Everyone has Problems,

1
Everyone has Problems,
but Chemists have Solutions
2

• A solution is defined as a homogeneous mixture of
  two or more substances.

What you are dissolving into
What you are dissolving
Types of Solutions Types of Solutions Types of Solutions
Solute Solvent Example
Gas Gas Air
Gas Liquid Carbonated soda
Liquid Liquid 3 Hydrogen peroxide
Solid Liquid Salt water
Solid Solid Brass (Cu/Zn)
3

• Solubility is a measure of how much solute will
  dissolve in a solvent at a specific temperature.
  
• The saying like dissolves like is helpful in
  predicting the solubility of a substance in a
  given solvent. What this expression means is
  that two substances with intermolecular forces of
  similar type and magnitude are likely to be
  soluble in each other.

4
Like Dissolves Like

• Will bromine be more soluble in water or in
  carbon tetrachloride?

H2O
Bromine (Br2) is nonpolar, therefore it will be
more soluble in the nonpolar carbon tetrachloride
CCl4
5

• FYI-When eating hot peppers or hot spicy foods
  has left you "breathing fire, drinking water
  will not relieve you, because the water cannot
  dissolve the oils which give the spicy taste. A
  chemist might suggest that the ideal would be to
  drink a chaser of a non-polar liquid. But most
  such liquids are toxic, and even gargling with
  benzene or toluene doesnt sound very pleasant. A
  better solution is to eat greasy foods like meats
  which will dissolve the oils. Alternatively, food
  like pasta or bread, that can absorb the oils
  will help "put out the fire." The smell and taste
  of onions and garlic are also due to oils that
  are not easily washed away by water and are not
  absorbed well.

6

• Two liquids are said to be miscible if they are
  completely soluble in each other in all
  proportions.
• Oil and water are immiscible.

7

• Chemists characterize solutions by their capacity
  to dissolve a solute.
• An unsaturated solution contains less solute than
  it has the capacity to dissolve.
• A saturated solution contains the maximum amount
  of a solute that will dissolve in a given solvent
  at a specific temperature
• A supersaturated solution contains more solute
  than is present in a saturated solution. The
  supersaturated solution is not very stable. In
  time, some of the solute will come out of
  solution and form crystals. The remaining
  solution will then be _________.

saturated
8
Concentration Units

• Chemists use several different concentration
  units, each of which has advantages as well as
  limitations. The four most common units of
  concentration percent by mass, mole fraction,
  molarity and molality.
• The choice of concentration unit is based on the
  purpose of the measurement.

9

• Mole Fraction (X)

Moles of A Moles of A Moles of
B Moles of C
XA

• mole fraction is appropriate for calculating
  partial pressures of gases and for dealing with
  vapor pressures of solutions

10
Calculate the mol fraction ethanol (C2H5OH) if
25.0 g ethanol is mixed with 50.0 grams water.
Moles of A Moles of A Moles of
B Moles of C
XA
25.0 g C2H5OH x 1 mol C2H5OH .543 mol
C2H5OH 46.07 g C2H5OH
50.0 g H2O x 1 mol H2O 2.77 mol H2O
18.02 g H2O
.543 .543 2.77
XEthanol
.164
11

• Molarity (M)

• the advantage of molarity is that it is generally
  easier to measure the volume of a solution using
  precisely calibrated volumetric flasks, than to
  mass the solvent

12
Calculate the molarity of an aqueous solution
containing 34.2 grams of potassium chlorate in
225 mL of solution.
34.2 g KClO3 x 1 mol KClO3 .279 mol KClO3
122.55 g KClO3
.279 mol KClO3 .225 L
1.24 M
13

• Molality (m)

• molality is independent of temperature (the
  volume of a solution typically increases with
  increasing temperature, so that a solution that
  is 1.0 M at 25 oC may become 0.97 M at 45 oC
  because of the increase in volume)

14
What is the molality of a solution containing
14.5 g of ferric nitrate in 285 mL of water?
14.5 g Fe(NO3)3 x 1 mol Fe(NO3)3 .0599 mol
Fe(NO3)3 241.88 g
Fe(NO3)3
.210 m
Dwater 1 g 1 mL 1 cm3
15

• Percent by Mass (also called percent by weight or
  weight percent)
• Mass of solute
• Mass of solution

X 100

• like molality, percent by mass is independent of
  temperature
• we do not need to know the molar mass of the
  solute in order to calculate the percent by mass

16
What is the mass percent table salt in a solution
containing 25.0 grams of table salt dissolved in
100.0 mL of water?
25 125
X 100
20.0
Dwater 1 g 1 mL 1 cm3
17
The effect of temperature on solubility

• The solubility of gases in water usually
  decreases with increasing temperature.
• In most, but not all cases, the solubility of a
  solid substance increases with temperature.

