Chapter 11
Solutions and Their Properties
Solutions: Definitions
• A solution is a homogeneous mixture.
• A solution is composed of a solute dissolved in a solvent.
– When two compounds make a solution, the solute is the lesser
quantity, and the solvent is the greater quantity.
• Solutions can exist in all three physical states
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Solution Formation
There are three types interactions among particles in solutions:
Solvent – solvent
Solute – solute
Solute – solvent
Like Dissolves Like
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Miscible & Immiscible
• Two liquids that completely
dissolve in each other are
miscible liquids.
• Two liquids that are not
miscible in each other are
immiscible liquids.
• Polar water and nonpolar oil are
immiscible liquids and do not
mix to form a solution.
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Polar & Nonpolar Solvents
• A liquid composed of polar molecules is a polar solvent.
– Water and ethanol are polar solvents.
• A liquid composed of nonpolar molecules is a nonpolar
solvent.
– Hexane is a non-polar solvent
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Formation of Solutions
• When a soluble crystal is placed into a solvent, it
begins to dissolve.
• When a sugar crystal is placed
in water, the water molecules
attack the crystal and begin
pulling part of it away and into
solution.
• The sugar molecules are held
within a cluster of water
molecules called a solvent cage.
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Dissolving of Ionic Compounds
• In an ionic compound, the water molecules pull individual ions off of
the crystal.
• When a sodium chloride crystal is
placed in water, the water molecules
attack the edge of the crystal.
• The anions (Cl-) are surrounded by the
positively charged hydrogens on
water.
• The cations (Na+) are surrounded by
the negatively charged oxygen on
water.
• This process is called solvation
(hydration when the solvent is water),
which diminished the attraction of the
Na+ for the ClChapter 11
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Energetics of Solution Formation:
Introduction to Entropy
• We have stated that chemical and physical processes occur
spontaneously only if they go “downhill” energetically (give off
energy) so that the final state is more stable and lower in energy
than the initial state.
• However, we know that reactions that require energy (endothermic
reactions) will also occur.
• This is possible because energy is more than just heat. There are
other forms of energy present in reactions that allow it to be
spontaneous despite absorbing heat.
• Another factor that effects spontaneity of a reaction is the molecular
disorder of the reaction.
–The amount of molecular disorder (or randomness) in a system is called
the system’s Entropy (S)
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Entropy
• Entropy has units of J/K (Joules per Kelvin).
∆S = Sfinal – Sinitial
Positive value of ∆S indicates increased disorder.
Negative value of ∆S indicates decreased disorder.
Second Law of
Thermodynamics:
Reactions proceed in the
direction that increases the
entropy of the system plus
surroundings.
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Entropy and Enthalpy
• To decide whether a process is spontaneous,
both enthalpy and entropy changes must be
considered:
Spontaneous process: Decrease in enthalpy (–∆H).
Increase in entropy (+∆S).
Nonspontaneous process: Increase in enthalpy (+∆H).
Decrease in entropy (–∆S).
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Introduction to Gibbs Free Energy
• Gibbs Free Energy Change (∆G) weighs the relative
contributions of enthalpy and entropy to the overall
spontaneity of a process.
∆G = ∆H – T∆S
∆G < 0
∆G = 0
∆G > 0
Spontaneous
Equilibrium
Non-spontaneous
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Gibbs Free Energy
∆G = ∆H – T∆S
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Energetics of Solution Formation (∆Ssoln)
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Energetics of Solution Formation (∆Hsoln)
Exothermic ∆Hsoln
• The solute–solvent
interactions are stronger
than solute–solute or
solvent–solvent.
• Favorable process
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Energetics of Solution Formation (∆Hsoln)
Endothermic ∆Hsoln
• The solute–solvent
interactions are weaker
than solute–solute or
solvent–solvent.
• Unfavorable process.
