Solutions

1
Solutions

• Chapter 14

2
Types of Solution

• A solution is a homogeneous mixture of two or
  more substances.
• Solvent medium in which the solute dissolves
• Usually a liquid
• Usually the most abundant species
• 10 grams of H2O(l) in 25 grams of CH3OH(l)
• Solute substance that is dissolved
• Discuss (review) homogeneous and heterogeneous
  mixtures

3
Spontaneity of the Dissolution Process

• Dissolving a substance in a solvent (usually a
  liquid)
• Dissolving with a reaction
• 2Na(s) 2H2O(l) ? 2Na(aq) 2OH-(aq) H2(g)
• Dissolving without a reaction
• We will focus on the latter type of dissolution.

4
Spontaneity of the Dissolution Process

• Two major factors determine the dissolution of
  solutes
• Change in energy, ?Hsolution
• An _____ process (a decrease in energy) favors
  dissolution.
• An _____ process (an increase in energy) does not
  favor dissolution.
• Change in disorder or randomness, ?Smixing
• ______ in disorder favors dissolution.
• ______ in disorder does not favor dissolution.
• When a substance dissolves the disorder of the
  system almost always increases.

5
Spontaneity of the Dissolution Process

• Heat of solution, ?Hsolution, primarily depends
  on the strength of ____________ between solute
  and solvent particles. A negative value
  indicates that heat is _____ favoring
  dissolution. The larger the magnitude of the
  value, the more dissolution is favored.
• Factors affecting ?Hsolution
• Solute-solute attractions
• Weak attractions favor solubility (why?)
• Solvent-solvent attractions
• Weak attractions favor solubility (why?)
• Solvent-solute attractions
• Strong attractions favor solubility (why?)

6
Spontaneity of the Dissolution Process
Input of energy is required to separate
solute-solute (step a) and solvent-solvent (step
b) attractions. Energy is released due to
solute-solvent attractions (step c). If the
amount of energy released in step c is greater
than the energy absorbed in steps a and b the
process is exothermic and favored for
dissolution. What if it is less?
7
Spontaneity of the Dissolution Process

• Many solids, however, will dissolve in liquids by
  endothermic processes. The increase in disorder
  of the system upon mixing (magnitude of ?Smixing)
  is enough to outweigh the endothermic process.
• The solute particles go from highly ordered in a
  crystalline solid to highly disordered in
  solution.
• Most dissolving processes involve an overall
  increase in disorder.
• Discuss mixing of gases

8
Dissolution of Solids in Liquids

• The crystal lattice energy is the energy change
  accompanying the formation of one mole of formula
  units in the crystalline state from constituent
  particles in the gaseous state.
• The crystal lattice energy is always ______ (i.e.
  the process is exothermic).
• M(g) X-(g) ? MX(s) energy
• The amount of energy released increases with the
  strength of attraction between particles (e.g.
  ions in the solid)
• Energy released increases with increasing charge
  density. Why?

9
Dissolution of Solids in Liquids

• Solute-solute attractive interactions present in
  the solid must be overcome for the solid to
  dissolve in the liquid.
• This can be related to the crystal lattice energy
• MX(s) energy ? M(g) X-(g)
• If the magnitude of the crystal lattice energy is
  small only a small amount of energy is needed to
  start the dissolution process.
• This can be thought of as step a in Figure 14-1.
• Is this step exothermic or endothermic?

10
Dissolution of Solids in Liquids

• Energy is also required to break up or expand the
  solvent molecules (i.e. solvent-solvent
  interactions). This can be fairly large if the
  solvent is water. Why?
• This can be thought of as step b in Figure 14-1.
• Energy is released, however, due to attractive
  solute-solvent interactions.
• This can be thought of as step c in Figure 14-1.
• Solvation is the process by which solvent
  molecules interact and surround solute particles
  or ions.
• Hydration refers to when the solvent molecules
  are water.

11
Dissolution of Solids in Liquids

• Solvation energy is the energy change involved in
  the solvation of one mole of gaseous ions.
• Process is almost always exothermic.
• Equivalent to the sum of steps b and c in Figure
  14-1.
• Hydration energy is the energy change when water
  is the solvent.
• Hydration is highly exothermic for ionic or polar
  covalent compounds. Why?
• ?Hsolution (solvation energy) (crystal
  lattice energy)
• If ?Hsolution is negative the dissolution
  process is exothermic.

