# The Solution Process by OznN8nG9

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```									The Solution Process

Chapter 13
Brown-LeMay
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
forms.
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
exothermic
• If delta H of solvation is negative or
exothermic it is energetically favorable or
spontaneous
• 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
changes
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
100g
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
g/mol)
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
solute
• At higher temperatures usually more solute
can be dissolved and solubility is higher
b. Supersaturated Solutions
• Solute is dissolved at high temperature to
saturation
• Solution is carefully cooled to a lower
temperature
• 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
place
• 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
solutes
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
process
• 5. polar liquids tend to dissolve in polar
solvents
• 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
miscible
• 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
solubility
• 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
solvent
• Deviations – hydrogen bonding or when
solute solvent interactions are extremely
strong
Vapor pressure lowering expression

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

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