1. 11.3 Factors Affecting Solubility
The Vapor Pressures of Solutions
Boiling-Point Elevation and Freezing-Point Depression
11.6 Osmotic Pressure
11.7 Colligative Properties of Electrolyte Solutions
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2. Solution Composition
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3. Solution = Solvent + Solute So,
Mole Fraction
Solution = Solvent + Solute
So,
Mole solution = mole solvent + mole solute
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4. Exercise
A solution of phosphoric acid was made by dissolving 8.00 g of H3PO4 (molar mass 98.0 g/mol) in mL of water. Calculate the mole fraction of H3PO4.. Assume water has a density of 1.00 g/mL.
0.0145
8.00 g H3PO4 × (1 mol / g H3PO4) = mol H3PO4
100.0 mL H2O × (1.00 g H2O / mL) × (1 mol / g H2O) = 5.55 mol H2O
Mole Fraction (H3PO4) = mol H3PO4 / [ mol H3PO mol H2O] =
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5. Molality
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6. Exercise
A solution of phosphoric acid was made by dissolving 8.00 g of H3PO4 (molar mass 98.0 g/mol) in mL of water. Calculate the molality of the solution. (Assume water has a density of 1.00 g/mL.)
0.816 m
8.00 g H3PO4 × (1 mol / g H3PO4) = mol H3PO4
100.0 mL H2O × (1.00 g H2O / mL) × (1 kg / 1000 g) = kg H2O
Molality = mol H3PO4 / kg H2O] = m
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7. Affecting aqueous solutions
Structural Effects:
Polarity
Pressure Effects:
Henry’s law
Temperature Effects:
Affecting aqueous solutions
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8. Pressure Effects
Henry’s law: C = kP
C = concentration of dissolved gas
k = constant
P = partial pressure of gas solute above the solution
Amount of gas dissolved in a solution is directly proportional to the pressure of the gas above the solution.
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9. A Gaseous Solute
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10. Temperature Effects (for Aqueous Solutions)
Although the solubility of most solids in water increases with temperature, the solubilities of some substances decrease with increasing temperature.
Predicting temperature dependence of solubility is very difficult.
Solubility of a gas in solvent typically decreases with increasing temperature.
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11. The Solubilities of Several Solids as a Function of Temperature
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12. The Solubilities of Several Gases in Water vs. Temperature
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13. Vapor pressure results from dynamic equilibrium between liquid state and gas state An Aqueous Solution and Pure Water in a Closed Environment: Eventually, pure solvent be transferred into solution state via evaporation/condensation equilibrium
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14. Vapor Pressures of Solutions
Mixing of nonvolatile solute reduces the chance of solvent molecule entering gas state, thus reducing vapor pressure. Nonvolatile solute lowers the vapor pressure of a solvent.
Raoult’s Law:
Psoln = observed vapor pressure of solution
solv = mole fraction of solvent
Psolv° = vapor pressure of pure solvent
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15. Practice: Vapor pressure for solution
The vapor pressure of a solution containing 107. g glycerin (molar mass 92.1 g/mol) in 266. g ethanol (molar mass 46.1 g/mol) is 104. torr at 35°C. Calculate the vapor pressure of pure ethanol at 35°C, assuming glycerin is nonvolatile nonelectrolyte in ethanol.
Mole fraction of solvent = 0.832
Vapor pressure of pure ethanol = 104/0.832 = 125 torr
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16. A Solution Obeying Raoult’s Law
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17. Liquid-liquid solutions where both components are volatile.
Nonideal Solutions
Liquid-liquid solutions where both components are volatile.
Modified Raoult’s Law:
Nonideal solutions behave ideally as the mole fractions approach 0 and 1, but deviate between 0 and 1.
