General Chemistry
Principles and Modern Applications
Petrucci • Harwood • Herring
8th Edition
Chapter 14 (13 in 9th Ed.): Solutions and Their
Physical Properties
Chem 1050
Chapter 14 Solutions
Dr. A. Ghumman
Contents
14-1
14-2
14-3
14-4
14-5
14-6
14-7
14-8
14-9
Types of Solutions: Some Terminology
Solution Concentration
Intermolecular Forces and the Solution Process
Solution Formation and Equilibrium
Solubilities of Gases
Vapor Pressure of Solutions
Osmotic Pressure
Freezing-Point Depression and Boiling-Point Elevation of
Nonelectrolyte solutions.
Solutions of Electrolytes
Focus on Chromatography
Chem 1050
Chapter 14 Solutions
Dr. A. Ghumman
Types of Solution:
Some Terminology
• Solutions: (solvent +solute)- homogeneous mixtures.
– Uniform throughout.
• Solvent.
– Determines the state of matter in which solution exists.
– the largest component of solution.
• Solute.
– Other solution component said to be dissolved in the
solution.
• Concentrated and dilute solution
• Gaseous and solid solutions
Chem 1050
Chapter 14 Solutions
Dr. A. Ghumman
1
Common solutions
Chem 1050
Chapter 14 Solutions
Dr. A. Ghumman
Solution Concentration.
• Concentration: a measure of quantity of solute in a given
quantity of solvent (or solution).
– Mass %
(m/m) widely used in industry
– Volume %
(v/v)
– e.g. 25% CH3OH solution is 25.0mLCH3OH for every
100mL of aqueous solution.
– Mass/volume percent (m/v): used in medicine and
pharmacy
• Isotonic saline is prepared by dissolving 0.9 g of NaCl in
100 mL of water solution and is said to be 0.9% NaCl
(mass/volume)
Chem 1050
Chapter 14 Solutions
Dr. A. Ghumman
Solution Concentration.
10% ethanol solution (V/V)
Chem 1050
Chapter 14 Solutions
Dr. A. Ghumman
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ppm, ppb and ppt
•
Very low solute concentrations are expressed as:
ppm: parts per million
ppm
ppb: parts per billion
ppt: parts per trillion
0.001g/1000g solution
µg/g = mg/L = 1x 10-3 g/1000g
ng/g = µg/L = 1x 10-6 g/1000g
pg/g = ng/L = 1x 10-9 g/1000g
* Note that for very dilute solutions, density of solution is equal to the
density of water so one liter of solution is equal to 1000g (1.0 Kg) of
solution.
1.0 L x 1.0 g/mL = 1000 g
ppm, ppb, and ppt are m/m or v/v.
Chem 1050
Chapter 14 Solutions
Dr. A. Ghumman
Mole Fraction and Mole Percent
χi=
Amount of component i (in moles)
Total amount of all components (in moles)
χ1 + χ2 + χ3 + …χn = 1
Mole % i = χi x 100%
Chem 1050
Chapter 14 Solutions
Dr. A. Ghumman
Molarity and Molality
Molarity (M) = Amount of solute (moles)
Volume of solution (liters)
Molality (m) = Amount of solute (moles)
Mass of solvent (kilograms)
Chem 1050
Chapter 14 Solutions
Dr. A. Ghumman
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Practice Example B
• A 11.3 mL solution of methanol(d=0.793 g.mL-1) is
dissolved in enough water to produce 75.0 mL of a
solution with a density of 0.980 g.mL-1. What is the
solution concentration expressed as
• (a) Volume % of methanol
• (b) Mass % of methanol
• (c) mass/volume%
• (d) mole fraction of H2O
• (e) molarity of CH3OH
• (f) molality of CH3OH?
Chem 1050
Chapter 14 Solutions
Dr. A. Ghumman
Intermolecular Forces and the Solution Process
∆Hb
∆Ha
∆Hc
∆Hsoln = ∆Ha + ∆Hb+ ∆Hc
Chem 1050
Chapter 14 Solutions
Dr. A. Ghumman
Intermolecular Forces in Mixtures
• Magnitude of ∆Ha, ∆Hb, and ∆Hc depend on
intermolecular forces.
• Ideal solution;
– Forces are similar between
all combinations of components.
•
-∆Hc = ∆Ha+ ∆Hb
∆Hsoln = 0
Chem 1050
Chapter 14 Solutions
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Ideal Solution
“Like dissolves like”
Chem 1050
Chapter 14 Solutions
Dr. A. Ghumman
Non-ideal Solutions
• Adhesive Forces: between different kind of molecules
• Cohesive forces: between like molecules
1) Adhesive forces are greater
than cohesive forces
e.g. CHCl3 and CH3COCH3
∆Hsoln < 0 (exothermic)
Chem 1050
Chapter 14 Solutions
Dr. A. Ghumman
Non-ideal Solutions
2) Adhesive forces are less than cohesive forces.
