Solutions: Properties and Behavior
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Transcript of Solutions: Properties and Behavior
Solutions: Properties and Behavior
Chapter 11 Solutions: Properties and Behavior States of Matter
Solid- Particles moving about a fixed point
Liquid-Particles moving about a moving point Gas-Particles filling
the volume of the container with complete random motions. Particle
Forces Affect
Solubility Vapor Pressures Freezing Points Boiling Points 3
Particle Forces Intramolecular forces (Relative strength =
100)
Ionic bonding Covalent bonding Interparticle forces Ion-dipole
forces Dipole-dipole (Polar molecules) (relative Strength = 1)
London Forces (Dispersion forces)( Nonpolar molecules) (relative
strength = 1) Hydrogen Bonding (Relative strength = 10) Ion-Ion
Interactions Coulombs law states that the energy (E) of the
interaction between two ions is directly proportional to the
product of the charges of the two ions (Q1 and Q2) and inversely
proportional to the distance (d) between them. Predicting Forces of
Attraction
Coulombs Law indicates the increases in the charges of ions will
cause an increase in the force of attraction between a cation and
an anion. Increases in the distance between ions will decrease the
force of attraction between them. Size of Ions Lattice Energy M+(g)
+ X-(g) ---> MX(s)
The lattice energy (U) of an ionic compound is the energy released
when one mole of the ionic compound forms from its free ions in the
gas phase. M+(g)+X-(g)--->MX(s) Comparing Lattice Energies
Lattice Energies of Common Ionic Compounds Compound U(kJ/mol) LiF
-1047 LiCl -864 NaCl -790 KCl -720 KBr -691 MgCl2 -2540 MgO -3791
Practice Determine which salt has the greater lattice energy.
MgO and NaF MgO and MgS Lattice Energy Using Hesss Law Electron
Affinity Cl(g) + e-(g) ---> Cl-(g)
Electron affinity is the energy changeoccurring when one mole of
electronscombines with one mole of atoms or ion inthe gas phase.
Step 4 in diagram on the last slide. Cl(g)+e-(g)---> Cl-(g) HEa
= -349 kj/mole Calculating U Na+(g) + e-(g) ---> Na(g)
-HIE1
Na(g)--->Na(s)-Hsub Cl-(g)--->Cl(g)+e-(g)-HEA
Cl(g)--->1/2Cl2(g)-1/2HBE Na(s)+1/2Cl2(g)--->NaCl(s) Hf
Na+(g)+Cl-(g)--->NaCl(s)U U = Hf - 1/2HBE - HEA - Hsub - HIE1
Lattice energy for NaCl.
U Interactions Involving Polar Molecules
An ion-dipole interaction occurs between anion and the partial
charge of a molecule witha permanent dipole. The cluster of water
molecules that surroundan ion in aqueous medium is a sphere
ofhydration. Illustrates of Ion-Dipole Interaction The Solution
Process Bond Breaking Processes
Break solute particle forces (expanding the solute), endothermic
Break solvent particle forces (expanding the solvent), endothermic
The Solution Process Attractive Forces
Energy released when solute solvent are attracted, exothermic
Energy is released due to new attractions Ion dipole if the solute
is ionic and the solvent polar. London-Dipole for nonpolar solute
and polar solvent Dipole-dipole for polar solute and polar solvent
The Solution Process Theromodynamics Enthalpy
Entropy (Perfect crystal, assumed to be zero) Gibbs free energy The
Solution Process Oil dissolving in water
London forces holding the oil moleculestogether are large do to the
large surface areaof the oil The hydrogen bonds holding water
moleculestogether are large The forces of attraction of between
nonpolaroil and polar water are weak at best Thus the overall
process is highly endothermicand not allowed thermo chemically The
Solution Process Oil dissolving in water
Entropy should be greater than zero Free energy should be greater
than zero, sincethe process is highly endothermic Thus the overall
process is nonspontaneous The Solution Process Sodium chloride
dissolving in water
Large amount of energy is required to break the ionic lattice of
the sodium chloride (expand solute) Large amount of energy is
required to separate the water molecules to expand the solvent
breaking hydrogen bonds Formation of the ion dipole forces releases
a large amount of energy, strong forces (why?) The sum of the
enthalpies is about +6 kJ (slightly endothermic), which is easily
overcome by the entropy of the solution formation. Water as a
Solvent Water most important solvent, importantto understand its
solvent properties Most of the unusual solvent properties ofwater
stem from it hydrogen bondingnature Consider the following S of
solution KCl 75j/K-mole LiF-36j/K-mole CaS-138j/K-mole Water as a
Solvent We would expect S>0 for all solutions, right?