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18. Vapor Pressure for a Solution of Two Volatile Liquids
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19. Summary of the Behavior of Various Types of Solutions
Interactive Forces Between Solute (A) and Solvent (B) Particles
Hsoln
T for Solution Formation
Deviation from Raoult’s Law
Example
A  A, B  B  A  B
Zero
None (ideal solution)
Benzene-toluene
A  A, B  B < A  B
Negative (exothermic)
Positive
Negative
Acetone-water
A  A, B  B > A  B
Positive (endothermic)
Ethanol-hexane
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20. Colligative Properties
Boiling-point elevation: Because of depressed vapor pressure of solvent in a solution, the boiling point of solution is higher than pure solvent.
Colligative Properties
Depend only on the number, not on the identity, of the solute particles in an ideal solution:
Boiling-point elevation
Freezing-point depression
Osmotic pressure
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21. Boiling-Point Elevation
Nonvolatile solute elevates the boiling point of the solvent.
ΔT = Kbmsolute
ΔT = boiling-point elevation
Kb = molal boiling-point elevation constant
msolute= molality of solute
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22. Freezing-Point Depression
When a solute is dissolved in a solvent, the freezing point of the solution is lower than that of the pure solvent.
ΔT = Kfmsolute
ΔT = freezing-point depression
Kf = molal freezing-point depression constant
msolute= molality of solute
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23. Changes in Boiling Point and Freezing Point of Water
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24. 100.35 °C ΔT = m Kb = (0.51 °C·kg/mol)(0.6938 mol/kg) = 0.35 °C
Exercise
A solution was prepared by dissolving g glucose in g water. The molar mass of glucose is g/mol. What is the boiling point of the resulting solution (in °C)? Glucose is a molecular solid that is present as individual molecules in solution. Kb = 0.51 °C·kg/mol.
mol glucose = mol
The change in temperature is ΔT = Kbmsolute. Kb is 0.51 °C·kg/mol. To solve formsolute, use the equation m = moles of solute/kg of solvent.
Moles of solute = (25.00 g glucose)(1 mol / g glucose) = mol glucose
Kg of solvent = (200.0 g)(1 kg / 1000 g) = kg water
msolute = ( mol glucose) / ( kg water) = mol/kg
ΔT = (0.51 °C·kg/mol)( mol/kg) = 0.35 °C.
The boiling point of the resulting solution is °C °C = °C.
Note: Use the red box animation to assist in explaining how to solve the problem.
Molality m = mol/ kg = m
ΔT = m Kb = (0.51 °C·kg/mol)( mol/kg) = 0.35 °C °C
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25. Osmotic pressure (): The pressure needed to stop the flow of solvent
Osmosis – flow of solvent into the solution through a semipermeable membrane.
Osmotic pressure (): The pressure needed to stop the flow of solvent
 = MRT (atm)
M = molarity of the solution
R = gas law constant
T = temperature (Kelvin)
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27. Salt causes minced water-rich plants/fruits to dehydrate
Osmosis in life
Salt causes minced water-rich plants/fruits to dehydrate
What happens when salt is applied to live tissues/organs/animals with thin membrane?
High concentration of salt from exterior of membrane causes water (solvent) leaving the interior of membrane.
The following videos are for scientific observation only, DO NOT imitate!!!
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28. Exercise: Determine molar mass using Colligative Properties
When 33.4 mg of a compound is dissolved in 10.0 mL of water at 25°C, the solution has an osmotic pressure of 558 torr. Calculate the molar mass of this compound.
111 g/mol
 = MRT
Find molarity of solution: M= /(RT)
The molar mass is 111 g/mol.
Note: Use the red box animation to assist in explaining how to solve the problem.
0.734/( *298) = M
Mol unknown = M x L = 3.00E-4 mol
Molar mass = mass/mol =3.34E-2 g/3.00E-4 mol = 111 g/mol
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29. Ionic compound dissociates in solution, resulting more particles.
van’t Hoff Factor, i
Ionic compound dissociates in solution, resulting more particles.
Example: 1 mole NaCl  1 mol Na+ + 1 mol Cl-
1 mole Fe(NO3)3  1 mol Fe mol NO3-
These “extra” ions enhance the colligative properties, thus a calibration factor i was introduced.
