Transcript Chapter one - gen. obc - Southington Public Schools

Solutions
Ch. 15 (p. 501-529)
4-1
Solutions
Solution:
A homogeneous mixture of two
or more components in a single
physical state
Sugar in water
Air
Dental fillings
Saline
4-2
Characteristics of a Solution
• The solute can’t be filtered out.
• The solute always stays mixed.
• Particles are always in motion.
• Volumes may not be additive.
• A solution will have different
properties than the solvent.
A solution consists of two components:
solvent - component in the greater extent
solute - component in the lesser extent
(There may be more than one.)
4-3
Physical states of solutions
Solutions can be made that exist in any of the
three states. (See Fig. 15-3, p. 503)
Solid solutions
dental fillings, 14K gold, sterling silver
Liquid solutions
saline, liquor, antifreeze, vinegar
Gas solutions
air, anesthesia gases
4-4
Solution Definitions
Alloy: solution of two or more metals
Soluble: able to dissolve in another substance
Insoluble: not able to dissolve in another
substance
Miscible: pairs of liquids that mix in any
amount (Ex. ethanol and water)
Immiscible: pairs of liquids that do not mix in
any amount (Ex. oil and water)
Aqueous: solutions with water as the solvent
4-5
Amount of Solute in Solution
Concentration: the amount of solute per
quantity of solvent or solution
There are many systems - we will cover four.
•
•
•
•
Molarity (M)
Mass percent
Mole fraction (X)
Molality (m)
4-6
Example: Molarity
Calculate the molarity of a 2.0 L solution that
contains 10 moles of NaOH.
moles solute
Molarity (M) =
liters of solution
MNaOH =
=
mol
L
10 mol / 2.0 L
=
5.0 M
4-7
Mass Percent
Mass %
=
Mass Solute x 100
Total Mass
Use the same units for both
If a ham contained 5 grams of fat in 200 g
of ham, what is the mass percent?
5 g / 200g * 100
=
2.5 %
On the label, it would say 97.5 % fat
free.
4-8
Molality
moles solute
Molality (m) =
kilograms of solvent
=
mol
kg
Example: Calculate the molality of a 10.0 g of
KCl in 80.0 grams of water.
 1 mole 
10 g

74.6 g 

molality 
 1.68 m
0.080 kg
4-9
Mole fraction
Mole fraction
The moles of solute, expressed as a fraction
of the total number of moles in the solution.
nA
XA = n + n + n + . . . .
A
B
C
Because the units in the numerator and
denominator are the same, mole fraction is
a unitless quantity.
The sum of all components must equal one.
4 - 10
Example: Mole Fraction
What is the mole fraction of sulfur dioxide (SO2) in an
industrial exhaust gas containing 128.0 g of sulfur
dioxide dissolved in 1500. g of carbon dioxide.
XSO2
 1mole 
128 g

