matter
impure
pure
at least two
substances
one substance
Elements
compounds
mixture
mono-atomic: He
Diatomic: O2
Polyatomic: Fe
ionic
consist of ions
NaCl
molecular
consist of molecules
H2O
matter
impure
pure
at least two
substances
one substance
mixture
homogeneous
one phase
Tab water
heterogeneous
two phases
oil/water
milk
region of physical uniformity
all parts of that space have the same physical properties
color, density, conductivity, magnetism, …
mixture
homogeneous
heterogeneous
two phases
oil/water
milk
one phase
Tab water
solvent
Solute(s)
Less abundant or other
more abundant
component(s) of mixture
component of mixture
salts in tab water
water in tab water
nitrogen in air
Water always solvent even in 98% H2SO4
amount of solute to solvent
amount of solute in solution
NaCl
Percent by Mass
What is the percent by mass of NaCl in a solution
consisting of 12.5 g of NaCl and 75.0 g water?
mass of solute
percent by mass 
 100%
mass of solution
wt %NaCl
12.5 g

 100
(12.5  75.0)g
wt %NaCl  14.3% NaCl
7
Molality (m)
– Number of moles of solute per kilogram solvent
mol of solute
molality  m 
kg of solvent
If you prepare a solution by dissolving 25.38 g of I2 in 500.0
g of water, what is the molality (m) of the solution?
8
What is the molarity (M) of this solution? The density
of this solution is 1.00 g/mL.
nC H OH 
2
5
mC H OH
2
M C H OH
2
M C H OH 
2
5
nC H OH
2
5
5
Vsolution
C2 H 5OH % 
5
1g
2


2
.
17

10
mol
1
46.07 g.mol
2.17  10 2 mol
mol

 0.215
 0.215 M
0.101 L
L
mC H OH
2
5
msolution
 100 % 
mC H OH
2
5
mC H OH  mwater
2
5
1g
C2 H 5OH % 
 100 %  0.99%
1 g  100 g
 100 %
xC H OH 
2
nC H OH 
2
5
5
mC H OH
2
M C H OH
2
nwater 
xC H OH 
2
5
2
5
5
nC H OH
2
5
wtsolvent/ kg
2
5
nsolution

nC H OH
2
5
nC H OH  nwater
2
5
1g
2


2
.
17

10
mol
1
46.07 g.mol
mwater
100 g

 5.56 mol
1
M water 18.0 g.mol
nC H OH  nwater
mC H OH 
5
5
nC H OH
2
2
5
nC H OH
2.17  10 2 mol

