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					                           Properties of Solutions

                                              Terminology


           Solvation                          Compositions                      Solubility Rules


                                            Heats of Solution


Structure/Intermolecular Forces               Temperature                          Pressure


                   Henry's Law              Vapor Pressures             Raoult's Law


                                          Colligative Properties


                       B. pt. Elevation    Fr. pt. Depression      Osomotic Pressure
               Terminology

• Solution = A homogeneous mixture.
• Solute = a substance dissolved in a solvent
  to form a solution; usually the smaller
  portion.
• Solvent = The dissolving medium of a
  solution; usually the greater portion.
• Solubility = Amount of substance dissolved.
• Dilute , Concentrated , Saturated .
The Solution Process
Figure 13.2: The Solution Process
     Hydration or Solvation
Figure 13.3:
Three Steps of
Solution Formation
           The Solution Process
                Solution Formation, Spontaneity, and
                                                  Disorder
• If the process leads to a greater state of disorder, then
  the process is spontaneous.
• Example: a mixture of CCl4 and C6H14 is less ordered
  than the two separate liquids. Therefore, they
  spontaneously mix even though Hsoln is very close to
  zero.
• There are solutions that form by physical processes and
  those by chemical processes.
Figure 13.6: The Solution Process
      Solution Formation, Spontaneity, and
                                 Disorder
        Factors Affecting Solubility

                             Solute-Solvent Interaction
• Miscible liquids: mix in any proportions.
• Immiscible liquids: do not mix. [ Oil and Vinegar ]
• Generalization: “like dissolves (in) like”.

• Polar liquids tend to dissolve in polar solvents.
Factors Affecting Solubility

             Solute-Solvent Interaction
      Ways of Expressing Concentration

                                     Mass Percentage
• All methods involve quantifying amount of solute per
  amount of solvent (or solution).
• Generally amounts or measures are masses, moles or
  liters.


                      mass of component in solution
mass % of component                                100
                          total mass of solution
     Ways of Expressing Concentration
• Mole Fraction, Molarity, and Molality

                             moles of component in solution
Mole fraction of component 
                                 total moles of solution

                            moles solute
              Molarity 
                         liters of solution

                             moles solute
               Molality, m 
                             kg of solvent
              Solution Compositions
• s = solute ; A = solvent; V = Tot. Vol. of solution.
• Weight %:                          ws
                                ws %              x 100
                                         ws  wA
• Mole Fraction:
                                   ns
                         s 
                                ns  nA
• Molarity:
                         ns
                    Ms 
                         V

• Molality:
                          ns
                    ms 
                         kg A
    Example of Solution Compositions
• A solution is prepared by mixing 78.9 g of ethanol
  (C2H5OH) with 100.0 g of water to give 190.5 mL of
  solution. Calculate the solution compositions.


• The electrolyte in automobile lead storage batteries is
  a 3.75 M H2SO4 solution that has a density of 1.230
  g/mL. Calculate mass %, molality, and mole fraction
  in terms of H2SO4 .
• [Hint: Assume exactly one liter of solution.]
• [Answers: 29.9% , 4.35 molal, 0.0727 ]
   s = ethanol (solute);      A = water (solvent);

            78 .9 g
  ws %             x 100  44 .1%
           178 .9 g

                     78 .9 
                            
s 
        ns
                    46 .07 
                                   
                                     1.71
                                            0.236
     ns  n A  78 .9 100 .0  1.71  5.55
                46 .07  18 .02 
                                

       ns 1.713 mol
   Ms              8.99 mol L1
       V 190 .5 mL


         ns     1.713 mol
 ms                         17 .1 mol kg 1
        kg A 0.1000 kg water
Exam ple 2 (Solution Com positions)

The e lectrolyte in autom obile le ad s torage batterie s is a 3.75 M HSO 4 solution that has a
                                                                      2
                       -1
de nsity of 1.230 g m L . Calculate w eight%, m olality, and m ole fraction in te rm s of H SO 4 .
                                                                                           2


Hint: As sum e exactly one liter of solution in your calculations.
De fine:     s = H2 SO 4     A=H                              2O


 In one liter of solution, from 3.75 mol/L solution, there are 3.75 moles of H       2 SO 4   .
 In one liter of solution, from d=1.230 g/mL, there are 1230 grams of solution.