12
Dissolution of Solids in Liquids

• A nonpolar solid such as naphthalene, C10H8, does
  not dissolve in polar solvents such as H2O. It
  does, however, dissolve in nonpolar solvents.
  Why?
• like dissolves like
• Lets look at the dissolving process of a soluble
  salt such as NaCl (Figure 14-2).
• The individual ions become solvated (hydrated if
  the solvent is water) due to electrostatic
  interactions between the ions and water
  molecules.
• Most cations (e.g. Na) are surrounded by 4 to 9
  H2O molecules.

13
Dissolution of Solids in Liquids

• An increasing charge density (charge/radius
  ratio) increases the hydration energy (heat of
  hydration). Table 14-1

14
Dissolution of Solids

• An increasing charge density, however, also
  increases the magnitude of the crystal lattice
  energy.
• For many salts that possess low-charge species
  (e.g. NaCl) the hydration energy and crystal
  lattice energy nearly cancel each other.
• Demo The dissolution of NH4NO3 is endothermic.
  What can you say about the heat of solvation and
  crystal lattice energy? The dissolution of
  Ca(CH3COO)2 is exothermic.
• With large charge densities the magnitude of the
  crystal lattice energy increases more than the
  hydration energy.
• This is the reason why the dissolution process
  for many solids that contain highly charged ions
  (e.g. Cr2O3) is endothermic. Many are insoluble.
  Look at equation.

15
Dissolution of Liquids in Liquids

• Miscibility is the ability of one liquid to
  dissolve in another. If two liquids are miscible
  one liquid completely dissolves in the other.
• Types of attractive forces to consider when
  determining miscibility
• Solute-solute attractions (step a)
• Solvent-solvent attractions (step b)
• Solute-solvent attractions (step c)
• When will two liquids be the most miscible?

16
Dissolution of Liquids in Liquids

• Like dissolves like determines miscibility
• Polar liquids tend to dissolve in other polar
  liquids
• H2O and CH3OH
• Discuss in terms of the types of interactions.
• What are some other liquids that will dissolve in
  water?

17
Dissolution of Liquids in Liquids

• Will hexane, C6H14, dissolve in water, H2O? Why
  or why not? Explain in terms of types of
  interactions.
• Will hexane dissolve in gasoline which is
  nonpolar? Why or why not? What are the
  strengths of attractive interactions?

18
Dissolution of Gases in Liquids

• Like dissolves like holds fairly well for gases
  dissolving in liquids.
• Polar gases dissolve in water
• Polar gases can also react with water

19
Dissolution of Liquids in Liquids

• Nonpolar gases (e.g. O2) dissolves to a limited
  extent in H2O due to dispersion forces
• Dissolved O2 is responsible for keeping fish
  alive
• CO2, a nonpolar gas, dissolves in water
  appreciable because it reacts with H2O
• CO2(g) H2O(l) H2CO3(aq)
• H2CO3(aq) H(aq) HCO3-(aq)
• HCO3-(aq) H(aq) CO32-(aq)
• What happens to the acidity of water when CO2(g)
  is dissolved ?
• Because gases have such weak solute-solute
  attractions, gases dissolve in liquids
  exothermically.

20
Rates of Dissolution and Saturation

• Rate of dissolution of a solid increases if the
  size of the solid particles is decreased (e.g.
  ground to a powder). Why?
• Sugar cubes versus granulated sugar
• The amount of solid in a liquid will increase
  until the rate of dissolution equals the rate of
  crystallization.
• The opposing processes are in dynamic
  equilibrium. The solution is saturated. It is
  holding all the solute it can at a given
  temperature.
• DEMO NaCl crystals increasing in size in a
  saturated solution.

21
Rates of Dissolution and Saturation

• Saturated solutions have an established
  equilibrium between dissolved and undissolved
  particles.
• NaCl(s) Na(aq) Cl-(aq)
• The forward and reverse rates are equal.
• Supersaturated solutions contain
  higher-than-saturated concentrations of solute.
  How is this possible?
• A supersaturated solution is in a metastable
  state. There needs to be mechanism to start the
  crystallization.
• Hot packs, sodium acetate in H2O

22
Effect of Temperature on Solubility

• LeChateliers Principle states that a chemical
  system responds in a way that best relieves the
  stress or disturbance.
• Exothermic dissolution
• Endothermic dissolution

23
Effect of Temperature on Solubility

• What will happen to the solubility if the
  temperature is changed? The system will respond
  according to LeChateliers Principle.
• Adding heat to an exothermic process ______
  dissolution. Why?
• Adding heat to an endothermic process _____
  dissolution. Why?
• For most solids dissolving in liquids the process
  is endothermic.
• Most gases dissolve in liquids by exothermic
  processes. What does this mean if the
  temperature is increased?
• Demo Dissolve NaCH3COO. Is this process
  endothermic or exothermic? What will happen if
  the temp. is increased? Cool the mixture. What
  type of solution results?

