Properties of Solutions
Chapter 11
Solutions
. . . the components of a mixture
are uniformly intermingled
(the mixture is homogeneous).
Solution Composition
moles of solute
1. Molarity (M) =
liters of solution
mass of solute
(100%)
2. Mass (weight) percent =
mass of solution
molesA
3. Mole fraction (A) =
total moles in solution
moles of solute
4. Molality (m) =
kilograms of solvent
Molarity Calculations
 1.00 gethanol  1000mL  1mol 
  0.215M



 101mL  1L  46.07 g 
Mass % Calculations
 massethanol 
%C 2 H 5OH  
100% 
 masssolution 
 1.00 g 
100% 
 
 101.0 g 
 0.990%C 2 H 5OH
Mole Fraction




n
ethanol
MoleFractionethanol  nethanolnH 2 O 


2.17 x10  2 mol

 
2
 2.17 x10 mol  5.56mol 
 2.17 x10  2 

 
 5.58 
 0.00389
Molality Calculations
 1.00 gethanol  1000 g  1mol 



  0.217m
 100.0 g  1kg  46.07 g 
Molarity & Molality
For dilute solutions, molarity (M) and
molality(m) are very similar.
In previous example, M = 0.215 M and
m = 0.217 m.
Normality
 equivalentssolute 
Normality ( N )  

