Transcript Chapter 6: Solutions, Acids, And Bases Goss

Warm-up
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
Define the term solution in your own
words.
What are three examples of solutions?
Solutions (ch.16)
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Solution – a homogeneous mixture
of pure substances

The SOLVENT is the medium in
which the SOLUTES are dissolved.
(The solvent is usually the most
abundant substance.)
– Example:
• Solution: Salt Water
• Solute: Salt
• Solvent: Water
READ ONLY SECTION:
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HOW DO SUBSTANCES DISSOLVE?
– “According to the kinetic theory, the water
molecules in each glass of tea are always
moving. Some moving water molecules
collide with sugar crystals. When this
happens, energy is transferred to the sugar
molecules at the surface of the crystal.”
(Holt, p. 192)
The process of dissolution (solute
dissolving in solvent) is favored
by:
1)
A decrease in the energy of the
system (exothermic-releasing energy)
2) An increase in the disorder of the
system (entropy)
Liquids Dissolving in Liquids
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Liquids that are soluble in one another
(“mix”) are called MISCIBLE.
– “LIKE dissolves LIKE”
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POLAR liquids are generally soluble in
other POLAR liquids.
NONPOLAR liquids are generally
soluble in other NONPOLAR liquids.
Factors affecting rate of
dissolution:
How can you speed up dissolution?
– Think of a cube of sugar dissolving in water
Factors affecting rate of
dissolution: think iced tea vs. hot tea &
the type of sugar you use: cubes or granulated
1) Surface area / particle size
– The greater the surface area, the faster it
dissolves
2) Temperature
– Most solids dissolve faster at higher
temperatures
3) Agitation
– Stirring/shaking will speed up dissolution
Saturation:
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Write a definition for the following terms using
the terms solute and solvent
Unsaturated solution –
Saturated solution –
Supersaturated solution –
Saturation: a solid solute dissolves in a solvent
until the soln is SATURATED
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Unsaturated solution – is able to dissolve
more solute
Saturated solution – has dissolved the
maximum amount of solute
Supersaturated solution – has dissolved
excess solute (at a higher temperature).
Solid crystals generally form when this
solution is cooled.
Demo: Supersaturated NaC2H3O2
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Crystallization of a supersaturated
solution:
Have more solute dissolved than a soln
can theoretically hold.
Can initiate the crystallization by using a
very small crystal (seed crystal) of
solute.
Crystallization is RAPID & draws excess
solute to crystallize
http://www.youtube.com/watch?v=BLq5
NibwV5g
ROCK CANDY, YUM!!
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Applying Concepts QUESTION
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When the crystallization has stopped,
will the solution be saturated or
unsaturated?

 answer
ANSWER: SATURATED
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Solution has the maximum amount of
solute for a given quantity of solvent at
a constant temperature and pressure.
SOLUBILITY
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Solubility = the amount of solute that will
dissolve in a given amount of solvent
Factors Affecting Solubility
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The nature of the solute and solvent:
different substances have different solubilities
Temperature: many solids substances
become more soluble as the temp of a
solvent increases; however, gases are less
soluble in liquids at higher temps.
Pressure: Only affects the solubility of
gases. As pressure increases, the solubility
of gases increases.
Gases: solubility
Temp and Pressure (think: flat soda)
 mentos in Diet Coke
– Nucleation site: the following factors that
contribute to the bubble formation:
Diet coke
– carbon dioxide is what makes the
bubbles form in the first place
– in synthetic mixtures aspartame, caffeine
and potassium benzoate where shown
give better fountains
Mentos
– the most important property is the rough surface which
provides plenty of nucleation sites for bubble formation
– the density makes them sink which is ideal as the bubbles
formed at the bottom of the bottle help expel much more soda
– mentos contains gelatin and gum arabic which could also
reduce surface tension
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NOTES DRAWINGS
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Write each concept/term in a square (small)
Explain the concept and draw an image to help you
remember the meaning:
•Solution
•Solute
•Solvent
•Entropy
•Saturated
•Supersaturated
•Solubility
•Factors affecting rate
•Demo: heat pack
•Unsaturated
•Demo: coke & mentos
•Factors affecting solubility
Concentration of Solution
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Concentration refers to the amount of
solute dissolved in a solution.
*MOLARITY
mol solute
Molarity (M) 
L sol'n
Example: Describe how you would prepare 2.50
L of 0.665 M Na2SO4 solution starting with:
b) solid Na2SO4
mol solute
Molarity (M) 
L sol'n
x
0.665 M 
2.50 L
Dissolve 236 g
of Na2SO4 in
enough water
to create 2.50 L
of solution.
x  1.6625 mol
1.6625mol Na 2SO 4 142.04 g

 236g
1 mol
Example Problem from Textbook
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Pg. 481
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#9
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#8
MOLARITY BY DILUTION
 When
you dilute a solution,
you can use this equation:
M1  V1  M2  V2
Example: Describe how you would prepare 2.50
L of 0.665 M Na2SO4 solution starting with:
a) 5.00 M Na2SO4
M1  V1  M2  V2
(5.00 M)(V1 )  (0.665 M)(2.50 L)
V1  0.333 L  333 mL
Add 0.333 L of Na2SO4 to 2.17 L of water.
MASS PERCENT
mass solute
mass % 
 100
total mass of sol' n
MASS PERCENT