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Formation of Solutions
1. Predict the relative solubilities in the following cases:
(a) Br2 in benzene (C6H6) and in water,
(b) KCl in carbon tetrachloride and in liquid ammonia,
(c) urea (NH2)2CO in carbon disulfide and in water.
2. Is iodine (I2) more soluble in water or in carbon disulfide
(CS2)?
3. Which would have the largest (most negative) hydration
energy and which should have the smallest? Al3+, Mg2+,
Na+
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Solubility
• The solubility of a compound is used to describe the
maximum amount of solute that can dissolve in a given
amount of solvent.
• Many factors affect a solute’s solubility:
•
•
•
•
Type of solute
Type of solvent
Temperature
Pressure (for solutions with gases)
• There are three terms we use to describe solubility:
Saturated
Unsaturated
Supersaturated
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Saturation of Solutions
• A solution containing exactly the
maximum amount of solute at a
given temperature is a saturated
solution.
• A solution that contains less than the
maximum amount of solute is an
unsaturated solution.
• Under certain conditions, it is
possible to exceed the maximum
solubility of a compound. A solution
with greater than the maximum
amount of solute is a supersaturated
solution.
Solid solute
Chapter 11
dissolves
crystallizes
Saturate Solution
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Concentration of Solutions
• The concentration of a solution tells us how much solute is
dissolved in a given quantity of solution or solvent.
• We often hear imprecise terms such as a “dilute solution” or a
“concentrated solution”.
• Four precise ways to express the concentration of a solution:
Mass percent (m %)
Mole Fraction (X)
Molality (m)
Molarity (M) (See Chapter 3)
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Mass Percent Concentration
• Mass percent concentration comparea the amount (g) of
solute to the amount of solution (g).
• The mass percent (m %) concentration is expressed as
the mass of solute dissolved in 100 g of solution.
mass of solute
× 100% = m %
mass of solution
g solute
g solute + g solvent
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× 100% = m %
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Mass Percent Concentration:
Parts per Million (ppm)
Mass of component
Parts per million (ppm) = Total mass of solution x 10
6
= % mass x 104
• One ppm gives 1 gram of solute per 1,000,000 g
or one mg per kg of solution. For dilute aqueous
solutions this is about 1 mg per liter of solution.
• What is a Part per Billion (ppb)?
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Other Concentration Units
• Mole Fraction (X): X A =
Moles of A
Total number of moles
• Molarity (M): Molarity =
• Molality (m): Molality =
Chapter 11
Moles of solute
Liters of SOLUTION
Moles of solute
Kilograms of SOLVENT
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Working Concentration Problems
• When working these types of problems you MUST first
identify the parts of your solution.
• Make a list of the solute, solvent and solution with any
numerical values as given in the problem.
• This will make your preparation of any conversion factors
much easier!
For Example:
A student prepares a 102.3 g of a solution of 5.62 g of
MgCl2 dissolved in water.
Identify the solute, solvent and solution
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Working Concentration Problems
• A sample of 0.892 g of potassium chloride (KCl) is
dissolved in 54.6 g of water. What is the percent by
mass of KCl in this solution?
• An aqueous solution is 5.50% H2SO4. How many
moles of sulfuric acid (MM = 98.08 g/mol) are
dissolved in 250.0 g of the solution?
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Working Concentration Problems
• Molality from Mass: Calculate the molality of a
sulfuric acid solution containing 24.4 g of
sulfuric acid in 198 g of water. The molar mass
of sulfuric acid is 98.08 g.
• Molality from Molarity: Calculate the molality
of a 5.86 M ethanol (C2H5OH) solution whose
density is 0.927 g/ml.
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Working Concentration Problems
• Mole Fraction from Molality: An aqueous
solution is 0.258 m in glucose (MM = 180.2
g/mol). What is the mole fraction of the glucose?
• Mass from Molality: What mass (in grams) of
a 0.500 m aqueous solution of urea [(NH2)2CO,
MM = 60.1 g/mol] would you use to obtain
0.150 mole of urea?