18

• Solubility curves, which are used to illustrate
  how the solubility of a solute is affected by
  temperature, are determined experimentally

19

• Because the solubility of solids generally
  decrease with decreasing temperatures fractional
  crystallization, the process of separating a
  mixture of substances into pure components on the
  basis of their differing solubilities, can be
  used to purify substances.

20
Crystallization is the process in which dissolved
solute comes out of solution and forms crystals.

• This occurs because supersaturated solutions are
  not very stable. In time, some of the solute
  will come out of a supersaturated solution as
  crystals.

21

• A 90.0 g sample of KNO3 is contaminated with a
  small amount (less than 5.0 g) of NaCl. Explain
  how you would purify the KNO3.
• The goal is not to necessarily get all 90.0 g of
  KNO3, but to get as much pure KNO3 as possible.

22
Add water to the crystals (100.0 g at least).
Heat the solution until all of the crystals
dissolve (excess of __ oC).
50
Slowly cool the solution until it reaches 0 oC.
At that temperature, about _____ of KNO3 is
still soluble and ____ of the NaCl is still
soluble.
13 g
5 g
That means that ____ of KNO3 will have formed a
pure crystal free from NaCl.
77 g
23

• Mixtures of ions can be separated by
  precipitation. Precipitation is when an
  insoluble solid forms and separates from a
  solution.
• Note that both precipitation and crystallization
  describe the separation of excess solid substance
  from a supersaturated solution.

24
The Effect of Pressure on the Solubility of Gases

• For all practical purposes, external pressure has
  no influence on the solubilities of liquids and
  solids, but it does greatly affect the solubility
  of gases. The quantitative relationship between
  gas solubility and pressure is given by Henrys
  Law, which states the the solubility of gas in a
  liquid is proportional to the pressure of the gas
  over the solution.

25

• c kP

Here c is the molar concentration (mol/L) of the
dissolved gas P is the pressure (in atm) of the
gas over the solution (if several gases are
present , P is the partial pressure of the
specific gas of interest) and k is a constant
(for that specific gas) that depends only on
temperature. (For example, at a given
temperature, CO2 has the same k value, AS LONG AS
THE TEMP STAYS CONSTANT.) The constant k has the
units mol/L . atm.
26

• The solubility of pure nitrogen gas at 25 oC and
  1 atm is 6.8 x 10-4 mol/L. What is the
  concentration of nitrogen dissolved in water
  under atmospheric conditions? The partial
  pressure of nitrogen gas in the atmosphere is
  0.78 atm.

Step 1 Solve for k (which remains constant for
this gas at this temp)
c kP
Now you can use this same k value at a different
pressure, as long as the temp stays the same.
6.8 x 10-4 mol/L k . (1 atm) K 6.8 x 10-4
mol/L . atm
Step 2 Solve for the solubility of nitrogen gas
at 0.78 atm
c kP
c (6.8 x 10-4 mol/L . atm)(0.78 atm) c 5.3 x
10-4 M
27

• Explain how the effervescence of a soft drink
  when the cap is removed is a demonstration of
  Henrys Law.

Before the bottle is sealed, it is pressurized
with a mixture of air and CO2 and water vapor.
Because of the high partial pressure of CO2, the
amount dissolved in the soft drink is many times
the amount that would dissolve under normal
atmospheric conditions. When the cap is removed,
the pressurized gases escape, and the excess CO2
comes out of solution causing the effervescence.
28

• Most gases obey Henrys law, but there are some
  important exceptions
• Dissolved gases that react with water have higher
  solubilities
• NH3 H2O ? NH4 OH-
• CO2 H2O ? H2CO3
• Normally, oxygen gas is only sparingly soluble in
  water, however, its solubility in blood is
  dramatically greater because hemoglobin can bind
  up to four oxygen molecules

29
Colligative Properties

• Colligative properties (or collective properties)
  are properties that depend only on the number of
  solute particles in solution and NOT on the
  nature of the solute particles. The colligative
  properties are vapor-pressure lowering, boiling
  point elevation, freezing point depression, and
  osmotic pressure.