But two are negative, why? Obviously, something must be
happeningfor the increased order. Ion-dipole forces are ordering
the watermolecules around the ions, thus causingmore order in water
i.e. less positions forwater than in the pure liquid state Water as
a Solvent Smaller ions, have stronger ion dipole forces, thus
pulling water closer, therefore less positions Also, ions with a
charge greater than one will attract to water stronger than a one
plus charge, thus more order due to less space between particles
Dipole-Dipole Interactions
Dipole-dipole interactions are attractive forces between polar
molecules. An example is the interaction between water molecules.
The hydrogen bond is a special class of dipole-dipole interactions
due to its strength. Dipole-Dipole Forces H Cl H Cl H Cl
Dipole-dipole (Polar molecules) +
Alignment of polar molecules to two electrodes charged+ and Forces
compared to ionic/covalent are about 1 in strength compared to a
scale of 100, thus 1% + + + H Cl H Cl H Cl Dipole Dipole
Interactions Hydrogen Bonding Hydrogen bonding a stronger
intermolecular force involving hydrogen and usually N, O, F, and
sometimes Cl Stronger that dipole-dipole, about 10 out of 100, or
10 Hydrogen needs to be directly bonded to the heteroatom Since
hydrogen is small it can get close to the heteroatom Also, the
second factor is the great polarity of the bond. Hydrogen Bonding
in HF(g) Hydrogen Bonding in Water
Chemistry 140 Fall 2002 Hydrogen Bonding in Water Solid ice has
lower density than liquid water.H-bonding holds the ice in a rigid
but open structure. Maximum density of water at 3.98 C. around a
molecule in the solid in the liquid Boiling Points of Binary
Hydrides Interacting Nonpolar Molecules
Dispersion forces (London forces) are intermolecular forces caused
by the presence of temporary dipoles in molecules. A temporary
dipole (or induced dipole) is a separation of charge produced in an
atom or molecule by a momentary uneven distribution of electrons.
Illustrations Strength of Dispersion Forces
The strength of dispersion forces depends on the polarizability of
the atoms or molecules involved. Poarizability is a term that
describes the relative ease with which an electron cloud is
distorted by an external charge. Larger atoms or molecules are
generally more polarizable than small atoms or molecules. London
Forces (Dispersion)
Induced dipoles (Instantaneous ) Strength is surface area dependent
More significant in larger molecules All molecules show dispersion
forces Larger molecules are more polarizable Instantaneous and
Induced Dipoles Molar Mass and Boiling Points of Common
Species.