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Rate of Dissolving
• There are three ways we can speed up the rate
of dissolving for a solid compound.
• Heating the solution:
– This increases the kinetic energy of the solvent and
the solute is attacked faster by the solvent molecules.
• Stirring the solution:
– This increases the interaction between solvent and
solute molecules.
• Grinding the solid solute:
– There is more surface area for the solvent to attack.
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Solubility and Temperature
• For most solids,
solubility increases as
the temperature
increases.
• For most gases,
solubility in a liquid
solution decreases as
the temperature
increases.
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Solubility and Pressure
• Henry’s Law: The solubility of a gas (c) is directly
proportional to the pressure (P) of the gas over the solution.
c = kP
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Solubility and Pressure
• Calculate the molar concentration of O2 in water at 25°C
for a partial pressure of 0.22 atm. The Henry’s law
constant for O2 is 3.5 x 10–4 mol/(L·atm).
• The solubility of CO2 in water is 3.2 x 10–2 M at 25°C
and 1 atm pressure. What is the Henry’s law constant for
CO2 in mol/(L·atm)?
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Colligative Properties
• Colligative Properties: Depend only on the number of
solute particles in solution. These affect properties of the
solvent.
• There are four main colligative properties:
Vapor pressure lowering
Freezing point depression
Boiling point elevation
Osmotic pressure
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Colligative Properties: Vapor Pressure
• Vapor Pressure is the pressure exerted on the surface of a liquid
by the vapor above it.
• When a nonvolatile solute (i.e. no vapor pressure of its own) is
mixed with a solvent, the solute molecules displace solvent
molecules at the surface of the solution.
• As a result, the vapor pressure decreases since fewer gas
molecules are needed to equalize the escape rate and capture
rates at the liquid surface.
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Colligative Properties: Vapor Pressure
• Raoult’s Law describes the relationship between the vapor
pressure of the solution (Psoln) and the vapor pressure of the
solvent (Psolv):
Psoln = Psolv • Xsolv
• The change in the vapor pressure (∆Psoln) is dependent on the
amount of solute that has been added to the solvent:
∆Psoln = Psolv • Xsolute
∆Psoln = Psolv – Psoln
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Colligative Properties: Vapor Pressure
• The vapor pressure of a glucose (C6H12O6) solution is
17.01 mm Hg at 20°C, while that of pure water is 17.25
mm Hg at the same temperature. Estimate the molality of
the solution.
• How many grams of NaBr must be added to 250 g of
water to lower the vapor pressure by 1.30 mm Hg at
40°C? The vapor pressure of water at 40°C is 55.3 mm
Hg.
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Colligative Properties: Vapor Pressure
• When you mix two volatile liquids, the overall vapor
pressure (Ptot) is the sum of the vapor pressure of the
individual components (Dalton’s Law!)
Ptot = PA + PB
• The individual vapor pressures (PA and PB) are
determined by Raoult’s Law:
Ptot = PA + PB = (PA • XA) + (PB • XB)
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Colligative Properties: Vapor Pressure
• Two miscible liquids, A and B, have vapor pressures of 250
mm Hg and 450 mm Hg, respectively. They were mixed in
equal molar amounts. What is the total vapor pressure of the
mixture and what are their mole fractions in the vapor
phase?
• What is the vapor pressure (in mm Hg) of a solution
prepared by dissolving 25.0 g of ethanol (C2H5OH) in 100.0
g of water at 25.0°C? The vapor pressure of pure water is
23.8 mm Hg and the vapor pressure of ethanol is 61.2 mm
Hg at 25.0°C.
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Colligative Properties: Boiling Point Elevation
• The vapor pressure
decreases when a solution
is formed.
• What do you think should
happen to the boiling
point?
