Properties of Solutions

1
Properties of Solutions

• Chapter 13

2
Three Steps to forming a solution

• DH1 Separate the solute
• DH2 Separate the solvent
• DH3 Form imfs

3

• DHsoln DH1 DH2 DH3
• Exothermic Reactions
• Solute-solvent imf is stronger than the imfs
  holding solute and solvent together (DH3)
• Gets warmer
• Endothermic Reactions
• Imf is weaker than the imfs holding solute and
  solvent together (DH3)
• Gets colder

4
Role of Disorder

• Disorder (Entropy) is also a factor
• Solutions mix to form maximum disorder

5
Two Ways to Form Solutions

• 1. Physical Dissolving
• NaCl(s) ? NaCl(aq)
• C12H22O11(s) ? C12H22O11(aq)
• Can evaporate water/solvent to get original
  compound back

6
(No Transcript)
7
Types of Reactions
8

• Chemical reaction
• Ni(s) 2HCl(aq) ? NiCl2(aq) H2(g)
• Evaporating solvent gives the products

9
Saturated Solutions

• Solution Homogeneous mixture
• Solvent Present in the greater amount
• Solute Present in the lesser amount
• Examples
• Kool-Aid
• Grape Juice from Conc.
• Brass

10

• 5. Solubility Maximum amount of a solute that
  can dissolve in 100 mL of a solution
• Ex NaCl 35.7 g/100mL
• Saturated solution Contains the max. amount of
  solute with some undissolved solid,
• Unsaturated more solute will dissolve.
• Supersaturated More than the max is dissolved
  by heating and slowly cooling.

11
Like Dissolves Like

• Polar dissolves polar (dipole-dipole Forces)
• Water and Ammonia
• Non-Polar dissolves non-polar (London Forces)
• Soap and grease

12

• Would acetone (shown below) dissolve in water?

O CH3CCH3
Acetone
13

• Using you knowledge of like dissolves like,
  explain the following trends in solubility.

14
(No Transcript)
15
Pressure Effects

• Solubilty of a gas increases with pressure of gas
  over the liquid (soda bottle)
• Henrys Law
• Cg kPg
• Cg concentration of the gas in the solution
• k Henrys Law Constant for the gas
• Pg Pressure in atm

16
Henrys Law Ex 1

• Calculate the concentration of CO2 in a soft
  drink with a partial pressure of 4.0 atm over a
  liquid at 25oC. The Henrys Law constant for CO2
  is 3.1 X 10-2 mol/L atm.
• Cg kPg
• Cg (3.1 X 10-2 mol/L atm)(4.0 atm)
• Cg 0.12 M

17
Henrys Law Ex 2

• Calculate the concentration of CO2 once the
  bottle is opened and the partial pressure is now
  only 3.0 X 10-4 atm.
• (ANS 9.3 X 10-6 M)

18
Temperature Effects

• Solubility of most solids increases with
  temperature
• Solubility of most gases decreases with
  temperature (warm soda)
• Warm water is deoxygenated
• Problem with thermal pollution of lakes

19
Ways of Expressing Concentration

• Mass mass of compound in soln X 100
• total mass of soln
• Parts Per Million
• ppm mass of component X 106
• total mass of soln

20
Concentration Ex 1

• 13.5 g of C6H12O6 is dissolved in 0.100 kg of
  water. Calculate the mass percentage.
• mass 13.5 g X 100 11.9
• (100 g 13.5 g)

21
Concentration Ex 2

• A 2.5 g sample of groundwater is found to contain
  5.4 mg of Zn2. What is the concentration of the
  Zn2 ion.
• 5.4 mg 1X10-6g 5.4 X 10-6 g
• 1 mg
• ppm mass of component X 106
• total mass of soln
• ppm 5.4 X 10-6 g X 106 2.2 ppm
• 2.5g

22
Concentration Ex 3

• Calculate the mass percentage of NaCl in a
  solution containing 1.50 g of NaCl in 50.0 g of
  water.
• ANS 2.91

23
Concentration Ex 4

• Bleach is 3.62 NaOCl. What mass of NaOCl is
  contained in 2500 g of bleach?
• ANS 90.5 g NaOCl

24
Mole Fraction

• Mole Fraction
• X moles of component
• total moles of all components
• What is the mole fraction of HCl if 36.5 grams is
  dissolved in 144 grams of water?
• ANS 0.111

25
Molality

• Molality moles of solute
• kilograms of solvent
• Why not use Molarity?
• Molarity varies with temperature
• Total volume of a solution changes with
  temperature (liquid expands)
• Mass does not change with temperature

26
Molality Ex 1

• A solution is made by dissolving 4.35 grams of
  C6H12O6 in 25.0 mL of water. Calculate the
  molality of the glucose.
• 4.35 g 1 mol 0.0241 mol
• 180.2 g
• Molality 0.0241 mol 0.964 m
• 0.0250 kg

27
Molality Ex 2

• Calculate the molality of a solution made by
  dissolving 36.5g C10H8 in 425 grams of toluene
  (solvent).
• ANS 0.671 m

