Chapter 12

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Chapter 12
Solutions
Solutions
solute is the dissolved substance
◦ seems to “disappear”
◦ “takes on the state” of the solvent
 solvent is the substance solute dissolves in
◦ does not appear to change state
 when both solute and solvent have the same state, the
solvent is the component present in the highest percentage
 solutions in which the solvent is water are called aqueous
solutions

2
Solution Concentration Molarity



moles of solute per 1 liter of solution
used because it describes how many molecules of solute
in each liter of solution
if a sugar solution concentration is 2.0 M, 1 liter of
solution contains 2.0 moles of sugar, 2 liters = 4.0 moles
sugar, 0.5 liters = 1.0 mole sugar
moles of solute
molarity =
liters of solution
3
Molarity and Dissociation





the molarity of the ionic compound allows you to
determine the molarity of the dissolved ions
CaCl2(aq) = Ca+2(aq) + 2 Cl-1(aq)
A 1.0 M CaCl2(aq) solution contains 1.0 moles of CaCl2 in each liter
of solution
◦ 1 L = 1.0 moles CaCl2, 2 L = 2.0 moles CaCl2
Because each CaCl2 dissociates to give one Ca+2 = 1.0 M Ca+2
◦ 1 L = 1.0 moles Ca+2, 2 L = 2.0 moles Ca+2
Because each CaCl2 dissociates to give 2 Cl-1 = 2.0 M Cl-1
◦ 1 L = 2.0 moles Cl-1, 2 L = 4.0 moles Cl-1
4
Solution ConcentrationMolality, m


moles of solute per 1 kilogram of solvent
◦ defined in terms of amount of solvent, not solution
 like the others
does not vary with temperature
◦ because based on masses, not volumes
moles of solute
molality, m 
kg of solvent
5
Percent


parts of solute in every 100 parts solution
mass percent = mass of solute in 100 parts solution by mass
◦ if a solution is 0.9% by mass, then there are 0.9 grams of
solute in every 100 grams of solution
 or 0.9 kg solute in every 100 kg solution
Mass of Solute, g
Mass Percent 
100%
Mass of Solution, g
Mass of Solute  Mass of Solvent  Mass of Solution
6
Using Concentrations as
Conversion Factors

concentrations show the relationship between the amount
of solute and the amount of solvent
◦ 12%(m/m) sugar(aq) means
◦ 5.5%(m/v) Ag in Hg means
◦ 22%(v/v) alcohol(aq) means
7
Preparing a Solution


need to know amount of solution and concentration of
solution
calculate the mass of solute needed
◦ start with amount of solution
◦ use concentration as a conversion factor
 5% by mass 5 g solute  100 g solution
◦ “Dissolve the grams of solute in enough solvent to total
the total amount of solution.”
8
Example

A solution is prepared by mixing 15.0 g of Na2CO3 and
235 g of H2O. Calculate the mass percent (% m/m) of
the solution.

What volume of 10.5% by mass soda contains 78.5 g of
sugar? Density of solution is 1.04 g/ml
Solution Concentration PPM



grams of solute per 1,000,000 g of solution
mg of solute per 1 kg of solution
1 liter of water = 1 kg of water
◦ for water solutions we often approximate the kg of the
solution as the kg or L of water
grams solute x 106 mg solute
mg solute
L solution
kg solution grams solution
10
Solution Concentrations Mole
Fraction, XA




the mole fraction is the fraction of the moles of one
component in the total moles of all the components of the
solution
total of all the mole fractions in a solution = 1
unitless
the mole percentage is the percentage of the moles of one
component in the total moles of all the components of the
solution
 = mole fraction x 100%
mole fraction of A = XA = moles of components A
total moles in the solution
11
Example

What is the molarity of a solution prepared by mixing 17.2 g
of C2H6O2 with 0.500 kg of H2O to make 515 mL of
solution?

What is the percent by mass of a solution prepared by mixing
17.2 g of C2H6O2 with 0.500 kg of H2O to make 515 mL of
solution?
Example

What is the mole fraction of a solution prepared by
mixing 17.2 g of C2H6O2 with 0.500 kg of H2O to make
515 mL of solution?

A water sample is found to contain the pollutant
chlorobenzene with a concentration of 15 ppb (by mass).
What volume of this water contains 5.00 x 102 mg of
chlorobenzene? Assume density of 1.00 g/ml
Spontaneous Mixing
When the barrier is removed, spontaneous mixing occurs,
producing a solution of uniform concentration.
14
Mixing and the Solution Process
Entropy

formation of a solution does not necessarily
lower the potential energy of the system
◦ the difference in attractive forces between atoms of
two separate ideal gases vs. two mixed ideal gases is
negligible
◦ yet the gases mix spontaneously
the gases mix because the energy of the system
is lowered through the release of entropy
 entropy is the measure of energy dispersal
throughout the system
 energy has a spontaneous drive to spread out
over as large a volume as it is allowed

15
Energy changes and the solution
process







Simply put, three processes affect the
energetics of the process:
_ Separation of solute particles
ΔH1( this is always endothermic)
_ Separation of solvent particles ΔH2 (
this too is always endothermic)
_ New interactions between solute and
solvent ΔH3 ( this is always
exothermic)
The overall enthalpy change associated
with these three processes :
ΔHsoln = ΔH1 + ΔH2+ ΔH3 (Hess’s
Law)
Intermolecular Forces and the Solution
Process Enthalpy of Solution
The solute-solvent
interactions are greater
than the sum of the
solute-solute and solventsolvent interactions.
The solute-solvent
interactions are less
than the sum of the
solute-solute and solventsolvent interactions.
Intermolecular Attractions
18
Relative Interactions and Solution Formation
Solute-to-Solvent
Solute-to-Solvent
Solute-to-Solvent