 Weight % of H2 SO 4 Calculation:

           98.07gm 
                 
        
   3.75mol            368gm H2SO 4
            1 mol 

                       368 gm
          Wt %                  100           Wt %  29.9
                          
                       1230gm




 Molality of H2 SO 4 Calculation :


  In one liter of solution => 1230 gm soln - 368 gm H          2 SO 4   = 862 gm H 2 O

                    
          n s  3.75mol                            
                                         kgA  0.862kg


               ns                                  mol
        ms                             ms  4.35
              kgA                                  kg




 Mole Fraction of H2 SO 4 Calculation :

                 862
       n A          mol
                18.015
                  ns
        s                              s  0.0727
                ns  nA
     Concentrations of Solutions

    In the Dilution process of a more concentrated
                                               solution:
• The number of moles are the same in diluted and
  concentrated solutions.
• So:
          MdiluteVdilute = moles = MconcentratedVconcentrated
   General Properties of Aqueous Solutions


                              Electrolytic Properties
• Three types:
   • Strong electrolytes,
   • Weak electrolytes, and
   • Nonelectrolytes.
   General Properties of Aqueous Solutions

                         Strong and Weak Electrolytes
• Strong electrolytes: completely dissociated in solution.
  e.g.
             HCl(aq)           H+(aq) + Cl-(aq)

• Weak electrolytes: produce small concentration of ions
  when dissolved.
• e.g.

      HC2H3O2(aq)             H+(aq) + C2H3O2-(aq)
Precipitation Reactions
          Precipitation Reactions

                    Exchange (Metathesis) Reactions
• Metathesis reactions involve swapping ions in solution:
                  AX + BY  AY + BX.
            Precipitation Reactions
                                         Ionic Equations
• Ionic equation: used to highlight reaction between ions.

• Molecular equation: all species listed as molecules:
     AgNO3(aq) + NaI(aq)  AgI(s) + NaNO3(aq)

• Complete ionic equation(CIE): lists all ions:

Ag+(aq) + NO3-(aq) + Na+(aq) + I-(aq)  AgI(s) + Na+(aq) + NO3-(aq)

• Net ionic equation: cancel spectator ions from CIE
                Ag+(aq) + I-(aq)  AgI(s)
Figure 13.14: Factors Affecting Solubility


  Intermolecular Forces


  Pressure


  Temperature
Figure 13.14: Factors Affecting Solubility

                            Pressure Effects
       Factors Affecting Solubility

                                             Pressure Effects
• If Sg is the solubility of a gas, k is a constant, and Pg is
  the partial pressure of a gas, then Henry’s Law gives:
                          S g  k  Pg
• Carbonated beverages are bottled with a partial pressure
  of CO2 >1 atm, ( pressure inside can ~4 atm above liq).
• As bottle is opened, partial pressure of CO2 decreases and
  solubility of CO2 decreases.
• Therefore, bubbles of CO2 escape from solution.
                                              CyberChem Diving Gases
      Factors Affecting Solubility

              Temperature Effects: Solids in Liquids
• Generally, as temperature increases, solubility of solids
  generally increases, BUT
• Sometimes, solubility decreases as temperature increases
  (e.g. Ce2(SO4)3).
Figure 13.17
       Factors Affecting Solubility

              Temperature Effects: Gases in Liquids
• Gases get less soluble as temperature increases.

• Thermal pollution: if lakes get too warm, CO2 and O2
  become less soluble and are not available for plants or
  animals.
Figure 13.18
              Colligative Properties

• Colligative properties depend on quantity of solute
  molecules. (e.g. freezing point depression and boiling
  point elevation.)
                              Lowering Vapor Pressure
• Non-volatile solutes reduce the ability of the surface
  solvent molecules to escape the liquid.
• Therefore, vapor pressure is lowered.
• The amount of vapor pressure lowering depends on the
  amount of solute.
Figure 11.22: Vapor Pressure
 Vapor Pressure on the Molecular Level
Figure 11.24
Figure 11.26: Phase Diagrams
Figure 11.27: Phase Diagrams

    The Phase Diagrams of H2O and CO2
Figure 13.20: Colligative Properties

                 Lowering Vapor Pressure
              Colligative Properties

                             Lowering Vapor Pressure
• Raoult’s Law:
                    PA   A  P A
• Where: PA = vapor pressure with solute,
        • PA = vapor pressure without solute (pure
          solvent), and
        • A = mole fraction of A.
            Vapor Pressure Examples
• Calculate the expected vapor pressure at 25oC for a
  solution prepared by dissolving 158.0 g of common
  table sugar (sucrose – MW=342.3) in 643.5 mL of
  water. At 25oC, the density of water is 0.9971 g/mL
  and the vapor pressure is 23.76 torr.
                                          PA   A  P A
• A solution was prepared by adding 20.0 g of urea to
  125 g of water at 25oC, a temperature at which pure
  water has a vapor pressure of 23.76 torr. The
  observed vapor pressure of the solution was found to
  be 22.67 torr. Calculate the molecular weight of
  urea.
  [Answer: 60. g/mol ]
 Example 1:            Sugar in Water
                                                                                                      1
s = sugar             A = w ater           ws  158.0
                                                     gm                            
                                                                       MW s  342.3gm mol
                  3                               3                                                            1
          
