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Physical Properties
of Solutions
Chapter 12
Solution
Stoichiometry
end of Chapter 4
12.1
A solution is a homogenous mixture of 2 or
more substances
The solute is(are) the substance(s) present in the
smaller amount(s)
The solvent is the substance present in the larger
amount
An electrolyte is a substance that, when dissolved in
water, results in a solution that can conduct electricity.
A nonelectrolyte is a substance that, when dissolved,
results in a solution that does not conduct electricity.
nonelectrolyte weak electrolyte strong electrolyte
4.1
A saturated solution contains the maximum amount of a
solute that will dissolve in a given solvent at a specific
temperature.
An unsaturated solution contains less solute than the
solvent has the capacity to dissolve at a specific
temperature.
A supersaturated solution contains more solute than is
present in a saturated solution at a specific temperature.
Sodium acetate crystals rapidly form when a seed crystal is
added to a supersaturated solution of sodium acetate.
12.1
12.2
Three types of interactions in the solution process:
• solvent-solvent interaction
• solute-solute interaction
• solvent-solute interaction
DHsoln = DH1 + DH2 + DH3
“like dissolves like”
Two substances with similar intermolecular forces are likely
to be soluble in each other.
• non-polar molecules are soluble in non-polar solvents
CCl4 in C6H6
• polar molecules are soluble in polar solvents
C2H5OH in H2O
• ionic compounds are more soluble in polar solvents
NaCl in H2O or NH3 (l)
12.2
Concentration Units
The concentration of a solution is the amount of solute
present in a given quantity of solvent or solution.
Percent by Mass
% by mass = x 100%
mass of solute
mass of solute + mass of solvent
= x 100%
mass of solute
mass of solution
12.3
Mole Fraction (X)
XA =
moles of A
sum of moles of all components
Concentration Units Continued
M =
moles of solute
liters of solution
Molarity (M)
Molality (m)
m =
moles of solute
mass of solvent (kg)
12.3
What is the molality of a 5.86 M ethanol (C2H5OH)
solution whose density is 0.927 g/mL?
m =
moles of solute
mass of solvent (kg)
M =
moles of solute
liters of solution
Assume 1 L of solution:
5.86 moles ethanol = 270 g ethanol
927 g of solution (1000 mL x 0.927 g/mL)
mass of solvent = mass of solution – mass of solute
= 927 g – 270 g = 657 g = 0.657 kg
m =
moles of solute
mass of solvent (kg)
=
5.86 moles C2H5OH
0.657 kg solvent
= 8.92 m
12.3
Temperature and Solubility
Solid solubility and temperature
solubility increases with
increasing temperature
solubility decreases with
increasing temperature
12.4
Fractional crystallization is the separation of a mixture of
substances into pure components on the basis of their differing
solubilities.
Suppose you have 90 g KNO3
contaminated with 10 g NaCl.
Fractional crystallization:
1. Dissolve sample in 100 mL of
water at 600C
2. Cool solution to 00C
3. All NaCl will stay in solution
(s = 34.2g/100g)
4. 78 g of PURE KNO3 will
precipitate (s = 12 g/100g).
90 g – 12 g = 78 g
12.4
Temperature and Solubility
Gas solubility and temperature
solubility usually
decreases with
increasing temperature
12.4
Pressure and Solubility of Gases
The solubility of a gas in a liquid is proportional to the
pressure of the gas over the solution (Henry’s law).
c = kP
c is the concentration (M) of the dissolved gas
P is the pressure of the gas over the solution
k is a constant (mol/L•atm) that depends only
on temperature
low P
low c
high P
high c
Chemistry In Action: The Killer Lake
Lake Nyos, West Africa
8/21/86
CO2 Cloud Released
1700 Casualties
Trigger?
• earthquake
• landslide
• strong Winds
Solution Stoichiometry (Chapter 4)
The concentration of a solution is the amount of solute
present in a given quantity of solvent or solution.
M = molarity =
moles of solute
liters of solution
What mass of KI is required to make 500. mL of
a 2.80 M KI solution?
volume KI moles KI grams KI
M KI M KI
500. mL = 232 g KI
166 g KI
1 mol KI
x
2.80 mol KI
1 L soln
x
1 L
1000 mL
x
4.5
4.5
Dilution is the procedure for preparing a less concentrated
solution from a more concentrated solution.
Dilution
Add Solvent
Moles of solute
before dilution (i)
Moles of solute
after dilution (f)
=
MiVi MfVf
=
4.5
How would you prepare 60.0 mL of 0.2 M
HNO3 from a stock solution of 4.00 M HNO3?