64.1 g 
moles of SO2




total moles
 1mole 
 1mole 
128 g
  1500 g

 64.1 g 
 44.0 g 
1.999 moles

 0.05539
1.999 moles  34.09 moles
XCO2 = 1 – XSO2 = 1 – 0.05539 = .9446
4 - 11
Saturation
When a solution contains as much solute as it
can at a given temperature.
Unsaturated
Can still dissolve more.
Saturated
Have dissolved all you can.
Supersaturated Temporarily have dissolved
too much.
Precipitate
Excess solute that falls out
of solution.
4 - 12
Properties of Aqueous Solutions
There are two general classes of solutes.
Electrolytes
• ionic compounds in polar solvents
• dissociate in solution to make ions
• conduct electricity
• may be strong (100% dissociation) or
weak (less than 100%)
Nonelectrolytes
• do not conduct electricity
• solute is dispersed but does not
dissociate
4 - 13
Colligative Properties
“Bulk” properties that change when you add
a solute to make a solution.
• Based on how much you add but not
what the solute is.
• Effect of electrolytes is based on number
of ions produced.
Colligative properties
• vapor pressure lowering
• freezing point depression
• boiling point elevation
• osmotic pressure
4 - 14
Vapor Pressure Lowering
Volatile: has a measurable vapor pressure
The introduction of a nonvolatile solute will
reduce the vapor pressure of the solvent in
the resulting solution.
• The vapor pressure of a nonvolatile
component is essentially zero.
• It does not contribute to the vapor pressure
of the solution.
• However, the solution’s vapor pressure is
dependent on the solute mole fraction.
4 - 15
Demo: Vapor Pressure Lowering
Water will end up in the ‘salt’ solution because it’s
vapor pressure is lower than the pure water.
4 - 16
Vapor Pressure (A Dynamic Process)
vapor
liquid
4 - 17
Boiling Point Elevation
When you add a nonvolatile solute to a
solvent, the boiling point goes up. This is
because the vapor pressure has been
lowered.
DTb = Kb m
The boiling point will continue to be
elevated as you add more solute until you
reach saturation.
Examples
Cooking pasta in salt water
Antifreeze
4 - 18
Example: Boiling point elevation
Example
Determine the boiling point for a 0.222 m
aqueous solution of sucrose.
Kb = 0.512 oC/m for water.
DTb = 0.512 oC/m (0.222 m)
= 0.114 oC
Tb = 100.00 oC + 0.114 oC = 100.11 oC
4 - 19
Phase Diagram (Boiling Point Elevation)
Diagram: www.sparknotes.com
4 - 20
Freezing Point Depression
When you add a solute to a solvent, the
freezing point goes down.
DTf = Kf m
The more you add, the lower it gets.
This will only work until you reach saturation.
Examples
“Salting” roads in winter
Making ice cream
4 - 21
Example: Freezing Point Depression
Example
Determine the freezing point for a 0.222 m
aqueous solution of sucrose.
Kf = 1.86 oC/m for water.
DTf = -1.86 oC/m (0.222 m)
= -0.413 oC
Tf
= 0.00 oC - 0.413 oC = -0.41 oC
4 - 22
Example Constants (p. 522/526)
Kb
Solvent
Water
100.0
+0.512
0.0
-1.86
Benzene
80.1
+2.53
5.5
-5.12
Camphor
207.4
+5.61
178.8
-39.7
78.3
+1.22
-117.3
-1.99
Ethanol
Normal
oC/m
Tf, oC
Kf
Normal
Tb, oCoC/m
4 - 23
Phase Diagram (Freezing Point Depression)
Diagram: www.sparknotes.com
4 - 24
Electrolytes and Colligative Properties
Ionic substances have a greater effect per
mole than covalent.
• 1 mol/kg of water for glucose = 1 molal
• 1 mol/kg of water for NaCl
= 2 molal ions
• 1 mol/kg of water for CaCl2
= 3 molal ions
Effects are based on the number of particles!
4 - 25
Van’t Hoff Factor
Van’t Hoff factor (i): number of ions created.
• Glucose (nonelectrolyte)—i = 1
• NaCl (Na+ + Cl-)—i = 2
• CaCl2(Ca2+ + 2 Cl-)—i = 3
Equations become:
DTb = iKbm
DTf = iKfm
4 - 26
Example: Electrolytes
Example
Determine the boiling point for a 1.35 m
aqueous solution of MgCl2.
i = 3 (Mg2+ + 2 Cl-)
Kb = 0.512 oC/m for water.
DTb = (3)0.512 oC/m (1.35 m)
= 2.07oC
Tb = 100.00 oC + 2.1 oC = 102.1 oC
4 - 27
Determining Molar Mass
EXAMPLE: A 10.0 gram sample of an
unknown nonvolatile, nonelectrolyte
compound is dissolved in 100 g of water.
The boiling point of the solution is elevated
to 0.433 ◦C. What is the molar mass of the
unknown sample? (Kb for water is 0.52 ◦C/m)
Given:
munk=10.0 g
msolv=100 g = .100 kg
DTb = 0.433 ◦C
i=1
Kb=0.52 ◦C/m
4 - 28
Determining Molar Mass (page 2)
Given:
munk=10.0 g
msolv=100 g = .100 kg
DTb = 0.433 ◦C
Kb = 0.52 ◦C/m
Equation:
DTb = Kbm
m= DTb/Kb = 0.433/0.52 = 0.83 m
By Definition:
(Step 1)
m = mol solute/kg solvent
Rearranging: mol solute = m x kg solvent (Step 2)
= 0.83 m * .100 kg = 0.083 mol
Conclusion: 10 g/0.083 mol = 120 g/mole
(Step 3)
4 - 29
How Solutions Form
When an ionic solid is placed in a polar
solvent, the outer ions are exposed to the
polar molecules.
Water will pull the ions from the solid and
surround them - solvate them.
Solvation of ions is an exothermic process
which helps overcome the lattice energy
that holds the crystal together. Hydration is
this same process—with water as the
solvent. (Solvation is entropically favorable,
as well.)
4 - 30
Solution of solids
4 - 31
Factors Affecting Solubility
1) Nature of Solvent and Solute
2) Temperature
3) Pressure
4 - 32
Nature of Solvent & Solute
“Like dissolves like.”
Materials with similar polarity are soluble in
each other. Dissimilar ones are not
Ionic substances are not soluble in nonpolar
solvents like hexane.
• A large amount of energy is need to
separate the ions.
• A nonpolar solvent can’t solvate ions so
there is no solvation energy to offset the
lattice energy.
4 - 33
Pressure and solubility of
gases
cg = kpgas
This law is accurate to
within 1-3% for slightly
soluble gases and
pressures up to one
atmosphere.
Solubility
(g/100g water)
Henry’s Law
At constant temperature, the solubility of a
gas is directly proportional to the pressure
of the gas above the solution.
0.010
O2
0.005
N2
He
0.000
0
1
2
Pressure (atm)
4 - 34
Solubilities of solids
Ionic substances are not soluble in nonpolar
solvents like hexane.
• A large amount of energy is need to
separate the ions.
• A nonpolar solvent can’t solvate ions so
there is no solvation energy to offset the
lattice energy.
Predicting the solubility of ionic solids in
water is difficult because a number of
competing factors are involved.
4 - 35
Solution of solids
While covalent compounds do not dissociate,
they are solvated in solution.
4 - 36
Saturated Solutions
At saturation, the solute is in dynamic
equilibrium. The concentration is constant.
Solute species are
constantly in
motion, moving
in and out of
solution.
4 - 37
Why do some
curves
decrease with
increasing
temperature
and some
increase?
4 - 38