VA  643.5cm                            
                             d A  0.9971gm cm                     
                                                          PoA  23.76t orr                      
                                                                                    MW A  18.015gm mol

           VA  d A
                                                                 643.5cm  
                                                                                  0.9971gm 
                                                                                                          
                                                                          3                          1 mol
  n A 
            MW A
                                   n A  35.62mol         nA         
                                                                                      3
                                                                                                 
                                                                                                 18.015gm 
                                                                                                           
                                                                                    cm        
            ws
                                                                    158.0gm               
  n s                            n s  0.4616mol                                   1 mol
           MW s                                             ns                            
                                                                                   342.3gm 
                                                                                        


              nA
   A                             A  0.9872                          35.62                35.62
            ns  nA                                         A
                                                                     0.4616 35.62            36.08



    PA   A  PoA                PA  23.46t orr             PA                
                                                                      0.9872( 23.76t orr)
    Example 2:               Urea Molecular Weight


•                                                                                              o
           A solution was prepared by adding 20.0 g of urea to 125 g of water at 25 C, a temp erature
         at which pure water has a vapor pressure of 23.76 torr. T he observed vapor p ressure of the
         solution was found to be 22.67 torr. Calculate the molecular weight of urea.
         [Answer:        60. g/mol ]




s = urea                 A = w ater

         
ws  20.0gm                 wA  125 gm                         
                                                        PoA  23.76t orr                       
                                                                                      PA  22.67t orr

                                               PA                                                        1
    PA      A  PoA                   A                  A  0.9541                         
                                                                                    MW A  18.015gm mol
                                               PoA


              nA                               wA
    A                                n A                        n A  6.939mol
           ns  nA                             MW A




               nA                 solving for n has solution of:
                                               s                                      A  1
    A                                                                       n A 
             ns  nA                                                                   A




     n s 
                    
              nA  1   A                     n s  0.334mol
                        A


               ws                               ws                                    1
     ns                               MW s                       MW s  60gm mol
              MW s                              ns
Figure 13.22
               Colligative Properties

                               Boiling-Point Elevation
• Molal boiling-point-elevation constant, Kb, expresses
  how much Tb changes with molality, mS :

                      Tb  K b  mS
• Decrease in freezing point (Tf) is directly proportional
  to molality (Kf is the molal freezing-point-depression
  constant):
                      Tf  K f  mS
Colligative Properties
   Applications of Colligative Properties
• A solution was made by dissolving 18.00 g of glucose
  in 150.0 g of water. The resulting solution was found to
  have a boiling point of 100.34oC. Calculate the
  molecular weight of glucose.
                                    Tb  K b  mS
• How many kg of ethylene glycol (C2H6O2, MW=62.1),
  antifreeze, must be added to 10.0 L of water to produce
  a solution for use in a car’s radiator that freezes at
  –23.3oC? Assume that the density of water is 1.00
  g/mL.
  [Answer: 7.78 kg (with d = 1.18 g/mL => 6.59 L)]
• A solution was made by dissolving 18.00 g of glucose in 150.0 g of water. The resulting solution
was found to have a boiling point of 100.34oC. Calculate the molecular weight of glucose.

                                                                                Tb  K b  mS
How many kg of ethylene glycol (C2H6O2, MW=62.1), antifreeze, must be added to 10.0 L
of water to produce a solution for use in a car’s radiator that freezes at –23.3oC? Assume
that the density of water is 1.00 g/mL.
                                                                   Tf  K f  mS
Free zing Point De pre ssion Example

•      How many kg of ethylene glycol (C2H6O2, M W=62.1), antifreeze, must be added to 10.0 L
    of water to produce a solution for use in a car’s radiator that freezes at           –23.3 oC? Assume
    that the density of water is 1.00 g/mL.
    [Answer:         7.78 kg (with d = 1.18 g/mL => 6.59 L)]

       Let:     s = ethylene glycol            A=water


                                                                      1 gm    1 mL    10  kg 
                                                                                                 3
                    1
Kf  1.86K kg mol
                                   Tf  23.3       kg_A  10.0                                 
                                                                      1 mL  10 3 L   1 gm 
                                              K                  L
                                                                                        