MiVi = MfVf
Mi = 4.00 Mf = 0.200 Vf = 0.06 L Vi = ? L
4.5
Vi =
MfVf
Mi
=
0.200 x 0.06
4.00
= 0.003 L = 3 mL
3 mL of acid + 57 mL of water = 60 mL of solution
Gravimetric Analysis
4.6
1. Dissolve unknown substance in water
2. React unknown with known substance to form a precipitate
3. Filter and dry precipitate
4. Weigh precipitate
5. Use chemical formula and mass of precipitate to determine
amount of unknown ion
Titrations
In a titration a solution of accurately known concentration is
added gradually added to another solution of unknown
concentration until the chemical reaction between the two
solutions is complete.
Equivalence point – the point at which the reaction is complete
Indicator – substance that changes color at (or near) the
equivalence point
Slowly add base
to unknown acid
UNTIL
the indicator
changes color
4.7
What volume of a 1.420 M NaOH solution is
Required to titrate 25.00 mL of a 4.50 M H2SO4
solution?
4.7
WRITE THE CHEMICAL EQUATION!
volume acid moles acid moles base volume base
H2SO4 + 2NaOH 2H2O + Na2SO4
4.50 mol H2SO4
1000 mL soln
x
2 mol NaOH
1 mol H2SO4
x
1000 ml soln
1.420 mol NaOH
x
25.00 mL = 158 mL
M
acid
rx
coef.
M
base
Chemistry in Action: Metals from the Sea
CaCO3 (s) CaO (s) + CO2 (g)
Mg(OH)2 (s) + 2HCl (aq) MgCl2 (aq) + 2H2O (l)
CaO (s) + H2O (l) Ca2+ (aq) + 2OH (aq)
-
Mg2+ (aq) + 2OH (aq) Mg(OH)2 (s)
-
Mg2+ + 2e- Mg
2Cl- Cl2 + 2e-
MgCl2 (l) Mg (l) + Cl2 (g)
Now back to Chapter 12…
Colligative Properties of Nonelectrolyte Solutions
Colligative properties are properties that depend only on the
number of solute particles in solution and not on the nature of
the solute particles.
Vapor-Pressure Lowering
Raoult’s law
If the solution contains only one solute:
X1 = 1 – X2
P 1
0
- P1 = DP = X2 P 1
0
P 1
0
= vapor pressure of pure solvent
X1 = mole fraction of the solvent
X2 = mole fraction of the solute
12.6
P1 = X1 P 1
0
PA = XA P A
0
PB = XB P B
0
PT = PA + PB
PT = XA P A
0
+ XB P B
0
Ideal Solution
12.6
PT is greater than
predicted by Raoults’s law
PT is less than
predicted by Raoults’s law
Force
A-B
Force
A-A
Force
B-B
< &
Force
A-B
Force
A-A
Force
B-B
> &
12.6
Fractional Distillation Apparatus
12.6
Boiling-Point Elevation
DTb = Tb – T b
0
Tb > T b
0
DTb > 0
T b is the boiling point of
the pure solvent
0
T b is the boiling point of
the solution
DTb = Kb m
m is the molality of the solution
Kb is the molal boiling-point
elevation constant (0C/m)
12.6
Freezing-Point Depression
DTf = T f – Tf
0
T f > Tf
0
DTf > 0
T f is the freezing point of
the pure solvent
0
T f is the freezing point of
the solution
DTf = Kf m
m is the molality of the solution
Kf is the molal freezing-point
depression constant (0C/m)
12.6
12.6
What is the freezing point of a solution containing 478 g
of ethylene glycol (antifreeze) in 3202 g of water? The
molar mass of ethylene glycol is 62.01 g.
DTf = Kf m
m =
moles of solute
mass of solvent (kg)
= 2.41 m
=
3.202 kg solvent
478 g x
1 mol
62.01 g
Kf water = 1.86 0C/m
DTf = Kf m = 1.86 0C/m x 2.41 m = 4.48 0C
DTf = T f – Tf
0
Tf = T f – DTf
0
= 0.00 0C – 4.48 0C = -4.48 0C
12.6
Colligative Properties of Nonelectrolyte Solutions
Colligative properties are properties that depend only on the
number of solute particles in solution and not on the nature of
the solute particles.
