The Solution Process by OznN8nG9


									The Solution Process

      Chapter 13
           I. Solution Forces
• Solution = Solvent + Solute
• Attractions exist between
  A. solvent and solute
  B. solute and solute
  C. solvent an solvent
• If solute by solute attraction is greater that
  solute by solvent (water) then a precipitate
           I. Solution Forces
• Lattice Energy – describes the attractive
  forces between solute molecules or ions
• Solvation – solvent – solute attraction is
  strong enough so the solute dissolves.
• Hydration – when water is the solvent.
      II. Enthalpy of Solvation
•   H (sol,total)= H1 + H2 + H3
•   H1 = Energy of dispersing solute
        particles (+) endothermic
•   H2 = Energy of dispersing solvents
        particles (+) endothermic
•   H3 = Energy of solvated solute
         molecules (-) exothermic
      II. Enthalpy of Solvation

• If the energy released from solvation is
  greater than that required for dispersion
  (delta 1 + delta 2) then the process is
• If delta H of solvation is negative or
  exothermic it is energetically favorable or
• Although some delta H solvations are
  endothermic they may proceed spontaneously
  1) the increase in disorder is the driving force
  that can overcome slightly positive enthalpy
  2) remember disorder is measured in degrees of
  entropy delta S
   a) high entropy = disorder
   b) low entropy = ordered
   c) systems tend to be disordered move from
      low to high
    III. Ways to Express Conc.
• A. Dilute vs Concentration
  1) Dilute – weak– few solute particles vs
                  solvent particles
  2) Concentrated– strong- greater solute vs
                  solvent particles
  B. Mass % - mass of solute X 100
               mass of solution
Ex: A 100g NaCl solution evaporated to dryness
  weighed 5g. What was the mass Percent of the
  solution.     5g/100g X 100 = 5%
   III. Ways to Express Conc.
C. Parts Per Million
ppm = mass g of solute x 106
        mass g of solution
1) a mass % of 1 would equal 10,000 ppm
      1g/100g x 106
2) 1 ppm would be equal to 0.0001%
    1% = xg x 106
    1 = xg  100 = 106x = 0.0001%
  106 100g
• A 100ml sample of water is evaporated to
  dryness and as 75ug Pb2+. of How many parts
  per million of lead II does it have?
(d water is 100g per milliliter, 1ug=1x106g) 75ug
  (1g/1x106ug) = 7.5 x 10-5g Pb2+
 100g of water x 1 x 106 = 7.5 x 10-1ppm
 7.5 x 10-5g Pb2+
          D. Mole Fraction
• X = moles of solute
     moles of solution
X Pb2+ = sum of moles fractions in a solution
  (including the solution itself) must equal one.
• E. Molarity = Moles of solute
       (M)     liters of solution
• F. Molality = Moles of solute
      (m)       kg of solvent
Ex. 1 a certain beverage contains 7% ethanol
  C2H5OH by mass. Calculate the mole fraction,
  molarity, and molality. (mm ethanol = 46.1
7g ethanol            7g ethanol
100g(C2H5OH+ H2O) (93g H2O+7g C2H5OH)

X eth = 0.152 mol eth            = 0.0286
       (0.152 mol eth + 5.17 mol H20)
• Molarity = moles of solute
      (M) liters of solution
7g ethanol
100g solution(1g/ml) assume ethanol has no
  effect on density
7g(1mole/46.1g) 0.152 moles = 1.52 M
 100ml = 0.1 liter 0.1 liter
Molality= moles of solute 0.152
        Kg of solvent 93g(1kg/1000g)
                      = 1.63 m
  Saturated solutions and solubility

• A. Two opposing forces
 1. dissolving – hydration of individual ions
 2. collision of ions – ions unite and increase
  crystal mass (crystallization)
  Solute + solvent dissolve  solution
               crystallize  solution
IV saturated solutions and solubility

• If dissolution>crystallization then the
  crystals in the solvent get smaller
• If crystallization>dissolution crystals in the
  solvent get larger
• If crystallization = dissolution the system is
  in dynamic equilibrium (saturated) no more
  solute will be dissolved
IV saturated solutions and solubility

• The concentration of solute present at
  saturated is known as the solubility of the
• At higher temperatures usually more solute
  can be dissolved and solubility is higher
    b. Supersaturated Solutions
• Solute is dissolved at high temperature to
• Solution is carefully cooled to a lower
• At the lower temperature the concentration
  of solute is higher than at the equilibrium
  concentration at that temperature
• The introduction of a “seed” crystal will
  stimulate rapid crystal formation
   V. Factors Affecting Solubility

• A. Gases in Water
  1. solute-solvent interactions
  gases have weak I.F. primarily London
  dispersion L.D. forces
  L.D. forces increase with increased
  molecular weight – solubility of a gas
  typically decreases with increasing mass of
  a gas molecule
          Gases in Water
• If a gas molecule appears to be more
  soluble than its mass would indicate a
  chemical reaction may have taken
• B. Polar solutes in polar solvents
  1. interactions between polar solutes
  are typically dipole-dipole (or
  hydrogen bonding)
  2. interactions between molecules of
  polar solvents are the same as the
  B. Polar solutes in polar solvents

• 3. thus the energy associated with disrupting
  solute-solute and solvent-solvent
  interactions are aprox. Equal
• 4. entropic forces drive the dissolution
• 5. polar liquids tend to dissolve in polar
• Note: liquids that mix are miscible- don’t
  mix are immiscible
              c. Alcohol’s
• Short chained alcohol’s are miscible in
  water - polar - D.D. and hydrogen bonding
• Long chained alcohol’s tend not to be
• The hydrocarbon chain is held together by
  C-H bonds that are primarily L.D. forces.
  Single OH groups at the end are not enough
  to make the liquids miscible.
               c. Alcohol’s
• In general “like dissolves like” substances
  with the same or similar intermolecular
  forces tend to be soluble in one another
• D. Effect of pressure on gases and solubility
  – increasing the pressure at constant temp.
  results in more collisions of gas molecules,
  per unit time with the surface of the solvent
  resulting in greater solubility
       E. Temperature effects
• Solvated gas molecules with enough K.E.
  can escape from the surface of a liquid
• K.E. increases with temperature
• Increased temperature reduces the solubility
  of gas molecules in a solvent
F. Temp. effects on Solid solutes and
• Insolubility – inability of solvent to
  overcome the solute – solute attraction
• Increase temp increases K.E. of solvent and
  solute molecules
• Both separate more readily and the effect of
  the solvent is increased
• Increasing temperature increases the
  solubility of solid solutes
      Colligative Properties Vapor
           Pressure lowering
                            • Vapor pressure - The
• Depend upon the             pressure of the gas that
  collective number of        collects above a liquid
  particles in a solution     in a closed container
• The physical bases for
  the behavior is the
  effect of the solute
  upon the vapor
  pressure of the water
     Colligative Properties Vapor
          Pressure lowering
Raoult's Law - The vapor pressure of the
 solvent above a solution is equal to the
 product of the mole fraction of the solvent
 and the vapor pressure of the pure solvent:
 Pa = vp solution (cont. solute) Pao= pure
 solvent –
     Colligative Properties Vapor
          Pressure lowering
• Ideal Solutions – solutions that obey
  Raoult’s law – generally for dilute solutions
  where the mole fraction is closer to one,
  also where I.F. are similar among solute and
• Deviations – hydrogen bonding or when
  solute solvent interactions are extremely
Vapor pressure lowering expression

• Given Raoult’s law holds true
•   P =Pao Xb(mole fraction solute)

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