chapter 11. solutions and their properties
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Chapter 11. Solutions and Their Properties. Solutions. Solute: material present in least amount Solvent: material present in most amount Solution = solvent + solute(s). What criteria must be satisfied in order to form a solution?. An artist’s conception of how a salt dissolves in water. - PowerPoint PPT PresentationTRANSCRIPT
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Chapter 11. Solutions and Their Properties
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Solutions
Solute: material present in least amount
Solvent: material present in most amount
Solution = solvent + solute(s)
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What criteria must be satisfied in order to form a solution?
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An artist’s conception of how a salt dissolves in waterHow could we identify the cation and anion?
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Dissolving of sugar
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Solution energy = M+X- (s) + H2O M+(aq) + X-(aq)
Lattice energy = M+(g) + X-(g) M+X- (s)
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Soluteaggregated
SolventaggregatedEn
thal
py, H
A Exothermic solution process
Hinitial
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Soluteaggregated
Solventaggregated
SolutionHsoln < 0
Enth
alpy
, H
A Exothermic solution processHfinal
Hinitial
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Solventseparated
Soluteaggregated
Solventaggregated
SolutionHsoln < 0
Enth
alpy
, H
Hso
lven
t
A Exothermic solution processHfinal
Hinitial
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Solventseparated
Soluteaggregated
Solventaggregated
Soluteseparated
SolutionHsoln < 0
Enth
alpy
, H
Hso
lute
Hso
lven
t
A Exothermic solution processHfinal
Hinitial
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Solventseparated
Soluteaggregated
Solventaggregated
Soluteseparated
SolutionHsoln < 0
Enth
alpy
, H
Hso
lute
Hsolute
+Hsolvent
Hso
lven
t
A Exothermic solution processHfinal
Hinitial
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Solventseparated
Soluteaggregated
Solventaggregated
Soluteseparated
SolutionHsoln < 0
Enth
alpy
, H
Hso
lute
Hsolute
+Hsolvent
Hmix
Hso
lven
t
A Exothermic solution processHfinal
Hinitial
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Soluteaggregated
Solventaggregated
Enth
alpy
, H
Hinitial
B Endothermic solution process
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Soluteaggregated
Solventaggregated
Enth
alpy
, H
Hfinal
Hinitial
Hsoln > 0
Solution
B Endothermic solution process
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Solventseparated
Soluteaggregated
Solventaggregated
Enth
alpy
, H
Hfinal
Hinitial
Hsoln > 0
Solution
B Endothermic solution process
Hso
lven
t
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Solventseparated
Soluteaggregated
Solventaggregated
Soluteseparated
Enth
alpy
, H
Hfinal
Hinitial
Hsoln > 0
Solution
B Endothermic solution process
Hso
lute
Hso
lven
t
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Solventseparated
Soluteaggregated
Solventaggregated
Soluteseparated
Enth
alpy
, H
Hfinal
Hinitial
Hsolute
+Hsolvent
Hsoln > 0
Solution
B Endothermic solution process
Hso
lute
Hso
lven
t
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Solventseparated
Soluteaggregated
Solventaggregated
Soluteseparated
Enth
alpy
, H
Hfinal
Hinitial
Hsolute
+Hsolvent
Hmix
Hsoln > 0
Solution
B Endothermic solution process
Hso
lute
Hso
lven
t
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Interactions that must be overcome and those that form
Why do salts with a positive enthalpy of solution form solutions?
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Variation of solubility of solids and liquids with temperature
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Solubility of gases with temperature
Solvent : H2O
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Solubility of gases as a function of pressure
Le Chatelier’s principle
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Pure solvent ASolution of A and B
Vapor pressure of a pure componentB is a non-volatile component
air + vapor
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Mole fraction of a = na/(na +nb +nc + ...)
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in the presence of a non-volatile solute
Colligative property: a physical property that depends on how the amount present
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Distillation of volatile materials
Raoult’s Law: PA = xAPAo; PB = xBPBo
PT obs = xA PAo + xBPBo
for two volatile components where PAo and PBo are the vapor pressures of the pure components
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Raoult’s Law: PA = XAPAo; PB = XBPB
o PT obs = XA PAo + XBPB
o
for two volatile components
25 °C
mol fraction
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760
30
0
450
76
0
100
0
93 °C
200
560
Boiling occurs when the total pressure = 760 mm
For a 1:1 mol mixture at 93 ° C
PA = XAPAo; PB = XBPBo PT obs = XA PAo + XBPBo
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Two volatile liquids
For an initial 1:1 mixture of benzene-toluene, the initial composition of the vapor based on its vapor pressure is approximately
Ptoluene = 200 mm
Pbenzene = 560 mm
Treating the vapor as an ideal gas, if we were to condense the vapor:
PbV/PtV = nbRT/ntRT
Pb/Pt = nb/nt
nb/nt = 56/20; the vapor is enriched in the more volatile componentxB = 56/(56+20) = 0.73
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Starting with a 50:50 mixture: (the composition of the first drop)
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A Volatile Liquid and Non-volatile Solid
A recipe for making maple syrup requires adding two cups sugar (sucrose, mw 342 g/mol) for every cup of water. Assuming a cup of water contains 300 mL of water and a cup of sugar contains 300 g of sucrose (C12H22O11), at what temperature would you expect the syrup to boil given the following vapor pressures of pure water?
