Physical Properties
of Solutions
Chapter 13
Factors Affecting Solubility
Fig 13.12 Structure and solubility
• Glucose (which has
hydrogen bonding) is
very soluble in water
• Cyclohexane (which
only has dispersion
forces) is not
Pressure Effect on Gases in Solution
• Solubility of liquids and solids does not change
appreciably with pressure
• Solubility of a gas in a liquid is directly proportional to its
pressure
Fig 13.14 Effect of pressure on gas solubility
Henry’s Law
Sg = kPg
where
• Sg ≡ solubility of the gas
• k ≡ the Henry’s Law
constant for that gas in that
solvent
• Pg ≡ partial pressure of the
gas above the liquid
Fig 13.15 Solubility decreases
as pressure decreases
Temperature Effect on Solids and Liquids
Fig 13.17 Solubilities of several ion compounds
as a function of temperature
• Generally, the solubility
of solid solutes in liquid
solvents increases with
increasing temperature
Temperature Effect on Gases
Fig 13.18 Variation of gas solubility with temperature
• The opposite is true of gases:
•
Carbonated soft drinks are
more “bubbly” if stored in the
refrigerator
• Warm lakes have less O2
dissolved in them than cool
lakes
Concentration Units
Concentration - amount of solute present in a given quantity
of solvent or solution:
• Mass percentage
• Mole fraction
• ppm and ppb
• Molarity
• Molality
Concentration Units
Mass percentage (w/w)
% by mass =
mass of solute
mass of solute + mass of solvent
mass of solute
=
mass of solution
x 100% (w/w)
Mole Fraction (X)
moles of A
XA =
sum of moles of all components
x 100%
Practice Exercise p 543
(a) Calculate the mass percentage of NaCl in a solution
containing 1.50 g of NaCl in 50.0 g of water.
(b) A commercial bleaching solution contains 3.62 mass %
sodium hypochlorite, NaOCl. What is the mass of NaOCl
in a bottle containing 2.50 kg of bleaching solution?
Answer:
(a) 2.91%
(b) 90.5 g of NaOCl
Sample Exercise 13.6 Calculation of Mole Fraction
An aqueous solution of hydrochloric acid contains 36%
HCl by mass. Calculate the mole fraction of HCl in the
solution.
Assume we have 100. g of solution:
Concentration Units Continued
Parts per million (ppm)
moles solute
ppm 
x 10 6
L soln
Parts per billion (ppb)
moles solute
ppm 
x 10 9
L soln
Concentration Units Continued
Molarity (M)
moles of solute
M =
liters of solution
Molality (m)
moles of solute
m =
mass of solvent (kg)
Practice Exercise p 544
What is the molality of a solution made by dissolving 36.5 g
of naphthalene (C10H8) in 425 g of toluene (C7H8)?
Answer: 0.670 m
Conversion of Concentration Units
Fig 13.19 Calculating molality and molarity
• If we know the density
of the solution, we can
calculate the molality
from the molarity and
vice versa.
What is the molality of a 5.86 M ethanol (C2H5OH)
solution whose density is 0.927 g/mL?
moles of solute
moles of solute
m =
M =
mass of solvent (kg)
liters of solution
Assume 1 L of solution:
5.86 moles ethanol = 270 g ethanol
927 g of solution (1000 mL) (0.927 g/mL)
mass of solvent = mass of solution – mass of solute
= 927 g – 270 g = 657 g = 0.657 kg
moles of solute
m =
mass of solvent (kg)
=
5.86 moles C2H5OH
0.657 kg solvent
= 8.92 m