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Physical Properties of Solutions
Chapter 12
A solution is a homogenous mixture of 2 or
more substances
The solute is(are) the substance(s) present in the
smaller amount(s)
The solvent is the substance present in the larger
amount
12.1
A saturated solution contains the maximum amount of a
solute that will dissolve in a given solvent at a specific
temperature.
An unsaturated solution contains less solute than the
solvent has the capacity to dissolve at a specific
temperature.
A supersaturated solution contains more solute than is
present in a saturated solution at a specific temperature.
Sodium acetate crystals rapidly form when a seed crystal is
added to a supersaturated solution of sodium acetate.
12.1
Three types of interactions in the solution process:
• solvent-solvent interaction
• solute-solute interaction
• solvent-solute interaction
Hsoln = H1 + H2 + H3
12.2
“like dissolves like”
Two substances with similar intermolecular forces are likely
to be soluble in each other.
•
non-polar molecules are soluble in non-polar solvents
CCl4 in C6H6
•
polar molecules are soluble in polar solvents
C2H5OH in H2O
•
ionic compounds are more soluble in polar solvents
NaCl in H2O or NH3 (l)
12.2
The Concentration of a solute can be
quantitatively expressed in several ways:
 Molarity
 Mass Percentage of solute
 Molality
 Mole Fraction
Solve this?
• A 650.0 ml solution contains 19.6 g H2SO4.
Calculate molarity and molality of the
solution
• Determine the number of equivalents of
H2SO4
Concentration Units
The concentration of a solution is the amount of solute
present in a given quantity of solvent or solution.
Percent by Mass
mass of solute
x 100%
% by mass =
mass of solute + mass of solvent
mass of solute x 100%
=
mass of solution
Mole Fraction (X)
moles of A
XA =
sum of moles of all components
12.3
Concentration Units Continued
Molarity (M)
M =
moles of solute
liters of solution
Molality (m)
m =
moles of solute
mass of solvent (kg)
12.3
What is the molality of a 5.86 M ethanol (C2H5OH)
solution whose density is 0.927 g/mL?
moles of solute
moles of solute
m =
M =
mass of solvent (kg)
liters of solution
Assume 1 L of solution:
5.86 moles ethanol = 270 g ethanol
927 g of solution (1000 mL x 0.927 g/mL)
mass of solvent = mass of solution – mass of solute
= 927 g – 270 g = 657 g = 0.657 kg
moles of solute
m =
mass of solvent (kg)
=
5.86 moles C2H5OH
= 8.92 m
0.657 kg solvent
12.3
Problem #1
A 3.0 g sample of ground water contains 3.5 ppm of
Arsenic ion. Calculate;
(A) How many grams Arsenic ion are in this
sample? What is the weight percentage of
Arsenic ion?
(B) If 0.25 L aqueous solution with density of 1.00
g/ml contains 13.7 g of pesticide, express the
(C) concentration in ppm and ppb.
Temperature and Solubility
Solid solubility and temperature
solubility increases with
increasing temperature
solubility decreases with
increasing temperature
12.4
Fractional crystallization is the separation of a mixture of
substances into pure components on the basis of their differing
solubilities.
Suppose you have 90 g KNO3
contaminated with 10 g NaCl.
Fractional crystallization:
1. Dissolve sample in 100 mL of
water at 600C
2. Cool solution to 00C
3. All NaCl will stay in solution
(s = 34.2g/100g)
4. 78 g of PURE KNO3 will
precipitate (s = 12 g/100g).
90 g – 12 g = 78 g
12.4
Temperature and Solubility
O2 gas solubility and temperature
solubility usually
decreases with
increasing temperature
12.4
Pressure and Solubility of Gases
The solubility of a gas in a liquid is proportional to the
pressure of the gas over the solution (Henry’s law).
c is the concentration (M) of the dissolved gas
c = kP
P is the pressure of the gas over the solution
k is a constant for each gas (mol/L•atm) that
depends only on temperature
low P
high P
low c
high c
12.5
Chemistry In Action: The Killer Lake
8/21/86
CO2 Cloud Released
1700 Casualties
Trigger?
•
earthquake
•
landslide
•
strong Winds
Lake Nyos, West Africa
Colligative Properties of Nonelectrolyte Solutions
Colligative properties are properties that depend only on the
number of solute particles in solution and not on the nature of
the solute particles.
Vapor-Pressure Lowering
P1 = X1 P
0
1
Raoult’s law
P 10 = vapor pressure of pure solvent
X1 = mole fraction of the solvent
If the solution contains only one solute:
X1 = 1 – X2
P 10 - P1 = P = X2 P 10
X2 = mole fraction of the solute
12.6
Ideal Solution
PA = XA P A0
PB = XB P 0B
PT = PA + PB
PT = XA P A0 + XB P B0
12.6
PT is greater than
predicted by Raoults’s law
PT is less than
predicted by Raoults’s law
Force
Force
Force
< A-A & B-B
A-B
Force
Force
Force
> A-A & B-B
A-B
12.6
Solve this problem
• . Dimethyl gloxime, DMG, is an organic
molecule used to test for aqueous nickel (II) ions.