Complete mixing may occur e.g.
∆Hsoln > 0
• Endothermic
CH3COCH3
CS2
3) Adhesive forces are very very weak than cohesive forces.
• The components remain segregated e.g. H2O and octane(C8H18)
– At the limit these solutions are heterogeneous.
Chem 1050
Chapter 14 Solutions
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Ionic Solutions
•Ions surrounded
by cluster of water
molecules.
•Hydration of ions.
•Ion-dipole forces.
Chem 1050
Chapter 14 Solutions
Dr. A. Ghumman
Hydration Energy
Hypothetical three step process to describe
dissolution of NaCl
1.
Dissociation NaCl(s) → Na+(g) + Cl-(g)
2.
Hydration
Na+(g) + x H2O(l) → Na+(aq)
∆Hhydration < 0
3.
Hydration
Cl-(g) + x H2O(l) → Cl-(aq)
∆Hhydration < 0
∆Hlattice > 0
∆Hsoln = ∆H1 + ∆H2 + ∆H3 = +5 kJ mol-1
∆Hsoln > 0
Chem 1050
but Entropy ∆Gsolution < 0
(Chem 1051)
Chapter 14 Solutions
Dr. A. Ghumman
Solution Formation and Equilibrium
Formation of Saturated solution.
Solubility: the concentration of the saturated solution is called
solubility.
Chem 1050
Chapter 14 Solutions
Dr. A. Ghumman
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Solubility Curves
•Solubility as a function of
temperature.
Supersaturated
Unsaturated
•We can obtain a supersaturated
solution by cooling a solution.
•Supersaturated solution
Contains quantity of solute
greater than in a saturated
solution.
•Unstable
Chem 1050
Chapter 14 Solutions
Dr. A. Ghumman
Solubility of gases; Effect of Temperature
•
For most gases (N2,O2 and air) solubility in water decreases as
temperature increases.
• In organic solvents the reverse is
often true ( gases may become
more soluble at higher temperature).
For Noble gases solubility
curves are complex.
• Dissolved air is released as water is heated,
even well below the boiling point of water.
Chem 1050
Chapter 14 Solutions
Dr. A. Ghumman
Solubility of gases; Effect of pressure
• Henry’s Law: Solubility of a gas increases with
increasing pressure.
C = k Pgas
C = solubility of a gas in a particular solvent at fixed
temperature.
Pgas= partial pressure of the gas above the solution
k= Henry’s law constant (units) mol. L-1.atm-1
Chem 1050
Chapter 14 Solutions
Dr. A. Ghumman
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Solubility of gases; Effect of pressure
• Henry’s Law
C = k Pgas
The aqueous solubility of N2 at 0C and 1.0 atm is 23.54 mL N2 L-1and k
can be determined as follows
C
23.54 mL/L = 23.54 ml N .L-1.atm-1
2
=
Pgas
1.00 atm
To increase the solubility of N2 to 100.0 mL at 0C, pressure has to
k=
be increased to
Pgas =
C
k
=
100 mL N2 L-1
= 4.25 atm
23.54 mL N2 .L-1.atm1
Chem 1050
Chapter 14 Solutions
Dr. A. Ghumman
Henry’s Law
•Why does a freshly opened
can of soda pop fizz?
•By maintaining the high
pressure above the soda pop
more CO2 can be dissolved in
it.
•The concentration of
dissolved gas is proportional to
the pressure of the gas above
the solution.
•Divers bend.
Chem 1050
Chapter 14 Solutions
Dr. A. Ghumman
Vapour Pressures of Solutions
• F. M. Roault, (1880s).
– Dissolved solute lowers vapour pressure of the solvent.
– The partial pressure exerted by solvent vapour above
an ideal solution is the product of the mole fraction of
solvent χA in the solution and the vapour pressure of
the pure solvent P°A at the given temperature.
PA = χA P°A
• PA= vapour pressure of solvent molecules above solution.
• χA= mole fraction of pure solvent in the solution.
• P° = vapour pressure of pure solvent
Chem 1050
Chapter 14 Solutions
Dr. A. Ghumman
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Vapour Pressures of Solutions
• In an ideal solution of two liquids,
– Each liquid lowers the vapour pressure of the other
liquid.
– Ptot = PA+ PB (Dalton Law)
= χA P° + χB P°
–
– Where χA + χB = 1
• Limitations;
– Raoult’s Law applies only to ideal solutions.