Halogen M(g/mol) Bp(K) Noble Gas He 2 4 F2 38 85 Ne 20 27 Cl2 71
239 Ar 40 87 Br2 160 332 Kr 84 120 I2 254 457 Xe 131 165 Rn 211
Hydrocarbon Alcohol Molecular Formula Molar Mass Bp (oC) CH4
16.04
-161.5 CH3CH3 30.07 -88 CH3OH 32.04 64.5 CH3CH2CH3 44.09 -42
CH3CH2OH 46.07 78.5 CH3CH(CH)CH3 58.12 -11.7 CH3CH(OH)CH3 60.09 82
CH3CH2CH2CH3 -0.5 CH3CH2CH2OH 97 The Effect of Shape on Forces
Practice Rank the following compound in order of increasing boiling
point.CH3OH, CH3CH2CH2CH3, andCH3CH2OCH3 Practice Rank the
following compound in order of increasing boiling point.CH3OH,
CH3CH2CH2CH3, andCH3CH2OCH3 MM IM Forces CH3OH 32.0 London and
H-bonding CH3CH2CH2CH3 58.0 London, only CH3CH2OCH3 60.0 London and
Dipole-dipole Practice Rank the following compound in order of
increasing boiling point.CH3OH, CH3CH2CH2CH3, andCH3CH2OCH3 MM IM
Forces CH3OH 32.0 London and H-bonding CH3CH2CH2CH3 58.0 London,
only CH3CH2OCH3 58.0 London and Dipole-dipole The order is:
CH3CH2CH2CH3 < CH3CH2OCH3< CH3OH Polarity and
Solubility
If two or more liquids are miscible, they form a homogeneous
solution when mixed in any proportion. Ionic materials are more
soluble in polar solvents then in nonpolar solvents. Nonpolar
materials are soluble in nonpolar solvents. Like dissolves like
Solubility of Gases in Water
Henrys Law states that the solubility of asparingly soluble
chemically unreactive gas ina liquid is proportional to the partial
pressureof the gas. Cgas = kHPgas where C is the concentration
ofthe gas, kH is Henrys Law constant for thegas. Henrys Law
Constants for Gas
kH[mol/(Latm)] kH[mol/(kgmmHg)] He 3.5 x 10-4 5.1 x 10-7 O2 1.3 x
10-3 1.9 x 10-6 N2 6.7 x 10-4 9.7 x 10-7 CO2 3.5 x 10-2 5.1 x 10-5
Terms A hydrophobic (water-fearing)interaction repels water and
diminisheswater solubility.Polar vs. nonpolar A hydrophilic
(water-loving)interaction attracts water and promoteswater
solubility.Polar vs. polar, bestwith hydrogen bonding involved.
Types of Forces Cohesive Forces
Intermolecular forces between the same particles. Adhesive Forces
Intermolecular forces between the different particles. Cohesive
Forces Example Beading or wetting on a surface
Surface tension (resistance to increasing the surface area) Def: To
increase surface area molecules must move from the middle.This
requires energy j/m2 The stronger the IMF the stronger the surface
tension Needle or paper clip on top of water Beading or wetting on
a surface Rounded surface of liquid mercury in a tube Adhesive
Forces Examples.
Capillary rise water forms a meniscus since the forces between the
glass and water are stronger than between water and water.Both are
hydrogen bonds Cohesive and Adhesive Forces
The right tube shows cohesive forces, since mercury is nonpolar and
attracts more strongly to itself, rather than to the glass (SiO2)
The left test tube shows adhesive forces due to the attraction of
water solvent to the polar glass (SiO2)Hydrogen bonding, right? 51
Terms Capillary rise is the rise of a liquid up a narrowtube as a
result of adhesive forces between theliquid and the tube and
cohesive forces within theliquid. Viscosity is a measure of the
resistance to flowof a fluid. 52 Surface Tension 53 The Liquid
State Example Blood up a capillary Meniscus
Adhesive Forces Intermolecular forces between unlike molecules
Example Capillary rise Blood up a capillary Meniscus Capillary rise
is when the adhesive forces are stronger than the cohesive forces
Capillary rise when polar bonds are present in the container walls
like glass, SiO2 Mercury is an example where the cohesive forces
are stronger than the adhesive forces Intermolecular Forces
Slide 55 The Liquid State Viscosity (resistance to flow) How fast
liquids flow
Due in part to intermolecular forces, but also entanglement
Newtons/m2 called poise Change of State (Water)
deposition freezing Condensation H2O(s) H2O(l) H2O(g) Water
Thermodynamic Properties Hfus = 6.02 kj/mole Hvap= 40.7 kj/mole
melting evaporation sublimation Change of State (Water)
Heat capacity of water = j/g-C Water has a very large heat capacity
since it hydrogen bonds and a lot of energy is required to break
these bonds Why water is used in radiators Used to cool animals
Sublimation Hsub= Hfus + Hvap = -Hdeposition Slide 59 Some
Properties of Solids
Melting Point Freezing Point Super cooling Hfus(H2O) = kJ/mol Slide
60 of 35 Super Cooling and Heating
Super cooled, when a liquid exists below its freezing point. Super
cooling occurs when the rate of cooling is faster than it takes for
the molecules to rotate for correct alignment to form crystals.
