Chapter Twelve

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Chapter Twelve

• Physical Properties
• Of Solutions

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Contents

• Solutions
• (1) Some Types Of Solutions
• (2) Solution Concentrations
• (3) Energetics Of Solution Formation
• (4) Equilibrium In Solution Formation
• (5) The Solubilities Of Gases
• Colligative Properties Of Solutions
• (1) Vapor Pressures Of Solutions
• (2) Freezing Point Depression And Boiling Point
  Elevation
• (3) Osmotic Pressure
• (4) Solutions Of Electrolytes
• 3. Colloids

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(1) Some Types Of Solutions
1. Solutions

• A homogeneous mixture of two or more substances.
• Solute dispersed in a Solvent.

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(2) Solution Concentrations
1) Regular Concentration Units of
Solution a) Molarity (M) The amount of solute
(in moles) per one liter of solution (in
liters). b) Percent by volume (v/v) The volume
of solute (in liters) divided by the volume of
solution (in liters) and multiplied by
100. c) Mass/Volume percent (w/v) The amount of
solute (in grams) divided by the volume of
solution (in liters) and multiplied by
100. d) Percent by mass (w/w) The amount of
solute (in grams) divided by the amount of
solution (in grams) and multiplied by
100. e) Molality (m) The number of solute (in
moles) per one kilogram of solvent. a)c)
Temperature dependent, d)e) Temperature
independent.
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2) Solutions By Parts
a) Parts per million (ppm, 106) The number of
particles of solute per one-million particles of
solution, 1 ppm means 1 mg/L (liquid
solution) 1 mg/kg (solid solution) 1 µmol/ mol
or 1 µL/L (gaseous solution) b) Parts per billion
(ppb, 109) The number of particles of solute per
one-billion particles of solution. c) Parts per
trillion (ppt, 1012) The number of particles of
solute per one-trillion particles of
solution.
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3) Mole Fraction And Mole Percent
a) Mole fraction (xi) amount of
component i (mol) ni xi
total amount of solution
components (mol) ntotal b) The mole
percent of a solution component is its mole
fraction multiplied by 100.

• Temperature independent units
• To be independent of temperature, a concentration
  unit must be based on mass, not volume.

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Example 12.1 How would you prepare 750 g of an
aqueous solution that is 2.5 NaOH by mass?
Solution
Ans measure out 19 g NaOH and dissolve it in 731
g H2O.
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Example 12.2 At 20 C, pure ethanol has a density
of 0.789 g/mL and USP ethanol (95 ethanol on
volume basis) has a density of 0.813 g/mL. What
is the mass percent ethanol in USP ethanol?
Solution For 100 mL USP ethanol
Ans 92.3 on a mass basis
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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)?
Solution
Ans 0.598 m
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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?
Solution
Ans 0.805 g
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Example 12.6 An aqueous solution of ethylene
glycol 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?
Solution (a)
Ans 6.77 M
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(b)
Ans 10.7 m
(c)
Ans 0.162
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Energetics Of Solution Formation
1) Enthalpy of solution

• ?H1 gt 0, ?H2 gt 0, ?H3 lt 0
• ?Hsoln ?H1 ?H2 ?H3

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2) Intermolecular Forces in Solution

• If all intermolecular forces are of comparable
  strength, ?Hsoln 0 and ?Vsoln 0, this type
  called ideal solution.
• If solute-solvent interaction gt solvent-solvent
  interaction, ?Hsoln lt 0 and ?Vsoln lt 0, It is an
  exothermic process.
• If solute-solvent interaction lt solvent-solvent
  interaction, ?Hsoln gt 0 and ?Vsoln gt 0, It is an
  endothermic process.
• If solute-solvent interaction ltlt solvent-solvent,
  the solute does not dissolve in the solvent.

Rules of thumb Like dissolve like, Polar
dissolve in polar, Nonpolar dissolve in
nonpolar, Oil and water dont mix.
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3) Aqueous solutions of ionic compounds

• The forces causing an ionic solid to dissolve in
  water are iondipole forces
• Hydration energy becomes more negative with
  increasing ion charge and decreasing ion radius.

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

• Solution
• form a solution
• form a solution
• no dissolving
• no dissolving (immicible)

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Equilibrium In Solution Formation(for ionic
compounds)
1) Formation of a Saturated Solution
Saturated solution
Unsaturated solution
Rate of crystallization Rate of
dissolving dynamic equlibrium
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2) Solubility as a function of Temperature

• Solubility curve (left figure)
• - More than 95 of all ionic compounds have
  aqueous solubilities that increase with
  increasing temperature.
• - Solubility difference can be applied for
  selective crystallization

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3) Supersaturated Solution
until all of the excess solute has precipitated.
Solute immediately begins to crystallize
A single seed crystal of solute is added.
Supersaturated solution The solution contain
excess solute NOT dynamic equilibrium
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(5) The Solubilities of Gases
1) Temperature effect Most gases dissolved in
the liquids are exothermic process, thus,
according to Le Chatelier principle, increasing
temperature, decreasing solubility.
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2) Pressure effect According to Henry's law 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. C kPgas
C Concentration of the gas (M) in solution ,
M P Pressure of the gas (atm) over the
solution. k Henry's law constant
(mol?L-1?atm-1), depend on the particular gas and
temperature.
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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.
Solution
Partial pressure of argon in air
Determining k
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Solubility of argon at 0.0089 atm
Mass of argon in 225 g of water
Ans 0.12 mg
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2. Colligative Properties Of Solutions

• The property of solutions depends on the number
  of solute in solution, not the nature of the
  solute