Example: What is the percent of NaCl
in a solution made by dissolving 24 g of
NaCl in 152 g of water?
24 g
100 
176 g
13.6%  14%
*MOLALITY
mol solute
molality (m) 
kg solvent
MOLALITY
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mol solute
m
kg solvent
Example: What is the molality of a
solution that contains 12.8 g of C6H12O6
in 187.5 g water?
12.8g C6 H12 O6
1 mol

180.18g
 0.07104 mol
0.07104mol
m
 0.379 m
0.1875kg
mol solute
m
MOLALITY
kg solvent
 Example: How many grams of H2O must be
used to dissolve 50.0 g of sucrose to prepare
a 1.25 m solution of sucrose, C12H22O11?
50.0g C12 H 22 O11
1 mol

 0.1461 mol
342.34g
0.1461 mol
1.25 m 
x
x  0.1168kg  117gH2O
Colligative Properties of Solutions
(chapter 16)
Colligative properties = physical
properties of solutions that depend
on the # of particles dissolved,
not the kind of particle.
Colligative Properties
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Lowering vapor pressure
Raising boiling point
Lowering freezing point
Generating an osmotic pressure
2 to focus on…
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Lowering vapor pressure
Raising boiling point
Lowering freezing point
Generating an osmotic pressure
Boiling Point Elevation
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a solution that contains a nonvolatile
solute has a higher boiling pt than the
pure solvent; the boiling pt elevation is
proportional to the # of moles of solute
dissolved in a given mass of solvent.
Like when
adding salt to
a pot of boiling
water to make
pasta 
Boiling Point Elevation
 Tb
where:
= k bm
Tb = elevation of boiling pt
m = molality of solute (mol solute/kg solvent)
kb = the molal boiling pt elevation constant
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kb values are constants; see table 16.3 pg. 495
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kb for water = 0.52 °C/m
Ex: What is the normal boiling pt of a 2.50
m glucose, C6H12O6, solution?
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“normal” implies 1 atm of pressure
Tb = kbm
Tb = (0.52 C/m)(2.50 m)
Tb = 1.3 C
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Tb = 100.0 C + 1.3 C = 101.3 C
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Freezing/Melting Point Depression
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The freezing point of a solution is
always lower than that of the pure
solvent.
Like when
salting roads
in snowy
places so the
roads don’t ice
over or when
making ice
cream 
Freezing/Melting Point
Depression
 Tf
= kfm
where:
Tf = lowering of freezing point
m = molality of solute
kf = the freezing pt depression constant
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kf for water = 1.86 °C/m
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kf values are constants;
see table 16.2 pg. 494
Ex: Calculate the freezing pt of a
2.50 m glucose solution.
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Tf = kfm
Tf = (1.86 C/m)(2.50 m)
Tf = 4.65 C
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Tf = 0.00C - 4.65 C = -4.65C
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Textbook pg. 495 #34
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Calculate the freezing-point depression
(ΔTf) of a benzene solution containing
400. g of benzene and 200. g of
acetone, C3H6O (solute).
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Kf for benzene is 5.12 °C/m
Answer
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ΔTf = Kf x m
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ΔTf = (5.12 °C/m) x (m)
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m=
mol solute

kg solvent
1mol
58.09 g
 8.61m
0.400 kg
200.g x
So, ΔTf = (5.12 °C/m) x (8.61 m) = 44.1 °C
Do the following on your paper
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1. What is the boiling point of each solution?
a) 0.50 mol glucose in 1000. g water
b) 1.50 mol NaCl in 1000. g water
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2. What is the freezing point of each solution?
a) 1.40 mol Na2SO4 in 1750 g water
b) 0.060 mol MgSO4 in 100. g water
 answers
ANSWERS
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1a) 100.26 °C
1b) 100.78 °C
2a) -1.49 °C
2b) -1.1 °C
More calcs.
Ex: When 15.0 g of ethyl alcohol, C2H5OH, is
dissolved in 750 grams of formic acid, the freezing
pt of the solution is 7.20°C. The freezing pt of
pure formic acid is 8.40°C. Determine Kf for
formic acid.
15.0 g C2 H5OH
1 mol

 0.3255mol
46.08g
0.3255mol
 0.4340m
0.75 kg
Tf = kfm
1.20 C= (kf)( 0.4340 m)
kf = 2.76 C/m
EXTRA NOTES
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Important info for lab, etc.
Electrolytes and Colligative
Properties
• Colligative properties depend on the # of
particles present in solution.
• Because ionic solutes dissociate into ions,
they have a greater effect on freezing pt and
boiling pt than molecular solids of the same
molal conc.
Electrolytes and Colligative
Properties
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For example, the freezing pt of water is
lowered by 1.86°C with the addition of any
molecular solute at a concentration of 1 m.
– Such as C6H12O6, or any other covalent
compound
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However, a 1 m NaCl solution contains 2
molal conc. of IONS. Thus, the freezing pt
depression for NaCl is 3.72°C…double that
of a molecular solute.
– NaCl  Na+ + Cl- (2 particles)
Electrolytes - Boiling Point Elevation and
Freezing Point Depression
The relationships are given by the following equations:
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Tf = kf ·m·n
or
Tb = kb·m·n
Tf/b = f.p. depression/elevation of b.p.
m = molality of solute
kf/b = b.p. elevation/f.p depression constant
n = # particles formed from the dissociation of
each formula unit of the solute
Ex: What is the freezing pt of a 1.15 m sodium
chloride solution?
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NaCl  Na+ + Cl-
n=2
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Tf = kf·m·n
Tf = (1.86 C/m)(1.15 m)(2)
Tf = 4.28 C
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Tf = 0.00C - 4.28 C = -4.28C
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