Properties of Solutions
Chapter 11
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1
Overview
 Introduce student to solution composition and
energy of solution formation.
 Factor affecting solubilities will be discussed:
structure, pressure and temperature effects.
 Vapor pressure of solutions, boiling point,
freezing point affected by solute addition.
 Colligative properties of electrolyte solutions and
colloids.
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2
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
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3
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.
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4
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
4. Molality (m) =
moles of solute
kilograms of solvent
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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
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Concentration Units Continued
Molarity (M)
M =
moles of solute
liters of solution
Molality (m)
m =
moles of solute
mass of solvent (kg)
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7
Normality (N)
N =
Number of equivalent
mass
1
x
=
liters of solution
V
Equivalent mass
Number of equivalent
Acid-Base reaction: is the amount needed to accept one mole of H+
Equivalent mass of H2SO4 =
MM
2
MM
Equivalent mass of Ca(OH)2 =
2
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Redox reaction : is the amount needed to accept exactly one mole e-
MnO4- + 5e- + 8H+
Equivalent mass of KMnO4 =
Mn2+ + 4H2O
MM
5
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S.Ex 11.1
• 1.0 g C2H5OH dissolved in 100.0 g water and the
final volume of solution is 101 mL.
• Calculate M, mass%, mole fraction, molality
• Moles of C2H5OH = 1.00/46.07 = 2.17x10-2
• M = 2.17x10-2 mol/0.101L = 0.215 M
• Mass % = 1.00 x100% = 0.990 % C2H5OH
100+1
Mols of H2O = 100/18.0 = 5.56
χ C2H5OH = 2.17x10-2 / (2.17x10-2 + 5.56) = 0.00389
• m = 2.17x10-2mol/0.1000 kg = 0.217 m
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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
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12
S.Ex 11.2
• Lead storage battery has 3.75 M sulfuric acid
solution with a density of 1.230 g/mL.Calculate
mass %, molality, and Normality.
• Consider 1 L solution, …. 1230 g mass of solution
• mass of H2SO4 = 3.75 x 98.1 = 368 g
• mass of water = 1230 – 368 = 862 g
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13
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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The formation of a liquid solution can be divided into three steps:
(1) expanding the solute, (2) expanding the solvent, and (3)
combining the expanded solute and solvent to form the solution.
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Figure 11.2 a&b (a) Exothermic
and (b) Endothermic
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H < 0
Exothermic
Solution will occur
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17
The driving factor that favor a process of solution
formation is an increase in disorder
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“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)
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The molecular structures of (a) vitamin A (nonpolar, fat-soluble)
and (b) vitamin C (polar, water-soluble). The circles in the
structural formulas indicate polar bonds. Note that vitamin C
contains far more polar bonds than vitamin A.
Fat Soluble
Water Soluble
Hydrophobic
Hydrophilic
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21
Ex 11.3
• Liquid hexane C6H14, or liquid methanol
CH3OH is the most appropriate solvent for
grease C20H42 and KI
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Hexane,
Liquid
Methanol,
Grease
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23
Factor Affecting Solubility
1. Structure
2. Pressure
3. Temperature
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Pressure Effects
• Pressure has little effect on solids and
liquids.
• It increases the solubility of gases.
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(a) A gaseous solute in equilibrium with a solution. (b) The piston is pushed in,
increasing the pressure of the gas and number of gas molecules per unit volume. This
causes an increase in the rate at which the gas enters the solution, so the concentration
of dissolved gas increases. (c) The greater gas concentration in the solution causes an
increase in the rate of escape. A new equilibrium is reached.
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Henry’s Law
The amount of a gas dissolved in a solution is
directly proportional to the pressure of the gas
above the solution.
C = kP
P = partial pressure of gaseous solute above
the solution
C = concentration of dissolved gas
k = a constant
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Henry’s Law Applied for
• Dilute solutions
• Gases that do not dissociate or react with
solvent:
– O2/water
– HCl/water
Applied
Is not applied
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S.Ex 11.4
• A certain soft drink bottle at 25 oC contains CO2 gas at 5.0 atm
over the liquid. Assuming partial pressure of CO2 in the
atmosphere is 4.0x10-4 atm, calculate the equilibrium
concentration of CO2 before and after the bottle is opened.
