Chapter 11 Properties of Solutions

1
Chapter 11Properties of Solutions
2
Solution Compositions
Chapter 11 Section 1

• Homogeneous mixtures are usually called
  solutions.
• They can be liquid, gas, or solid.

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Solution Compositions
Chapter 11 Section 1

• Homogeneous mixtures are usually called
  solutions.
• They can be liquid, gas, or solid.
• A solute a substance being dissolved.
• A solvent a dissolving medium.
• Dilute a relatively small amount of solute
  present.
• Concentrated a relatively large amount of
  solute present.

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Solution Compositions
Chapter 11 Section 1

• Molarity mol of solute per liter M
• of solvent
• Mass percent of mass of solute in mass the
  solution
• Mole fraction ration of of moles of ?A one
  component to the total of moles.
• Molality mol of solute per kilo- m gram of
  solvent
• Normality of equivalents per liter N of
  solution.

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Calculation of Solution Compositions
Chapter 11 Section 1

• Sample Exercise 11.2
• In an automobile battery there is 3.75 M
  sulfuric acid solution with a density of
  1.230g/mL. Calculate mass percent, molarity, and
  normality of the sulfuric acid solution.

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Calculation of Solution Compositions
Chapter 11 Section 1

• Sample Exercise 11.2
• In an automobile battery there is 3.75 M
  sulfuric acid solution with a density of
  1.230g/mL. Calculate mass percent, molarity, and
  normality of the sulfuric acid solution.

7
Normality
Chapter 11 Section 1

• Normality number of equivalents per one liter of
  solution.
• Equivalents depend on the reaction taking place
  in the solution.
• Acid-base reactions.
• Oxidation-reduction reactions.
• For Acid-base reactions the equivalent is
  related to how many protons (H) or hydroxyl ions
  (OH) are generated in the reaction.
• Equivalent mass the molar mass of the acid (or
  base) divided by its equivalent.

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Normality
Chapter 11 Section 1

• For oxidation-reduction reactions
• MnO4 158 158/5 1M 5N
• MnO4 5e ? Mn2 4H2O

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Energies of Solution Formation
Chapter 11 Section 2

• Like dissolves like.
• Polar solvents (water, ethanol, liquid methanol,
  etc.) can dissolve polar/ionic solutes.
• Nonpolar solvents (benzene, hexane, chloroform
  CCl4) can dissolve nonpolar solutes.
• Dose water dissolve oil?
• Does water dissolve NaCl?

10
Energies of Solution Formation
Chapter 11 Section 2

• The formation of a liquid solution is assumed to
  take place in three distinct steps.
• ?H1 ?H2 ?H3 ?Hsoln (Enthalpy of solution)

11
Energies of Solution Formation
Chapter 11 Section 2

• ?H1 ?H2 ?H3 ?Hsoln (Enthalpy of solution)

• ?H1 and ?H2 are endothermic

• ?H3 is exothermic

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Energies of Solution Formation
Chapter 11 Section 2

• ?H1 ?H2 ?H3 ?Hsoln (Enthalpy of solution)

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Energies of Solution Formation
Chapter 11 Section 2

• ?H1 ?H2 ?H3 ?Hsoln (Enthalpy of solution)

• Oil in water
• ?H1 will be small.
• ?H2 will be large and ve.
• ?H3 will be small.

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Energies of Solution Formation
Chapter 11 Section 2

• NaCl in water
• ?H1 will be large and ve.
• ?H2 will be large and ve.
• ?H3 will be large and -ve.
• ?H2 ?H3 ?Hhyd (Enthalpy of hydration)
• ?H1 786 kJ/mol
• ?H2 ?H3 783 kJ/mol
• ?Hsoln 3 kJ
• Then, why salt is highly soluble in water??

NaCl(s)? Na(g) Cl(g)
Hydration H2O(l) Na(g) Cl(g) ? Na(aq)
Cl(aq)
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Energies of Solution Formation
Chapter 11 Section 2

• Substances naturally tend to higher probability.
• A factor that favors a process is an increase in
  probability.