– In nonideal solutions it often works well for the solvent
(not for solute) in very dilute solutions i.e.
–
xsolv > 0.98.
Chem 1050
Chapter 14 Solutions
Dr. A. Ghumman
Predicting vapour pressure of ideal solutions
Example 14-6 The vapour pressure pressures of pure benzene and toluene
at 25°C are 95.1 and 28.4 mm Hg, respectively. A solution is prepared
in which the mole fractions of benzene and toluene are both 0.500.
(a) What are the partial pressures of the benzene and toluene above this
solution?
(b) What is the total vapor pressure?
Balanced Chemical Equation:
•
Pbenzene = χbenzene P°benzene
•
Ptoluene = χtoluene P°toluene
•
Ptotal = Pbenzene + Ptoluene
Chem 1050
Chapter 14 Solutions
Dr. A. Ghumman
Calculating the Composition of Vapor in Equilibrium
with a Liquid Solution
Example: What
is the composition of the vapour in equilibrium
with the benzene-toluene solution in previous example?
Solution:
Partial pressure and mole fraction (recall chapter 6)
χbenzene = Pbenzene/Ptotal
χtoluene = Ptoluene/Ptotal
Chem 1050
Chapter 14 Solutions
Dr. A. Ghumman
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Liquid-Vapour Equilibrium for Benzene toluene
mixtures (ideal solution) at 25 °C
Phase diagram
Partial pressure and total
Pvap. are plotted as a
function of the solution and
vapour composition
Chem 1050
Chapter 14 Solutions
Dr. A. Ghumman
Osmotic Pressure
•
•
•
•
Colligative properties:- The properties whose values depend only on
concentration of solute particles in solution and not on the nature of
solute
– e.g. Osmotic pressure, vapour pressure lowering, freezing point
depression, boiling point elevation,.
Semipermeable membrane: A membrane that allows the flow of
smaller (solvent) water molecules through it but does not allow the
flow of bigger solute molecules.
Osmosis:The net flow of the solvent molecules through a
semipermeable membrane, from a more dilute solution into a more
concentrated solution.
Osmotic pressure: The necessary pressure that would have to be
applied to a solution to stop the osmotic flow through a semipermeable
membrane of solvent molecules from the pure solvent into the
solution(on the other side of the membrane).
Chem 1050
Chapter 14 Solutions
Dr. A. Ghumman
Osmotic Pressure
water molecules pass through the
membrane into more concentrated
solution where concentration of water
molecules is low.
Semipermeable membrane.
Chem 1050
Chapter 14 Solutions
Dr. A. Ghumman
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Osmotic Pressure
Osmotic pressure π of dilute solutions of non electrolytes
can be calculated by the following equation
πV = nRT ( Ideal Gas Law??)
π=
n
RT = C RT
V
π = osmotic pressure
C = concentration in mol L-1
R = gas constant and T= Temperature in K
•The magnitude of osmotic pressure depends only on the number
of solute particles per unit volume of the solution.
•Identity of solute is not important.
Chem 1050
Chapter 14 Solutions
Dr. A. Ghumman
Osmotic pressure
• Establishing a mass from a measurement of osmotic
pressure.
• Example: A 50.00mL sample of an aqueous solution
contains 1.08 g of human serum albumin, a blood plasma
protein. The solution has an osmotic pressure of 5.85
mmHg at 298 K. What is the molar mass of the albumin?
π=
n
RT
V
m/M
RT
V
mRT
M =
πV
π =
Chem 1050
Chapter 14 Solutions
Dr. A. Ghumman
Osmotic Pressure; Applications
•Red blood cell placed in concentrated salt
solution (NaCl) will shrink.
Hypertonic > 0.92% NaCl (m/V)
crenation
•Dehydration treatment.
Isotonic Saline 0.92% NaCl (m/V)
•Red blood cell placed in water will
rupture.
Hypotonic < 0.92% NaCl (m/V)
rupture
Chem 1050
Chapter 14 Solutions
Dr. A. Ghumman
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Reverse Osmosis - Desalination
Normal flow is from side
A to B.
•Desalination of sea water;
If we apply a pressure
>osmotic pressure of saltwater to side B, net flow of
water can be reversed.
•Result: pure drinking
water can be obtained
from seawater for emergency
situations.
•Cleaning of wastewater
Chem 1050
Chapter 14 Solutions
Dr. A. Ghumman
Freezing-Point depression and Boiling Point
elevation of Non-electrolyte Solutions
•
Vapour pressure of a solvent is lowered when a solute is
present in the solution. This results in
1. Freezing point depression (lowering)
∆Tf = -Kf x m
2. Boiling point elevation
∆Tb = Kb x m
Colligative properties
where m = molality of solute.
