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11.1 Solution Composition
11.2 The Energies of Solution Formation
11.3 Factors Affecting Solubility
11.4 The Vapor Pressures of Solutions
11.5 Boiling-Point Elevation and Freezing-Point Depression
11.6 Osmotic Pressure
11.7 Colligative Properties of Electrolyte Solutions
11.8 Colloids
Solution Properties
Various Types of Solutions
ExampleState of Solution
State of Solute
State of Solvent
Air, natural gas Gas Gas Gas
Vodka, antifreeze Liquid Liquid Liquid
Brass Solid Solid Solid
Carbonated water (soda) Liquid Gas Liquid
Seawater, sugar solution Liquid Solid Liquid
Hydrogen in platinum Solid Gas Solid
Solution Composition
AA
moles of soluteMolarity ( ) =
liters of solution
mass of soluteMass (weight) percent = 100%
mass of solution
molesMole fraction ( ) =
total moles of solution
moles of soluteMolality ( ) =
kilogram of s
M
molvent
Molarity
moles of soluteMolarity ( ) =
liters of solution M
Exercise #1
You have 1.00 mol of sugar in 125.0 mL of solution. Calculate the concentration in units of molarity.
8.00 M
Exercise #2
You have a 10.0 M sugar solution. What volume of this solution do you need to have 2.00 mol of sugar?
0.200 L
Exercise #3
Consider separate solutions of NaOH and KCl made by dissolving 100.0 g of each solute in 250.0 mL of solution. Calculate the concentration of each solution in units of molarity.
10.0 M NaOH
5.37 M KCl
Mass Percent
mass of soluteMass (weight) percent = 100%
mass of solution
Exercise #4
What is the percent-by-mass concentration of glucose in a solution made my dissolving 5.5 g of glucose in 78.2 g of water?
6.6%
Mole Fraction
AA
molesMole fraction ( ) =
total moles of solution
Exercise #5
A solution of phosphoric acid was made by dissolving 8.00 g of H3PO4 in 100.0 mL of water. Calculate the mole fraction of H3PO4.
(Assume water has a density of 1.00 g/mL.)
0.0145
Molality
moles of soluteMolality ( ) =
kilogram of solvent m
Exercise #6
A solution of phosphoric acid was made by dissolving 8.00 g of H3PO4 in 100.0 mL of water. Calculate the molality of the solution. (Assume water has a density of 1.00 g/mL.)
0.816 m
Formation of a Liquid Solution
1. Separating the solute into its individual components (expanding the solute).
2. Overcoming intermolecular forces in the solvent to make room for the solute (expanding the solvent).
3. Allowing the solute and solvent to interact to form the solution.
Steps in the Dissolving Process
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.
Enthalpy (Heat) of Solution
• Enthalpy change associated with the formation of the solution is the sum of the ΔH values for the steps:
ΔHsoln = ΔH1 + ΔH2 + ΔH3
• ΔHsoln may have a positive sign (energy absorbed) or a negative sign (energy released).
Enthalpy (Heat) of Solution
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.
The Energy Terms for Various Types of Solutes and Solvents
H1 H2 H3 Hsoln Outcome
Polar solute, polar solvent
Large Large Large, negative
Small Solution forms
Nonpolar solute, polar solvent
Small Large Small Large, positive
No solution forms
Nonpolar solute, nonpolar solvent
Small Small Small Small Solution forms
Polar solute, nonpolar solvent
Large Small Small Large, positive
No solution forms
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”.
• Structural Effects: Polarity – “like dissolves like”
• Pressure Effects: Henry’s law – for solubility of gases
• Temperature Effects: Affecting aqueous solutions
Factors Affecting Solubility
Pressure Effects
• Henry’s law: c = kPc = 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 partial pressure of gas above the solution.
A Gaseous Solute
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.
The Solubilities of Several Solids as a Function of Temperature
The Solubilities of Several Gases in Water
Ideal Solution: One that obeys Raoult’s Law
Ideal Solutions Consisting of Two Volatile Liquids
• Two volatile Liquids form ideal solution if: – they are structurally very similar, and
– molecular interactions between nonidentical molecules were relatively similar to identical molecules.
• The vapor of each liquid obeys Raoult’s Law:
PA = XAPoA; PB = XBPo
B
PT = PA + PB = XAPoA + XBPo
B
(X : mole fraction; Po : vapor pressure of pure liquid)
Summary of the Behavior of Various Types of Solutions of Two Volatile Liquids
Interactive Forces Between Solute (A) and
Solvent (B) ParticlesHsoln
T for Solution
Formation
Deviation from
Raoult’s Law
Example
A A, B B A B Zero ZeroNone (ideal
solution)
Benzene-toluene
A A, B B < A BNegative
(exothermic)Positive Negative
Acetone-water
A A, B B > A BPositive
(endothermic)Negative Positive
Ethanol-hexane
Vapor Pressure for a Solution of Two Volatile Liquids
Laboratory Fractional Distillation Apparatus
Fractional Distillation Towers in Oil Refinaries
Refined Crude Oil Mixtures
Concept Check
For each of the following solutions, would you expect it to be relatively ideal (with respect to Raoult’s Law), show a positive deviation, or show a negative deviation?
a) Hexane (C6H14) and chloroform (CHCl3)
b) Ethyl alcohol (C2H5OH) and water
c) Hexane (C6H14) and octane (C8H18)
Exercise #7
• A solution of benzene (C6H6) and toluene (C7H8) contains 50.0% benzene by mass. The vapor pressures of benzene and pure toluene at 25oC are 94.2 torr and 28.4 torr, respectively. Assuming ideal behavior, calculate the following:
(a) The mole fractions of benzene and toluene;
(b) The vapor pressure of each component in the mixture, and the total vapor pressure above the solution.