 Litersolut ion 
Acid-Base Equivalents = (moles) (total (+)
charge)
Redox Equivalents = (moles)(# e- transferred)
Normality Calculations
.250 M H3PO4 =______N
N = M(total(+) charge)
N = (0.250)(3)
N = 0.750 N H3PO4
Concentration & Density
Calculations
See Example 11.2 on pages 517518.
Know how to do this problem!!
Steps in Solution Formation
Step 1 - Expanding the solute (endothermic)
Step 2 - Expanding the solvent (endothermic)
Step 3 - Allowing the solute and solvent to
interact to form a solution (exothermic)
Hsoln = Hstep 1 + Hstep 2 + Hstep 3
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Step 1
Step 2
 H1
 H2
Solute
Expanded solute
Direct
formation
of solution
Step 3
Solvent
H soln
Expanded solvent
H3
Three steps of a liquid solution: 1) expanding the solute,
2) expanding the solvent, & 3) combining the expanded
solute and solvent to form the solution.
a) Hsoln is negative and solution process is exothermic.
b) Hsoln is positve and solution process is endothermic.
Processes that require large amounts of energy tend not to
occur. Solution process are favored by an increase in entropy.
Structure & Solubility
Like dissolves like.
Hydrophobic --water-fearing. Fat soluble
vitamins such as A, D, E, & K.
Hydrophilic --water-loving. Water soluble
vitamins such as B & C.
Hypervitaminosis--excessive buildup of
vitamins A, D, E, & K in the body.
Henry’s Law
The amount of a gas dissolved in a solution is
directly proportional to the pressure of the gas
above the solution.
P = kC
P = partial pressure of gaseous solute above
the solution
C = concentration of dissolved gas
k = a constant
11_274
300
Sugar
(C12 H 22 O 11 )
Solubility (g solute/100 g H2O)
260
KNO 3
220
180
140
NaNO 3
NaBr
100
KBr
Na 2 SO 4 KCl
60
20
0
Ce 2 (SO 4 )3
0
20
40
60
80
100
Temperature (C)
Solubility of several solids as a function of temperature.
The solubility of various gases at different
temperatures.
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Water
Vapor
Water
(a)
Aqueous
solution
Aqueous
solution
(b)
When an aqueous solution and pure water are in a closed
environment, the water is transferred to the solution
because of the difference in vapor pressure.
Raoult’s Law
The presence of a nonvolatile solute lowers
the vapor pressure of a solvent.
Psoln = solventPsolvent
Psoln = vapor pressure of the solution
solvent = mole fraction of the solvent
Psolvent = vapor pressure of the pure solvent
Raoult’s Law Calculations
Sample Exercise 11.6 on page 532.
Na2SO4 forms 3 ions so the number of moles
of solute is multiplied by three.
Psoln = waterPwater
Psoln = (0.929)(23.76 torr)
Psoln = 22.1 torr
11_279
Vapor pressure
of solution
Vapor pressure
Vapor pressure of pure B
Vap
o
Pa
rtia
l
r pr
ess
ure
of s
pr
olu
es
tion
su
r
eB
u
ess
l pr
a
i
t
Par
re A
A
B
(a)
Vapor pressure
of solution
Vapor pressure
of pure A
A
B
(b)
A
B
(c)
Vapor pressure for a solution of two volatile liquids.
a) Ideal(benzene & toluene) -- obeys Raoult’s Law,
b) Positive deviation (ethanol & hexane) from Raoult’s
Law, & c) Negative deviation (acetone & water).
Negative deviation is due to hydrogen bonding.
Liquid-Liquid Solutions
Ptotal = PA + PB
= APoA + BPoB
Raoult’s Law Calculations
Sample Exercise 11.7 on page 535.
A= nA/(nA+nC)
A= 0.100 mol/(0.100 mol + 0.100 mol)
A = 0.500  C = 0.500
Ptotal = APoA + CPoC
Ptotal = (0.500)(345 torr) + (0.500)(293 torr)
Ptotal = 319 torr
Colligative Properties
Depend only on the number, not on the
identity, of the solute particles in an ideal
solution.
- Boiling point elevation
- Freezing point depression
- Osmotic pressure
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atm
Pressure (atm)
Vapor pressure
of pure water
Vapor pressure
of solution
Freezing point
of water
Freezing
point of
solution
Boiling point
of water
Tf
Boiling point
of solution
T b
Temperature (C)
Phase diagrams for pure water and for an aqueous
solution containing a nonvolatile solute -- liquid range
is extended for the solution.
Boiling Point Elevation
A nonvolatile solute elevates the boiling point
of the solvent. The solute lowers the vapor
pressure of the solution.
T = Kbmsolutei
Kb = molal boiling point elevation constant
m = molality of the solute
i = van’t Hoff factor ( # ions formed)
Boiling Point Calculations
Sample Exercise 11.8 on page 537.
T = Kbmsolutei
msolute = T/(Kbi)
msolute = (0.34 Co)/[(0.51 Cokg/mol)(1)]
msolute = 0.67 mol/kg
Boiling Point Calculations
(Continued)
msolute = nsolute/ kgsolvent
nsolute = msolute kgsolvent
nsolute = (0.67 mol/kg)(0.1500 kg)
nsolute = 0.10 mol
Boiling Point Calculations
(Continued)
n = m/M
M = m/n
M = 18.00 g/0.10 mol
M = 180 g/mol
Freezing Point Depression
A nonvolatile solute depresses the freezing
point of the solvent. The solute interferes with
crystal formation.
T = Kfmsolutei
Kf = molal freezing point depression constant
m = molality of the solute
i = van’t Hoff factor ( # ions formed)
Freezing Point Calculations
Sample Exercise 11.10 on page 539.
T = Kfmsolutei
msolute = T/(Kfi)
msolute = (0.240 Co)/[(5.12 Cokg/mol)(1)]
msolute = 4.69 x 10-2 mol/kg
Freezing Point Calculations
(Continued)
msolute = nsolute/ kgsolvent
nsolute = msolute kgsolvent
nsolute = (4.69 x 10-2 mol/kg)(0.0150 kg)
nsolute = 7.04 x 10 -4 mol
Freezing Point Calculations
(Continued)
n = m/M
M = m/n
M = .546 g/7.04 x 10-4 mol
M = 776 g/mol
Osmotic Pressure
Osmosis: The flow of solvent into the
solution through the semipermeable
membrane.
Osmotic Pressure: The excess hydrostatic
pressure on the solution compared to the
pure solvent.
Due to osmotic pressure,
the solution is diluted by
water transferred through
the semipermeable
membrane. The diluted
solution travels up the
thistle tube until the
osmotic pressure is
balanced by the
gravitational pull.
Osmosis
The solute particles interfere with the
passage of the solvent, so the rate of
transfer is slower from the solution
to the solvent than in the reverse
direction.
a) The pure solvent travels at a greater rate into the solution than
solvent molecules can travel in the reverse direction.
b) At equilibrium, the rate of travel of solvent molecules in both
directions is equal.
Osmotic Pressure
 = MRT
 = osmotic pressure (atm)
M = Molarity of solution
R = 0.08206 Latm/molK
T = Kelvin temperature
Osmotic Pressure
Calculations
Sample Exercise 11.11 on pages 541-542.
 = MRT
M = /RT
M = (1.12 torr)(1 atm/760 torr)/[(0.08206
Latm/molK)(298K)]
M = 6.01 x 10-5 mol/L
Osmotic Pressure Calculations
Continued
Molar Mass = (1.00 x 10 -3g/1.00 mL)(1000
mL/1 L)(1 L/6.01 x 10-5 mol) =
1.66 x 104 g/mol protein
Crenation & Lysis
Crenation-solution in which cell is bathed is
hypertonic (more concentrated)-cell
shrinks. Pickle, hands after swimming in
ocean. Meat is salted to kill bacteria and
fruits are placed in sugar solution.
Lysis-solution in which cell is bathed is
hypotonic (less concentrated)-cell
expands. Intravenous solution that is
hypotonic to the body instead of isotonic.
If the external pressure is larger
than the osmotic pressure, reverse
osmosis occurs.
One application is desalination of
seawater.
Colligative Properties of
Electrolyte Solutions
van’t Hoff factor, “i”, relates to the number of
ions per formula unit.
NaCl = 2, K2SO4 = 3
moles of particles in solution
i =
moles of solute dissolved
T = mKi
 = MRTi
Electrolyte Solutions
The value of i is never quite what is expected
due to ion-pairing. Some ions stay linked
together--this phenomenon is most
noticeable in concentrated solutions.
Osmotic Pressure Calculation for
Electrolyte
Sample Exercise 11.13 on page 548.
Fe(NH4)2(SO4)2 produces 5 ions.
 = MRTi
i  /MRT
i = 10.8 atm/[(0.10 mol/L)(0.08206
Latm/molK)(298 K)]
i = 4.4
Colloids
Colloidal Dispersion (colloid): A suspension
of tiny particles in some medium.
aerosols, foams, emulsions, sols
Coagulation: The addition of an electrolyte,
causing destruction of a colloid. Examples
are electrostatic precipitators and river deltas.
The eight types of colloids and examples of each.
Tyndall Effect
The scattering of light by particles of a colloid
is called the Tyndall Effect. Which of the
glasses below contains a colloid?
Calorimeter Problem
Add this problem to the Chapter 11 set of
problems. KNOW how to work this
problem--show the appropriate formula!!
When 8.50 g of sodium nitrate is dissolved in
600. g of water, the temperature of the
solution rises 0.817 Co. What is the molar
heat of solution for sodium nitrate?