30
Electrolytes and Nonelectrolytes

• All solutes that dissolve in water fit into one
  of two categories electrolytes and
  nonelectrolytes.
• An electrolyte is a substance that, when
  dissolved in water, results in a solution that
  can conduct electricity.
• A nonelectrolyte does not conduct electricity
  when dissolved in water

31
A solutions ability to conduct electricity
depends on the number of ions it contains.
The molar amounts of the dissolved solutes are
equal in all three cases.
32
Vapor Pressure

• When a liquid evaporates, its gaseous molecules
  exert a vapor pressure. Vapor pressure is defined
  as the pressure of the vapor of a substance in
  contact with its liquid or solid phase. Vapor
  pressure changes with temperature the higher the
  temperature, the higher the vapor pressure.

33

• At the boiling point, bubbles form within a
  liquid. The pressure inside the bubble is due
  solely to the vapor pressure of the liquid. The
  pressure exerted on the bubble is largely
  atmospheric.
• When the vapor pressure (internal) equals the
  external (atmospheric) pressure, the bubble rises
  to the surface of the liquid and bursts.

34
Phase Diagram

• A phase diagram shows the conditions at which a
  substance exists as a solid, liquid or gas. This
  is a phase diagram for water.
• At 1 atm (760 mm Hg) the vapor pressure of water
  (inside the bubble) is equal to the external
  atmospheric pressure at 100 oC, the point at
  which water boils. (See your Water Vapor
  Reference Sheet to obtain waters vapor pressure
  at different temperatures.)

35

• If you live at higher altitudes, the atmospheric
  pressure is lower, therefore, the liquid boils at
  a

lower
36

• Do you need to boil your eggs for less time or
  more time if you live in Denver, Colorado?
  Explain.

It takes a certain amount of heat (total kinetic
energy) to cook a hard-boiled egg. Because water
boils at a lower temperature in Denver, (approx
95 oC) it is supplying less heat to the egg,
therefore the egg needs to be boiled for
longer. Remember, liquid water in Denver will
not get hotter than 95 oC if that is its boiling
point!!!
37
Vapor-pressure Lowering

• If a solute is nonvolatile (therefore the solute
  does not have a measurable vapor pressure), the
  vapor pressure of the solution will always be
  less than that of the pure solvent. Thus, the
  relationship between solution vapor pressure and
  solvent vapor pressure depends on the
  concentration of the solute in the solution.
  This relationship is expressed by Raoults law
• P1 X1 . P01 , where
• P1 partial pressure of the solvent over a
  solution
• P01 vapor pressure of the pure solvent (torr,
  mm Hg, atm or kPa)
• X1 mole fraction of the solvent in the solution

38

• Calculate the new vapor pressure when 10.0 mL
  glycerol (C3H8O3), a nonvolatile solute, is added
  to 500.0 mL water at 50 oC. At this temperature,
  the vapor pressure of pure water is 92.5 torr and
  its density is 0.988 g/mL. The density of
  glycerol is 1.26 g/mL.

We must first must calculate the mole fraction of
the solvent (water).
Now we can use Raoults law to calculate the new
vapor pressure
Psolution (.995) (92.5 torr) 92.0 torr
Psolution Xwater . P0water
39

• In a solution containing only one solute, X1 1
  X2, where X1 is the mole fraction of the
  solvent, and X2 is the mole fraction of the
  solute. The previous equation can therefore be
  rewritten substituting 1 X2 for X1

P1 (1 X2) . P01
P1 P01 X2P01
-P01 P1 - X2P01
P01 - P1 X2P01
Using this formula will give you the CHANGE in
vapor pressure.
DP X2P01
40

• Lets solve the previous problem again, this time
  solving for change (decrease) in vapor pressure
  using this derived formula.
• Calculate the vapor pressure lowering when 10.0
  mL glycerol (0.137 mol) is added to 500.0 mL
  (27.4 mol) water at 50 oC. At this temperature,
  the vapor pressure of pure water is 92.5 torr

We must first must calculate the mole fraction of
the solute (glycerol).
Now we can use Raoults law to calculate how much
the vapor pressure was lowered by the addition of
the glycerol
This shows that the vapor pressure would be
lowered by .461 torr (92.5 torr 0.461 torr
92.0 torr), the same answer we obtained two
slides ago!
DP X2P01
DP Xglycerol . P0water
DP (.00498) (92.5 torr) 0.461 torr
41
Why is the vapor pressure of a solution less than
that of the pure solvent?