∆Tb = Tbsoln – Tbsolv
Chapter 11
∆Tb = Kb • m
Kb is the molal boiling-point
elevation constant. 37
Colligative Properties: Freezing Point Depression
• The vapor pressure
decreases when a solution
is formed.
• What do you think should
happen to the freezing
point?
∆Tf = Tfsolv – Tfsoln
Chapter 11
∆Tf = Kf • m
Kf is the molal Freezing-point
depression constant. 38
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Boiling Point Elevation &
Freezing Point Depression
• How many grams of ethylene glycol antifreeze, CH2(OH)CH2(OH), must you
dissolve in one liter of water to get a freezing point of –20.0°C. The molar
mass of ethylene glycol is 62.01 g. For water, Kf = 1.86 (°C·kg)/mol. What
will be the boiling point?
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The van’t Hoff Factor (i)
• For ionic compounds, dissolving results in ions in solution.
• The van’t Hoff Factor (i) generally equals the number of ions
produced from each molecule of a compound upon dissolving.
Use the subscripts to determine the number of ions.
i = 1 for CH3OH
i = 3 for CaCl2
i = 2 for NaCl
i = 5 for Ca3(PO4)2
• However, this dissolving is not always 100% meaning that you
cannot determine the amount of each ion in solution merely from
the balanced equation. Therefore, the van’t Hoff Factor has also
been determined experimentally for many ionic compounds.
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The van’t Hoff Factor (i)
• Whenever you are working a problem that involves an
ionic compound, you MUST include the van’t Hoff
factor in your calculations:
∆Tb = i • Kb • m
∆Tf = i • Kf • m
∆Psoln = i • Psolv • Xsolute
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Boiling Point Elevation &
Freezing Point Depression
• What is the molality of an aqueous solution of KBr
whose freezing point is –2.95°C? Kf for water is 1.86
(°C·kg)/mol.
• What is the freezing point (in °C) of a solution prepared
by dissolving 7.40 g of K2SO4 in 110 g of water? The
value of Kf for water is 1.86 (°C·kg)/mol.
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Boiling Point Elevation &
Freezing Point Depression
• A 7.85 g sample of a compound with the
empirical formula C5H4 is dissolved in 301 g
of benzene. The freezing point of the solution
is 1.05°C below that of pure benzene. What
are the molar mass and molecular formula of
this compound?
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Osmosis and Osmotic Pressure (Π)
• Osmosis is the selective passage of solvent molecules through a
porous membrane from a dilute solution to a more concentrated
one.
• Osmotic pressure (Π) is
the pressure required to
stop osmosis.
Π = i•(MM)•R•T
R = 0.08206 (L⋅atm)/(mol⋅K)
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Osmosis and Osmotic Pressure (Π)
• Isotonic: Solutions have equal concentration of solute,
and so equal osmotic pressure.
• Hypertonic: Solution with higher concentration of
solute.
• Hypotonic: Solution with lower concentration of
solute.
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Osmosis and Osmotic Pressure (Π)
• The average osmotic pressure of seawater is
about 30.0 atm at 25°C. Calculate the molar
concentration of an aqueous solution of urea
[(NH2)2CO] that is isotonic with seawater.
• What is the osmotic pressure (in atm) of a 0.884
M sucrose solution at 16°C?
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Osmosis and Osmotic Pressure (Π)
• A 202 ml benzene solution containing 2.47 g of an
organic polymer has an osmotic pressure of 8.63 mm
Hg at 21°C. Calculate the molar mass of the polymer.
• What is the molar mass of sucrose if a solution of 0.822
g of sucrose in 300.0 mL of water has an osmotic
pressure of 149 mm Hg at 298 K?
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Uses of Colligative Properties
Desalination
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Uses of Colligative Properties
• Fractional Distillation is the separation of volatile liquid
mixtures into fractions of different composition.
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Uses of Colligative Properties
• Fractional distillation can be represented on a phase
diagram by plotting temperature against composition.
Chapter 11
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