• Kco2 = 3.1x10-2 mol/L.atm at 25 oC
» CCO2 = KCO2. PCO2
Unopened bottle; CCO2 = 3.1x10-2 mol/L.atm x 5atm
= 0.16 mol/L
Opened bottle; CCO2 = 3.1x10-2 mol/L.atmx 4.0x10-4 atm
= 1.2x10-5 mol/L
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Temperature Effect on Gases
The solubilities of
several gases in
water as a function
of temperature at a
constant pressure
of 1 atm of gas
above the solution.
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Temperature Effect on Solids
The solubilities
of several solids
as a function of
temperature.
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The Vapor Pressure of Solutions
• Solutions have different physical properties
from pure solvent.
• Solutions of nonvolatile solutes differ from
solutions of volatile solvents.
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An aqueous solution and pure water in a closed
environment. (a) Initial stage. (b) After a period of time,
the water is transferred to the solution.
Vapor pressure of pure water is higher
VP of solution is lower
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Raoult’s Law
The presence of a nonvolatile solute lowers
the vapor pressure of a solvent.
Psoln = solvent Psolvent
Psoln = vapor pressure of the solution
solvent = mole fraction of the solvent
Psolvent = vapor pressure of the pure solvent
solvent
=
Moles of solvents
Moles of solvent + Moles of solute
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For a solution
that obeys
Raoult's law, a
plot of Psoln
versus xsolvent
gives a straight
line.
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Note
• Solutions that obey Raoult’s Law are called
Ideal Solutions.
• Can be used to determine the molar mass of
unknown.
• For ionic compounds you should multiply
by the total number of ions per molecule
• e.g. Na2SO4
n = 3 x Na2SO4
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11.5
• 158 g sugar (MM=342.3 g/mol) dissolved in 643.5 ml
of water. At 25 oC, density of water= 0.9971 g/ml and
v.p. = 23.76 torr.
• Calculate v.p. of the sugar solution.
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11.6
• 35.0 g solid Na2SO4 (MM= 142 g/mol) in
175 g water at 25 oC. Po(water). = 23.76 torr
Calc v.p. of solution.
• The lowering of v.p. depends on the number of
solute particles present in the solution.
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Non-Ideal Solutions
When a solution
contains two volatile
components, both
contribute to the total
vapor pressure.
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PA = XA P A0
PB = XB P 0B
PT = PA + PB
PT = XA P A0 + XB P 0B
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(a) ideal liquid-liquid solution by Raoult's law. (b) This solution
shows a positive deviation from Raoult's law. (c) This solution
shows a negative deviation from Raoult's law.
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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
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11.7
• 5.81 g acetone
+ 11.9 g Chloroform
• MM=58.1
• Po = 345 torr
• n = 0.100 mol
MM = 119.4
Po = 293 torr
n = 0.100 mol
• At 35 oC, total V.P. of solution = 260 torr
• Is this an ideal solution ?
• Xac = 0.5
Xch = 0.5
• Raoult’s Law; PT = XA PA + XB PB
• Ptotal = 0.5 x 345 torr + 0.5 x 293 torr = 319 torr
• 260 torr < 319 (ideal) ;
Negative deviation
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Colligative Properties of Non-Electrolyte
Solutions
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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Boiling Point Elevation
A nonvolatile solute elevates the boiling
point of the solvent.
