Physical Properties of Solutions

1
Physical Properties of Solutions
Chapter Twelve
2
Some Types of Solutions
Solution Solute dispersed in a solvent.
3
Solution Concentration
Amount of solute
Most concentration units are expressed as
Amount of solvent or solution

• Molarity moles of solute/liter of solution
• Percent by mass grams of solute/grams of
  solution (then multiplied by 100)
• Percent by volume milliliters of
  solute/milliliters of solution (then multiplied
  by 100)
• Mass/volume percent grams of solute/milliliters
  of solution (then multiplied by 100)

4

• Example 12.1
• How would you prepare 750 g of an aqueous
  solution that is 2.5 NaOH by mass?
• Example 12.2
• At 20 C, pure ethanol has a density of 0.789
  g/mL and USP ethanol has a density of 0.813 g/mL.
  What is the mass percent ethanol in USP ethanol?

5
Solution Concentration (contd)
Amount of solute
Most concentration units are expressed as
Amount of solvent or solution

• Parts per million (ppm) grams of solute/grams of
  solution (then multiplied by 106 or 1 million)
• Parts per billion (ppb) grams of solute/grams of
  solution (then multiplied by 109 or 1 billion)
• Parts per trillion (ppt) grams of solute/grams
  of solution (then multiplied by 1012 or 1
  trillion)

• ppm, ppb, ppt ordinarily are used when expressing
  extremely low concentrations (a liter of water
  that is 1 ppm fluoride contains only 1 mg F!)

6

• Example 12.3
• The maximum allowable level of nitrates in
  drinking water in the United States is 45 mg
  NO3/L. What is this level expressed in parts per
  million (ppm)?

7
Solution Concentration (contd)
Amount of solute
Most concentration units are expressed as
Amount of solvent or solution

• Molality (m) moles of solute/kilograms of
  solvent.
• Molarity varies with temperature (expansion or
  contraction of solution).
• Molality is based on mass of solvent (not
  solution!) and is independent of temperature.
• We will use molality in describing certain
  properties of solutions.

8

• Example 12.4
• What is the molality of a solution prepared
  by dissolving 5.05 g naphthalene C10H8(s) in
  75.0 mL of benzene, C6H6 (d 0.879 g/mL)?
• Example 12.5
• How many grams of benzoic acid, C6H5COOH,
  must be dissolved in 50.0 mL of benzene, C6H6 (d
  0.879 g/mL), to produce 0.150 m C6H5COOH?

9

• Example 12.6
• An aqueous solution of ethylene glycol
  HOCH2CH2OH used as an automobile engine coolant
  is 40.0 HOCH2CH2OH by mass and has a density of
  1.05 g/mL. What are the (a) molarity, (b)
  molality, and (c) mole fraction of HOCH2CH2OH in
  this solution?
• Example 12.7 An Estimation Example
• Without doing detailed calculations,
  determine which aqueous solution has the greatest
  mole fraction of CH3OH (a) 1.0 m CH3OH, (b)10.0
  CH3OH by mass, or (c) xCH3OH 0.10.

10
Solution Concentration (contd)
Amount of solute
Concentration expressed as
Amount of solvent or solution

• Mole fraction (xi) moles of component i per
  moles of all components (the solution).
• The sum of the mole fractions of all components
  of a solution is ____.
• Mole percent mole fraction times 100.

11
Enthalpy of Solution

• Solution formation can be considered to take
  place in three steps
• Move the molecules of solvent apart to make room
  for the solute molecules. DH1 gt 0 (endothermic)
• Separate the molecules of solute to the distances
  found between them in the solution. DH2 gt 0
  (endothermic)
• Allow the separated solute and solvent molecules
  to mix randomly. DH3 lt 0 (exothermic)
• DHsoln DH1 DH2 DH3

12
Visualizing Enthalpy of Solution
For dissolving to occur, the magnitudes of DH1
DH2 and of DH3 must be roughly comparable.
13
Intermolecular Forcesin Solution Formation

• An ideal solution exists when all intermolecular
  forces are of comparable strength, DHsoln 0.
• When solutesolvent intermolecular forces are
  somewhat stronger than other intermolecular
  forces, DHsoln lt 0.
• When solutesolvent intermolecular forces are
  somewhat weaker than other intermolecular forces,
  DHsoln gt 0.
• When solutesolvent intermolecular forces are
  much weaker than other intermolecular forces, the
  solute does not dissolve in the solvent.
• Energy released by solutesolvent interactions is
  insufficient to separate solute particles or
  solvent particles.