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Energies of Solution Formation
Chapter 11 Section 2

• Two opposing factors
• The tendency of system to achieve higher
  probabilities.
• The large amount of energy needed to carry out
  that process.
• A process that requires higher energy tends not
  to occur. However, for a small amount of required
  energy, the process tends to happen due to the
  increase in probability.

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Energies of Solution Formation
Chapter 11 Section 2
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Factors Affecting Solubility
Chapter 11 Section 3

• Structure effect.
• Pressure effect.
• Temperature effect.

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Factors Affecting Solubility
Chapter 11 Section 3

• Structure effect.
• Structure ? Polarity ? Solubility
• Vitamins
• Fat-soluble vitamins (vitamins A, D, E, and K)
• Water-soluble vitamins (vitamins B and C)
• The different structures of these two types of
  vitamins explain the solubility differences.

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Factors Affecting Solubility
Chapter 11 Section 3
Vitamin A
Vitamin C
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Factors Affecting Solubility
Chapter 11 Section 3

• Structure effect.
• Structure ? Polarity ? Solubility
• Vitamins
• Fat-soluble vitamins (vitamins A, D, E, and K)
• Water-soluble vitamins (vitamins B and C)
• The different structures of these two types of
  vitamins explain the solubility differences.
• Nonpolar materials are usually called hydrophobic
  (water-fearing).
• Polar materials are usually called hydrophilic
  (water-loving).

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Factors Affecting Solubility
Chapter 11 Section 3

• Pressure effect.
• It greatly affects the solubility of gases.
• In Carbonated drinks (Pepsi, Coca Cola, etc.),
• CO2 is dissolved in the soda (high concentration
  under high pressure).
• Henrys Law
• The amount (concentration) of the dissolved gas
  in a solution is directly proportional to the
  pressure of the gas above the solution.
• C k P

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Henrys Law
Chapter 11 Section 3
Equilibrium being disturbed
Equilibrium being returned
Equilibrium
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Henrys Law
Chapter 11 Section 3

• Henrys law is only applied when there is no
  chemical reaction between the solute and solvent.
• Henrys law is hold when oxygen is dissolved in
  water.
• Henrys law is NOT hold when HCl(g) is dissolved
  in water (dissociation taking place).
• HCl(g) H(aq) Cl(aq)

H2O
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Application of Henrys Law
Chapter 11 Section 3

• Sample Exercise 11.4
• A certain soft drink is bottled at 25ºC with CO2
  gas at a pressure of 5.0 atm over the liquid. If
  the partial pressure of CO2(g) in the atmosphere
  is 4.010-4 atm, calculate the equilibrium
  concentration for CO2(g) in the soda both before
  and after the bottle is opened. The Henrys law
  constant for CO2 is 3.110-2 mol/Latm _at_ 25ºC.

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Factors Affecting Solubility
Chapter 11 Section 3

• Temperature effect (in aqueous solution).
• Solids The solubility of most of the solids
  increases with the increase of temperature. For
  some solids, such as Na2SO4 and Ce2(SO4)3 , the
  solubility decreases with increasing temperature.

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Factors Affecting Solubility
Chapter 11 Section 3

• Temperature effect (in aqueous solution).
• Solids The solubility of most of the solids
  increases with the increase of temperature. For
  some solids, such as Na2SO4 and Ce2(SO4)3 , the
  solubility decreases with increasing temperature.
• Gases The solubility of gases in water decreases
  with temperature.
• Thermal pollution
• Boiler scale

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Vapor Pressure of Solutions
Chapter 11 Section 4

• Generally, nonvolatile solutes tend not to escape
  from the solvents. Consequently, adding a solute
  (nonvolatile) to a solvent lowers the vapor
  pressure of the solvent.

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Raoults Law
Chapter 11 Section 4

• Adding volatile solutes to the solvent decreases
  the number of solvent molecules per unit volume.
  Thus, the rate of solvent molecules trying to
  escape from the surface of the solution will be
  lower, leading to a lower vapor pressure.