• One of the driving forces in physical and
  chemical processes is an increase in disorder
  the greater the disorder, the more favorable the
  process. Vaporization increases the disorder of
  a system because molecules in a vapor are not as
  closely packed and therefore have less order than
  those in a liquid. Because a solution is more
  disordered that a pure solvent, the difference in
  disorder between a solution and a vapor is less
  than that between a pure solvent and a vapor.
  Thus solvent molecules in a solution have less of
  a tendency to leave the solution than to leave
  the pure solvent to become vapor, and the vapor
  pressure of a solution is less than that of the
  pure solvent.

42
Nature tends towards disorder (entropy)
The pure solvent is highly ordered, therefore
many of the particles will leave the solvent
because it GREATLY increases disorder, this
results in high vapor pressure.
Nonvolatile solute
This solution is already somewhat disordered,
therefore less of the solvent particles will
leave the solvent, this results in lower vapor
pressure.
Pure solvent
43

• If both components of a solution are volatile
  (that means they both have measurable vapor
  pressure), the vapor pressure of the solution is
  the sum of the individual partial pressures.
• Raoults law holds equally well in this case.
• PT XAP0A XBP0B
• Where PT is equal to the total pressure (PA PB)

44
Fractional Distillation

• Solution vapor pressure has a direct bearing on
  fractional distillation, a procedure for
  separating liquid components of a mixture based
  on their different boiling points.

45

• Fractional distillation is somewhat analogous to
  fractional crystallization.
• When you boil a mixture containing two substances
  with appreciably different boiling points (80.1
  oC and 110.6 oC), the vapor formed will be
  somewhat richer in the more volatile compound
  (the one that vaporizes/boils at the lower temp).
  If the vapor is condensed in a separate
  container and that liquid is boiled again, a
  still higher concentration of the more volatile
  substance will be obtained in the vapor phase.
  By repeating this process many times, it is
  possible to obtain a pure sample of the more
  volatile substance.

46
Boiling-point elevation

• Because the presence of a nonvolatile solute
  lowers the vapor pressure of a solution, it must
  also affect the boiling point of the solution.

47
Phase Diagram of a pure substance. Normal
boiling and freezing points are indicated.
48
Phase diagram of a pure substance and curves
representing the effect of adding a solute.
For each temperature the solute decreases the
vapor pressure of the solution. Notice that the
solution boils at a higher temperature than the
pure substance (boiling occurs when the vapor
pressure is equal to atmospheric pressure in
this case 1 atm - so now the solution needs to be
hotter to reach a vapor pressure of 1 atm.)
49

• Boiling-point elevation (DTb) is defined as the
  CHANGE in the boiling point (boiling point of the
  solution minus the boiling point of the pure
  solvent).
• The value of DTb is proportional to the
  concentration (molality) of the solution. That
  is,
• DTb kbm
• where kb is the molal boiling point elevation
  constant. The units of kb are oC/m.
• (You can find a table of molal boiling-point
  elevation and freezing-point depression constants
  of several common liquids on your reference
  sheet.)

50
What is the new boiling point when 175 grams of
sucrose, C12H22O11, are added to 285 mL of water?
DTb kbm
175 g C12H22O11 x 1 mol C12H22O11 .511 mol
C12H22O11 342.30 g
C12H22O11
1.79 m
DTb (.52 oC/m)(1.79 m) .93 oC
.93 oC 100.00 100.93 oC
51
Freezing-point depression

• Freezing involves a transition from the
  disordered state to the ordered state. For this
  to happen, energy must be removed from the
  system. Because a solution has greater disorder
  than the solvent, more energy needs to be removed
  from it to create order than in the case of the
  pure solvent. Therefore, it needs to get colder
  to freeze a solution, giving the solution a lower
  freezing point than the solvent.

52

• Freezing-point depression (DTf) is defined as the
  CHANGE in the freezing point (freezing point of
  the pure solvent minus the freezing point of the
  solution). Note DTf is ALWAYS positive.

The value of DTf is proportional to the
concentration (molality) of the solution. That
is, DTf kfm where kf is the molal freezing
point depression constant.
53
An aqueous solution of glucose freezes at 2.56
oC. How many grams of glucose must have been
added to 150.0 mL of water to form this solution?
DTf (1.86 oC/m)(m) 2.56 oC
m 1.38
1.38 mol C6H12O6 1 kg water
X mol C6H12O6 .150 kg water
X .207 mol

.207 mol C6H12O6 x 180.15 g C6H12O6 37.3 g
C6H12O6 1 mol
C6H12O6
54

• Explain why salt is placed on frozen roads and
  sidewalks.