24
Effect of Temperature on Solubility

• Increasing the temperature increases the
  solubility of most solids in H2O. Are these
  processes exothermic or endothermic?
• Na2SO4 is the only exothermic process

25
Effect of Pressure on Solubility

• Changing the pressure has no appreciable effect
  on the solubilities of solids or liquids in
  liquids.
• Pressure changes have large effects on the
  solubilities of gases in liquids.
• Carbonated beverages
• Scuba divers get the bends

26
Effect of Pressure on Solubility

• Henrys Law expresses that the concentration of
  dissolved gas is directly related to the pressure
  of the gas above the solution.
• Pgas kCgas
• Pgas pressure of the gas above the sollution
• k is a constant for a particular gas and solvent
  at a specific T
• Cgas the concentration of dissolved gas
  (molarity)

27
Molality and Mole Fraction

• Previously, we covered molarity and weight
  precent to express concentration. What are they?
• Molality number of moles of solute per kilogram
  of solvent
• What is the molality of a solution that contains
  142 grams of CH3OH in 315 grams of water?

28
Molality and Mole Fraction

• Calculate the molality and the molarity of an
  aqueous solution that is 10.0 glucose, C6H12O6.
  The density of the solution is 1.04 g/mL. 10.0
  glucose solution has several medical uses. 1 mol
  C6H12O6 180 g
• Calculate the molality of a solution that
  contains 7.25 g of benzoic acid C6H5COOH, in 200
  mL of benzene, C6H6. The density of benzene is
  0.879 g/mL. 1 mol C6H5COOH 122 g

29
Molality and Mole Fraction

• Mole fraction is the number of moles of one
  component per moles of all the components.
• The sum of the mole fractions (XA XB) 1
• What are the mole fraction for methanol and water
  in the previous problem?
• What are the mole fractions of glucose and water
  in a 10.0 glucose solution?

30
Colligative Properties of Solutions

• Colligative properties depend solely on the
  number of particles dissolved in the solution and
  not the kinds of particles dissolved.
• Physical property of solutions
• Four kinds that will be discussed
• Vapor pressure lowering
• Freezing point depression
• Boiling point elevation
• Osmotic pressure

31
Lowering of Vapor Pressure and Raoults Law

• Addition of a nonvolatile solute to a solution
  lowers the vapor pressure of the solution
• Fewer solvent molecules present at the surface
  since some solute molecules occupy the space.
• As a result, molecules evaporate at a slower rate
• Raoults Law describes this effect in ideal
  solutions
• Where Xsolvent represents the mole fraction,
  is the vapor pressure of the pure solvent,
  and Psolvent is the vapor pressure of the solvent
  in the solution.
• Figure 14-9

32
Lowering of Vapor Pressure and Raoults Law

• The change in vapor pressure of the solvent in
  solution can be expressed in terms of the solute
  mole fraction.
• Where ?Psolvent is the change in vapor pressure
  of the solvent in the solution.
• This relationship assumes ideal solutions and
  that the solute is nonvolatile (i.e. has not
  vapor pressure). This is Raoults Law.

33
Lowering of Vapor Pressure and Raoults Law

• Determine the vapor pressure of a solution, at
  25?C, that is made by dissolving 5.00 grams of
  sucrose, C6H12O6, in 15.0 grams of water. The
  vapor pressure of pure water at 25?C is 23.8 torr
  (Appendix E).
• Determine the vapor pressure of a 5.25 molal
  aqueous sucrose solution at 47?C.

34
Determining the Vapor Pressure of a Two-Component
System

• Both components are considered as volatile and
  contribute to the total vapor pressure.
• A solution of hexane and heptane
• Each component behaves as if it were pure.
  Therefore, the vapor pressure would be a sum of
  its components.
• Ptotal PA PB
• From Raoults Law
• Therefore,

35
Determining the Vapor Pressure when a
Two-Component System

• The left-hand side corresponds to pure B (XB1)
  and the right-hand side to pure A (XA1). Which
  is more volatile?
• Notice that the black line is always equal to the
  sum of PA and PB.

