properties of solutions. terminology solution = a homogeneous mixture. solute = a substance...
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S olva tion
S truc ture /Inte rm olecula r F orces
Henry's L aw
B. pt. E leva tion F r. p t. D epress ion O som otic Pressure
C oll iga tive Properties
V apor Pressures Raoult's L aw
T em pera ture Pressure
Hea ts of S olution
C om pos itions S olubi l i ty Rules
T erm inology
Properties of SolutionsProperties of Solutions
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TerminologyTerminology
• Solution = A homogeneous mixture.• Solute = a substance dissolved in a solvent to
form a solution; usually the smaller portion.• Solvent = The dissolving medium of a
solution; usually the greater portion.• Solubility = Amount of substance dissolved.• Dilute , Concentrated , Saturated .
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The Solution ProcessThe Solution Process
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Figure 13.2: The Solution ProcessFigure 13.2: The Solution ProcessHydration or SolvationHydration or Solvation
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Solution Formation, Spontaneity, and Disorder
• If the process leads to a greater state of disorder, then the process is spontaneous.
• Example: a mixture of CCl4 and C6H14 is less ordered than the two separate liquids. Therefore, they spontaneously mix even though Hsoln is very close to zero.
• There are solutions that form by physical processes and those by chemical processes.
The Solution ProcessThe Solution Process
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Solute-Solvent Interaction• Miscible liquids: mix in any proportions.• Immiscible liquids: do not mix. [ Oil and Vinegar ]• Generalization: “like dissolves (in) like”.
• Polar liquids tend to dissolve in polar solvents.
Factors Affecting SolubilityFactors Affecting Solubility
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Mass Percentage• All methods involve quantifying amount of solute per
amount of solvent (or solution).• Generally amounts or measures are masses, moles or
liters.
Ways of Expressing ConcentrationWays of Expressing Concentration
100solution of mass total
solution incomponent of masscomponent of % mass
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• Mole Fraction, Molarity, and Molality
Ways of Expressing ConcentrationWays of Expressing Concentration
solution of moles totalsolutionin component of moles
component offraction Mole
solution of literssolute moles
Molarity
solvent of kgsolute moles
Molality, m
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Solution CompositionsSolution Compositions
• s = solute ; A = solvent; V = Tot. Vol. of solution.• Weight %:
• Mole Fraction:
• Molarity:
• Molality:
100% xww
ww
As
ss
As
ss nn
n
V
nM s
s
Akg
nm s
s
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Example of Solution CompositionsExample of Solution Compositions
• A solution is prepared by mixing 78.9 g of ethanol (C2H5OH) with 100.0 g of water to give 190.5 mL of solution. Calculate the solution compositions.
• The electrolyte in automobile lead storage batteries is a 3.75 M H2SO4 solution that has a density of 1.230 g/mL. Calculate mass %, molality, and mole fraction in terms of H2SO4 .
• [Hint: Assume exactly one liter of solution.]• [Answers: 29.9% , 4.35 molal, 0.0727 ]
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%1.441009.178
9.78% x
g
gws
s = ethanol (solute); A = water (solvent);
236.04549.5371.1
371.1
02.180.100
07.469.78
07.469.78
As
ss nn
n
199.85.190
371.1 LmolmL
mol
V
nM s
s
11.171000.0
371.1 kgmolwaterkg
mol
Akg
nm s
s
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In the Dilution process of a more concentrated solution:
• The number of moles are the same in diluted and concentrated solutions.
• So:MdiluteVdilute = moles = MconcentratedVconcentrated
Concentrations of SolutionsConcentrations of Solutions
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Electrolytic Properties• Three types:
• Strong electrolytes,• Weak electrolytes, and• Nonelectrolytes.
General Properties of Aqueous SolutionsGeneral Properties of Aqueous Solutions
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Strong and Weak Electrolytes• Strong electrolytes: completely dissociated in solution.
e.g.
• Weak electrolytes: produce small concentration of ions when dissolved.