14
Intermolecular Forces in Solution
For a solute to dissolve, the strength of
solventsolvent forces
must be comparable to solutesolvent forces.
and solutesolute forces
15
Non-Ideal Solutions
when mixed, give less than 100 mL of solution.
50 mL of ethanol
and 50 mL of water
In this solution, forces between ethanol and
water are _____er than other intermolecular
forces.
16
Aqueous Solutions of Ionic Compounds

• The forces causing an ionic solid to dissolve in
  water are iondipole forces, the attraction of
  water dipoles for cations and anions.
• The attractions of water dipoles for ions pulls
  the ions out of the crystalline lattice and into
  aqueous solution.
• The extent to which an ionic solid dissolves in
  water is determined largely by the competition
  between
• interionic attractions that hold ions in a
  crystal, and
• iondipole attractions that pull ions into
  solution.

17
IonDipole Forces in Dissolution
Negative ends of dipoles attracted to cations.
Positive ends of dipoles attracted to anions.
18

• Example 12.8
• Predict whether each combination is likely to
  be a solution or a heterogeneous mixture
• (a) methanol, CH3OH, and water, HOH
• (b) pentane, CH3(CH2)3CH3, and octane,
  CH3(CH2)6CH3
• (c) sodium chloride, NaCl, and carbon
  tetrachloride, CCl4
• (d) 1-decanol, CH3(CH2)8CH2OH, and water, HOH

19
Some Solubility Terms

• Liquids that mix in all proportions are called
  miscible.
• When there is a dynamic equilibrium between an
  undissolved solute and a solution, the solution
  is saturated.
• The concentration of the solute in a saturated
  solution is the solubility of the solute.
• A solution which contains less solute than can be
  held at equilibrium is unsaturated.

20
Formation of a Saturated Solution
Eventually, the rates of dissolving and of
crystallization are equal no more solute appears
to dissolve.
Solid begins to dissolve.
As solid dissolves, some dissolved solute begins
to crystallize.
Longer standing does not change the amount of
dissolved solute.
21
Solubility as a Functionof Temperature

• Most ionic compounds have aqueous solubilities
  that increase significantly with increasing
  temperature.
• A few have solubilities that change little with
  temperature.
• A very few have solubilities that decrease with
  increasing temperature.
• If solubility increases with temperature, a hot,
  saturated solution may be cooled (carefully!)
  without precipitation of the excess solute. This
  creates a supersaturated solution.
• Supersaturated solutions ordinarily are unstable

22
A Supersaturated Solution
Solute immediately begins to crystallize
until all of the excess solute has precipitated.
A single seed crystal of solute is added.
23
Some Solubility Curves
What is the (approx.) solubility of KNO3 per 100
g water at 90 C? At 20 C?
24
Selective Crystallization
When KNO3(s) is crystallized from an aqueous
solution of KNO3 containing CuSO4 as an impurity,
CuSO4 (blue) remains in the solution.
KNO3 crystallized from a hot, saturated solution
is virtually pure.
25
The Solubilities of Gases

• Most gases become less soluble in liquids as the
  temperature increases. (Why?)
• At a constant temperature, the solubility (S) of
  a gas is directly proportional to the pressure of
  the gas (Pgas) in equilibrium with the solution.
• S k Pgas
• The value of k depends on the particular gas and
  the solvent.
• The effect of pressure on the solubility of a gas
  is known as Henrys law.

