Two component mixtures - comments and preview of material to be covered
  gas/gas  -  always homogeneous, mix in all proportions, completely
                   miscible

  solid/solid - alloys such as brass, bronze, steel

  gas/liquid - solute/solvent, limited solubility, increase in partial pressure of the gas
                                        over liquid increases the solubility of the gas ==> Henry's Law  C_g = kP_g

  liquid/liquid - polarity of the liquid compounds determine miscibility,
                    "likes dissolve likes", vapor pressure of each pure liquid and the composition
                    of the solution determines the vapor pressure of the solution ==> Raoult's Law = P_A=X_AP^o_A.

  solid/liquid - usually solids in water - need to define some concentration units - weight percent,
                                                        ppm, molality, molarity, mole fraction.

The solution process involves the "breaking" and making of interactions between
particles in solution: ion-ion, ion-dipole, dipole-dipole, dipole-induced dipole,
induced dipole- induced dipole.  The "breaking" of interactions requires energy and
the making of interactions releases energy

Energies involved in the solution process (solvent-solvent, solute-solute and solute-solvent)
will determine whether energy will be released or absorbed when mixing two compounds
together to form a solution.

 Separating the solute molecules requires energy --- DH₁

 Separating the solvent molecules requires energy --- DH₂

 Mixing the solute and solvent molecules releases energy --- DH₃

  If   DH₁ +  DH₂  <   DH₃   then solution process is exothermic
  If  DH₁ +  DH₂  >   DH₃   then solution process is endothermic

Concentration units:
    Weight percent and ppm -- mass of solute/total mass gives you the mass
                                              fraction of solute

             Multiply the mass fraction of solute by 100 to get percent and by 10⁶ to get ppm

                  e.g.  30 g of KBr (119) in 120 g of water
                         30/150 = 0.20 20%      2.0 x 10⁵ ppm

Molality --- moles of solute per kg of solvent temperature independent unit
                   convenient to determine colligative property

                    use example above  {30/119}/0.120 kg = 2.1m

Molarity --- moles of solute per liter of solution (KBr in water)  convenient
                   for titrations and dilution of more concentrated solutions, ie,
                   volumetric analysis

                    use example above and given that the density of the solution at
                    room temperature is 1.25 g/mL

                    then the volume of the solution is 150/1.25 = 120 mL
                    so the molarity is {30/119}/0.12 = 2.10M

Mole fraction --- moles of solute/(moles of solute + moles of solvent)

  Use the example above X_KBr = 30/119/{30/119 + 120/18}

Factors affecting solubility
         Solvent-solute interactions
Gases dissolve in water due primarily to induced dipole-induced dipole interactions (London dispersion forces)
Under 1 atm pressure the solubility of the following gases increases (N₂, CO, O₂, Ar, Kr) as the molar mass increases
Exceptions are Cl₂, HCl, CO₂, etc. - chemical reaction occurs

        Hydrogen bonding interactions between solute and solvent often lead to high solubility -
                consider low molecular weight alcohols and many sugars

        "Likes dissolve likes" - substances with similar intermolecular attractive forces tend to be
                                              soluble in one another - polar with polar and non-polar with non-polar.
                                              Oil in water does not work because this represents non-polar with polar.

        Pressure and temperature affect solubility
               Increase pressure of gas above solution increases gas solubility
               Increase temperature of solution - solubility of gases decrease
                                                                   and, in most cases,  solubility of solids increase

Colligative properties
         Pure solvents have a melting point, boiling point and vapor pressure.  When a solute is added to
         a solvent these properties are altered - there is a vapor pressure lowering (DP), a boiling point
         elevation (DT_b), a melting point depression (DT_f) - all called colligative properties

    The colligative property is directly proportional to the number of independent solute particles --
        i.e. DP = Xsolute Po , DT_f  = K_f m, DT_b = K_b m where X_solute is the mole fraction and m is the
        molality of independent solute particles.  If the solute is an electrolyte such as KBr then one mole
        of KBr produces two moles of particles (ions) and at relatively high concentrations these ions will
        not act independently -- there are interionic attractions between these ions.

 Let's consider a solution that is prepared by adding 100 g of ethylene glycol, C₂H₆O₂, (a non-electrolyte)
        to 200 g water.  P^o = 23 torr

              P_W = X_WP_W^o  P_W = (200/18)/{100/62 + 200/18) x 23 = 0.873x23 = 20.1 torr
          or P_W = X_WP_W^o  P_W = (1-Xsolute) P_W^o = P_W^o - X_soluteP_W^o
          and rearranging P_W^o - P_W = X_soluteP_W^o  or
                DP = X_solute P^o = vapor pressure lowering

 Vapor pressure lowering at 25^oC
            DP = X_solute P^o  =  (100/62)/{100/62 + 200/18} x 23 = 0.127(23) = 2.9 torr
           The vapor pressure of the solution is 23.0 - 2.9 = 20.1 torr

 Boiling point elevation = DT_b = K_b m = 0.52(100/62)/.200 = 4.19^oC

 Freezing point depression = DT_f = K_f m = 1.86(100/62)/.200 = 15.0 ^oC

For non-electrolytes the concentration of the independent particles is the concentration of the solute.

For electrolytes one needs to consider the number of ions produced when dissolved in water.

Consider 25 g of sodium sulfate, Na₂SO₄, in 350 g of water.  If we assume no interionic attractions,
        then the moles of independent particles is three times the concentration of the solute.
        Moles of Na₂SO₄ is 25/142 = 0.176 moles  and this produces 3(0.176)= 0.528 moles of particles.
            DP = X_solute P^o  =  0.528/{0.528 + 350/18)  x 23 = 0.42 torr

            Boiling point elevation = DT_b = K_b m = 0.52(0.176)/.350 x 3 = 0.784^oC

            Freezing point depression = DT_f = K_f m = 1.86(0.176)/.350 x 3 = 2.81^oC

There is one additional colligative property -- osmotic pressure, p.  Certain materials called semipermeable membranes when in contact with a solution allow the passage of small solvent molecules but not relatively large solute molecules or ions.
 

The membranes of red blood cells are semipermeable.  When replacing body fluids or nutrients intervenously (feed nutrients directly into the veins), the concentration of glucose and/or salt in this solution must be the same as that in the red blood cell - otherwise the cells will rupture (hemolysis) or shrivel (crenation)  The solution is:
         Hypertonic - if the solution has a higher conc. - lower osmotic pressure
         Hypotonic - if the solution has a lower conc. - higher osmotic pressure
         Isotonic - if the solution has the same conc. - same osmotic pressure

Colloids and surfactants
 Dispersions of finely divided particles (10 to 2000 A in diameter) that remain suspended in a solvent such as water are said to be colloidal dispersions or colloids.  Examples are milk, paint, soap solution, etc. The suspended particles are large enough to scatter light from a light beam - called the Tyndall effect.

 Hydrophobic and Hydrophilic colloids.

Hydrophilic (water loving) colloids are easily suspended in water.  Large molecules such as enzymes fold in such a way that the hydrophobic groups of the molecule are folded in away from the water molecules while the hydrophilic, polar groups are found on the surface of the molecule.

Hydrophobic (water hating) colloids can be prepared in water only if they are stabilized.  Oil droplets can be suspended upon the addition of a soap such as sodium stearate.