Solutions and Chemical
Equilibrium
Preparation for College Chemistry
Columbia University
Department of Chemistry
Chapter Outline
Concentration of Solutions
Colligative Properties
Osmosis and Osmotic Pressure
Chemical Equilibrium
Ion Product of Water
Types of Solutions
Phase
Solute
Solvent
Example
Gas
gas
gas
air
Liquid
Liquid
Liquid
Solid
gas
liquid
solid
gas
liquid
liquid
liquid
solid
Coke
antifreeze
Coke
H2 in Pt
Solid
solid
solid
alloys
Concentration of Solutions
Units
Mass percent
Symbol
Definition
% m/m
(msolute/msolution ) x100
Volume percent
Mass/volume percent
% v/v
% m/v
(vsolute/vsolution ) x100
(msolute/vsolution ) x100
Parts per million
Parts per billion
ppm
ppb
mgsolute/Lsolution
µgsolute/Lsolution
M
molsolute/Lsolution
molsolute/kgsolvent
Molarity
Molality
m
Mass % Solute
Mass % solute =
ppm solute =
mass solute
Total mass solution
mass solute
total mass solution
x 100
x 10 6
When concentration is so low that the d ~ dwater:
ppm solute(aqueous solutions) = mg solute / Lsolution
ppb solute = mass % x 10 9
ppb solute (aqueous solutions) = µg solute / Lsolution
http://pubs.acs.org:80/hotartcl/est/99/oct/oct-news5.html
Molarity
moles solute
=
[solute] = M =
Liter solution
mol
L
What volume of a 0.035 M AgNO3 solution can
be made from 5.0 g AgNO3 ?
L
5.0 g AgNO3 x 1 mole AgNO3 x
= 840 mL
169.91g
0.035 mole AgNO3
Dilution Equation
Ci Vi = Cf Vf
C ,mol  L V , L  C  V ,mol  chemical amount of solute
i
i
i
i
1
Only solvent is added
250mL
Cf = Ci Vi
Vf
Preparing a dilute solution of specified concentration
Raoult’s Law
0
P1  X1 P1
0
P1
n1
X1 
n1  n2  n3  ...
P1
Positive
Negative
Ideal
0
X1
1
Basis for four properties
of DILUTE SOLUTIONS
Colligative Properties
Depend on the concentration of solute species
and not on its nature
 Freezing point depression. Kf (°C kgsolvent mol -1solute)
 Boiling point elevation Kb (°C kgsolvent mol -1solute)
 Vapor-pressure lowering (atm)
 Osmotic Pressure (atm)
Vapor-pressure of liquids
Pressure exerted by a vapor in equilibrium with its liquid
For water:
Atmospheric pressure
800
600
boiling point
Vapor pressure (torr)
1000
400
200
0
20
40
60
80
100
Temperature(°C)
120
Vapor-pressure lowering
For a two component system : solvent 1, solute 2:
X1  1 X2
Raoult Law:
P1  X1 P1
0
The vapor pressure lowering is
0
1
0
1 1
0
1
0
2 1
P1  P1  P  X P  P   X P
The change in vapor pressure of the solvent is
proportional to the mole fraction of the solute (< 0)
Boiling Point Elevation (∆Tb)
Vapor Pressure lowering (∆P1)
P0solvent
∆Tb
Solvent vapor pressure
1 atm
∆P1
P0solution
Tb T’b
Temperature
∆Tb and ∆Tf
∆Tb =Tb’ - Tb = Kbm
∆Tf = Tf ’ - Tf = -Kfm
mol solute
mol solute
m

kg solvent 1000g solvent
∆Tb = b.p. elevation
Kb = b.p. elevation constant
∆Tf = f.p. depression
Kf = f.p. depression constant
Kf and Kb (°C kgsolvent mol -1solute)
Solvent
m.p (°C)
Kf
b.p.(°C) Kb
Water
0.00
1.86
100.0
0.512
Acetic acid
16.6
3.90
118.5
3.07
Benzene
5.5
5.1
80.1
2.53
Camphor
178
40
208.1
5.95
Osmosis and Osmotic Pressure, p
Jacobus van’t Hoff in 1887
p  cRT
c = M; R = universal gas constant; T absolute temperature
Pure water
Solution
Semipermeable membrane
Chemical Equilibrium
T
2NO2
N 2O 4
T
N2O4
2NO2
N2O4
2NO2
REVERSIBLE REACTION:
forward
PRODUCTS
REACTANTS
reverse
Kinetics. Rates of Reaction
Reaction rate
RATEforward
A+B
C+D
RATEforward = RATEreverse
Equilibrium
RATEreverse
C+D
A+B
Time
Chemical Equilibrium
Saturated solution
NaCl (s)
Na+(aq) + Cl -(aq)
Weak electrolyte dissociation
HC2H3O2 (aq)
H3O+(aq) + C2H3O2 -(aq)
Complex ion formation
Fe 3+ (aq) + SCN - (aq)
Fe(SCN)2+ ( aq)
Le’Chatelier’s Principle
“A system in equilibrium that is subjected to a stress
will react in ways that counteract the stress”
Four ways to stress a chemical system:
• Concentration Change
• Volume Change
• Temperature Change
• Presence of a Catalyst
Equilibrium Constants, Keq
aA + bB
Keq =
aA(g) + bB(g)
Keq =
cC + dD
[C] c [D] d
Kc
[A] a[B] b
cC(g) + dD(g)
(PC) c (PD) d
Kp
(PA) a (PB) b
LAW OF MASS ACTION. Guldberg and Waage. 1867
Reaction quotient
Reaction Quotient, Q
RATEforward
Q =
Q>K
[C] c [D] d
[A] a[B] b
Q=K
Q<K
Keq =
[C] c [D] d
[A] a[B] b
RATEreverse
Time
Writing Equilibrium Constants
• Gases enter equilibrium expressions as partial
pressures in atm
• Dissolved species enter as concentrations in M
• Pure solids and pure liquids are represented by the
number 1 (unity)
• A solvent in a chemical reaction is represented by 1,
provided the solution is diluted
Ion Product Constant for Water, KW
Autoionization of Water
H3O+(aq) + OH -(aq)
H2O + H2O
Keq =
[H2O] =
1kg H2O
1L H2O
[H3O +] [OH -]
[H2O] 2
x
103 g H2O
1kg H2O
x
1 mol H2O
18 g H2O
= 55.5 M
Keq [H2O] 2 = Kw = [H3O +] [OH -] = 1x 10 -14
[H3O +] = [OH -] = 1x 10 -7 M