 0.00389
2
2.17  10 mol  5.56 mol
2.17  10 2 mol
mol

 0.217
 0.217 m
0.100 kg
kg water
3.75 M sulfuric acid solution has a density of 1.230
g/mL. Calculate the mass percent and molality of
H2SO4 in solution.
368 g H2SO4; 862 g H2O; 29.9% H2SO4; 4.35 m
Why do Solutions Form?
1. Energetic effects:
Intermolecular Interaction forces
2. Entropy effects:
Tendency of nature towards disorder
Hsoln = H1solute + H2solvent + H3solvation
14
• Hsoln < 0 (negative)
– Energy given off when
solution is made
– Exothermic
–  PE of system
• Hsoln > 0 (positive)
– Costs energy to make
solution
– Endothermic
–  PE of system
Ideal Solution
– Hsoln = 0
Ex. Benzene/CCl4
– All London
forces
– Hsoln ~ 0
• Step 1 + Step 2 = –Step 3
Hsolute + Hsolvent = –Hsolvation
16
Why oil doesn’t dissolve in water?
Hsoln = Hsolute + Hsolvent + Hsolvation
Hsoln = H1 + H2 + Hsolvation
Hsolute > 0
expanding oil (bringing oil molecules apart)
but small (small London forces, nonpolar)
Hwater> 0
Expanding water (bringing water molecules apart)
very large (hydrogen bond!)
Hsolv< 0
Attraction oil molecules-water molecules
Attraction: Ep decrease
small (polar-nonpolar)
Hsoln strongly positive: strongly endo
Large amounts of energy needed to form
solution: usually can not be afforded
Immiscible Liquids
•
•
•
•
Two insoluble liquids
Do not mix
Get two separate phases
Strengths of IMFs are different in solute
and solvent
• Different polarity
Ex. Benzene and water
Benzene;
no polar bond
H
H
H
C
C
C
C
H
18
C
C
H
H
Rule of Thumb
• “Like dissolves Like”
– Use polar solvent for polar solute
– Use Nonpolar solvent for nonpolar solute
19
Hydration of Solid Solute
• At edges, fewer
oppositely charged
ions around
– H2O can come in
– Ion-dipole forces
– Remove ion
• New ion at surface
– Process continues
until all ions in
solution
• Hydration of ions
– Completely
surrounded
by solvent
20
Dissolving NaBr in H2O
Hsoln = Hsolute + Hsolvent + Hsolvation
Hsolute > 0
NaBr(s) →
Na+
(g)
+
Br-(g)
Lattice energy
Hhydration = Hsolvent + Hsolvation
Na+
(g)
+
Br-(g)
H2O
→ Na+(aq) + Br-(aq)
Hsoln = Hlattice + Hhydration
NaBr(s)
H2O
→ Na+(aq) + Br-(aq)
Hlattice
Dissolving NaBr in H2O
Hlattice (NaBr) = 728 kJ mol–1
Hhydration (NaBr) = –741 kJ mol–1
Hsolution = Hlatt + Hhydr
Hsoln = 728 kJ mol–1
– 741 kJ mol–1
Hsoln = – 13 kJ mol–1
 Formation of NaBr(aq) is
exothermic
22
Dissolving KI in H2O
Hlattice (KI) = 632 kJ mol–1
Hhydration (KI) = – 619 kJ mol–1
Hsolution = Hlatt + Hhydr
Hsoln = 632 kJ mol–1
– 619 kJ mol–1
Hsoln = +13 kJ mol–1
• Formation of KI(aq) is
endothermic
Still: KI dissolves in water!!
Why?
23
Spontaneous Mixing
• 2 gases mix
spontaneously
– Due to random motions
– Mix without outside work
– Never separate
spontaneously
• Tendency of system left
to itself, to
become increasingly
disordered
– Entropy effect
24
Gas A
Gas B
separate
mixed
saturated
Contains the maximum
amount of solute that can
be disssolved
unsaturated
Contains less than the
maximum amount of solute
that can be disssolved
Maximum amount of NaCl that can be dissolved in 100 g
water at 25oC is 36 g.
If 50 g added to water: only 36 g can dissolve, 14 g settle down.
If only 20 g NaCl are dissolved in 100 g H2O at 25oC:
unsaturated: can take more.
Solubility
• Mass of solute that forms saturated solution with given
mass of solvent at specified temperature
g solute
solubility 
100 g solvent
• If extra solute added to saturated solution, extra solute will
remain as separate phase
•
Solubility (NaCl)=36 g/100 g H2O at 25oC
26
Solubility of Most Substances Increases with
Temperature
• Most substances
become more
soluble as T 
• Amount solubility 
– Varies considerably
– Depends on
substance
27
Effect of T on Gas Solubility in Liquids
• Solubility of gases usually  as T 
28
Case Study: Dead Zones
• During the industrial revolution, factories were
built on rivers so that the river water could be used
as a coolant for the machinery. The hot water was
dumped back into the river and cool water
recirculated. After some time, the rivers began to
darken and many fish died. The water was not
found to be contaminated by the machinery.
What was the cause of the mysterious fish kills?
Increased temperature, lowered
amounts of dissolved oxygen
29
Effect of Pressure on Gas Solubility
A. At some P, equilibrium exists between vapor phase
and solution
– ratein = rateout
B.  in P
–  frequency of collisions so ratein > rateout
– More gas molecules dissolve than are leaving
solution
C. More gas dissolved
– Rateout will  until Rateout = Ratein and equilibrium
restored
30
Effect of Pressure on Gas Solubility
• Solubility  as P 
solubility  as P 
– Soda in can
31
Henry’s Law
“Concentration of gas in liquid at any given temperature is
directly proportional to partial pressure of gas over
solution”
Cgas = kH Pgas
(T is constant)
Cgas = concentration of gas in the liquid phase
Pgas = partial pressure of gas above solution
kH = Henry's Law constant
»Unique to each gas
»Tabulated
32
Henry’s Law
• True only at low concentrations and pressures
where gases do NOT react with solvent
• Alternate form
C1 C2

