Chapter 12: Solutions

1
Chapter 12 Solutions

• Solubility and Concentration Units
• Temperature and Pressure Effects on Solubility
• Vapor Pressure Lowering
• Boiling Point Elevation
• Freezing Point Depression
• The vant Hoff Factor

2
Solutions Homogeneous Mixtures

• Most stuff we find in our daily lives are
  mixtures of substances called solutions
• Tap Water (water, iron, calcium, nitrate)
• Air (N2, O2, Ar)
• O2(g) in H2O(l) O2(aq)
• This stuff has consistent properties throughout,
  so they are homogenous
• In most cases, one of the substances that makes
  up a solution is in great excess
• Lets call the excess component the solvent
• Everything else is called a solute
• For example, air is 78 nitrogen, so nitrogen
  would be the solvent while oxygen and argon would
  be the solutes.
• Notice that more than one solute is possible, but
  only one solvent is ever present
• 14kt Gold is 58.3 Au mixed with other metals
• When mixing the solutes into the solvent, we say
  they are dissolved. The two substances are
  termed soluble, since they mix to form a solution
• We are assuming they dont react, in this case
• If two things dont mix to make a solution, we
  say they are insoluble

3
Miscibility

• Insoluble stuff you know about
• Oil and water
• Rocks and water
• Macaroni and cheese

• When discussing liquids, the term immiscible is
  often used when the liquids are insoluble
• Here we see how liquid molecules behave for
  immiscible (left) and miscible (right)
  substances.
• On the left, the molecules would rather interact
  with their own kind then with those from the
  other liquid
• This screams to us intermolecular forces, since
  they hold all stuff together

4
Intermolecular Forces Solubility

• In mixtures of gases, IM forces are not very
  important, and all gases are soluble with each
  other
• Unless they react chemically, that is

• Solids and liquids form solutions only when the
  IM forces are strong between the components of
  the solution
• In LiF(aq), the ion-dipole forces are strong and
  we have complete solubility

• The ion-dipole interaction is more favorable than
  the crystal lattice interactions
• Favorable does not mean stronger. Instead, it
  encompasses many factors including the strength
  of the bond/IM force and the number of
  interactions

5
Intermolecular Forces Solubility II

• For liquid-liquid mixtures, we either see
• One liquid dissolving into the other to give a
  homogeneous mixture
• Less dense liquid settles on top of the other
  liquid to give a heterogeneous mixture
• Below, the B-B interactions are more favorable
  than the A-B interactions, and
• The A-A interactions are more favorable than the
  A-B interactions

• Note there is always some mixing, no matter how
  poor the A-B interactions are!!!
• Note 2 we see a boundary region called the
  meniscus between the two layers!

6
Likes Dissolve Likes

• A general rule on whether or not an interaction
  is favorable is that likes-dissolve-likes
• This means that molecules with similar polarity
  will tend to dissolve in each other
• Thus, the polar molecules below are more likely
  soluble in mixtures with other polar molecules
• And, the nonpolar molecules are more likely
  soluble in mixtures with other nonpolar molecules

7
Alcohol Dissolution in Water

• Problem 1 Examine the data below for a series
  of alcohols and their solubility in water. Which
  of the answers below best explains the trend.

• The hydroxide group on the long chain molecules
  interacts more with the chain than the H2Os
• The longer the hydrocarbon chain, the more the
  dipole in the OH is reduced, making the molecule
  less and less polar as the chain increases
• The added hydrocarbon chain length increases the
  viscosity of the alcohol, reducing its ability to
  dissolve
• Small hydrocarbon chains ionize more readily than
  long hydrocarbon chains

8
Dissolution and Dissociation

• Dissolution breaking of IM forces of a solid or
  liquid in order to make solvent-solute IM forces
• Sugar(s) H2O --gt sugar(aq)
• CH3CN (l) H2O --gt CH3CN(aq)
• Not 100 true, more on this in EQ
• Dissociation breaking of ionic bonds of
  dissolved substance to make solvent-ion IM forces
• NaCl(s) H2O --gt Na(aq) Cl-(aq)
• HCl(l) H2O --gt H3O(aq) Cl-(aq)
• Usually, both dissolution and dissociation
• Are spontaneous
• Give off heat
• Change the properties of the solvent
• Depend on the temperature and pressure
• Problem 2 Which of the following soluble
  substances will dissociate in pure water?
• CH3OH C. HNO3 E. Both A and C
• C2H6 D. CaCl2 F. Both C and D
• G. All of these