(c) The composition of the vapor in mole percent.
Exercise #8
• A solution composed of 24.3 g acetone (CH3COCH3) and 39.5 g of carbon disilfide (CS2) has a measured vapor pressure of 645 torr at 35oC.
(a) Is the solution ideal or nonideal?
(b) If not, does it deviate positively or negatively from Raoult’s law?
(c) What can you say about the relative strength of carbon disulfide-acetone interactions compared to the acetone-acetone and carbon disulfide-carbon disulfide interaction?
(Vapor pressures at 35oC of pure acetone and pure carbon disulfide are 332 torr and 515 torr, respectively.)
An Aqueous Solution and Pure Water in a Closed Environment
Liquid/Vapor Equilibrium
Vapor Pressure Lowering: Addition of a Solute
Vapor Pressures of Solutions of Nonvolatile Solutes
• Nonvolatile solute lowers the vapor pressure of solvent.
• Raoult’s Law:
Psoln = vapor pressure of solution
solv = mole fraction of solvent
= vapor pressure of pure solvent
soln solv solv = P P
solvP
Lowering of solvent vapor pressure Freezing-point depression Boiling-point elevation Osmotic pressure
Colligative properties depend only on the number, not on the identity, of the solute particles in an ideal solution.
Colligative Properties
Lowering of Solvent Vapor Pressure
• The presence of nonvolatile solute particles lowers the number of solvent molecules in the vapor that is in equilibrium with the solution.
• The solvent vapor pressure is lowered;• Assuming ideal behavior, the lowering of vapor
pressure is proportional to the mole fraction of solute:
P = Xsolute.Posolvent (for nonelectrolytes)
= iXsolute.Posolvent (for electrolytes)
(i is the van’t Hoff’s factor, which approximately relates to the number of ions per formula unit of the compound)
Changes in Boiling Point and Freezing Point of Water
• When a solute is dissolved in a solvent, the freezing point of the solution is lower than that of the pure solvent.
• ΔT = Kfmsolute (for nonelectrolytes)
= iKfmsolute (for electrolytes)
ΔT = freezing-point depression
Kf = freezing-point depression constant
msolute = molality of solute
Freezing-Point Depression
Freezing Point Depression: Solid/Liquid Equilibrium
Freezing Point Depression: Addition of a Solute
Freezing Point Depression: Solid/Solution Equilibrium
• Nonvolatile solute elevates the boiling point of the solvent.
• ΔT = Kbmsolute
ΔT = boiling-point elevation
Kb = boiling-point elevation constant
msolute = molality of solute
Boiling-Point Elevation
Boiling Point Elevation: Liquid/Vapor Equilibrium
Boiling Point Elevation: Addition of a Solute
Boiling Point Elevation: Solution/Vapor Equilibrium
Exercise #9
• What mass of ethylene glycol (C2H6O2), in grams, must be added to 1.50 kg of water to produce a solution that boils at 105oC?
(Boiling point elevation constant for water is Kb = 0.512oC/m)
At what temperature will the solution freeze?
(Freezing point depression constant for water is Kf = 1.86oC/m)
• Osmosis – flow of solvent into the solution through a semipermeable membrane.
• = MRT
= osmotic pressure (atm)
M = molarity of the solution
R = gas law constant
T = temperature (Kelvin)
Osmotic Pressure
Osmosis
• The relationship between the moles of solute dissolved and the moles of particles in solution is usually expressed as:
van’t Hoff Factor, i
moles of particles in solution =
moles of solute dissolvedi
Modified Equations for the Colligative Properties of Electrolytes
= T imK
= iMRT
Ion Pairing• At a given instant a small percentage of the
sodium and chloride ions are paired and thus count as a single particle.
• Ion pairing is most important in concentrated solutions.
• As the solution becomes more dilute, the ions are farther apart and less ion pairing occurs.
• Ion pairing occurs to some extent in all electrolyte solutions.
• Ion pairing is most important for highly charged ions.
Ion Pairing
Exercise #10
A solution was prepared by dissolving 25.00 g glucose in 200.0 g water. The molar mass of glucose is 180.16 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.
100.35 °C
Exercise #11
You take 20.0 g of a sucrose (C12H22O11) and NaCl mixture and dissolve it in 1.0 L of water. The freezing point of this solution is found to be -0.426°C. Assuming ideal behavior, calculate the mass percent composition of the original mixture, and the mole fraction of sucrose in the original mixture.
72.8% sucrose and 27.2% sodium chloride;
mole fraction of the sucrose is 0.313
Exercise #12
A plant cell has a natural concentration of 0.25 m. You immerse it in an aqueous solution with a freezing point of – 0.246°C. Will the cell explode/expand, shrivel, or do nothing?
Exercise #13
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
• 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 i = 2 KNO3 i = 2
Na3PO4 i = 4
Examples
Colloidal Mixtures
• 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.
Scattering of Light by Colloid Particles
Tyndall Effect of Colloidal Mixture
Tyndall Effect of Morning Mist
Types of Colloids
Micelle – A Colloidal Suspension
Micelle in Soap Bubbles
• Destruction of a colloid.
• Usually accomplished either by heating or by adding an electrolyte.
Coagulation