12.6
Vapor-Pressure Lowering P1 = X1 P 1
0
Boiling-Point Elevation DTb = Kb m
Freezing-Point Depression DTf = Kf m
Osmotic Pressure (p) p = MRT
Colligative Properties of Electrolyte Solutions
12.7
0.1 m NaCl solution 0.1 m Na+ ions & 0.1 m Cl- ions
Colligative properties are properties that depend only on the
number of solute particles in solution and not on the nature of
the solute particles.
0.1 m NaCl solution 0.2 m ions in solution
van’t Hoff factor (i) =
actual number of particles in soln after dissociation
number of formula units initially dissolved in soln
nonelectrolytes
NaCl
CaCl2
i should be
1
2
3
Which would you use for the streets of
Bloomington to lower the freezing point
of ice and why? Would the temperature
make any difference in your decision?
a) sand, SiO2
b) Rock salt, NaCl
c) Ice Melt, CaCl2
Change in Freezing Point
Boiling-Point Elevation DTb = i Kb m
Freezing-Point Depression DTf = i Kf m
Osmotic Pressure (p) p = iMRT
Colligative Properties of Electrolyte Solutions
12.7
Change in Freezing Point
Common Applications
of Freezing Point
Depression
Propylene glycol
Ethylene
glycol –
deadly to
small
animals
Change in Boiling Point
Common Applications of
Boiling Point Elevation
At what temperature will a 5.4 molal
solution of NaCl freeze?
Solution
∆TFP = Kf • m • i
∆TFP = (1.86 oC/molal) • 5.4 m • 2
∆TFP = 20.1 oC
FP = 0 – 20.1 = -20.1 oC
Freezing Point Depression
The Cleansing Action of Soap
12.8
Osmotic Pressure (p)
12.6
Osmosis is the selective passage of solvent molecules through a porous
membrane from a dilute solution to a more concentrated one.
A semipermeable membrane allows the passage of solvent molecules but
blocks the passage of solute molecules.
Osmotic pressure (p) is the pressure required to stop osmosis.
dilute
more
concentrated
High
P
Low
P
Osmotic Pressure (p)
p = MRT
M is the molarity of the solution
R is the gas constant
T is the temperature (in K) 12.6
A cell in an:
isotonic
solution
hypotonic
solution
hypertonic
solution
12.6
Chemistry In Action:
Desalination
A colloid is a dispersion of particles of one substance
throughout a dispersing medium of another substance.
Colloid versus solution
• collodial particles are much larger than solute molecules
• collodial suspension is not as homogeneous as a solution
12.8
Colloids
• Tyndall Effect
• Brownian motion
Suspensions

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Ch 12 Physical Properties of Solutions.ppt

  • 1. Physical Properties of Solutions Chapter 12 Solution Stoichiometry end of Chapter 4
  • 2. 12.1 A solution is a homogenous mixture of 2 or more substances The solute is(are) the substance(s) present in the smaller amount(s) The solvent is the substance present in the larger amount
  • 3. An electrolyte is a substance that, when dissolved in water, results in a solution that can conduct electricity. A nonelectrolyte is a substance that, when dissolved, results in a solution that does not conduct electricity. nonelectrolyte weak electrolyte strong electrolyte 4.1
  • 4. A saturated solution contains the maximum amount of a solute that will dissolve in a given solvent at a specific temperature. An unsaturated solution contains less solute than the solvent has the capacity to dissolve at a specific temperature. A supersaturated solution contains more solute than is present in a saturated solution at a specific temperature. Sodium acetate crystals rapidly form when a seed crystal is added to a supersaturated solution of sodium acetate. 12.1
  • 5. 12.2 Three types of interactions in the solution process: • solvent-solvent interaction • solute-solute interaction • solvent-solute interaction DHsoln = DH1 + DH2 + DH3
  • 6. “like dissolves like” Two substances with similar intermolecular forces are likely to be soluble in each other. • non-polar molecules are soluble in non-polar solvents CCl4 in C6H6 • polar molecules are soluble in polar solvents C2H5OH in H2O • ionic compounds are more soluble in polar solvents NaCl in H2O or NH3 (l) 12.2
  • 7. Concentration Units The concentration of a solution is the amount of solute present in a given quantity of solvent or solution. Percent by Mass % by mass = x 100% mass of solute mass of solute + mass of solvent = x 100% mass of solute mass of solution 12.3 Mole Fraction (X) XA = moles of A sum of moles of all components
