chapter 11. solutions and their propertieschickosj/chem11/lecture/lecture13.pdf · solutions and...
TRANSCRIPT
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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
SolutionΔHsoln < 0
Enth
alpy
, H
A Exothermic solution processHfinal
Hinitial
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Solventseparated
Soluteaggregated
Solventaggregated
SolutionΔHsoln < 0
Enth
alpy
, H
ΔH
solv
ent
A Exothermic solution processHfinal
Hinitial
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Solventseparated
Soluteaggregated
Solventaggregated
Soluteseparated
SolutionΔHsoln < 0
Enth
alpy
, H
ΔH
solu
te
ΔH
solv
ent
A Exothermic solution processHfinal
Hinitial
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Solventseparated
Soluteaggregated
Solventaggregated
Soluteseparated
SolutionΔHsoln < 0
Enth
alpy
, H
ΔH
solu
te
ΔHsolute+
ΔHsolvent
ΔH
solv
ent
A Exothermic solution processHfinal
Hinitial
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Solventseparated
Soluteaggregated
Solventaggregated
Soluteseparated
SolutionΔHsoln < 0
Enth
alpy
, H
ΔH
solu
te
ΔHsolute+
ΔHsolvent
ΔHmix
ΔH
solv
ent
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
ΔHsoln > 0
Solution
B Endothermic solution process
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Solventseparated
Soluteaggregated
Solventaggregated
Enth
alpy
, H
Hfinal
Hinitial
ΔHsoln > 0
Solution
ΔH
solv
ent
B Endothermic solution process
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Solventseparated
Soluteaggregated
Solventaggregated
Soluteseparated
Enth
alpy
, H
Hfinal
Hinitial
ΔHsoln > 0
Solution
ΔH
solu
te
ΔH
solv
ent
B Endothermic solution process
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Solventseparated
Soluteaggregated
Solventaggregated
Soluteseparated
Enth
alpy
, H
Hfinal
Hinitial
ΔHsolute+
ΔHsolvent
ΔHsoln > 0
Solution
ΔH
solu
te
ΔH
solv
ent
B Endothermic solution process
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Solventseparated
Soluteaggregated
Solventaggregated
Soluteseparated
Enth
alpy
, H
Hfinal
Hinitial
ΔHsolute+
ΔHsolvent
ΔHmix
ΔHsoln > 0
Solution
ΔH
solu
te
ΔH
solv
ent
B Endothermic solution process
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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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Variation of the solubility of gas liquids with pressure
Amount
Increasing Pressure
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Le Chatelier’s principle
• If a stress is applies to a system in equilibrium, the system reacts in a way to try to minimize this stress
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Pure solvent ASolution of A and B
Vapor pressure of a pure componentB is a non-volatile component
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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
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Distillation of two ideal volatile materials
Raoult’s Law: PA = xAPAo; PB = xBPBo
PT obs = xA PAo + xBPBofor two volatile components
where PAo and PBo are the vapor pressures of the pure components
What is the boiling temperature of a 1:1 mixture of toluene/benzene?
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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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300
4
50
760
1
000
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 component
xB = 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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Distillation
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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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ΔT = Kf m; Kf = boiling point elevation constant
m = molality; moles /1000g of solvent
Kf (water) = 0.51 °C/m
Boiling Point Elevation for A Volatile Liquid and Non-volatile Solid
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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?
sucrose, 600/342 g/mol = 1.75 mol
ΔT = Kf m; ΔT = 0.51*5.83m; ΔT = 3 °C
Boiling point of water = 103 °C
1.75 mol/300 g H2O = x mol/1000gH2O
x = 5.83 m molality of sugar
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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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4% starch 10% starch
7% starch7% starch
A semipermeable membrane separates a 4% starch solution from a 10% starch solution. Starch cannot pass through the membrane, but water can. What happens?
Osmotic pressure
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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)
The molarity of the solution refers to the total concentration of all the particles in the solution
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Red blood cells have cell membranes that are semipermeable
they maintain an osmotic pressure that cannot change without damage occuring
they must maintain an equal flow of water between the cell and its surrounding environment
An isotonic solution exerts the same osmotic pressure as red blood cells
5.0% (m/v) glucose or 0.9% (m/v) NaCl is used medically because each has a solute concentration equal to the osmotic pressure equal to red blood cells
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A hypotonic solution has a lower osmotic pressure than red blood cells (RBCs)
containes fewer dissolvedparticles than blood serum
causes water to flow into RBCs
causes hemolysis (RBCsswell and may burst)
A hypertonic solutionhas a higher osmotic pressure than RBCs
containes more dissolvedparticles than blood serum
causes water to flow out of RBCs
causes crenation (RBCsshrink in size)
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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 andpure 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)*310 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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Π
A B C
Solution
Semipermeablemembrane
SolutemoleculesSolventmolecules
Osmoticpressure
Π
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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Elemental analysis of β-carotene, a dietary source of vitamin A, shows that it containes 10.51 % H and 89.49% C. Dissolving 0.025 g of β-carotene in 1.5 g of camphor gives a freezing point depression of 1.17 °C. What are the molar mass and formula of β-carotene? [Kf for camphor is 37.7 (°C kg/mol)]
ΔT = Kfm; 1.17 = 37.7*m; m = 1.17/37.7 = 0.031 mol/1000g
0.025g/mw β-carotene/1.5 g camphor = x/1000;
x = 16.67g β-carotene/1000 g camphor; 16.67/mw = 0.031 mol
mw = 537.7 g/mol
89.49/12 = 7.458 C 7.458/7.458 = 1
10.51/1 = 10.51 H 10.51/7.458 = 1.409 C1H1.409
n(12 + 1.409) = 537.7
13.409n = 537.7; n = 40.0 C40H56
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C40H56 β-carotene
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Met-enkephalin is one of the so called endorphins, a class of naturaly occuring morphine like chemicals in the brain. What is the molecular ass of met-enkephalin if 20.0 mL of an aqueous solution containing 15 .0 mg of met-enkephalin at 298 K supports a column of water 32.9 cm high? The density of Hg is 13.534 g/mL
pure H2O
H2O + met-enkephalin
32.9 cm H2O
32.9cm/13.434 = 2.43 cm Hg
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Met-enkephalin is one of the so called endorphins, a class of naturaly occuring morphine like chemicals in the brain. What is the molecular mass of met-enkephalin if 20.0 mL of an aqueous solution containing 15 .0 mg of met-enkephalin at 298 K supports a column of Hg 2.43 cm high?
π = MRT = 2.45/76cm/atm = M (0.0821Latm/(K mol)*298K
M = 0.03198/(.0821*298); M = 0.001307 mole/L
0.015g/20 mL = x/1000mL; x = 0.75 g
0.75/mw = 0.001307 mol; mw = 573.8
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mw = 573.66; calc 573.8; C27H35N5O7S