Ch13 properties of-solutions

Solutions
Chapter 13
Properties of Solutions
Section 13.1-13.5 are covered
Chemistry, The Central Science, 11th edition
Theodore L. Brown; H. Eugene LeMay, Jr.;
and Bruce E. Bursten
1

Solutions
13.1) The Solution Process
• Solutions are homogeneous mixtures of two
or more pure substances.
• In a solution, the solute is dispersed uniformly
throughout the solvent.
2

Solutions
© 2009, Prentice-Hall, Inc.
Solutions
The intermolecular
forces between solute
and solvent particles
must be strong enough
to compete with those
between solute particles
and those between
solvent particles.

Solutions
How Does a Solution Form?
As a solution forms, the solvent pulls solute
particles apart and surrounds, or solvates,
them.

Solutions
How Does a Solution Form
If an ionic salt is
soluble in water, it is
because the ion-
dipole interactions
are strong enough
to overcome the
lattice energy of the
salt crystal.

Solutions
Energy Changes in Solution
• Simply put, three
processes affect the
energetics of solution:
– separation of solute
particles,
– separation of solvent
particles,
– new interactions
between solute and
solvent.

Solutions
Energy Changes in Solution
The enthalpy
change of the
overall process
depends on H for
each of these steps.
The formation of a
solution can be
either exothermic or
endothermic.
Example: dissolution of NaOH in water; exothermic
Hsoln = -44.48 kJ/mol

Solutions
Why Do Endothermic
Processes Occur?
Things do not tend to
occur spontaneously
(i.e., without outside
intervention) unless
the energy of the
system is lowered.

Solutions
Why Do Endothermic Processes
Occur?
Yet we know that in
some processes,
like the dissolution
of NH4NO3 in water,
heat is absorbed,
not released.
i.e. is endothermic,
Hsoln = 26.4 kJ/mol
Ammonium nitrate has been used to make
instant ice packs.

Solutions
Enthalpy Is Only Part of the Picture
The reason is that
increasing the disorder
or randomness (known
as entropy) of a system
tends to lower the
energy of the system.

Solutions
Enthalpy Is Only Part of the Picture
So even though
enthalpy may increase,
the overall energy of
the system can still
decrease if the system
becomes more
disordered.

Solutions
Student, Beware!
Just because a substance disappears when it
comes in contact with a solvent, it doesn’t
mean the substance dissolved.

Solutions
Student, Beware!
• Dissolution is a physical change — you can get back the
original solute by evaporating the solvent.
• If you can’t, the substance didn’t dissolve, it reacted.

Solutions
3.2) Saturated Solutions and Solubility
• Saturated
 Solvent holds as much
solute as is possible at that
temperature.
 Dissolved solute is in
dynamic equilibrium with
solid solute particles.
 The amount of solute
needed to form a saturated
solution in a given quantity
of solvent at a specified
temperature is known as
the Solubility of that solute.
Solute + Solvent Solution
dissolve
crystallize

Solutions
Types of Solutions
• Unsaturated
Less than the
maximum amount of
solute for that
temperature is
dissolved in the
solvent.

Solutions
Types of Solutions
• Supersaturated
Solvent holds more solute than is normally
possible at that temperature.
These solutions are unstable; crystallization can
usually be stimulated by adding a ―seed crystal‖ or
scratching the side of the flask.

Solutions
17
For example, the solubility of NaCl in water at 0 C is 35.7 g per 100 ml of water.
This is the maximum amount of NaCl (35.7 g) that can be dissolved in 100 ml of
water to give a stable equilibrium solution at 0 C.
If we dissolve less than 35.7 g of NaCl in 100 ml of water at 0 C, this will form
unsaturated solution.
If we dissolve greater than 35.7 g of NaCl in 100 ml of water at 0 C, this will
form supersaturated solution.

Solutions
13.3) Factors Affecting Solubility
• Chemists use the axiom
―like dissolves like‖:
Polar substances tend to
dissolve in polar solvents.
Nonpolar substances tend
to dissolve in nonpolar
solvents.
The extent to which one substance dissolves in another depends
on the nature of both the solute and solvent (solute-solvent
interactions). It also depends on Temperature and,
at least for gases, on Pressure.

Solutions
Factors Affecting Solubility
The more similar the
intermolecular
attractions, the more
likely one substance
is to be soluble in
another.

Solutions
Glucose (which has
hydrogen bonding)
is very soluble in
water, while
cyclohexane (which
only has dispersion
forces) is not.