  • 8. Concentration Units Continued M = moles of solute liters of solution Molarity (M) Molality (m) m = moles of solute mass of solvent (kg) 12.3
  • 9. What is the molality of a 5.86 M ethanol (C2H5OH) solution whose density is 0.927 g/mL? m = moles of solute mass of solvent (kg) M = moles of solute liters of solution Assume 1 L of solution: 5.86 moles ethanol = 270 g ethanol 927 g of solution (1000 mL x 0.927 g/mL) mass of solvent = mass of solution – mass of solute = 927 g – 270 g = 657 g = 0.657 kg m = moles of solute mass of solvent (kg) = 5.86 moles C2H5OH 0.657 kg solvent = 8.92 m 12.3
  • 10. Temperature and Solubility Solid solubility and temperature solubility increases with increasing temperature solubility decreases with increasing temperature 12.4
  • 11. Fractional crystallization is the separation of a mixture of substances into pure components on the basis of their differing solubilities. Suppose you have 90 g KNO3 contaminated with 10 g NaCl. Fractional crystallization: 1. Dissolve sample in 100 mL of water at 600C 2. Cool solution to 00C 3. All NaCl will stay in solution (s = 34.2g/100g) 4. 78 g of PURE KNO3 will precipitate (s = 12 g/100g). 90 g – 12 g = 78 g 12.4
  • 12. Temperature and Solubility Gas solubility and temperature solubility usually decreases with increasing temperature 12.4
  • 13. Pressure and Solubility of Gases The solubility of a gas in a liquid is proportional to the pressure of the gas over the solution (Henry’s law). c = kP c is the concentration (M) of the dissolved gas P is the pressure of the gas over the solution k is a constant (mol/L•atm) that depends only on temperature low P low c high P high c
  • 14. Chemistry In Action: The Killer Lake Lake Nyos, West Africa 8/21/86 CO2 Cloud Released 1700 Casualties Trigger? • earthquake • landslide • strong Winds
  • 15. Solution Stoichiometry (Chapter 4) The concentration of a solution is the amount of solute present in a given quantity of solvent or solution. M = molarity = moles of solute liters of solution What mass of KI is required to make 500. mL of a 2.80 M KI solution? volume KI moles KI grams KI M KI M KI 500. mL = 232 g KI 166 g KI 1 mol KI x 2.80 mol KI 1 L soln x 1 L 1000 mL x 4.5
  • 16. 4.5
  • 17. Dilution is the procedure for preparing a less concentrated solution from a more concentrated solution. Dilution Add Solvent Moles of solute before dilution (i) Moles of solute after dilution (f) = MiVi MfVf = 4.5
  • 18. How would you prepare 60.0 mL of 0.2 M HNO3 from a stock solution of 4.00 M HNO3? MiVi = MfVf Mi = 4.00 Mf = 0.200 Vf = 0.06 L Vi = ? L 4.5 Vi = MfVf Mi = 0.200 x 0.06 4.00 = 0.003 L = 3 mL 3 mL of acid + 57 mL of water = 60 mL of solution
  • 19. Gravimetric Analysis 4.6 1. Dissolve unknown substance in water 2. React unknown with known substance to form a precipitate 3. Filter and dry precipitate 4. Weigh precipitate 5. Use chemical formula and mass of precipitate to determine amount of unknown ion
  • 20. Titrations In a titration a solution of accurately known concentration is added gradually added to another solution of unknown concentration until the chemical reaction between the two solutions is complete. Equivalence point – the point at which the reaction is complete Indicator – substance that changes color at (or near) the equivalence point Slowly add base to unknown acid UNTIL the indicator changes color 4.7
  • 21. What volume of a 1.420 M NaOH solution is Required to titrate 25.00 mL of a 4.50 M H2SO4 solution? 4.7 WRITE THE CHEMICAL EQUATION! volume acid moles acid moles base volume base H2SO4 + 2NaOH 2H2O + Na2SO4 4.50 mol H2SO4 1000 mL soln x 2 mol NaOH 1 mol H2SO4 x 1000 ml soln 1.420 mol NaOH x 25.00 mL = 158 mL M acid rx coef. M base
  • 22. Chemistry in Action: Metals from the Sea CaCO3 (s) CaO (s) + CO2 (g) Mg(OH)2 (s) + 2HCl (aq) MgCl2 (aq) + 2H2O (l) CaO (s) + H2O (l) Ca2+ (aq) + 2OH (aq) - Mg2+ (aq) + 2OH (aq) Mg(OH)2 (s) - Mg2+ + 2e- Mg 2Cl- Cl2 + 2e- MgCl2 (l) Mg (l) + Cl2 (g) Now back to Chapter 12…