A solution prepared by dissolving 65.0 g of DMG
in 375 g of ethanol boils at 80.3 oC.
•
What is the molar mass of DMG?
•
Kb = 1.22 oC/m, and boiling point
of pure ethanol = 78.5 oC.
Problem
• A solution is formed by dissolving 10.0 g of KCl in 500.0
g of H2O.
•
(a) What is the vapor pressure of the solution at
25 oC if the vapor pressure of pure H2O is
•
23.8 mmHg at 25 oC?
•
(b) What is the boiling-point elevation? Kb
for water is 0.52 oC/m.
•
(c) What is the freezing-point depression?
Kf for water is 1.86 oC/m.
•
(d) What is the osmotic pressure of the
solution? Assume that the volume of the solution is
•
0.500 L.
Fractional Distillation Apparatus
12.6
Boiling-Point Elevation
Tb = Tb – T 0b
T b0 is the boiling point of
the pure solvent
T b is the boiling point of
the solution
Tb > T b0
Tb > 0
Tb = Kb m
m is the molality of the solution
Kb is the molal boiling-point
elevation constant (0C/m)
for a given solvent
12.6
Freezing-Point Depression
Tf = T 0f – Tf
T
0
Tf
f
is the freezing point of
the pure solvent
is the freezing point of
the solution
T 0f > Tf
Tf > 0
Tf = Kf m
m is the molality of the solution
Kf is the molal freezing-point
depression constant (0C/m)
for a given solvent
12.6
12.6
What is the freezing point of a solution containing 478 g
of ethylene glycol (antifreeze) in 3202 g of water? The
molar mass of ethylene glycol is 62.01 g.
Tf = Kf m
Kf water = 1.86 0C/m
moles of solute
m =
mass of solvent (kg)
478 g x
1 mol
62.01 g
=
= 2.41 m
3.202 kg solvent
Tf = Kf m = 1.86 0C/m x 2.41 m = 4.48 0C
Tf = T 0f – Tf
Tf = T 0f – Tf = 0.00 0C – 4.48 0C = -4.48 0C
12.6
Problem#2
• An Antifreeze mixture is 11.0 molal
ethylene glycol solution. Water is the
solvent and it
•
has a mass of 800 g. What is the
mass of the solution? The molar mass of
ethylene glycol
•
is 62.0 g/mol
Osmotic Pressure ()
Osmosis is the selective passage of solvent molecules through a porous
membrane from a dilute solution to a more concentrated one.
A semipermeable membrane allows the passage of solvent molecules but
blocks the passage of solute molecules.
Osmotic pressure () is the pressure required to stop osmosis.
dilute
more
concentrated
12.6
Osmotic Pressure ()
High
P
Low
P
 = MRT
M is the molarity of the solution
R is the gas constant
T is the temperature (in K)
12.6
A cell in an:
isotonic
solution
hypotonic
solution
hypertonic
solution
12.6
Colligative Properties of Nonelectrolyte Solutions
Colligative properties are properties that depend only on the
number of solute particles in solution and not on the nature of
the solute particles.
Vapor-Pressure Lowering
P1 = X1 P 10
Boiling-Point Elevation
Tb = Kb m
Freezing-Point Depression
Tf = Kf m
Osmotic Pressure ()
 = MRT
12.6
Colligative Properties of Electrolyte Solutions
0.1 m Na+ ions & 0.1 m Cl- ions
0.1 m NaCl solution
Colligative properties are properties that depend only on the
number of solute particles in solution and not on the nature of
the solute particles.
0.1 m NaCl solution
van’t Hoff factor (i) =
0.2 m ions in solution
actual number of particles in soln after dissociation
number of formula units initially dissolved in soln
i should be
nonelectrolytes
NaCl
CaCl2
1
2
3
12.7
Colligative Properties of Electrolyte Solutions
Boiling-Point Elevation
Tb = i Kb m
Freezing-Point Depression
Tf = i Kf m
Osmotic Pressure ()
 = iMRT
12.7
A colloid is a dispersion of particles of one substance
throughout a dispersing medium of another substance.
Colloid versus solution
•
collodial particles are much larger than solute molecules
•
collodial suspension is not as homogeneous as a solution
12.8
The Cleansing Action of Soap
12.8
Chemistry In Action:
Desalination
Problem
• What is the vapor pressure of a solution
made by dissolving 24.6 g of camphor
(C10H16O) in 98.5 g of
•
Benzene. Vapor pressure of
benzene is 100.0 mmHg. (Camphor is a
low-volatility solid).