Solutions
• Vitamin A is soluble in nonpolar compounds
(like fats).
• Vitamin C is soluble in water.

Solutions
Gases in Solution
• In general, the
solubility of gases in
water increases with
increasing mass.
• Larger molecules
have stronger
dispersion forces.

Solutions
Gases in Solution
• The solubility of
liquids and solids
does not change
appreciably with
pressure.
• The solubility of a
gas in a liquid is
directly proportional
to its pressure.

Solutions
Henry’s Law
Sg = kPg
where
• Sg is the solubility of
the gas;
• k is the Henry’s law
constant for that gas in
that solvent;
• Pg is the partial
pressure of the gas
above the liquid.

Solutions
Temperature
Generally, the
solubility of solid
solutes in liquid
solvents increases
with increasing
temperature.

Solutions
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.

Solutions
13.4) Ways of
Expressing
Concentration
27

Solutions
Mass Percentage
Mass % of A =
mass of A in solution
total mass of solution
 100%
28

Solutions
Parts per Million and
Parts per Billion
ppm =
 106
Parts per Million (ppm)
Parts per Billion (ppb)
ppb =
 109
29

Solutions
SAMPLE EXERCISE 13.4 Calculation of Mass-Related Concentrations
(a) A solution is made by dissolving 13.5 g of glucose (C6H12O6) in 0.100 kg of water. What is the mass
percentage of solute in this solution? (b) A 2.5-g sample of groundwater was found to contain 5.4g of
Zn2+ What is the concentration of Zn2+ in parts per million?
PRACTICE EXERCISE
(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 2500 g of bleaching solution?
Answers: (a) 2.91%, (b) 90.5 g of NaOCl
Comment: The mass percentage of water in this solution is (100 – 11.9)% = 88.1%.
(b) Analyze: In this case we are given the number of micrograms of solute. Because 1g is
1  10–6 g, 5.4g = 5.4  10–6 g.
30

Solutions
moles of A
total moles in solution
XA =
Mole Fraction (X)
• In some applications, one needs the
mole fraction of solvent, not solute—
make sure you find the quantity you
need!
31

Solutions
mol of solute
Volume of solution (in Liters)
M =
Molarity (M)
• You will recall this concentration
measure from Chapter 4.
• Because volume is temperature
dependent, molarity can change with
temperature.
32

Solutions
mol of solute
kg of solvent
m =
Molality (m)
Because both moles and mass do not
change with temperature, molality
(unlike molarity) is not temperature
dependent.
33

Solutions
SAMPLE EXERCISE 13.5 Calculation of Molality
A solution is made by dissolving 4.35 g glucose (C6H12O6) in 25.0 mL of water at 25°C. Calculate the
molality of glucose in the solution.
PRACTICE EXERCISE
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
Solution
Solve: Use the molar mass of glucose, 180.2 g/mol, to convert grams to moles:
Because water has a density of 1.00 g/mL, the mass of the solvent is
Finally, use Equation 13.9 to obtain the molality:
34

Solutions
SAMPLE EXERCISE 13.6 Calculation of Mole Fraction and Molality
An aqueous solution of hydrochloric acid contains 36% HCl by mass. (a) Calculate the mole fraction of
HCl in the solution. (b) Calculate the molality of HCl in the solution.
Solve: (a) To calculate the mole fraction of HCl, we convert the masses of HCl and H2O to moles
and then use Equation 13.7:
35

Solutions
PRACTICE EXERCISE
A commercial bleach solution contains 3.62 mass % NaOCl in water. Calculate (a) the molality and (b)
the mole fraction of NaOCl in the solution.
Answers: (a) 0.505 m, (b) 9.00  10–3
SAMPLE EXERCISE 13.6 continued
(b) To calculate the molality of HCl in the solution, we use Equation 13.9. We calculated the number of
moles of HCl in part (a), and the mass of solvent is 64 g = 0.064 kg:
36

Solutions
SAMPLE EXERCISE 13.7 Calculation of Molarity
A solution contains 5.0 g of toluene (C7H8) and 225 g of benzene and has a density of 0.876 g/mL.
Calculate the molarity of the solution.
Plan: The molarity of a solution is the number of moles of solute divided by the number of liters of
solution (Equation 13.8). The number of moles of solute (C7H8) is calculated from the number of grams
of solute and its molar mass. The volume of the solution is obtained from the mass of the solution
(mass of solute + mass of solvent = 5.0 g + 225 g = 230 g) and its density.
Solve:The number of moles of solute is
The density of the solution is used to convert the mass of the solution to its volume:
Molarity is moles of solute per liter of solution:
37

Solutions
Changing Molarity to Molality
If we know the
density of the
solution, we can
calculate the
molality from the
molarity, and vice
versa.
38

Solutions
13.5) Colligative Properties
• Changes in colligative properties
depend only on the number of solute
particles present, not on the identity of
the solute particles.
• Among colligative properties are
– Vapor pressure lowering
– Boiling point elevation
– Melting point depression
– Osmotic pressure

Solutions
Vapor Pressure
Because of solute-
solvent intermolecular
attraction, higher
concentrations of
nonvolatile solutes
make it harder for
solvent to escape to
the vapor phase.