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30. Examples
The expected value for i can be determined for a salt by noting the number of ions per formula unit (assuming complete dissociation and that ion pairing does not occur).
NaCl = Na+ + Cl- i = 2
NH4NO3 = NH4+ + NO3- i = 2
K3PO4 = 3 K+ + PO43- i = 4
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31. Considering the dissociation of ionic solute, freezing-point depression, boiling-point elevation, and osmotic pressure equations become
Because of the dissociation, ionic compounds are more effective than nonelectrolytes with similar concentration in causing freezing-point depression, boiling-point elevation, and enhancing osmotic pressure.
Homemade icecream so easy 
Salt helps melt snow/ice on the road
Salt causes freshly cut vegetables to “liquefy” (osmosis through cell membrane)
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32. Practice: Predict the freezing point depression
The old fashion to make homemade ice-cream is to churn the ingredient (milk, sugar, vanilla, etc) in a container placed in a mixture of ice and salt. Which solution, 500. mL 1.0 M NaCl solution or 250. mL 0.80 M CaCl2 solution, will give the lower temperature? Freezing point constant Kf for water is 1.86 °C/kg. The molarity of solution is treated approximately same as the molality of solution.
500. mL NaCl: T  2x1.0x1.86 = 3.72°C
250. mL CaCl2 : T  3x0.80x1.86 = 4.46°C
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33. Practice: Predict the osmotic pressure
Seawater contains 3.5% salt, roughly 0.60 M NaCl. Calculate the minimal pressure needed at 298 K to purify the water by osmosis. Assume NaCl is completely dissociated.
 = 2 x 0.60 x x 298 = 30 atm.
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34. Incomplete separation of Ions: Ion Pairing
In diluted solutions, the ions are farther apart (free ions).
Ion Pairing: As concentration increases or charges are high, counter ions are more likely to attract each other, leading to “pairing”.
Ion pairing occurs to some extent in all electrolyte solutions, but is most important in concentrated solutions, and/or for highly charged ions.
The result of ion pairing is van’t Hoff facotr i < ideal value. For example, i = 1.3 for 0.05 m MgSO4 solution (ideal value = 2.0)
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35. Ion Pairing
At a given instant a small percentage of the sodium and chloride ions are paired and thus count as a single particle.
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36. Formation of a Liquid Solution
Separating the solute into its individual components (expanding the solute).
Overcoming intermolecular forces in the solvent to make room for the solute (expanding the solvent).
Allowing the solute and solvent to interact to form the solution.
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37. Steps in the Dissolving Process
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38. Steps in the Dissolving Process
Steps 1 and 2 require energy, since forces must be overcome to expand the solute and solvent.
Step 3 usually releases energy.
Steps 1 and 2 are endothermic, and step 3 is often exothermic.
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39. Concept Check
Explain why water and oil (a long chain hydrocarbon) do not mix. In your explanation, be sure to address how ΔH plays a role.
Oil is a mixture of nonpolar molecules that interact through London dispersion forces, which depend on molecule size. ΔH1 will be relatively large for the large oil molecules. The term ΔH3 will be small, since interactions between the nonpolar solute molecules and the polar water molecules will be negligible. However, ΔH2 will be large and positive because it takes considerable energy to overcome the hydrogen bonding forces among the water molecules to expand the solvent. Thus ΔHsoln will be large and positive because of the ΔH1 and ΔH2 terms. Since a large amount of energy would have to be expended to form an oil-water solution, this process does not occur to any appreciable extent.
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40. Processes that require large amounts of energy tend not to occur.
In General
One factor that favors a process is an increase in probability of the state when the solute and solvent are mixed.
Processes that require large amounts of energy tend not to occur.
Overall, remember that “like dissolves like”.
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41. A suspension of tiny particles in some medium.
Tyndall effect – scattering of light by particles.
Suspended particles are single large molecules or aggregates of molecules or ions ranging in size from 1 to 1000 nm.
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42. Types of Colloids
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43. Destruction of a colloid.
Coagulation
Destruction of a colloid.
Usually accomplished either by heating or by adding an electrolyte.
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