When the crystals rotate and form inter-particle forces, heat is
released, thus raising the temperature up to the correct m.p. Super
heated Called bumping Use boiling stones, cannot reuse the stones
Hot vapor at bottom expands rapidly and bursts Vapor Pressure
Vaporization or evaporation is the transformation of molecules in
the liquid phase to the gas phase. Vapor pressure is the force
exerted at a given temperature by a vapor in equilibrium with its
liquid phase. Vapor Pressure What evaporates faster pure distilled
water in the beaker on the left, or seawater in the beaker on the
right??Both beakers are the same size and at the same temperature.
Intermolecular Forces Vapor Pressure Yes, pure distilled water
evaporates faster, since there are more water molecules on the
surface to evaporate? Vapor Pressure Physical properties that
depend on the number of particles, and not on the particle nature
are called colligative properties An Aqueous Solution and Pure
Water in a Closed Environment Vapor Pressure Clausius-Clapeyron
Equation Linear Which one is water?
(e) (d)(c) (b) (a) Clausius-Clapeyron Equation ( ) 1 T -Hvap Linear
Ln P = +B R S Which one is water? (Entropy of vaporization) No
Units! B = y-intercept = R Slide 68 ln((P1/P2) = -
H/R(T2-T1)/T1T2
Vapor Pressure Clasius Clapeyron Equation Assume data for two
different temperatures and pressures to generate two separate
equations By subtracting the equations the y-intercept component is
eliminated. ln P1 = - H/R(1/T1) +C -(ln P2 = - H/R(1/T2) + C)
ln((P1/P2) = -H/R(1/T2 1/T1) Another useful version of the two
point equation ln((P1/P2) = -H/R(T2-T1)/T1T2 Vapor Pressure As a
liquid evaporates in a closed container the concentration of vapor
increases, thus the rate of condensation increases As the rate of
condensation is increasing eventually it will equal the constant
rate of evaporation, then we have vapor in equilibrium with the
liquid The pressure of the vapor at equilibrium is called the
equilibrium vapor pressure Raoults Law Psolution = Xsolvent
(Psolvent) P - vapor pressure
X - mole fraction Xsolvent + Xsolute = 1 For a Solution that Obeys
Raoult's Law, a Plot fo Psoln Versus Xsolvent, Give a Straight Line
Vapor Pressure of Solvent and Solution Liquid-Vapor
Equilibrium
Chemistry 140 Fall 2002 Liquid-Vapor Equilibrium The vapor phase is
richer in the more volatile component. July 2009 General Chemistry:
Chapter 11 Slide 74 of 46 74 Two Volatile Liquids Ideal Solution
Positive deviation
Negative deviation Positive deviation exists when experimental
value is larger than calculated value, weaker solute solvent
attraction; more evaporation. Negative deviation exists when
experimental value is smaller than calculated value; stronger
solvent solute attraction; less evaporation Fractional
Distillation
July 2009 General Chemistry: Chapter 11 Slide 76 of 46 Fractional
Distillation
July 2009 General Chemistry: Chapter 11 Slide 77 of 46 Practice A
solution contains mL of water and mol of ethanol. What is the mole
fraction of water and the vapor pressure of the solution at 25oC,
if the vapor of pressure of pure water is 23.8 torr? Practice A
solution contains mL of water and mol of ethanol. What is the mole
fraction of water and the vapor pressure of the solution at 25oC,
if the vapor of pressure of pure water is 23.8 torr? 100.0mL
Practice A solution contains mL of water and mol of ethanol. What
is the mole fraction of water and the vapor pressure of the
solution at 25oC, if the vapor of pressure of pure water is 23.8
torr? 1.00 g mole 100.0mL mL 18.0 g Practice A solution contains mL
of water and mol of ethanol. What is the mole fraction of water and
the vapor pressure of the solution at 25oC, if the vapor of
pressure of pure water is 23.8 torr? 1.00 g mole 100.0mL = 5.56
mole mL 18.0 g 0.500 mole C2H6O 6.06 mole 5.56 XHOH = = 0.917 6.06
Practice A solution contains mL of water and mol of ethanol. What
is the mole fraction of water and the vapor pressure of the
solution at 25oC, if the vapor of pressure of pure water is 23.8
torr? 1.00 g mole 100.0mL = 5.56 mole mL 18.0 g 0.500 mole C2H6O
6.06 mole PHOH = 0.917(23.8 torr) PHOH = 21.8 torr 5.56 XHOH = =
0.917 6.06 Boiling Point Vs. Pressure Calculating Changes in
Boiling Point
Tb = Kbm Tb is the increase in Bp Kb is the boiling-point elevation
constant m is a new concentration unit called molality moles solute
Molality (m) = Kg solvent Practice Calculate the molality of a
solution containing mol of glucose (C6H12O6) in 1.5 kg of water.