  • 23. Colligative Properties of Nonelectrolyte Solutions Colligative properties are properties that depend only on the number of solute particles in solution and not on the nature of the solute particles. Vapor-Pressure Lowering Raoult’s law If the solution contains only one solute: X1 = 1 – X2 P 1 0 - P1 = DP = X2 P 1 0 P 1 0 = vapor pressure of pure solvent X1 = mole fraction of the solvent X2 = mole fraction of the solute 12.6 P1 = X1 P 1 0
  • 24. PA = XA P A 0 PB = XB P B 0 PT = PA + PB PT = XA P A 0 + XB P B 0 Ideal Solution 12.6
  • 25. PT is greater than predicted by Raoults’s law PT is less than predicted by Raoults’s law Force A-B Force A-A Force B-B < & Force A-B Force A-A Force B-B > & 12.6
  • 27. Boiling-Point Elevation DTb = Tb – T b 0 Tb > T b 0 DTb > 0 T b is the boiling point of the pure solvent 0 T b is the boiling point of the solution DTb = Kb m m is the molality of the solution Kb is the molal boiling-point elevation constant (0C/m) 12.6
  • 28. Freezing-Point Depression DTf = T f – Tf 0 T f > Tf 0 DTf > 0 T f is the freezing point of the pure solvent 0 T f is the freezing point of the solution DTf = Kf m m is the molality of the solution Kf is the molal freezing-point depression constant (0C/m) 12.6
  • 29. 12.6
  • 30. What is the freezing point of a solution containing 478 g of ethylene glycol (antifreeze) in 3202 g of water? The molar mass of ethylene glycol is 62.01 g. DTf = Kf m m = moles of solute mass of solvent (kg) = 2.41 m = 3.202 kg solvent 478 g x 1 mol 62.01 g Kf water = 1.86 0C/m DTf = Kf m = 1.86 0C/m x 2.41 m = 4.48 0C DTf = T f – Tf 0 Tf = T f – DTf 0 = 0.00 0C – 4.48 0C = -4.48 0C 12.6
  • 31. Colligative Properties of Nonelectrolyte Solutions Colligative properties are properties that depend only on the number of solute particles in solution and not on the nature of the solute particles. 12.6 Vapor-Pressure Lowering P1 = X1 P 1 0 Boiling-Point Elevation DTb = Kb m Freezing-Point Depression DTf = Kf m Osmotic Pressure (p) p = MRT
  • 32. Colligative Properties of Electrolyte Solutions 12.7 0.1 m NaCl solution 0.1 m Na+ ions & 0.1 m Cl- ions Colligative properties are properties that depend only on the number of solute particles in solution and not on the nature of the solute particles. 0.1 m NaCl solution 0.2 m ions in solution van’t Hoff factor (i) = actual number of particles in soln after dissociation number of formula units initially dissolved in soln nonelectrolytes NaCl CaCl2 i should be 1 2 3
  • 33. Which would you use for the streets of Bloomington to lower the freezing point of ice and why? Would the temperature make any difference in your decision? a) sand, SiO2 b) Rock salt, NaCl c) Ice Melt, CaCl2 Change in Freezing Point
  • 34. Boiling-Point Elevation DTb = i Kb m Freezing-Point Depression DTf = i Kf m Osmotic Pressure (p) p = iMRT Colligative Properties of Electrolyte Solutions 12.7
  • 35. Change in Freezing Point Common Applications of Freezing Point Depression Propylene glycol Ethylene glycol – deadly to small animals
  • 36. Change in Boiling Point Common Applications of Boiling Point Elevation
  • 37. At what temperature will a 5.4 molal solution of NaCl freeze? Solution ∆TFP = Kf • m • i ∆TFP = (1.86 oC/molal) • 5.4 m • 2 ∆TFP = 20.1 oC FP = 0 – 20.1 = -20.1 oC Freezing Point Depression
  • 38. The Cleansing Action of Soap 12.8
  • 39. Osmotic Pressure (p) 12.6 Osmosis is the selective passage of solvent molecules through a porous membrane from a dilute solution to a more concentrated one. A semipermeable membrane allows the passage of solvent molecules but blocks the passage of solute molecules. Osmotic pressure (p) is the pressure required to stop osmosis. dilute more concentrated
  • 40. High P Low P Osmotic Pressure (p) p = MRT M is the molarity of the solution R is the gas constant T is the temperature (in K) 12.6
  • 41. A cell in an: isotonic solution hypotonic solution hypertonic solution 12.6
  • 43. A colloid is a dispersion of particles of one substance throughout a dispersing medium of another substance. Colloid versus solution • collodial particles are much larger than solute molecules • collodial suspension is not as homogeneous as a solution 12.8