Temp (°C) Vapor pressure (mm) pure H2O100 760101 787102 815103 845104 875105 906
Raoult’s Law Pobs = xH2OPoH2O
Pobs = 760 mm for boiling to occur
300g/18 g/mol = 16.7 mol H2O
600g/342 g/mol = 1.75 mol sugar
xH2O =16.7/(16.7 + 1.75)= 0.91
PoH20 = Pobs/xH20; 760 /0.91 = 835 mm
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Freezing point depression and boiling point elevation (using non-volatile solutes)
T = Km
K is a constant characteristic of each substance
m = molality (mols/1000g solvent)
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T = Km m = molality, mol/1000g solvent
The boiling point elevation is for non-volatile solutes
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Boiling Point Elevation for A Volatile Liquid and Non-volatile Solid
A recipe for making maple syrup requires adding two cups sugar (sucrose, mw 342 g/mol) for every cup of water. Assuming a cup of water contains 300 mL of water and a cup of sugar contains 300 g of sucrose (C12H22O11), at what temperature would you expect the syrup to boil given the following vapor pressures of pure water?
T = Kf m; T = 0.51*5.83m; T = 3 °C
Boiling point of water = 103 °C
sucrose, 600/342 g/mol = 1.75 mol
1.75 mol/300 g H2O = x mol/1000gH2O
x = 5.83 m
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It is found that 1 g of naphthalene in 100 g of camphor depresses the freezing point of camphor by 2.945 °C. What is the molecular weight of naphthalene?
T = Kf m T= 37.7 °C kg/mol*m
2.945 °C = 37.7 °C /molal*m; m = 0.0781molal
If 1 g of naphthalene is dissolved in 10 g of the solvent camphor, 10 g of naphthalene is dissolved in 1000 g camphor
molality is equal to the number of moles/1000 g of solvent
therefore, 10g/MW = 0.0781 moles; MW = 10/0.0781 = 128 g /mol
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Osmosis and osmotic pressure
A simple example of osmosis: evaporation of water from a salt solution
What acts as the semipermeable membrane?
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Osmotic pressure
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Osmotic Pressure
π = MRT
where π is the osmotic pressure between a pure solvent and a solution containing that solvent; M is the molarity of the solution (mols /liter), R is the gas constant, and T = temperature (K)
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The total concentration of dissolved particle in red blood cells in 0.3 M. What would be the osmotic pressure between red cells and pure water if the blood cells were to be placed in pure water at 37 °C?
π = 0.3moles/liter*0.0821liter atm/(mol K)*(273.15+37 )K
π = 0.3 mol/l*0.0821l atm/(K mol)*298 K
π = 7.6 atm
In what direction would the pressure be directed?
What would happen to the cells?
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What is the osmotic pressure developed from a solution of 0.15 M NaCl in contact with a semipermeable membrane of pure water at T = 37 °C?
NaCl + H2O = Na+ + Cl-
π = nRT
π = 0.3 mol/l*0.0821l atm/(K mol)*310.2 K
π = 7.6 atm
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A B C
Solution
Semipermeablemembrane
SolutemoleculesSolventmolecules
Osmoticpressure
Applied pressure needed toprevent volume increase
Puresolvent
Netmovementof solvent
Reverse Osmosis
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De-salination of sea water
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What happens when polar molecules meet non-polar ones?
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What happens when there are incompatible interactions are attached to the same molecule?
A typical soap molecule:CH3CH2CH2CH2CH2CH2CH2CH2CH2CH2CH2CH2CH2CO2
- K+
How does soap clear your clothes?
CO2-
-O2C CO2-
-O2C CO2-
-O2C CO2-
-O2C CO2-
CO2-
micelle
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Soap in water forms micelles, small spherical objects suspended in water
- - - -
- -
- - - -
non-polar
interior
ionic exterior
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Other examples of self assembly:
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