28
Molality Ex 3

• A solution of HCl contains 36 percent HCl by
  mass. Calculate the mole fraction and molality
  of HCl.
• Pretend 100 g
• 36 g HCl
• 64 g H2O

29

• 36 g HCl 1 mol HCl 0.99 mol HCl
• 36.5 g HCl
• 64 g H2O 1 mol H2O 3.6 mol HCl
• 18 g H2O
• XHCl 0.99 mol HCl 0.22
• 0.99 mol 3.6mol
• Molality 0.99 mol HCl 15 m
• 0.064 kg H2O

30
Molality Ex 4

• A commercial bleach solutions contains 3.62
  percent NaOCl by mass. Calculate the mole
  fraction and molality of NaOCl.
• ANS XNaOCl 0.00900, 0.505 m

31
Molality Ex 5

• The density of a solution of 5.0 g of toluene
  (C7H8) and 225 g of benzene is 0.876 g/mL.
  Calculate the molality and molarity of the
  solution.
• Molality
• 5.0 g 1 mol 0.054 mol
• 92.0 g
• m 0.054 mol/ 0.225 kg 0.24 m

32
Molality Ex 6

• A solution containing equal masses of glycerol
  (C3H8O3) and water has a density of 1.10 g/mL.
  Calculate the molality and mole fraction of
  glycerol in the solution.
• ANS 10.9 m, XC3H8O3 0.163

33

• Molarity
• D mass/V
• V mass/D
• V 230 g 263 mL
• 0.876 g/ml
• M 0.054 mol 0.21 M
• 0.263 L

34
Colligative Properties Vapor Pressure Lowering

• Non-volatile solutes lower the vapor pressure of
  the solvent
• Raoults law
• PA XAPoA
• PA Vapor pressure
• XA Mole fraction of solvent
• PoA Pressure of pure solvent

35
Raoults Law Ex 1

• What is the vapor pressure of a solution made by
  adding 50.0 mL of glycerin (C3H8O3) to 500.0 mL
  of water? The density of glycerin is 1.26 g/mL
  and the vapor pressure of pure water is 23.8
  torr.
• MassC3H8O3 (50.0 mL)(1.26 g/mL ) 63.0 g
• MolesC3H8O3 500.0 g/92.1 g/mol 0.684 mol
• MolesH2O 500.0 g/18 g/mol 27.8 mol

36

• XH2O 27.8 mol 0.976
• (27.8 mol 0.684 mol)
• PA XAPoA
• PA (0.979)(23.8 torr) 23.2 torr

37
Raoults Law Ex 2

• The vapor pressure of water at 110oC is 1070
  torr. A solution of ethylene glycol and water
  has a vapor pressure of 1 atm at 110oC. What is
  the mole fraction of ethylene glycol in the
  solution?
• ANS 0.290

38
Colligative Properties Boiling Point Elevation

• Non-volatile solute raises the boiling point of a
  solution
• Shifts the phase diagram
• The pressure of the solution reaches atmospheric
  pressure at a higher temp.
• DTb iKbm

39
Colligative Properties Freezing Point Depression

• Solutions freeze at a lower temperature than pure
  solvent
• Salt water freezes lower (-2oC) than distilled
  water (0oC)
• DTf iKfm
• m molality of the nonvolatile solute

40

• The more ions produced, the greater the freezing
  point depression or boiling point elevation
• NaBr Produces two ions
• CaBr2 Produces three ions (more effect)

41
(No Transcript)
42
Colligative Ex 1

• Ethylene Glycol, C2H6O2, is used in antifreeze.
  What will be the freezing and boiling point of a
  25.0 mass percent solution of ethylene glycol and
  water?
• Pretend 100 grams of solution
• 25 grams of C2H6O2
• 75 grams of H2O (0.075 kg)

43

• 25 grams of C2H6O2 0.403 moles
• m 0.403 moles 5.37 m
• 0.075 kg H2O
• DTb iKbm (1)(0.52oC/m)(5.37 m) 2.8oC
• DTf iKfm (1)( 1.86oC/m)(5.37 m) 10.0oC
• Boiling Point 102.8oC
• Freezing Point -10.0oC

o
o
o
44
Colligative Ex 2

• Calculate the freezing point of a solution
  containing 0.600 kg of CHCl3 and 42.0 g of
  C10H18O. Kf for C10H18O is 4.68oC/m and the
  normal freezing point is -63.8oC.
• ANS -65.9oC

45
Colligative Ex 3

• Rank the following aqueous solutions in order of
  their expected freezing points
• 0.050 m CaCl2
• 0.15 m NaCl
• 0.10 m HCl
• 0.050 m HC2H3O2 (acetic acid)
• 0.10 m C12H22O11 (sugar)

46

• 0.050 m CaCl2 (0.15 m in particles)
• 0.15 m NaCl (0.30 m in particles)
• 0.10 m HCl (0.20 m in particles)
• 0.050 m HC2H3O2 (0.05 to 0.10 m)
• 0.10 m C12H22O11 (0.10 m in particles)
• Lowest FP Highest FP
• NaCl lt HCl lt CaCl2 lt C12H22O11 lt HC2H3O2