P1 P2
– C1 and P1 refer to an initial set of conditions
– C2 and P2 refer to a final set of conditions
33
Ex. 1 Using Henry’s Law
Calculate the concentration of CO2 in a soft drink that is
bottled with a partial pressure of CO2 of 5 atm over the
liquid at 25°C. The Henry’s Law constant for CO2 in water
at this temperature is 3.12  102 mol/L·atm.
CCO2  k H (CO 2 ) PCO2
= 3.12  102 mol/L·atm * 5.0 atm
= 0.156 mol/L  0.16 mol/L
When under 5.0 atm pressure
34
Ex. 1 Using Henry’s Law
Calculate the concentration of CO2 in a soft drink after the
bottle is opened and equilibrates at 25°C under a partial
pressure of CO2 of 4.0  104 ·atm.
C1 C 2

P1
P2
C2
P2 C1
C2 
P1

0.156 mol/L4.0  10 4 atm 

5.0atm
C2 = 1.2  104 · mol/L
When open to air
35
V
V
pi  xi  p
o
i
L
L
Raoult’s Law
Calculate the expected vapor pressure at 25oC for a solution
prepared by dissolving 158.0 g common table sugar (sucrose,
molar mass 342.3 g/mol) in 643.5 cm3 of water. At 25oC, the
density of water is 0.9971 g/cm3 and the vapor pressure is
23.76 torr.
pi  xi  p
o
i
psolution  pwater  xwater  p
o
water
nA
xA 
n A  nB
xwater
nwater

nwater  nsugar
nwater
mwater Vwater d water 643.5 cm3  0.9771 cmg



 35.6 mol
1
M water
M water
18.0 g.mol
3
msucrose
158.0 g
nsucrose 

 0.4616 mol
1
M sucrose 342.3 g.mol
xwater
nwater
35.6


 0.9873
nwater  nsugar 35.6  0.4616
o
psolution  pwater  xwater  pwater
 0.9873  23.76 torr
psolution  23.46 torr
Binary System
Solute volatile
psolution  p A  p B
V
psolution  x A p Ao  xB pBo
psolution  1  xB  p Ao  xB p Bo
psolution  p Ao  xB p Ao  xB pBo
L
A+B


psolution  p Ao  p Bo  p Ao xB
y
a 
b
x
psolution  p Ao   pBo  p Ao  xB
y
a 
b
x
P Total  PA  PB  X APA  X BPB
psolution  pA  pB
PA  X APA
PB 

X BPB
positive
deviation
negative
deviation
P > Pideal
P < Pideal
A solution is prepared by mixing 5.81 g acetone (C3H6O, molar mass
58.1 g/mol) and 11.9 g chloroform (HCCl3, molar mass 119.4 g/mol). At
35oC, this solution has a total vapor pressure of 260 torr. Is this an ideal
solution? The vapor pressures of pure acetone and pure chloroform at
35oC are 345 and 293 torr, respectively.
Does it obey Raoult’s Law?
? P = Pideal ?
psolution,ideal  pacetone  pchloroform
o
o
psolution,ideal  xacetone pacetone
 xchloroform pchloroform
nacetone 
macetone
5.81 g

 0.100 mol
1
M acetone 58.1 g.mol
nHCCl 3 
mHCCl 3
11.9 g

 0.100 mol
1
M HCCl 3 119.4 g.mol
nacetone
0.100
xacetone 

 0.500
nacetone  nHCCl 3 0.100  0.100
xHCCl 3  1  0.500  0.500
o
o
psolution,ideal  xacetone pacetone
 xchloroform pchloroform
psolution,ideal  0.5  345  0.5  293  319 torr
P < Pideal
260 < 319
negative
deviation
Phase diagram
pure water
Phase diagram
aq solution
Colligative Properties
solute
solute
solute
Non-volatile solutes
Kf and Kb
47
A solution was prepared by dissolving 18.00 g glucose in 150.0 g water.
The resulting solution was found to have a boiling point of 100.34oC.
Calculate the molar mass of glucose. Glucose is a molecular solid that is
present as individual molecules in solution.
Tb  K b  msolute
Tb  T  T  100.34 C  100 C  0.34 C
o
o
o
o
Tb
0.34 o C
msolute 