36
Determining the Vapor Pressure when a
Two-Component System

• At 45?C, the vapor pressure of pure heptane is
  112 torr and the vapor pressure of pure octane is
  36 torr. The solution contains two moles of
  heptane and three moles of octane. Calculate the
  vapor pressure of each component (heptane and
  octane) and the total vapor pressure above the
  solution. What is the composition (in mole
  fractions) of the vapor above the solution?
• The volume above the solution is equal to 1.50 L.
  Assuming that the gases behave ideally calculate
  the amount of heptane and octane in the gas
  phase.
• Also, do Example 14-5 for practice.

37
Boiling Point Elevation

• The boiling point of a liquid is the temperature
  at which its _____ ______ equals the external
  pressure. It was stated with Raoults law that
  the addition of a nonvolatile solute decreases
  the vapor pressure. Therefore, a _____
  temperature must be acquired to cause the liquid
  to boil (vapor pressure equals atmospheric
  pressure).
• The amount of boiling point elevation depends on
  the number of moles of solute particles dissolved
  in the solvent.

38
Boiling Point Elevation

• The boiling point change can be expressed as
• ?Tb Kbm
• ?Tb is the change in boiling point (add)
• Kb is the molal boiling point constant (Table
  14-2)
• Kb corresponds to the change in boiling point by
  a one-molal solution
• m is the molality of the solution
• What is the normal boiling point of a 2.50 m
  glucose, C6H12O6, solution?

39
Boiling Point Elevation

• Do this later. Predict the boiling point
  elevation if 33.0 grams of NaCl is added to 100
  grams of water? Note Calculate moles of solute
  particles not moles of solute.

40
Freezing Point Depression

• Addition of a nonelectrolyte lowers the freezing
  point according to the following expression
• ?TfKfm
• ?Tf is the change in the freezing point
  (subtract)
• Kf is the the molal freezing point depression
  constant (Table 14-2)
• m is the molality of the solution
• Upon freezing, the solvent solidifies as a pure
  substance. Solute molecules make it more
  difficult for the solvent molecules to come
  together and freeze. A lower temperature must be
  acquired to freeze the solvent.

41
Freezing Point Depression

• Calculate the freezing point of a 2.50 m aqueous
  glucose solution.
• CaCl2(s) is commonly added to melt snow on roads
  at lower temperatures. What is the freezing
  point when 20.0 grams of CaCl2 is dissolved in
  100 grams of H2O at 20?C? For comparative
  purposes, calculate the lowering of the freezing
  point when 20.0 grams of NaCl is dissolved in 100
  grams of H2O at the same temperature.
• Antifreeze (HOCH2CH2OH) addition to water is
  another example of freezing point depression.

42
Changes in Boiling and Freezing Point Temperatures

• The relationships for calculating the change in
  freezing and boiling points are very similar.
• It is essentially the same effect. The
  difference is in the size or magnitude of the
  effect which is indicated by the constants, Kf
  and Kb.

43
Determination of Molecular Weight from Freezing
Point Depression or Boiling Point Elevation

• These colligative properties (especially freezing
  point depression) can be used to determine
  molecular weight of a solute.
• Kf and the amount of solvent in kilograms is
  usually known. The freezing point lowering was
  measured. Therefore, the molality can be
  determined, and the moles of solute can be
  calculated. The molecular weight can then be
  determined.

44
Determination of Molecular Weight from Freezing
Point Depression or Boiling Point Elevation

• A 37.0 g sample of a new covalent compound, a
  nonelectrolyte, was dissolved in 200 g of water.
  The resulting solution froze at -5.580C. What is
  the molecular weight of the compound?
• Either ethylene glycol (C2H6O2) or propylene
  glycol (C3H8O2) can be used to make an antifreeze
  solution. A 15.0 gram sample of either ethylene
  glycol or propylene glycol dissolved in 50.0
  grams of water lowers the freezing point to
  7.33?C. Which glycol is added to the solution?
  The solutions behave ideally at these
  concentrations.

45
Colligative Properties and Dissociation of
Electrolytes

• Since a colligative property only depends on the
  number of solute particles in a given mass,
  electrolytes have a larger effect on boiling
  point elevation and freezing point depression.
• How many moles of solute particles will be
  produced from one mole of sucrose (C6H12O6)? How
  many from 1 mole of MgCl2?
• As a result, the boiling point elevation and
  freezing point depression are larger for a given
  molar quantity of a typical electrolyte.