Nonvolatile solutes dilute the solvent
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Raoults Law
Chapter 11 Section 4

• Psoln ?solvent Pºsolvent
• Applying this law to
• A solution of half solute and half solvent.
• A solution with three-fourth of its molecules are
  solvent.

Raoults law is a linear relationship
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Application of Raoults Law
Chapter 11 Section 4

• Sample Exercise 11.5
• A solution made by dissolving 158.0g of table
  sugar (MM342.3 g/mol) in 643.5 cm3 of pure water
  _at_ 25ºC. The density and vapor pressure of water _at_
  25ºC is 0.9771 g/cm3 and 23.76 torr,
  respectively. Calculate the vapor pressure of the
  solution _at_ 25ºC.

32
Application of Raoults Law
Chapter 11 Section 4

• Sample Exercise 11.6
• A solution prepared by dissolving 35.0g Na2SO4
  (MM142 g/mol) in 175g of pure water _at_ 25ºC. The
  vapor pressure of water _at_ 25ºC is 23.76 torr.
  Calculate the vapor pressure of the solution _at_
  25ºC.

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Nonideal Solutions
Chapter 11 Section 4

• For solutes that are relatively volatile, the
  partial pressure of the solution is given by
• Ptotal Psolvent Psolvent PA PB
• ?APºA ?BPºB
• Raoults law deals with ideal solutions that are
  never perfectly achieved (a situation similar to
  ideal gases).

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Enthalpy of solution
Chapter 11 Section 4

• ?H1 ?H2 ?H3 ?Hsoln (Enthalpy of solution)

?Hsoln is exothermic
?Hsoln is endothermic
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Nonideal Solutions
Chapter 11 Section 4
?

• ? ?H1 ?H2 ?H3
• ?Hsoln is nearly zero
• (Nearly ideal solution)

Benzene
Toluene
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Nonideal Solutions
Chapter 11 Section 4
?

• ? ?H1 ?H2 lt ?H3
• ?Hsoln is large and ve
• (Exothermic process)

d
d
Acetone
Water
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Nonideal Solutions
Chapter 11 Section 4
?

• ? ?H1 ?H2 gt ?H3
• ?Hsoln is large and ve
• (Endothermic process)

Ethanol
Hexane
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Deviations of Nonideal Solutions from Raoults Law
Chapter 11 Section 4

• Strong solute-solvent interaction gives a vapor
  pressure lower than that predicted by Raoults
  law.
• ?H1 ?H2 lt ?H3
• Strong solute-solute / solvent-solvent
  interactions give a vapor pressure higher than
  that predicted by Raoults law.
• ?H1 ?H2 gt ?H3

39
Boiling-Point Elevation and Freezing-Point
Depression
Chapter 11 Section 5

• Colligative properties
• Boling-point elevation.
• Freezing-point depression.
• Osmotic pressure.
• Colligaive properties for solutions depend on the
  number of the solute particles (solutes
  concentration) in an ideal solution, NOT on the
  identity of the solutes.

40
Boiling-Point Elevation
Chapter 11 Section 5

• Nonvolatile solutes cause the vapor pressure of
  the solvent to decrease. Thus, a higher
  temperature is needed to boil the solution as
  compared to the pure liquid state.
• A nonvolatile solute elevates the boiling point
  of the solvent.

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Boiling-Point Elevation
Chapter 11 Section 5

• The boiling point elevation depends on the
  concentration of the solute.
• ?T Kb msolute
• ?T Tb.p.(solution) Tb.p.(pure solvent)
• m molality of the solute in solution.
• Kb molal boiling-point elevation constant.

