Title: Chapter 6: Solutions, Acids, And Bases Goss
1SOLUTIONS
2Solutions (ch.16)
- 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
3READ ONLY SECTION
- 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)
4The process of dissolution is favored by
- A decrease in the energy of the system
(exothermic) - 2) An increase in the disorder of the system
(entropy)
5Liquids Dissolving in Liquids
- Liquids that are soluble in one another (mix)
are MISCIBLE. - LIKE dissolves LIKE
- POLAR liquids are generally soluble in other
POLAR liquids. - NONPOLAR liquids are generally soluble in other
NONPOLAR liquids.
LIKE DISSOLVES LIKE demo
6Factors affecting rate of dissolution think iced
tea vs. hot tea the type of sugar you use
cubes or granulated
- 1) Surface area / particle size
- Greater surface area, faster it dissolves
- 2) Temperature
- Most solids dissolve faster _at_ higher temps
- 3) Agitation
- Stirring/shaking will speed up dissolution
7Saturation a solid solute dissolves in a solvent
until the soln is SATURATED
- 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.
8ROCK CANDY, YUM!!
9Applying Concepts QUESTION
- When the crystallization has stopped, will the
solution be saturated or unsaturated? - ? answer
10ANSWER SATURATED
- Solution has the maximum amount of solute for a
given quantity of solvent at a constant
temperature and pressure.
11SOLUBILITY
- Solubility the amount of solute that will
dissolve in a given amount of solvent
12Factors Affecting Solubility
- 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.
13Gases solubility
- Temp and Pressure (think flat soda)
- DEMO 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 aspartam, 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
14STOP
15Concentration of Solution
- Concentration refers to the amount of solute
dissolved in a solution.
16MOLARITY
17Example Describe how you would prepare 2.50 L of
0.665 M Na2SO4 solution starting witha) solid
Na2SO4b) 5.00 M Na2SO4
Dissolve 236 g of Na2SO4 in enough water to
create 2.50 L of solution.
18MOLARITY BY DILUTION
- When you dilute a solution, you can use this
equation -
19Example Describe how you would prepare 2.50 L of
0.665 M Na2SO4 solution starting witha) 5.00 M
Na2SO4
Add 0.333 L of Na2SO4 to 2.17 L of water.
20Example Describe how you would prepare 2.50 L of
0.665 M Na2SO4 solution starting withb) solid
Na2SO4
Dissolve 236 g of Na2SO4 in enough water to
create 2.50 L of solution.
21MASS PERCENT
22MASS PERCENT
- Example What is the percent of NaCl in a
solution made by dissolving 24 g of NaCl in 152 g
of water?
23MOLALITY
24MOLALITY
- Example What is the molality of a solution that
contains 12.8 g of C6H12O6 in 187.5 g water?
25MOLALITY
- 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?
26Colligative Properties of Solutions (chapter 16)
- Colligative properties physical properties of
solutions that depend on the of particles
dissolved, not the kind of particle.
27Colligative Properties
- Lowering vapor pressure
- Raising boiling point
- Lowering freezing point
- Generating an osmotic pressure
282 to focus on
- Lowering vapor pressure
- Raising boiling point
- Lowering freezing point
- Generating an osmotic pressure
29Boiling Point Elevation
- 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 ?
30Boiling Point Elevation
- ?Tb kbm
- where ?Tb elevation of boiling pt
- m molality of solute (mol solute/kg solvent)
- kb the molal boiling pt elevation constant
- kb values are constants see table 16.3 pg. 495
- kb for water 0.52 C/m
31Ex What is the normal boiling pt of a 2.50 m
glucose, C6H12O6, solution?
- normal implies 1 atm of pressure
- ?Tb kbm
- ?Tb (0.52 ?C/m)(2.50 m)
- ?Tb 1.3 ?C
- Tb 100.0 ?C 1.3 ?C 101.3 ?C
32Freezing/Melting Point Depression
- 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 dont ice over or when making ice cream ?
33Freezing/Melting Point Depression
- ?Tf kfm
- where ?Tf lowering of freezing point
- m molality of solute
- kf the freezing pt depression constant
- kf for water 1.86 C/m
- kf values are constants
- see table 16.2 pg. 494
34Ex Calculate the freezing pt of a 2.50 m glucose
solution.
- ?Tf kfm
- ?Tf (1.86 ?C/m)(2.50 m)
- ?Tf 4.65 ?C
- Tf 0.00?C - 4.65 ?C -4.65?C
35Textbook pg. 495 34
- Calculate the freezing-point depression (?Tf) of
a benzene solution containing 400. g of benzene
and 200. g of acetone, C3H6O (solute). - Kf for benzene is 5.12 C/m
36Answer
- ?Tf Kf x m
- ?Tf (5.12 C/m) x (m)
- m
So, ?Tf (5.12 C/m) x (8.61 m) 44.1 C
37Do the following on your paper
- 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
- 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
38ANSWERS
- 1a) 100.26 C
- 1b) 100.78 C
- 2a) -1.49 C
- 2b) -1.1 C
39More calcs.
40Ex When 15.0 g of ethyl alcohol, C2H5OH, is
dissolved in 750 grams of formic acid, the
freezing pt of the solution is 7.20C. The
freezing pt of pure formic acid is 8.40C.
Determine Kf for formic acid.
- ?Tf kfm
- 1.20 ?C (kf)( 0.4340 m)
- kf 2.76 ?C/m
41EXTRA NOTES
- Important info for lab, etc.
42Electrolytes 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.
43Electrolytes and Colligative Properties
- For example, the freezing pt of water is lowered
by 1.86C with the addition of any molecular
solute at a concentration of 1 m. - Such as C6H12O6, or any other covalent compound
- However, a 1 m NaCl solution contains 2 molal
conc. of IONS. Thus, the freezing pt depression
for NaCl is 3.72Cdouble that of a molecular
solute. - NaCl ? Na Cl- (2 particles)
44Electrolytes - Boiling Point Elevation and
Freezing Point Depression
- The relationships are given by the following
equations - ?Tf kf mn or ?Tb kbmn
- ?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
45Ex What is the freezing pt of a 1.15 m sodium
chloride solution?
- NaCl ? Na Cl- n2
- ?Tf kfmn
- ?Tf (1.86 ?C/m)(1.15 m)(2)
- ?Tf 4.28 ?C
- Tf 0.00?C - 4.28 ?C -4.28?C