• e.g.
General Properties of Aqueous SolutionsGeneral Properties of Aqueous Solutions
HCl(aq) H+(aq) + Cl-(aq)
HC2H3O2(aq) H+(aq) + C2H3O2-(aq)
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Precipitation ReactionsPrecipitation Reactions
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Exchange (Metathesis) Reactions• Metathesis reactions involve swapping ions in solution:
AX + BY AY + BX.
Precipitation ReactionsPrecipitation Reactions
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Ionic Equations• Ionic equation: used to highlight reaction between ions.
• Molecular equation: all species listed as molecules:AgNO3(aq) + NaI(aq) AgI(s) + NaNO3(aq)
• Complete ionic equation(CIE): lists all ions:
Ag+(aq) + NO3-(aq) + Na+(aq) + I-(aq) AgI(s) + Na+(aq) + NO3
-(aq)
• Net ionic equation: cancel spectator ions from CIEAg+(aq) + I-(aq) AgI(s)
Precipitation ReactionsPrecipitation Reactions
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Figure 13.14: Factors Affecting SolubilityFigure 13.14: Factors Affecting Solubility
Intermolecular Forces
Pressure
Temperature
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Pressure Effects
Figure 13.14: Factors Affecting SolubilityFigure 13.14: Factors Affecting Solubility
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Pressure Effects
• If Sg is the solubility of a gas, k is a constant, and Pg is the partial pressure of a gas, then Henry’s Law gives:
• Carbonated beverages are bottled with a partial pressure of CO2 >1 atm, ( pressure inside can ~4 atm above liq).
• As bottle is opened, partial pressure of CO2 decreases and solubility of CO2 decreases.
• Therefore, bubbles of CO2 escape from solution.
Factors Affecting SolubilityFactors Affecting Solubility
gg PkS
CyberChem Diving Gases
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Temperature Effects: Solids in Liquids• Generally, as temperature increases, solubility of solids
generally increases, BUT• Sometimes, solubility decreases as temperature increases
(e.g. Ce2(SO4)3).
Factors Affecting SolubilityFactors Affecting Solubility
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Figure 13.18
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Temperature Effects: Gases in Liquids• Gases get less soluble as temperature increases.
• Thermal pollution: if lakes get too warm, CO2 and O2 become less soluble and are not available for plants or animals.
Factors Affecting SolubilityFactors Affecting Solubility
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Figure 13.18
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• Colligative properties depend on quantity of solute molecules. (e.g. freezing point depression and boiling point elevation.)
Lowering Vapor Pressure• Non-volatile solutes reduce the ability of the surface solvent
molecules to escape the liquid.• Therefore, vapor pressure is lowered.• The amount of vapor pressure lowering depends on the
amount of solute.
Colligative PropertiesColligative Properties
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Vapor Pressure on the Molecular Level
Figure 11.22: Vapor PressureFigure 11.22: Vapor Pressure
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Figure 11.24
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Figure 11.26: Phase DiagramsFigure 11.26: Phase Diagrams
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The Phase Diagrams of H2O and CO2
Figure 11.27: Phase DiagramsFigure 11.27: Phase Diagrams
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Lowering Vapor Pressure
Figure 13.20: Colligative PropertiesFigure 13.20: Colligative Properties
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Lowering Vapor Pressure• Raoult’s Law:
• Where: PA = vapor pressure with solute,
• PA = vapor pressure without solute (pure solvent), and
A = mole fraction of A.
Colligative PropertiesColligative Properties
AAA PP
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Vapor Pressure ExamplesVapor Pressure Examples
• Calculate the expected vapor pressure at 25oC for a solution prepared by dissolving 158.0 g of common table sugar (sucrose – MW=342.3) in 643.5 mL of water. At 25oC, the density of water is 0.9971 g/mL and the vapor pressure is 23.76 torr.