9
Just How Soluble?

• Experience shows that even soluble stuff cant
  forever go into solution
• Put enough salt in the water the salt will
  start to pile up on the bottom
• That is, the water can only hold so much
  dissolved substance
• An equilibrium is established, which we will
  quantify in chapter 18!
• Some salt has dissolved but the rest has not,
  creating a saturated solution
• The measure of how much stuff can dissolve in a
  solvent is its solubility

Solute (s) ? Solute (aq)
10
Concentration Units

• Problem 3 Solubility depends on many factors,
  including the T and P. So why is it a bad idea
  to use M (molarity) to express solubility?
• Thus, it would be best to have concentration
  units that are T P independent
• Molarity (as a reminder) is
• Mass Percent of Solute is defined as
• Molality (m) is defined as
• Mole Fraction (X) is defined as

11
Henrys Law

• Critical Thinking Questions
• Consider both carbon monoxide (C?O) and water
  (H-O-H)
• Does CO have a permanent dipole? If so, draw it
  and show the partial charges.
• Does HOH have a permanent dipole? If so, draw it
  and show the partial charges.
• What is the strongest type of intermolecular
  force (IMF) formed between CO and HOH? Draw one
  possible interaction between a single CO and a
  single HOH due to this IMF.

• Model 1

• Information
• Model 1 shows a sealed container of pure water to
  which carbon monoxide gas has been added (the
  gas is over the liquid). For simplicity, no
  vaporized water molecules are shown
• If a CO molecule in model 1 were to collide with
  the surface of the liquid water,the CO molecule
  could either
• bounce off the surface
• be captured by the IMF between CO and HOH

• Critical Thinking Questions
• a) What phase is the CO molecule after it
  bounces off the surface of the liquid water?
• What phase is the CO molecule after it is
  captured by the liquid water molecules?
• What kind of IMF is responsible for the capture
  of the CO molecule by the liquid water molecules?
• Would you characterize the captured CO
  molecules as being dissolved or dissociated?
  Explain.

12
Henrys Law

• Suppose a configuration like model 1 was set up,
  but oxygen molecules were used in place of CO.
• What type of IMF would be responsible for the
  dissolved oxygen in the water?
• If an equal number of moles of CO and O2 were
  placed above pure water in a closed container, do
  you expect more CO or O2 to dissolve in the
  water? Justify your choice.
• Information
• Henrys Law states that the partial pressure of a
  gas over a liquid is proportional to the amount
  of dissolved gas within the liquid. You will
  often see Henrys Law written as Sgas
  kHenryPgas, where
• Sgas is the solubility of the gas in appropriate
  units
• Pgas is the partial pressure of the gas over the
  liquid
• kHenry is the Henrys law constant
• The Henrys Law constant is specific for each gas
  over each liquid at a specific temperature.

• Critical Thinking Questions
• Is solubility (Sgas) a measure of the number of
  gas molecules over the liquid or dissolved in the
  liquid?
• If the pressure of a gas above a liquid is
  increased, would the number of collisions of the
  gas with the surface of the liquid increase?
• If the number of collisions between the gas
  molecules over a liquid and the surface of the
  liquid increased, would you expect the number of
  captured gas molecules to increase? Why or why
  not?

13
Henrys Law

• The mythical gas buburu is found to dissolve in
  water. At room temperature, 35mg of buburu
  dissolves in a half liter (0.500L) of pure water
  when the partial pressure above the water is 300
  torr. What is the value of the Henrys Law
  constant in mg per Latm?
• 0.888 atm of buburu is over pure water at room
  temperature. How many milligrams of buburu are
  dissolved in the water?
• Exercises
• 250 mg of buburu were bubbled through 1.50 L of
  pure water at room temperature. After all the
  buburu is allowed to bubble through the water

14
Temperature Effects on Solubility Ionic Solid
Solutes

• When a solid is dissolved in water, it must first
  be broken apart into individual molecules
• C12H22O11(s) C12H22O11(aq)
• MgCl2(s) MgCl2(aq)
• For sugar, this is as far as it goes (covalent
  bonds)
• For MgCl2, we know it breaks into ions!
• MgCl2(aq) Mg2(aq) 2Cl-(aq)
• If the temperature of the solvent is increased,
  the trend is that the solubility increases, as
  is seen in the graph to the right