T = Kbmsolute
Kb = molal boiling point elevation constant
m = molality of the solute
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Boiling-Point Elevation
Tb = Tb – T b0
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)
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11.8
18.00 g glucose in 150 g water. B.P. = 100.34 oC
Calc molar mass of glucose
ΔT = Kbmsolute
Kb = 0.51 for water
ΔT = 100.34 – 100.00 = 0.34 oC
msolute= ΔT /Kb = 0.34 oC
= 0.67 mol/kg
0.51 oC.kg/mol
msolute = nglucose/0.1500 kg
nglucose = 0.67 mol/kg x 0.1500 kg = 0.10 mol
MM = mass/mol = 18.00g / 0.10 mol = 180 g/mol
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Freezing Point Depression
A nonvolatile solute depresses the freezing
point of the solvent.
T = Kfmsolute
Kf = molal freezing point depression constant
m = molality of the solute
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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
Tf > 0
T 0f > Tf
Tf = Kf m
m is the molality of the solution
Kf is the molal freezing-point
depression constant (0C/m)
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• S.Ex. 11.9
• •What mass of glycol(MM=62.1 g/mol) must be
added to 10.0 L of water to produce a solution that
freezes at -23.3 oC? (d =1 g/ml)
• ∆T = 0.0 –(-23.3) = 23.3 oC= Kfmsolute
• msolute = 23.3 oC
= 12.5 mol/kg
1.86oC.kg/mol
• n = 12.5 mol/kg x 10.0 kg = 1.25x102mol
• Mass of glycol = 1.25x102mol x 62.1g/mol
=7.76x103g = 7.76 kg
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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.
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A tube with a bulb
on the
end that is covered
by a
semipermeable
membrane.
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Osmotic Pressure (p)
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 (p) is the pressure required to stop osmosis.
p = MRT
M is the molarity of the solution
R is the gas constant
T is the temperature (in K)
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• Ex 11.1
• 1.00x10-3 g of protein dissolved to make 1.00
mL solution. Osmotic pressure was found to be
1.12 torr at 25 oC. Calculate the MM of protein.
• π = MRT =1.12 torr/760 = 1.47x10-3 atm
• R = 0.08206 L atm/K.mol
• T = 25.0 + 273 = 298 K
• M = 1.47x10-3 atm
= 6.01x10-5 mol/L
(0.08206 L.atm/K.mol)(298 K)
• 1.00x10-3 g in 1 ml = 1.00 g /L
• MM = 1.00 g
x
L
–L
= 1.66x104 g/mol
6.01x10-5 mol
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(a) A pure solvent and its solution (containing a nonvolatile solute) are
separated by a semipermeable membrane through which solvent molecules
(blue) can pass but solute molecules (green) cannot. The rate of solvent transfer
is greater from solvent to solution than from solution to solvent. (b) The system
at equilibrium, where the rate of solvent transfer is the same in both directions.
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The normal flow of solvent
into the solution (osmosis)
can be prevented
by applying an external
pressure
to the solution. The
minimum pressure required
to stop the osmosis is equal
to the osmotic pressure of
the solution.
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If the external pressure is larger
than the osmotic pressure, reverse
osmosis occurs.
One application is desalination of
seawater.
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Reverse osmosis.
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A cell in an:
isotonic
Solution
(identical p)
hypotonic
Solution
(water in)
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hypertonic
Solution
(water out)
61
S. Ex. 11.12
• What concentration of NaCl in water is
needed to produce an aqueous solution
isotonic with blood (π = 7.70 atm at 25 oC)
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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 = imK
p = iMRT
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Colligative Properties of Electrolyte Solutions
Boiling-Point Elevation
Tb = i Kb m
Freezing-Point Depression
Tf = i Kf m
Osmotic Pressure (p)
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p = iMRT
64
In an aqueous solution a few ions
aggregate, forming ion pairs that
behave as a unit: Ion Pairing
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S. Ex 11.13
• The observed osmotic pressure for a 0.10 M
solution of Fe(NH4)2(SO4)2 at 25 oC is 10.8
atm. Compare the expected and
experimental value of i.
• Fe(NH4)2(SO4)2 → Fe2+ + 2NH4+ + 2SO42-
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The Cleansing Action of Soap
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The Cottrell
precipitator installed
in a smokestack. The
charged plates attract
the colloidal particles
because of their ion
layers and thus
remove them from
the smoke.
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