• A solution was prepared by adding 20.0 g of urea to 125 g of water at 25oC, a temperature at which pure water has a vapor pressure of 23.76 torr. The observed vapor pressure of the solution was found to be 22.67 torr. Calculate the molecular weight of urea. [Answer: 60. g/mol ]
AAA PP
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Calculate the expected vapor pressure at 25oC for a solution prepared by dissolving 158.0 g of common table sugar (sucrose – MW=342.3) in 643.5 mL of water. At 25oC, the density of water is 0.9971 g/mL and the vapor pressure is 23.76 torr.
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Example 1: Sugar in Water
s = sugar A = water ws 158.0gm MWs 342.3gm mol1
VA 643.5cm3 dA 0.9971gm cm
3 PoA 23.76torr MWA 18.015gm mol1
nA
VA dA
MWA nA 35.62mol nA 643.5cm
30.9971gm
cm3
1 mol
18.015gm
ns
ws
MWs ns 0.4616mol
ns 158.0gm1 mol
342.3gm
A
nA
ns nA A 0.9872 A
35.62
0.4616 35.6235.62
36.08
PA A PoA PA 23.46torr PA 0.9872 23.76torr( )
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Example 2: Urea Molecular Weight
• A solution was prepared by adding 20.0 g of urea to 125 g of water at 25oC, a temperature
at which pure water has a vapor pressure of 23.76 torr. The observed vapor pressure of thesolution was found to be 22.67 torr. Calculate the molecular weight of urea.[Answer: 60. g/mol ]
s = urea A = water
ws 20.0gm wA 125gm PoA 23.76torr PA 22.67torr
PA A PoA A
PA
PoA A 0.9541 MWA 18.015gm mol
1
A
nA
ns nAnA
wA
MWA nA 6.939mol
solving for ns has solution of:
A
nA
ns nAnA
A 1
A
ns
nA 1 A
A ns 0.334mol
ns
ws
MWsMWs
ws
ns MWs 60gm mol
1
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Figure 13.22
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Boiling-Point Elevation
• Molal boiling-point-elevation constant, Kb, expresses how much Tb changes with molality, mS :
• Decrease in freezing point (Tf) is directly proportional to molality (Kf is the molal freezing-point-depression constant):
Colligative PropertiesColligative Properties
Sff mKT
Sbb mKT
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Colligative PropertiesColligative Properties
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Applications of Colligative PropertiesApplications of Colligative Properties
• A solution was made by dissolving 18.00 g of glucose in 150.0 g of water. The resulting solution was found to have a boiling point of 100.34oC. Calculate the molecular weight of glucose.
• How many kg of ethylene glycol (C2H6O2, MW=62.1), antifreeze, must be added to 10.0 L of water to produce a solution for use in a car’s radiator that freezes at –23.3oC? Assume that the density of water is 1.00 g/mL. [Answer: 7.78 kg (with d = 1.18 g/mL => 6.59 L)]
Sbb mKT
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•A solution was made by dissolving 18.00 g of glucose in 150.0 g of water. The resulting solution was found to have a boiling point of 100.34oC. Calculate the molecular weight of glucose.
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• A solution was made by dissolving 18.00 g of glucose in 150.0 g of water. The resulting solution was found to have a boiling point of 100.34oC. Calculate the molecular weight of glucose.
Sbb mKT
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How many kg of ethylene glycol (C2H6O2, MW=62.1), antifreeze, must be added to 10.0 L of water to produce a solution for use in a car’s radiator that freezes at –23.3oC? Assume that the density of water is 1.00 g/mL.