Practice Calculate the molality of a solution containing mol of
glucose (C6H12O6) in 1.5 kg of water. 0.875 mole 1.5 kg Practice
Calculate the molality of a solution containing mol of glucose
(C6H12O6) in 1.5 kg of water. 0.875 mole = 0.58 m 1.5 kg Practice
Seawater contains M Cl- at the surface at 25oC.If the density of
sea water is g/mL, what is the molality of Cl- in sea water?
Practice Seawater contains M Cl- at the surface at 25oC.If the
density of sea water is g/mL, what is the molality of Cl- in sea
water? 103 mL solution 1.022 g = 1022 g solution mL Practice
Seawater contains M Cl- at the surface at 25oC.If the density of
sea water is g/mL, what is the molality of Cl- in sea water? 103 mL
solution 1.022 g = 1022 g solution mL 0.558 mole Cl- 45.45 g Cl- =
g Cl- mole Cl- Practice Seawater contains M Cl- at the surface at
25oC.If the density of sea water is g/mL, what is the molality of
Cl- in sea water? 103 mL solution 1.022 g = 1022 g solution mL
0.558 mole Cl- 45.45 g Cl- = g Cl- mole Cl- 1022 g solution g Cl- =
g H2O Practice Seawater contains M Cl- at the surface at 25oC.If
the density of sea water is g/mL, what is the molality of Cl- in
sea water? 103 mL solution 1.022 g = 1022 g solution mL 0.558 mole
Cl- 45.45 g Cl- = g Cl- mole Cl- 1022 g solution g Cl- = g H2O
0.558 mole Cl- 103 g = m 996.6 g H2O Kg Practice Cinnamon owes its
flavor and odor to cinnamaldehyde (C9H8O).Determine the
boiling-point elevation of a solution of 100 mg of cinnamaldehyde
dissolved in 1.00 g of carbon tetrachloride (Kb = 2.34oC/m).
Practice Cinnamon owes its flavor and odor to cinnamaldehyde
(C9H8O).Determine the boiling-point elevation of a solution of 100
mg of cinnamaldehyde dissolved in 1.00 g of carbon tetrachloride
(Kb = 2.34oC/m). 100 mg C9H8O 10-3 mole 103 g mmole = m 1.00 g CCl4
mg mmole Kg 2.34 C m = 1.77C m Freezing-point Depression
Tf = Kfm Kf is the freezing-point depression constant and m is the
molality. Practice The freezing point of a solution prepared by
dissolving X 102 mg of caffeine in 10.0 g of camphor is 3.07
Celsius degree lower than that of pure camphor (Kf = 39.7oC/m).What
is the molar mass of caffeine? The vant Hoff Factor Tb = iKbm &
Tf = iKfm
vant Hoff factor, i is thenumber of ions in oneformula unit The
vant Hoff Factor Used for ionic compounds, why not
osmolarity?