46
Colligative Properties and Dissociation of
Electrolytes

• Solute particles, however, are not randomly
  distributed in an ionic solution. This causes
  the boiling point elevation and freezing point
  depression to be not as large.
• Ionic solutions behave nonideally. The ions
  start to undergo association, and the number of
  solute particles decreases. The effective
  molality, therefore, is reduced.
• Especially noticeable at higher concentrations

47
Colligative Properties and Dissociation of
Electrolytes

• Due to association in ionic solutions, the
  dissociation (or ionization) is reduced. The
  extent of dissociation is measured by the vant
  Hoff factor, i.
• i has an ideal value of 2 for NaCl
• What is the ideal value for i with CaCl2?
• Look at Table 14-3 at values for i and notice
  that the actual values of i are closer to the
  ideal values of i at lower concentrations.

48
Colligative Properties and Dissociation of
Electrolytes

• The freezing point of 0.0100 m NaCl solution is
  -0.0360oC. Calculate the vant Hoff factor and
  apparent percent dissociation of NaCl in this
  aqueous solution.
• A 0.0500 m acetic acid solution freezes at
  -0.09480C. Calculate the percent ionization of
  CH3COOH in this solution.

49
Osmostic Pressure

• Osmosis is the net flow of solvent between two
  solutions separated by a semipermeable membrane.
• Solvent will travel from lower concentration
  solutions to higher concentration solutions
• Semipermeable membranges
• Skin, Saran wrap, and cells

50
Osmotic Pressure
Semipermeable membrane
The solvent molecules will pass through the
semipermeable membrane into the more concentrated
solution. The sugar cannot pass through the
membrane. The solvent passes through the
semipermeable membrane into the more concentrated
solution at a faster rate.
51
Osmotic Pressure

• Solvent molecules will pass through to the more
  concentrated solution until the pressure is
  sufficient to force molecules back through the
  membrane at the same rate. The pressure exerted
  to make the rates through the membrane equal is
  called the osmotic pressure of the solution.
• Look at Figure 14-6a
• Osmotic pressure depends on the number, and not
  the kind, of solute particles in solution.
• Colligative property

52
Osmotic Pressure

• Osmotic pressure follows the equation
• n number of moles
• R is the gas constant (0.082057 atm?L/mol?K)
• T is the temperature
• M molarity
• Osmotic pressure can be very significant.
• 1 M sugar solution has an osmotic pressure of
  22.4 atm or 330 p.s.i.

53
Osmotic Pressure

• What osmotic pressure would a 1.25 molal sucrose
  solution exhibit at 25?C? The density of the
  solution is 1.34 g/mL.
• A 1.00 g sample of a biological material was
  dissolved in enough water to give 100 mL of
  solution. The osmotic pressure of the solution
  was 2.80 torr at 250C. Calculate the molarity
  and approximate molecular weight of the material.

54
Colloids

• In a solution, the solute particles do not settle
  upon standing. Additionally, the mixture is at
  the molecular or ionic level.
• Colloids (or colloidal dispersions) are mixtures
  that have particle sizes between solutions and
  suspensions (Table 14-5). The different
  components do not separate upon standing.
• Fog, smoke, paint, and milk are examples
• The different components of a suspension do
  separate upon standing.

55
Colloids

• The Tyndall effect
• Colloidal particles will scatter light due to
  their size
• DEMO Laser through a container filled with
  smoke
• The Adsorption Phenomenon
• Colloids have very large surface areas and
  exhibit interesting surface chemistries
• They interact strongly with substances near their
  surfaces.

56
Hydrophilic and Hydrophobic Colloids

• Hydrophilic colloids (water loving)
• These colloids are polar (or possess a polar
  component) that will enable them to be well
  dispersed in polar solvents such as water.
• Examples are blood plasmas and proteins
• Hydrophobic colloids (water hating)
• These colloids are nonpolar and will not dissolve
  in polar solvents such as water. These colloids
  require emulsifiers which coat the colloids to
  enable dispersion in a polar solvent such as
  water.
• Emulsifier agents coats the particles of a
  dispersed phase and prevents coagulation (i.e.
  coming together of colloidal particles.

57
Hydrophobic Colloids and Emulsifiers

• Most oils and greases are long-chain hydrocarbons
  that are essentially nonpolar. Will they
  dissolve water?
• Soaps (and detergents) are excellent emulsifying
  agents. They are long-chain fatty acids that
  contain a polar head and a nonpolar tail.
  The nonpolar tail is attracted to the oil or
  grease and the polar head is compatible with
  the polar solvent (e.g. water).

58
Hydrophobic Colloids and Emulsifiers
These emulsifers (here it is sodium stearate)
coat or surround the entire nonpolar particle so
that it will become dispersed in the polar
solvent.
59
Hydrophobic Colloids and Emulsifiers
Look at Figure 14-21 in book. DEMO Sulfur on
surface of water. Add soap.