Solutions
Vapor Pressure
Therefore, the vapor
pressure of a solution
is lower than that of
the pure solvent.

Solutions
Raoult’s Law
PA = XAPA
where
– XA is the mole fraction of solvent (A), and
– PA is the normal vapor pressure of pure
solvent (A) at that temperature.
NOTE: This is one of those times when you
want to make sure you have the vapor
pressure of the solvent.
Quantitatively, the vapor pressure of solutions containing
Nonvolatile solutes is given by Raoult’s law:

Solutions
43
The solution that obeys Raoult’s law is called ideal solution.
Real solutions best approximate ideal behavior when:
1- the solute concentration is low, and
2- the solute and solvent have similar molecular sizes and
Similar types of intermolecular attractions.
Example; the vapor pressure of water is 17.4 torr at 20C.
Adding glucose to water so that the solution has Xwater = 0.80 and
Xglucose = 0.20 at constant temperature, calculate the vapor pressure
Over the solution.
Pwater = Xwater Pwater = 0.80 x 17.5 = 14 torr

Solutions
Boiling Point Elevation and
Freezing Point Depression
Nonvolatile solute-
solvent interactions
also cause solutions
to have higher boiling
points and lower
freezing points than
the pure solvent.

Solutions
Boiling Point Elevation
• The change in boiling
point is proportional to
the molality of the
solution:
Tb = Kb  m
where Kb is the molal
boiling point elevation
constant, a property of
the solvent.Tb is added to the normal
boiling point of the solvent.

Solutions
Boiling Point Elevation
• The change in freezing
point can be found
similarly:
Tf = Kf  m
• Here Kf is the molal
freezing point
depression constant of
the solvent.
Tf is subtracted from the normal
boiling point of the solvent.

Solutions
Boiling Point Elevation and
Freezing Point Depression
Note that in both
equations, T does
not depend on what
the solute is, but
only on how many
particles are
dissolved.
Tb = Kb  m
Tf = Kf  m

Solutions
Colligative Properties of
Electrolytes
Since these properties depend on the number of
particles dissolved, solutions of electrolytes (which
dissociate in solution) should show greater changes
than those of nonelectrolytes.

Solutions
Colligative Properties of
Electrolytes
However, a 1M solution of NaCl does not show
twice the change in freezing point that a 1M
solution of methanol does.

Solutions
van’t Hoff Factor
One mole of NaCl in
water does not
really give rise to
two moles of ions.

Solutions
van’t Hoff Factor
Some Na+ and Cl-
reassociate for a short
time, so the true
concentration of
particles is somewhat
less than two times the
concentration of NaCl.

Solutions
van’t Hoff Factor
• Reassociation is
more likely at higher
concentration.
• Therefore, the
number of particles
present is
concentration-
dependent.

Solutions
Osmosis
• Some substances form semipermeable
membranes, allowing some smaller
particles to pass through, but blocking
other larger particles.
• In biological systems, most
semipermeable membranes allow water
to pass through, but solutes are not free
to do so.

Solutions
Osmosis
In osmosis, there is net movement of solvent
from the area of higher solvent
concentration (lower solute concentration) to
the are of lower solvent concentration
(higher solute concentration).

Solutions
Osmotic Pressure
The pressure required to stop osmosis,
known as osmotic pressure, , is
n
V
 = ( )RT = MRT
where M is the molarity of the solution.
If the osmotic pressure is the same on both sides
of a membrane (i.e., the concentrations are the
same), the solutions are isotonic.

Solutions
Osmosis in Blood Cells
• If the solute
concentration outside
the cell is greater than
that inside the cell, the
solution is hypertonic.
• Water will flow out of
the cell, and crenation
results.

Solutions
Osmosis in Cells
• If the solute
concentration outside
the cell is less than
that inside the cell, the
solution is hypotonic.
• Water will flow into the
cell, and hemolysis
results.

Ch13 properties of-solutions

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