The value of i assumes that all of the salt dissolves and
dissociates in to its component ions This is not always true, for
example 0.10m NaClI is 1.87 Ion pairing often occurs in solutions
Ion pairing most important in concentrated solutions Ion pairing
important in highly charged solutions Values of vant Hoff Factors
Practice CaCl2 is widely used to melt frozen precipitation on
sidewalks after a winter storm.Could CaCl2 melt ice at -20oC?
Assume that the solubility of CaCl2 at this temperature is 70.0
g/100.0 g of H2O and that the vant Hoff factor for a saturated
solution of CaCl2 is 2.5 (Kf for water is C/m). Osmotic Pressure
Osmotic pressure () is the pressure that has to be applied across a
semipermeable membrane to stop the flow of solvent form the the
compartment containing pure solvent or a less concentratedsolution
towards a more concentrated solution. = iMRT where i is the vant
Hoff factor, M is molarity of solute, R is the idea gas constant (
latm/(molK)), and T is in Kelvin 101 Osmosis at the Molecular Level
Osmotic pressure Equation from the ideal gas law (pv = nRT) =
MRT
Semi permeable membrane Isotonic same concentration Cells placed in
lower concentration hypotonic, cellwill swell called hemolosis If
concentration on the outside of the cells isgreater then the
solution is called hypertonic andthe cells shrink called crenation
Osmosis In osmosis, solvent passes through a semipermeable
membrane
to balance the concentration of solutes in solution on both sides
of the membrane. Figure 10.30 THE END ChemTour: Lattice
Energy
Click to launch animation PC | Mac Students learn to apply Coulombs
law to calculate the exact lattice energies of ionic solids.
Includes Practice Exercises. ChemTour: Intermolecular Forces
Click to launch animation PC | Mac This ChemTour explores the
different types of intermolecular forces and explains how these
affect the boiling point, melting point, solubility, and
miscibility of a substance. Includes Practice Exercises. Click to
launch animation
ChemTour: Henrys Law Click to launch animation PC | Mac Students
learn to apply Henrys law and calculate the concentration of a gas
in solution under varying conditions of temperature and pressure.
Includes interactive practice exercises. ChemTour: Molecular
Motion
Click to launch animation PC | Mac Students use an interactive
graph to explore the relationship between kinetic energy and
temperature. Includes Practice Exercises. ChemTour: Raoults
Law
Click to launch animation PC | Mac Students explore the connection
between the vapor pressure of a solution and its concentration as a
gas above the solution.Includes Practice Exercises. ChemTour: Phase
Diagrams
Click to launch animation PC | Mac Students use an interactive
phase diagram and animated heating curve to explore how changes in
temperature and pressure affect the physical state of a substance.
ChemTour: Capillary Action
Click to launch animation PC | Mac In this ChemTour, students learn
that certain liquids will be drawn up a surface if the adhesive
forces between the liquid on the surface of the tube exceed the
cohesive forces between the liquid molecules. ChemTour: Boiling and
Freezing
Click to launch animation PC | Mac Students learn about colligative
properties by exploring the relationship between solute
concentration and the temperature at which a solution will undergo
phase changes. Interactive exercises invite students to practice
calculating the boiling and freezing points of different solutions.
ChemTour: Osmotic Pressure
Click to launch animation PC | Mac Students discover how a solute
can build up pressure behind a semipermeable membrane. This
tutorial also discusses the osmotic pressure equation and the vant
Hoff factor. Solubility of CH4, CH2Cl2, and CCl4
Which of the following three compounds is most soluble in water?
2008 W. W. Norton & Company Inc. All rights reserved. A)CH4(g)
B)CH2Cl2()C)CCl4() Solubility of CH4, CH2Cl2, and CCl4 Solubility
of CH4, CH2Cl2, and CCl4
Consider the following arguments for each answer and vote again: A
gas is inherently easier to dissolve in a liquid than is another
liquid, since its density is much lower. The polar molecule CH2Cl2
can form stabilizing dipole-dipole interactions with the water
molecules, corresponding to a decrease in Hsoln. The nonpolar
molecule CCl4 has the largest molecular mass, and so is most likely
to partially disperse into the water, corresponding to an increase
in Ssoln. Answer: B Solubility of CH4, CH2Cl2, and CCl4 The End