26
Effect of Temperature on Solubility of Gases
Thermal pollution as river/lake water is warmed
(when used by industry for cooling), less oxygen
dissolves, and fish no longer thrive.
27
Pressure and Solubility of Gases
thus more frequent collisions of gas molecules
with the surface
giving a higher concentration of dissolved gas.
Higher partial pressure means more molecules of
gas per unit volume
28

• Example 12.9
• A 225-g sample of pure water is shaken with air
  under a pressure of 0.95 atm at 20 C. How many
  milligrams of Ar(g) will be present in the water
  when solubility equilibrium is reached? Use data
  from Figure 12.14 and the fact that the mole
  fraction of Ar in air is 0.00934.

29
Colligative Properties of Solutions

• Colligative properties of a solution depend only
  on the concentration of solute particles, and not
  on the nature of the solute.
• Non-colligative properties include color, odor,
  density, viscosity, toxicity, reactivity, etc.
• We will examine four colligative properties of
  solutions
• Vapor pressure (of the solvent)
• Freezing point depression
• Boiling point elevation
• Osmotic pressure

30
Vapor Pressure of a Solution

• The vapor pressure of solvent above a solution is
  less than the vapor pressure above the pure
  solvent.
• Raoults law the vapor pressure of the solvent
  above a solution (Psolv) is the product of the
  vapor pressure of the pure solvent (Psolv) and
  the mole fraction of the solvent in the solution
  (xsolv)
• Psolv xsolv Psolv
• The vapor in equilibrium with an ideal solution
  of two volatile components has a higher mole
  fraction of the more volatile component than is
  found in the liquid.

31

• Example 12.10
• The vapor pressure of pure water at 20.0 C is
  17.5 mmHg. What is the vapor pressure at 20.0 C
  above a solution that has 0.250 mol sucrose
  (C12H22O11) and 75.0 g urea CO(NH2)2 dissolved
  per kilogram of water?

32

• Example 12.11
• At 25 C, the vapor pressures of pure benzene
  (C6H6) and pure toluene (C7H8) are 95.1 and 28.4
  mmHg, respectively. A solution is prepared that
  has equal mole fractions of C7H8 and C6H6.
  Determine the vapor pressures of C7H8 and C6H6
  and the total vapor pressure above this solution.
  Consider the solution to be ideal.
• Example 12.12
• What is the composition, expressed as mole
  fractions, of the vapor in equilibrium with the
  benzenetoluene solution of Example 12.11?

33
Fractional Distillation
The vapor here
is richer in the more volatile component than
the original liquid here
so the liquid that condenses here will also be
richer in the more volatile component.
34

• Example 12.13 A Conceptual Example
• Figure 12.16 (below) shows two different
  aqueous solutions placed in the same enclosure.
  After a time, the solution level has risen in
  container A and dropped in container B. Explain
  how and why this happens.

35
Vapor Pressure Lowering by a Nonvolatile Solute
the vapor pressure from the pure solvent.
Raoults Law the vapor pressure from a solution
(nonvolatile solute) is lower than
Result the boiling point of the solution
increases by DTb.
36
Freezing Point Depression and Boiling Point
Elevation
DTf Kf m DTb Kb m
37

• Example 12.14
• What is the freezing point of an aqueous
  sucrose solution that has 25.0 g C12H22O11 per
  100.0 g H2O?
• Example 12.15
• Sorbitol is a sweet substance found in fruits
  and berries and sometimes used as a sugar
  substitute. An aqueous solution containing 1.00 g
  sorbitol in 100.0 g water is found to have a
  freezing point of 0.102 C. Elemental analysis
  indicates that sorbitol consists of 39.56 C,
  7.75 H, and 52.70 O by mass. What are the (a)
  molar mass and (b) molecular formula of sorbitol?

