of Kf depends on the solvent
 the units of Kf are °C/m
Kf
Colligative Properties
Freezing Point Depression and Boiling Point Elevation
• the boiling point of a solution is higher than the boiling point of the
pure solvent
Colligative Properties
Freezing Point Depression and Boiling Point Elevation
• the difference between the bpt. of the solution and bpt. of the pure
solvent is directly proportional to the molal concentration of solute
particles
BPsolution – BPsolvent) = DTb = m∙Kb
• the proportionality constant is called the Boiling Point Elevation
Constant, Kb
 the value of Kb depends on the solvent
 the units of Kb are °C/m
 For water Kb = 0.512 °C/m
Colligative Properties
Osmosis
• osmosis is the flow of solvent through a semi-permeable membrane from
•
solution of low concentration to solution of high concentration
the amount of pressure needed to keep osmotic flow from taking place is
called the osmotic pressure
• the osmotic pressure, P, is directly proportional to the molarity of the solute
particles
 R = 0.08206 (atm∙L)/(mol∙K)
P = MRT
Where M = molarity, T = temperature
Colligative Properties of
Strong Electrolyte Solutions
• Ionic solutions:
DTf = i m∙Kf
DTb = i m∙Kb
P = i MRT
Use the van’t Hoff factor, i
e.g. which has the highest bpt?
1 M sugar solution, 1 M NaCl solution or 1 M MgCl2 solution?