47
Colligative Ex 4

• Rank the following in order of the increase in
  boiling point that they will produce in 1 kg of
  water
• 1 mol Co(NO3)2
• 2 mol KCl
• 3 mol C2H6O2 (a very, very weak electrolyte)

48

• 1 mol Co(NO3)2 (3 mol particles)
• 2 mol KCl (4 mol of particles)
• 3 mol C2H6O2 (3 mol of particles)
• Lowest BP Highest BP
• Co(NO3)2 lt C2H6O2 lt KCl

49
Freezing Pt Depression Ex 5

• What would be the molality of salt water if it
  freezes at 0 oF? Kf 1.86 oC/m.
• 0oF -17.8oC
• DTf iKfm
• m DTf/iKf
• m 17.8oC/(2X1.86oC/m) 4.78 m

50
Colligative Properties Osmotic Pressure

• Osmosis movement of solvent from high
  concentration to low concentration
• semipermeable membrane allows to passage of
  some particles but not others
• Cucumber Skin cell after
• in salt water soaking in a tub

51

• Note that solvent moves both ways
• Solute too large to pass through membrane
• Net movement is to try to dilute the side with
  solutes

52

• Osmotic Pressure (p) pressure required to
  prevent osmosis
• PV inRT
• V inRT
• p inRT
• V
• p iMRT
• M molarity

53
Osmotic Pressure Ex 1

• The average osmotic pressure of blood is 7.7 atm
  at 25oC. What concentration of glucose will be
  isotonic with blood?
• p iMRT
• M p/iRT
• M 7.7 atm 0.31 M
• (1)(0.0821 L-atm/mol-K)(298)

54
Osmotic Pressure Ex 2

• What is the osmotic pressure at 20oC of a 0.0020
  M sucrose, C12H22O11, solution? Express your
  answer both in atmosphere and in torr.
• ANS 0.048 atm, 37 torr

55
Molar Mass Ex 1

• A solution of an unknown nonelectrolyte was
  prepared by dissolving 0.250 g in 40.0 g of CCl4.
  The boiling point of the resulting solution was
  0.357oC higher than that of the pure solvent. Kb
  for CCl4 is 5.02 oC/m.
• DTb iKbm
• m DTb /iKb
• m (0.357oC)/(1 X 5.02 oC/m) 0.0711 m

56

• m mol of solute
• kilograms of solvent
• molsolute (m)(kg of solvent)
• molsolute (0.0711 m)(0.0400 kg) 0.00284 mol
• Molar Mass 0.250 g 88.0 g/mol
• 0.00284 mol

57
Molar Mass Ex 2

• Camphor, C10H16O, melts at 179.8oC and has a Kf
  of 40.0 oC/m. When 0.186 g of an unknown
  substance is dissolved in 22.01 g of liquid
  camphor, the freezing point is 176.7oC. What is
  the molar mass of the solute?
• ANS 110 g/mol

58
Molar Mass Ex 3

• A solution contains 3.50 mg of protein dissolved
  in water to form 5.00 mL of a solution. The
  osmotic pressure at 25oC was found to be 1.54
  torr. Calculate the molar mass of the protein.
• 1.54 torr 1 atm 0.00203 atm
• 760 torr

59

• p iMRT
• M p/iRT
• M 0.00203 atm 8.28X10-5 M
• (1)(0.0821 L-atm/mol-K)(298)
• M moles
• liter
• moles (M)(liters) (8.28X10-5 M)(0.00500L)
• moles 4.14 X 10-7 mol
• Molar mass 3.50 X 10-3 g 8454 g/mol
• 4.14 X 10-7 mol

60
Molar Mass Ex 4

• A 2.05 g sample of a plastic was dissolved in
  enough toluene to form 100 mL of solution. The
  osmotic pressure of this solution is 1.21 kPa at
  25oC. Calculate the molar mass of the plastic.
  (1 atm 101.325 kPa).
• ANS 42,000 g/mol

61
Colloids
62

• Examples
• Fog
• Smoke
• Whipped Cream
• Milk
• Tyndall effect scattering of light

63
Stabilization of Colloids

• Hydrophobic/hydrophilic imf
• Biomolecules

64

• Adsorption of ions
• Emulsifying agent
• Soap
• Sodium stearate(used to digest fats)

65
(No Transcript)
66

• 36a) 7.2 I2 b) 7.9 ppm Sr2
• 38a) 0.0285 b) 5.66 c) 0.638 m
• 40a) 0.125 M b) 0.140 M c) 0.630 M
• 42a) 4.34 m b) 3.1 g S8
• 44a) 27.7 b) 0.0377 c) 2.18 m
• d) 1.92 M
• 46a) 0.0439 b) 0.498 m c) 0.417 M
• 48a) 0.278 mol b) 6.25X10-5 mol c) 0.00329 mol
• 50a) 21.8 g b) 7.7 g/112.3g c) 209 g
• d) 11 mL of 6.0 M HCl
• 52) 15 M NH3