                                                            1
    kg_A  10.0kg                                
                                      MW s  62.1gm mol



                         ns                ws       
    Tf    Kf ms   Kf            Kf              
                         kg_A           MW s  kg_A 


                       Tf MW s  kg_A
               ws 
                             Kf




                                     1
                        
               23.3K 62.1 gm mol            
                                          10.0 kg
          ws
                                      1                                  ws  7.78kg
                          
                       1.86K kg mol




                                    1                     ws
                          
                d s  1.18gm mL                    Vs               Vs  6.59L
                                                           ds
               Colligative Properties
                                                Osmosis
• movement of a solvent from low solute concentration to
  high solute concentration across a semipermeable
  membrane.


Figure 13.23
              Colligative Properties

                                                  Osmosis
• Osmotic pressure, , is the pressure required to stop
  osmosis:
                  V  n  R  T
                         n
                          R T
                         V 


                M S  R T
 Application of Osmotic Pressure

• To determine the molecular weight of a
  certain protein, 1.00 mg of it was dissolved
  in enough water to make 1.00 mL of
  solution. The osmotic pressure of this
  solution was found to be 1.12 torr at 25.0oC.
  Calculate the molecular weight of the
  protein.
  [Answer:       1.66x104 g/mol]

                                 M S  R T
To determine the molecular weight of a certain protein, 1.00 mg of it was dissolved in enough water to
make 1.00 mL of solution. The osmotic pressure of this solution was found to be 1.12 torr at 25.0 oC.
Calculate the molecular weight of the protein.
                                                                           M S  R T
Colligativ e Prope rties: Osmotic Pressure Example


•      To determine the molecular weight of a certain protein, 1.00 mg of it was dissolved in
     enough water to make 1.00 mL of solution. T he osmotic pressure of this solution was found
     to be 1.12 torr at 25.0 oC. Calculate the molecular weight of the protein.
     [Answer:          1.66x104 g/mol]


Def ine:              s = protein         A = w ater

                3                        3                                                                          1     1
         
ws  1.0010        gm               
                             V  1.0010    L                 1.12 orr
                                                                    t                                
                                                                                          R  0.08206L at m mol       K


                                                    1.12t orr            
                                                                          1at m                                 3
      T  ( 25.0  273.15 K
                         )                                                                      1.474 10        at m
                                                                  760 t orr 

                             ns       ws 
              M s  R T      R T             R T
                            V         MW s  V
                                                  


    The only unknow n is the molecular w eight of the protein ( MW s ).




    MW s 
                ws  R T
                                        MW s 
                                                     
                                                  1.0010
                                                          3            
                                                              gm 0.08206L at m mol
                                                                                            1
                                                                                                 K
                                                                                                   1     298.15K
                                                                                                                 
                   V
                                                                   
                                                               1.47410
                                                                            3            
                                                                                             3
                                                                                  at m 1.0010    L  

                                    4          1
                 MW s  1.66 10 gm mol
             Colligative Properties

                                                  Osmosis
• Isotonic solutions: two solutions with the same 
  separated by a semipermeable membrane.

• Osmosis is spontaneous.
• Red blood cells are surrounded by semipermeable
  membranes.
      Figure 13.25: Colligative Properties
                                                    Osmosis




Crenation: hypertonic solution     Hemolysis: hypotonic solution


           IV(intravenous) fluids must be Isotonic.
                                    CyberChem: Desalination
                     Colloids

• Colloids are suspensions in which the suspended particles
  are larger than molecules but too small to drop out of the
  suspension due to gravity.
• Particle size: 10 to 2000 Å.
• Tyndall effect: ability of a Colloid to scatter light. The
  beam of light can be seen through the colloid.
Colloids
Figure 13.31: Colloids
                            Properties of Solutions
                                               Terminology


            Solvation                          Compositions                      Solubility Rules


                                             Heats of Solution


 Structure/Intermolecular Forces               Temperature                          Pressure


                    Henry's Law              Vapor Pressures            Raoult's Law
                                                                                           PA   A  PA
                                                                                                              o



                                           Colligative Properties                                      nA
                                                                                           A 
                                                                                                    nS  nA
                        B. pt. Elevation    Fr. pt. Depression      Osomotic Pressure
        nS
mS 
       kg A                  Tb, f  Kb, f  mS                        M S  R T           MS 
                                                                                                       nS
                                                                                                       V

				
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