• Vapor Pressures of Solutions
• 1) Raoults law
• Psolv xsolv Posolv

• Pi Partial pressure of the volatile solvent
  above the solution
• Poi Vapor pressure of the pure solvent
• xi Mole fraction of the solvent in the solution

2) For nonvolatile solute dissolved in a volatile
solvent The vapor pressure is lower than pure
solvent
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An Interesting Phenomenon
PH2O
PH2O xH2O PoH2O
Concentrated solution
Diluted Solution
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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
C12H22O11 (sucrose) and 75.0 g CO(NH2)2 (urea)
dissolved per kilogram of water?
Solution
moles of sucrose
moles of urea
moles of water
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Total number of moles of solute
Mole fraction of waters
Vapor pressure of water above the solution
Ans 17.0 mmHg.
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• For volatile solute dissolved in a volatile
  solvent
• Pi xi Pio, Pt SPi Sxi Pio
• The vapor in equilibrium with an ideal solution
  has a higher mole fraction of the more volatile
  component than is found in the liquid.
• Can be applied for fractional distillation

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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 C6H6
and C7H8. Determine the vapor pressures of C6H6
and C7H8 and the total vapor pressure above this
solution. Consider the solution to be ideal.
Solution
Ans 61.8 mmHg
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Freezing Point Depression And Boiling
PointElevation
1) Phase Diagram Comparison
Blue curve Pure solvent. Red curve A solution
with volatile solvent and nonvolatile solute.
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2) Freezing Point Depression

• At freezing point, liquid solvent molecules are
  solidified to an ordered crystal form.
• The solution contain solute which block the
  solvent molecule, therefore, slow down the
  collision frequency for the liquid state to the
  solid surface, that is the reason the freezing
  point is depressed.
• The freezing point depression calculation
• ?Tf Tf(solution) Tf(solvent) Kf m
• m molality
• Kf molal freezing-point depression constant
• For water, Kf is 1.86 oC/m

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3) Boiling Point Elevation

• The boiling point of liquid is the temperature at
  which its "vapor pressure" equals the "external
  atmospheric pressure"
• As the solution are made by adding a nonvolatile
  solute in a volatile solvent, vapor pressure
  lowering occurred.
• To make the "solution's pressure" equal to the
  external pressure, higher temperature (e.g. BP)
  is required.
• The boiling point elevation calculation
• ?Tb Tb(solution) Tb(solvent) Kb m
• m molality
• Kb molal boiling-point elevation constant
• For water, Kb is 0.512 oC/m

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4) Some FPD And BPE Constants
?Tf Kf m ?Tb Kb m
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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?
Solution (a)
Ans 182 g
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(b)
Ans C6H14O6
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(3) Osmotic Pressure
1) Semipermeable membrane The specific membrane
that allows solvent molecules to pass through,
but block the large solute particles 2) Osmosis
The net movement of solvent molecules through a
semipermeable membrane from a pure solvent (or a
dilute solution) to a more concentrated solution.

• When the flow of water through the membrane is
  the same in both directions, the hydrostatic
  pressure is now called the osmotic pressure.

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3) Osmotic pressure The pressure difference at
the equilibrium or the pressure required to stop
osmosis, it is expressed as p MRT p
osmotic pressure, atm M molarity of solute,
M T absolute temperature, K R gas constant,
0.08206 Latm/Kmol
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• Practical applications of osmosis
• For two solutions separated by semipermeable
  membrane

Isotonic solution equal concentration of the
both solutions Hypertonic solution the more
concentration solution Hypotonic solution the
more dilute solution

• Reverse osmosis The process of reversing the
  normal net flow of solvent molecules through a
  semipermeable membrane by applying to the
  solution a pressure exceeding the osmotic
  pressure can be applied for water purification.

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c) Determine the molar mass

• Osmotic pressure technique For higher molar mass
  and less soluble molecules (e.g., protein)
• Freezing point depression technique For
  smaller molar mass and more soluble molecules.

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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?
Solution (a)
Ans 4.00 x 104 g/mol
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(b) Assume that the molarity and molality of the
dilute solution are equal
Ans 0.000279 C (extremely difficult to
measure with precision and accuracy)
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(4) Colligative Properties Of Electrolytes

• The vant Hoff factor (i) is used to modify the
  colligative property equations for electrolytes
• ?Tf iKf m
• ?Tb iKb m
• p iMRT

• For electrolytes, the expected i equal to the
  number of ions for a substance dissociates into
  in solution. e.g., for NaCl, i 2 and for
  Pb(NO3)2, i 3.
• For nonelectrolyte solutes, i 1.

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2) The Experimental i (measured)
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3. Colloids

• Solution, Suspension and Colloid
• 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) dispersed
  particles are relatively large, and will settle
  from suspension.
• In a colloid dispersed particles are on the
  order of 11000 nm in size, larger than
  molecules/atoms/ions, colloidal particles are
  small enough to remain dispersed indefinitely.

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(2) Common Types Of Colloids
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• (3) More About collids
• 1) Tyndall effect
• The scattering of a visible light beam through a
  colloidal material is known as the Tyndall
  effect, which preferential scattering of blue
  light (shorter wavelength) than red light (longer
  wavelength).

Solution
Colloidal Fe(III) Oxide in water
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• Formation and Coagulation of a Colloid
• A high concentration of an inert electrolyte can
  cause a colloid to coagulate, or precipitate

After adding inert electrolyte e.g., Al2(SO4)3
Colloidal Iron(III) oxide
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End of Chapter 12