 0.67 mol
kg
o
K b 0.51 C. kg / mol
msolute 
nsolute
wtsolvent / kg
nsolute  msolute  wtsolvent / kg
nsolute  0.67 mol
kg  0.150 kg  0.10 mol
nsolute 
wtsolute
Mwtsolute
Mwtsolute 
wtsolute 18.00 g
g

 180 mol
nsolute 0.10 mol
What mass of ethylene glycol (C2H6O2, molar mass 62.1 g/mol), the
main component of antifreeze, must be added to 10.0 L water to produce
a solution for use in a car’s radiator that freezes at -23.3oC? Assume the
density of water is exactly 1 g/mL.
Tf  K f  msolute
Tf  Tf  Tf  23.3o C
o
T f
23.3o C
msolute 

 12.5 mol
kg
o
K f 1.86 C. kg / mol
nsolute
msolute 
wtsolvent / kg
nsolute  msolute  wtsolvent / kg
nsolute  12.5 mol
kg  10.0 kg  125 mol
wtsolute
nsolute 
Mwtsolute
wtsolute  nsolute  Mwtsolute  125  62.1  7760 g
Ex. Freezing Point Depression
Estimate the freezing point of a permanent type of antifreeze
solution made up of 100.0 g ethylene glycol, C2H6O2, (MM = 62.07)
and 100.0 g H2O (MM = 18.02).
T f  K f  msolute
msolute  mEG
nEG

wt EG / kg
nEG 
mEG
100 g

 1.611 mol
Mwt EG 62.07 g / mol
nEG
1.611
T f  K f 
 1.86
 30o C
wt EG / kg
0.10
T f  T fo  T f
T f  T fo  T f  0o C  30o C  30o C
50
Osmosis and Osmotic Pressure
A. Initially, Soln B separated from pure water, A, by osmotic
membrane (permeable to water). No osmosis occurred yet
B. After a while, volume of fluid in tube higher. Osmosis has
occurred.
51
Osmotic pressure (p): Pressure needed to stop the flow.
Flow of water
molecules
Net flow
Column rises
Pressure increases
Increase of flow from right to left
Finally:
Equilibrium established
Flow of water
molecules
Net flow = 0
Equation for Osmotic Pressure
• Assumes dilute solutions
p=iMRT
– p = osmotic pressure
– i = number of ions per formula unit
= 1 for molecules
– M = molarity of solution
• Molality, m, would be better, but M simplifies
• Especially for dilute solutions, where m  M
– T = Kelvin Temperature
– R = Ideal Gas constant
= 0.082057 L·atm·mol1K1
53
Eye drops must be at the same osmotic pressure as the human eye
to prevent water from moving into or out of the eye. A commercial
eye drop solution is 0.327 M in electrolyte particles. What is the
osmotic pressure in the human eye at 25°C?
p = MRT
T(K) = 25°C + 273.15
L  atm
p  0.327 M  0.08206
 298 K  8.00atm
K  mol
Using p to determine MM
The osmotic pressure of an aqueous solution of certain protein was
measured to determine its molar mass. The solution contained 3.50
mg of protein in sufficient H2O to form 5.00 mL of solution. The
measured osmotic pressure of this solution was 1.54 torr at 25 °C.
Calculate the molar mass of the protein.
 1atm 
1.54torr 

p
mol
 760torr 
M

 8.28 10 5
L  atm 
RT 
L
 0.08206
 298K
K  mol 



n  M V  8.28 105 M  5.00 103 L  4.14 107 mol
mass
3.50 10 3 g
3
Mwt 


8
.
45

10
g / mol
7
n
4.14 10 mol
solute
solute
p=iMRT
The observed osmotic pressure for a 0.10 M solution
of Fe(NH4)2(SO4)2 at 25oC is 10.8 atm. Compare the
expected and experimental values for i.
i=5