Salt depresses the freezing point of water, which
means that the ice on the roads and sidewalks
will melt into water, even at temperatures below
0 oC.
55
Osmotic Pressure

• The net movement of solvent molecules through a
  semipermeable membrane from a pure solvent of a
  dilute solution to a more concentrated solution
  is called osmosis. The osmotic pressure (p) of a
  solution is the pressure required to stop
  osmosis. This pressure can be measured directly
  from the difference in the final fluid levels.

56

• Like boiling point elevation and freezing point
  depression, osmotic pressure is directly
  proportional to the concentration of solution.
  If two solutions are of equal concentration
  (hence, equal osmotic pressure), they are said to
  be isotonic. If two solutions are of unequal
  osmotic pressures, the more concentrated solution
  is said to be hypertonic and the more dilute
  solution is described as hypotonic.

57
Using colligative properties to determine molar
mass

• The colligative properties of nonelectrolyte
  solutions provide a means of determining the
  molar mass of a solute. Theoretically, any of
  the four colligative properties is suitable for
  this purpose. In practice, however, only
  freezing-point depression and osmotic pressure
  are used because they show the most pronounced
  changes.

58

• A 7.85 g sample of a compound is dissolved in 301
  g of benzene. The freezing-point of the solution
  is 1.05 oC below that of pure benzene. What is
  the molar mass of this compound?

• Our first step is to calculate the molality of
  the solution.

• Since there is 0.205 mole of the solute in 1 kg
  of solvent, (0.250 m) , the number of moles of
  solute in 301 g, or 0.301 kg of solvent is

.0617 mol
59

• The colligative properties of electrolytes
  require a slightly different approach than the
  one used for the colligative properties of
  nonelectrolytes. The reason is that electrolytes
  dissociate into ions in solution, and so one unit
  of an electrolyte compound separates into two or
  more particles when it dissolves. (Remember, it
  is the number of solute particles that determines
  the colligative properties of a solution.)

60

• For example, each unit of NaCl dissociates into
  two ions Na and Cl-. Thus the colligative
  properties of a 0.1 m solution of NaCl should be
  twice as great as those of a 0.1 m solution
  containing a nonelectrolyte, such as sucrose.
  Similarly, we would expect a 0.1 m Ba(NO3)2
  solution to depress the freezing point by three
  times as much as a 0.1 m sucrose solution because
  Ba(NO3)2 produces three ions. To account for
  this effect we must modify the equations for
  colligative properties as follows
• DTb ikbm
• DTf ikfm
• The variable i is the vant Hoff factor which is
  equal to the number of particles in solution
  after dissociation. (Remember, only electrolytes
  dissociate!)

61
What is the new freezing point when 25 grams of
ammonium nitrate are added to 285 mL of water?
DTb ikbm
25 g NH4NO3 x 1 mol NH4NO3 .312 mol NH4NO3
80.05 g NH4NO3
1.10 m
DTb (2)(1.86 oC/m)(1.10 m) 4.09 oC
New freezing point -4.09 oC
62
Colloids

• The solutions we discussed so far are true
  homogeneous mixtures. Now consider what happens
  if we add fine sand to a beaker of water and
  stir. The sand particles are suspended at first
  but then gradually settle to the bottom. This is
  an example of a heterogeneous mixture. Between
  these two extremes is an intermediate state
  called a colloidal suspension, or simply, a
  colloid. A colloid is a dispersion of particles
  of one substance (the dispersed phase) throughout
  a dispersing medium made of another substance.

63
Types of Colloids
Dispersing Medium Dispersed phase Name Example
Gas Liquid Aerosol Fog, mist
Gas Solid Aerosol Smoke
Liquid Gas Foam Whipped cream
Liquid Liquid Emulsion Mayonnaise
Liquid Solid Sol Milk of magnesia
Solid Gas Foam Plastic foams
Solid Liquid Gel Jelly, butter
Solid Solid Solid gel Certain alloys (steel), gemstones
64

• One way to distinguish a solution from a colloid
  is by the Tyndall effect. When a beam of light
  passes through a colloid, it is scattered by the
  dispersed medium. No such scattering is observed
  with ordinary solutions because the solute
  molecules are too small to interact with visible
  light.
• Another demonstration of the Tyndall effect is
  the scattering of sunlight by dust or smoke in
  the air.

65

• The End