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Boiling-Point Elevation
Chapter 11 Section 5

• ?T Kb msolute

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Freezing-Point Depression
Chapter 11 Section 5

• The m.p. temperature occurs when the vapor
  pressures of the liquid and solid are identical
  (equilibrium condition).
• Adding a solute to a solvent decreases the vapor
  pressure of the solvent.
• In order to return to the equilibrium condition
  between the liquid and solid phases, the
  temperature will have to be lowered (depressed).
• For water with solutes (like ethylene glycol or
  salt), the freezing point temperature will be
  less than 0ºC.

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Freezing-Point Depression
Chapter 11 Section 5

• For H2O
• (a) At 0ºC, ice and water are at equilibrium.
• (b) For water with solute, in order to reduce
  the rate of molecules escaping the ice cube, the
  temperature should be lowered until the
  equilibrium is re-achieved.

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Freezing-Point Depression
Chapter 11 Section 5

• ?T Kf msolute
• ?T Tb.p.(solution) Tb.p.(pure solvent)
• m molality of the solute in solution.
• Kf molal freezing-point elevation constant.

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Sample Exercise 11.8
Chapter 11 Section 5

• A solution was prepared by dissolving 18.00g of
  glucose in 150.0g of water. The resulting
  solutions b.p was fount to be 100.34ºC.
  Calculate the molar mass of glucose.

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Osmotic Pressure
Chapter 11 Section 6

• A solution and pure solvent are separated by a
  semipermeable membrane (osmosis), which allows
  solvent and not solute molecules to pass through.
• The excess hydrostatic pressure on the
  solution is called the osmotic pressure.

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Osmotic Pressure
Chapter 11 Section 6

• In the U-tube shown, the normal flow of the
  solvent through the membrane can be prevented by
  applying an external pressure to the solution.
• Osmotic pressure of a solution is the minimum
  pressure that stops the osmosis.
• ? MRT
• ? The osmotic pressure.
• M Solution molarity.
• R Gas constant.
• T Kelvin temperature.

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Osmotic Pressure
Chapter 11 Section 6

• ? MRT
• ? The osmotic pressure.
• M Solution molarity.
• R Gas constant.
• T Kelvin temperature.

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Determining Molar Mass from Osmotic Pressure
Chapter 11 Section 6

• Sample Exercise 11.11
• 1.0010-3g of a certain protein 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_ 25ºC. Calculate the molar mass of the
  protein.

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Applications of Osmotic Pressure
Chapter 11 Section 6

• Dialysis is a special type of osmosis. In
  dialysis, the membrane allows transfer of solvent
  molecules and small solute molecules /ions.
• One important application of dialysis is the
  artificial kidney machine that is used to purify
  blood.

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Reverse Osmosis
Chapter 11 Section 6

• Reverse osmosis occurs when the external pressure
  is larger than the osmotic pressure of the
  solution.
• The membrane in this case acts as a molecular
  filter to remove solute particles from the
  solution.
• One major application is seawater desalination.

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Colligative Properties of Electrolyte Solutions
Chapter 11 Section 7

• Compare the freezing point depression, ?T Kf
  msolute ,for the following aqueous solutions
• 0.1 m glucose solution
• ?T 0.186ºC.
• 0.1 m sodium chloride solution
• ?T 0.37ºC.
• 0.1 m FeCl4 solution
• ?T is expected to be 0.74ºC.
• Vant Hoff factor
• i

Expected Observed
i 1.0 1.0
i 2.0 1.9
i 4.0 3.4
moles of particles in solution
moles of solute dissolves
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Colligative Properties of Electrolyte Solutions
Chapter 11 Section 7
55
Colligative Properties of Electrolyte Solutions
Chapter 11 Section 7

• Some observed is are in general smaller than the
  expected ones by vant Hoff equation due to the
  ion pairing effect.
• Observed i becomes less smaller than the expected
  i when the solution concentration is diluted.
• The colligative properties of electrolyte
  solutions can be described using the equations we
  studies earlier
• ?T iKf msolute
• ? iMRT

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Colligative Properties of Electrolyte Solutions
Chapter 11 Section 7
57
Colligative Properties of Electrolyte Solutions
Chapter 11 Section 7