∆Tf and ∆Tb = decrease in freezing point and increase in
boiling point.
Kf = freezing point depression constant, depends on mp,
∆Hfusand molar mass of solvent
Kb=boiling point elevation constant, depends on bp,
∆Hvapand molar mass of solvent.
The solution stays liquid over a wide range of temperature.
Chem 1050
Chapter 14 Solutions
Dr. A. Ghumman
Vapor Pressure Lowering; Phase diagram
∆Tf and ∆Tb = freezing
point depression and boiling
point elevation.
fp0 = normal fp of pure
solvent.
bp0= normal bp of pure
solvent.
fp and bp= freezing and
boiling points of solution.
Chem 1050
Chapter 14 Solutions
Dr. A. Ghumman
12
Major applications and Limitations of ∆Tb& ∆Tf
Applications:
– To establish, molecular mass for larger molecules.
– To establish molecular formula.
Limitations:
– Equations apply only to dilute solution of
nonelectrolyte, (usually less than 1m solution).
• ∆Tf = -Kf x m
• ∆Tb = Kb x m
Chem 1050
Chapter 14 Solutions
Dr. A. Ghumman
Example: Nicotine extracted from tobacco leaves, is a liquid
completely miscible with water at temperatures below 60
°C.
a). What is the molality of the nicotine in an aqueous solution
that starts to freeze at –0.450 °C?
b). If the solution is obtained by dissolving 1.921 g of
nicotine in 48.92 g H2O, what must be the molar mass of
nicotine?
c). Combustion analysis shows nicotine consist of 74.03% C,
8.70%H and 17.27% N by mass. What is the molecular
formula of nicotine?
Chem 1050
Chapter 14 Solutions
Dr. A. Ghumman
Practical Applications
De-icing of an aircraft with hot, high
pressure spray of propylene glycol
diluted with water.
Ethylene glycol in antifreeze mixture.
Chem 1050
Chapter 14 Solutions
Dr. A. Ghumman
Newfoundland’s Winter
Salt(NaCl) lowers the fp
of water on the roads in
winter.
13
Solutions of Electrolytes
•
Svante Arrhenius
– Nobel Prize 1903.
– Ions form when electrolytes dissolve in solution.
– Explained anomalous colligative properties.
•
Comparison of 0.0100 m aqueous urea (non electrolyte) to 0.0100 m
NaCl(aq).
– NaCl is strong electrolyte and one formula unit of NaCl produces
two moles of ions compared to urea.
– Expected ∆Tf for NaCl solution is
•
∆Tf = -Kf x m = -1.86°C m-1 x 0.0100 m = -0.0186°C
Actual Freezing point depression for NaCl is -0.0361°C
Chem 1050
Chapter 14 Solutions
Dr. A. Ghumman
Solution of electrolytes.
• Electrolytes in solution dissociate in to ions and produce
greater number of particles compared to non electrolytes.
– Ionic compounds dissociate to form ions in solution
while molecular compounds do not produce ions(urea,
glycerol, sucrose).
• Colligative properties like π, ∆Tf, and ∆Tb depend on the
concentration of particles in the solution.
– Electrolytes (increased no. of particles) increase the
values of colligative properties such as π, ∆Tf, and ∆Tb.
Chem 1050
Chapter 14 Solutions
Dr. A. Ghumman
van’t Hoff factor
Van’t Hoff’s factor i: is the ratio of the measured value of a colligative
property to the expected value. For NaCl(aq)
i=
measured ∆Tf
expected ∆Tf
=
-0.0361°C
= 1.98
-0.0186°C
For non electrolyte (urea) i = 1 and for NaCl i =2
Al2(SO4)3 i= 5
The previous equations has to be modified to take into account the greater
number of particles produced by electrolytes vs nonelectrolytes,
π = i x C x RT
∆Tf = -i x Kf x m
∆Tb = i x Kb x m
Chem 1050
Chapter 14 Solutions
Dr. A. Ghumman
14
Interionic Interactions
• Arrhenius theory does not correctly predict the
conductivity of concentrated electrolytes.
Chem 1050
Chapter 14 Solutions
Dr. A. Ghumman
Debye and Hückel; Interionic attractions
• Arrhenius’s theory could not explain low electrical
conductivities of concentrated solution of strong
electrolytes.
• strong electrolyte exists as ions in
aqueous solution, but
– Ions in solution do not behave independently.
– Each ion is surrounded by others of opposite
charge.
• Ion mobility is reduced by
the drag of the ionic atmosphere.
For NaCl the value of i is 1.94 (less than 2)
Chem 1050
Chapter 14 Solutions
Dr. A. Ghumman
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