15
Temperature Effects on Solubility Gaseous
Solutes

• For ionic solids, as T increases the solubility
  usually increases
• Increased KE from higher T makes overcoming the
  IM forces holding the molecules together and the
  ionic bonds holding the ions together easier for
  the solvent molecules
• For gases, as T increases the solubility usually
  decreases
• Weak dispersion forces of most gases are less and
  less effective as the average velocity of the
  solvent molecules increases due to the T increase
• The solution process can be either exothermic or
  endothermic, and the temperature change
  associated with making a solution is termed the
  heat of solution
• Problem 9 Which of the following would show a
  decrease in solubility as the temperature
  increased?
• NH3(aq)
• CO2(aq)
• K2SO4(aq)

16
Rauolts Law

• Information
• A nonvolatile substance is one that show no
  appreciable evaporation or sublimation at a
  normal temperatures
• A good example is a rock, which stays seemingly
  unchanged over years, as none of its molecule
  evaporate away
• A volatile substance will produce a vapor via
  evaporation or sublimation
• A volatile liquid will produce a vapor because
• At any temperature, some molecules in the system
  have enough energy to overcome the IMFs
• At any temperature, some molecules with enough
  energy will be at the interface between the
  atmosphere and the liquid water (the surface of
  the water).
• The pressure produced by the vapor (the vapor
  pressure, of Pvap) from a volatile substance
  becomes constant when the rate of evaporation
  equals the rate of condensation at the surface of
  the liquid.
• The left panel of model 1 shows the liquid A and
  its vapor. The vapor pressure for the left
  panel is 300 torr.
• The right panel of model 1 displays the solution
  formed byadding the nonvolatile substance B to
  the liquid A. The solvent for this solution is
  A, which is in great excess.
• Critical Thinking Questions
• What is the vapor pressure of pure A at 25ºC?
• Why will there be no vapor phase (gas phase) B
  molecules in the right panel of model 1?
• According to model 1, are the chances of
  evaporation better or worse for liquid A
  molecules after the addition of the nonvolatile
  B? Explain.

17
Raoults Law

• Based on your answer to CTQ3, do you expect more
  or less A molecules to be in the vapor phase
  after the addition of B to the liquid A?
  Explain.
• Is the following statement true or false.
  Justify your choice.
• The vapor pressure of A will be lower after the
  addition of the B molecules because some B
  molecules have evaporated and take up space over
  the liquid, blocking the exit path for any
  evaporating A molecules.
• Information
• Raoults Law states that, for a dilute solution,
  the drop in vapor pressure due to the addition of
  a nonvolatile solute is directly proportional to
  the composition of the solution (the liquid
  phase).
• Note what happens in the gas phase is a result
  of the composition of the condensed phase
• Model 2 shows a mathematical representation of
  Raoults Law
• The composition, or concentration unit, utilized
  in RaoultsLaw is the mole fraction of the
  solvent

Model 2 Raoults Law
mole fraction of solvent
Pvap of pure solvent (A)
Pvap of solution (B dissolved in A)
18
Raoults Law

• Critical Thinking Questions
• Based on the depiction of model 1, what is the
  mole fraction of A in the right panel? Assume
  that every instance of an A or B represents the
  same number of moles of each species.
• Use Raoults Law to calculate the vapor pressure
  of A for the right panel of model 1. Draw in the
  appropriate number of gas phase As to complete
  model 1.
• Calculate the vapor pressure of water over a
  solution made of 50.0 g of sucrose (MW342) and
  150.0 g of water at 45ºC. Pvap 71.88 torr for
  pure water at this temp.

19
Colligative Properties

• Information
• The boiling point of a pure liquid is related to
  the vapor pressure of the liquid. A highly
  volatile substance will boil at a lower
  temperature than a substance of low volatility.
• Boiling occurs when the vapor pressure of the
  liquid equals or exceeds the pressure of the
  atmosphere surrounding the system.
• Model 1 represents a solution of A that is
  boilingprior to the addition of the nonvolatile
  solute B.
• The atmospheric pressure in model 1 is 0.900 atm.
• Critical Thinking Questions
• What is the vapor pressure of A on the left
  handpanel of model 1. Explain how you know
  this.
• Is the reduction in vapor pressure by the
  addition of B a true reflection of Raoults Law?
• Is the solution created by adding B boiling? Why
  or why not?