Sff mKT
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Freezing Point Depression Example
• How many kg of ethylene glycol (C2H6O2, MW=62.1), antifreeze, must be added to 10.0 L
of water to produce a solution for use in a car’s radiator that freezes at –23.3 oC? Assumethat the density of water is 1.00 g/mL.[Answer: 7.78 kg (with d = 1.18 g/mL => 6.59 L)]
Let: s = ethylene glycol A=water
Kf 1.86K kg mol1 Tf 23.3K kg_A 10.0L
1 gm1 mL
1 mL
103
L
10
3kg
1 gm
kg_A 10.0kg MWs 62.1gm mol1
Tf Kf ms Kf
ns
kg_A
Kf
ws
MWs kg_A
ws
Tf MWs kg_A
Kf
ws23.3K 62.1 gm mol
1 10.0 kg
1.86K kg mol1
ws 7.78kg
ds 1.18gm mL1 Vs
ws
ds Vs 6.59L
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Colligative Properties: Homework due for Friday Dec. ?, 201?.Colligative Properties: Homework due for Friday Dec. ?, 201?.
What is the molecular weight (g/mol) of a non-volatile solute if 7.32 kg of this solute dissolved in 10.0 L of water produced a solution with a freezing point of -23.3C? Assume density of water as 1.00 g/mL.
molgMWS /4.58
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Osmosis• movement of a solvent from low solute concentration to
high solute concentration across a semipermeable membrane.
Colligative PropertiesColligative Properties
Figure 13.23
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Osmosis• Osmotic pressure, , is the pressure required to stop
osmosis:
Colligative PropertiesColligative Properties
TRV
n
TRnV
TRM S TRM S
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Application of Osmotic PressureApplication of Osmotic Pressure
• To determine the molecular weight of a certain protein, 1.00 mg of it was dissolved in enough water to make 1.00 mL of solution. The osmotic pressure of this solution was found to be 1.12 torr at 25.0oC. Calculate the molecular weight of the protein.
[Answer: 1.66x104 g/mol]
TRM S TRM S
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To determine the molecular weight of a certain protein, 1.00 mg of it was dissolved in enough water to make 1.00 mL of solution. The osmotic pressure of this solution was found to be 1.12 torr at 25.0 oC.
Calculate the molecular weight of the protein. TRM S TRM S
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Colligative Properties: Osmotic Pressure Example
• To determine the molecular weight of a certain protein, 1.00 mg of it was dissolved in
enough water to make 1.00 mL of solution. The osmotic pressure of this solution was found
to be 1.12 torr at 25.0oC. Calculate the molecular weight of the protein.[Answer: 1.66x104 g/mol]
Define: s = protein A = water
ws 1.00103 gm V 1.0010
3 L 1.12torr R 0.08206L atm mol1 K
1
T 25.0 273.15( ) K 1.12 torr1atm
760torr
1.474 103 atm
Ms R Tns
V
R Tws
MWs V
R T
The only unknown is the molecular weight of the protein ( MWs ).
MWs
ws R T
V MWs
1.00103 gm 0.08206L atm mol
1 K1 298.15 K
1.474103 atm 1.0010
3 L
MWs 1.66 104 gm mol
1
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Osmosis• Isotonic solutions: two solutions with the same
separated by a semipermeable membrane.
• Osmosis is spontaneous.• Red blood cells are surrounded by semipermeable
membranes.
Colligative PropertiesColligative Properties
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Osmosis
Figure 13.25: Colligative PropertiesFigure 13.25: Colligative Properties
Crenation: hypertonic solution Hemolysis: hypotonic solution
IV(intravenous) fluids must be Isotonic.CyberChem: Desalination
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• Colloids are suspensions in which the suspended particles are larger than molecules but too small to drop out of the suspension due to gravity.
• Particle size: 10 to 2000 Å.• Tyndall effect: ability of a Colloid to scatter light. The
beam of light can be seen through the colloid.
ColloidsColloids
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ColloidsColloids
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Figure 13.31: ColloidsFigure 13.31: Colloids
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Solvation
Structure/Interm olecular Forces
Henry's Law
B. pt. E levation Fr. pt. Depression Osom otic Pressure
Colligative Properties
Vapor Pressures Raoult's Law
T em perature Pressure
Heats of Solution
Com positions Solubility Rules
T erm inology
Properties of SolutionsProperties of Solutions
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nm S
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