38
Osmotic Pressure

• A semipermeable membrane has microscopic pores,
  through which small solvent molecules can pass
  but larger solute molecules cannot.
• During osmosis, there is a net flow of solvent
  molecules through a semipermeable membrane, from
  a region of lower concentration to a region of
  higher concentration.
• The pressure required to stop osmosis is called
  the osmotic pressure (p) of the solution.
• p (nRT/V) (n/V)RT M RT

This equation should look familiar
39
Osmosis and Osmotic Pressure
The solution increases in volume until
the height of solution exerts the osmotic
pressure (p) of the solution.
Net flow of water from the outside (pure H2O) to
the solution.
40

• Example 12.16
• An aqueous solution is prepared by dissolving
  1.50 g of hemocyanin, a protein obtained from
  crabs, in 0.250 L of water. The solution has an
  osmotic pressure of 0.00342 atm at 277 K. (a)
  What is the molar mass of hemocyanin? (b) What
  should the freezing point of the solution be?

41
Practical Applications of Osmosis
Ordinarily a patient must be given intravenous
fluids that are isotonichave the same osmotic
pressure as blood.
External solution is hypertonic produces osmotic
pressure gt pint. Net flow of water out of the
cell.
Red blood cell in isotonic solution remains the
same size.
External solution is hypotonic produces osmotic
pressure lt pint. Net flow of water into the cell.
42
Practical Applications of Osmosis (contd)
Pressure greater than p is applied here

• Reverse osmosis (RO) reversing the normal net
  flow of solvent molecules through a semipermeable
  membrane.
• Pressure that exceeds the osmotic pressure is
  applied to the solution.
• RO is used for water purification.

water flows from the more concentrated
solution, through the membrane.
43
Solutions of Electrolytes

• Whereas electrolytes dissociate, the number of
  solute particles ordinarily is greater than the
  number of formula units dissolved. One mole of
  NaCl dissolved in water produces more than one
  mole of solute particles.
• The vant Hoff factor (i) is used to modify the
  colligative-property equations for electrolytes
• DTf i (Kf) m
• DTb i Kb m
• p i M RT
• For nonelectrolyte solutes, i 1.
• For electrolytes, we expect i to be equal to the
  number of ions into which a substance dissociates
  into in solution.

44
At very low concentrations, the theoretical
values of i are reached.
At higher concentrations, the values of i are
significantly lower than the theoretical values
ion pairs form in solution.
45

• Example 12.17 A Conceptual Example
• Without doing detailed calculations, place the
  following solutions in order of decreasing
  osmotic pressure
• 0.01 M C12H22O11(aq) at 25 C
• 0.01 M CH3CH2COOH(aq) at 37 C
• 0.01 m KNO3(aq) at 25 C
• a solution of 1.00 g polystyrene (molar mass 3.5
  105 g/mol) in 100 mL of benzene at 25 C

46
Colloids

• In a solution, dispersed particles are molecules,
  atoms, or ions (roughly 0.1 nm in size). Solute
  particles do not settle out of solution.
• In a suspension (e.g., sand in water) the
  dispersed particles are relatively large, and
  will settle from suspension.
• In a colloid, the dispersed particles are on the
  order of 11000 nm in size.
• Although they are larger than molecules/atoms/ions
  , colloidal particles are small enough to remain
  dispersed indefinitely.

47
Why are there no gas-in-gas colloids?
48
The Tyndall Effect
Light scattered by the (larger) colloidal
particles of Fe2O3 makes the beam visible.
The dissolved Fe3 ions are not large enough to
scatter light the beam is virtually invisible.
49
A Suspension and a Colloid
Each colloidal particle of SiO2 (Ludox) attains
a () charge, which repels other colloidal
particles.
Suspended SiO2 (sand) settles very quickly.
50
Formation and Coagulationof a Colloid
When a strong electrolyte is added to colloidal
iron oxide, the charge on the surface of each
particle is partially neutralized
and the colloidal particles coalesce into a
suspension that quickly settles.
51

• Cumulative Example
• A 375-mL sample of hexane vapor in equilibrium
  with liquid hexane C6H14 (d 0.6548 g/mL), at
  25.0 C is dissolved in 50.0 mL of liquid
  cyclohexane, C6H12, at 25.0 C (d 0.7739 g/mL,
  vp 97.58 Torr). Use information found elsewhere
  in the text (such as Example 11.3) to calculate
  the total vapor pressure above the solution at
  25.0 C. How reliable is this calculation?