Model 1 T 111ºC for both containers
20
Colligative Properties

• Is the following statement true or false?
  Justify your choice.
• The solution in model 1 is not boiling because
  the vapor pressure of the solution is less than
  that of pure A.
• What would you do to the solution on the right
  panel of model 1 to make it boil?
• At which temperature is the solution on the right
  panel of model 1 most likely to boil 101ºC,
  111ºC, 121ºC? Justify your choice.
• Information
• The boiling point of a solution is higher than
  that of the pure solvent, when nonvolatile
  solutes are used.
• In general, the more solute particles you add to
  the solution, the more interface positions
  between the liquid and vapor phase are occupied
  by nonvolatile molecules. Thus, increasing the
  concentration of solute particles will result in
  a lowering of the vapor pressure of the solution
  as compare to the pure solvent. To achieve
  boiling, the solutions temperature must be
  raised by adding energy that

21
Colligative Properties

• 3.00 g of methyl iodide are dumped into 20.g of
  benzene. What change in boiling point do we
  expect for benzene if the Kb for benzene is 2.53
  ºC kg/mol?

• Exercises
• Do we add salt to food for the taste or to ensure
  our water is warm enough to cook our food? You
  can check by supposing you placed a 1g clump of
  salt into a liter of water (around 1000g), and
  determining the temperature change.

22
Colligative Properties

• Information
• When water freezes, the ice produced is nearly
  pure water. The act of freezing purifies the
  water because the water freeze undergoes a phase
  change at its freezing point, but the solutes in
  water do not. That is, the crystal formed is
  made of molecules that are in the act of
  freezing, which are water molecules in an aqueous
  solution.
• When pure, liquid water freezes, there is a
  collision between water molecules in the liquid
  phase with those in the solid phase. If the
  colliding liquid molecule is of low enough
  energy, it will become part of the crystal
  structure of the solid phase. Otherwise, a
  liquid molecule with relatively high energy will
  simply bounce off the crystal.
• The lower the temperature, the more molecules in
  the liquid that are of low energy and can freeze
  upon collision.
• Critical Thinking Questions
• Two students are discussing the possible effect
  of having a solute in pure water on the freezing
  point. Do you agree with student 1, student 2,
  both or neither? Justify your choice
• Student 1 The presence of the solute blocks
  the interface between the liquid molecules and
  the solid molecules. So, there ends up being
  less collisions between liquid molecules and the
  crystal forming in the container. Less
  collisions means less chance of freezing
  occurring. In the end, the solution will freeze
  at a lower temperature because by removing some
  energy, more of the collisions that do happen are
  successful at freezing instead of bouncing off.
• Student 2 Actually, the solute molecules
  absorb interfere with the energy transfer in the
  solution. So, instead of the solvent molecules
  becoming colder and slowing down (so they can be
  captured by the crystal), the solute molecules
  slow down. Unfortunately, the solute molecules
  dont crystallize with the water, meaning that we
  have to cool the solution even further to make
  sure the solvent molecules get cold enough to
  freeze, too.

23
Colligative Properties

• Information
• When a nonvolatile solute is added to a liquid,
  the freezing point decreases. The relationship
  for dilute solutions is equivalent to the boiling
  point elevation equation.. That is..
• The lowering in freezing point can be calculated
  via DT Kfmsolution
• Where Kf is the freezing point depression
  constant, which is specific for every substance
• Critical Thinking Questions
• An aqueous solution of 0.10 m CaCl2 is cooled to
  freezing. At what T will this occur? Kf for
  water is 1.86 ºC/m.

• Information
• A colligative property is one that is directly
  related to the number of solute particles
  involved in the process.
• A particle can be an atom, a molecule, or even
  ions.
• vant Hoff adjusted the freezing point lowering
  and boiling point elevation equations to include
  the theoretical number of particles produced via
  dissociation of ionic substances in water.
• DT iKbmsolution and DT iKfmsolution
• Here, i is the vant Hoff factor

24
Colligative Properties
Table 1

• Critical Thinking Questions
• What does the symbol i stand for in the3rd
  column of Table 1?
• Fill in the missing components of Table 1.
• Confirm that a better answer to CTQ9would be
  -0.56ºC.
• Which of the following solutions will raise the
  BP of water the most? Explain your choice.
• 0.010 m CaCl2
• 0.015 m KCl
• 0.020 m Sugar
• 0.025 m NH3