Solute-to-Solute +
>
Solvent-to-Solvent
Solute-to-Solute +
=
Solvent-to-Solvent
Solute-to-Solute +
<
Solvent-to-Solvent
Solution Forms
Solution Forms
Solution May or
May Not Form
when the solute-to-solvent attractions are weaker than the sum of
the solute-to-solute and solvent-to-solvent attractions, the solution
will only form if the energy difference is small enough to be
overcome by the entropy
19
Will It Dissolve?

Chemist’s Rule of Thumb –
Like Dissolves Like


a chemical will dissolve in a solvent if it has a similar
structure to the solvent
when the solvent and solute structures are similar, the
solvent molecules will attract the solute particles at least as
well as the solute particles to each other
20
Classifying Solvents
Solvent
Class
Structural
Feature
Water, H2O
polar
O-H
Methyl Alcohol, CH3OH
Ethyl Alcohol, C2H5OH
Acetone, C3H6O
polar
polar
polar
O-H
O-H
C=O
Toluene, C7H8
Hexane, C6H14
Diethyl Ether, C4H10O
nonpolar
nonpolar
nonpolar
C-C & C-H
C-C & C-H
C-C, C-H & C-O,
(nonpolar > polar)
Carbon Tetrachloride
nonpolar
C-Cl, but symmetrical
Tro, Chemistry: A Molecular Approach
21
Example 12.1a  predict whether the
following vitamin is soluble in fat or water
OH
The 4 OH groups make the
molecule highly polar and it will
also H-bond to water.
Vitamin C is water soluble
OH
H 2C
C
H
H
C
O
C
C
O
C
HO
OH
Vitamin C
22
Example 12.1b  predict whether the
following vitamin is soluble in fat or water
The 2 C=O groups are polar, but
their geometric symmetry
suggests their pulls will cancel
and the molecule will be
nonpolar.
Vitamin K3 is fat soluble
O
H
C
C
CH3
HC
C
C
HC
C
CH
C
H
C
O
Vitamin K3
Tro, Chemistry: A Molecular Approach
23
Heats of Hydration

for aqueous ionic solutions, the energy added to
overcome the attractions between water molecules
and the energy released in forming attractions
between the water molecules and ions is
combined into a term called the heat of
hydration
◦ attractive forces in water = H-bonds
◦ attractive forces between ion and water = iondipole
◦ DHhydration = heat released when 1 mole of
gaseous ions dissolves in water
Tro, Chemistry: A Molecular Approach
24
Ion-Dipole Interactions


when ions dissolve in water they become hydrated
each ion is surrounded by water molecules
25
Solution Equilibrium
26
Solubility Limit



a solution that has the maximum amount of solute
dissolved in it is said to be saturated
◦ depends on the amount of solvent
◦ depends on the temperature
 and pressure of gases
a solution that has less solute than saturation is said to be
unsaturated
a solution that has more solute than saturation is said to be
supersaturated
27
Example

Example: The solubility of NaNO3 in water at 50oC is
110g/100g of water. In a laboratory, a student use 50.0 g of
NaNO3 with 200 g of water at the same temperature
◦ How many grams of NaNO3 will dissovle?
◦ Is the solution saturated or unsaturated?
◦ What is the mass, in grams, of any solid NaNO3 on the
bottom of the container?
Temperature Dependence of Solubility
of Solids in Water





Solubility
depends on temperature
most solids increases as temperature increases.
◦ Hot tea dissolves more sugar than does cold tea because the
solubility of sugar is much greater in higher temperature
When a saturated solution is carefully cooled, it becomes a
supersaturated solution because it contains more solute than the
solubility allowssolubility is generally given in grams of solute that
will dissolve in 100 g of water
for most solids, the solubility of the solid increases as the
temperature increases
◦ when DHsolution is endothermic
29
Solubility Curve
solubility curves can be
used to predict whether a
solution with a particular
amount of solute dissolved
in water is saturated (on
the line), unsaturated
(below the line), or
supersaturated (above the
line)
Temperature Dependence of Solubility
of Gases in Water




solubility is generally given in moles of solute that will
dissolve in 1 Liter of solution
generally lower solubility than ionic or polar covalent solids
because most are nonpolar molecules
for all gases, the solubility of the gas decreases as the
temperature increases
◦ the DHsolution is exothermic because you do not need to
overcome solute-solute attractions
the solubility of gases in water increases with increasing
mass as the attraction between the gas and the solvent
molecule is mainly dispersion forces
◦ Larger molecules have stronger dispersion forces.
31
Henry’s Law


the solubility of a gas in a liquid is directly related to the
pressure of that gas above the liquid.
at higher pressures, more gas molecules dissolve in the
liquid.
Soap Action
most dirt and grease is made of nonpolar molecules – making it
hard for water to remove it from the surface
 soap molecules form micelles around the small oil particles with the
polar/ionic heads pointing out
 this allows the micelle to be attracted to water and stay suspended
 The polar heads of the micelles attract them to the water, and
simultaneously repel other micelles so they will not coalesce and
settle out.

33
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