1. “Everything around us is made up of chemicals, including ourselves
“Everything around us is made up of chemicals, including ourselves. Some things are made up of a single chemical, or substance. Orange juice, however, contains different kinds of molecules and is a mixture of substances. Vinegar is a solution with water acting as a solvent in which the other substances are dissolved. These substances can be separated physically from one another. Most foods are mixtures. Salad dressing is a mixture, called an emulsion, containing oil and vinegar. They do not mix properly, but separate out into the lighter oily layer and the heavier water layer containing the vinegar. Other kinds of mixtures may be solids, like coins, or gases, like the air around us. Toothpaste is a mixture called a suspension, in which fine particles are suspended in a liquid, and they do not dissolve. It is often necessary for chemist to separate out the different chemicals to find out what they are.”
Eyewitness Science “Chemistry” , Dr. Ann Newmark, DK Publishing, Inc., 1993, pg 14
Solutions

2. Chemical Stewardship
Be responsible in how you dispose of and use chemicals.
Chemical pollution can travel far – and harm organisms.
Rachel Carson, environmentalist wrote the book “Silent Spring” (Sept. 27, 1962), which warned everyone of the hazards associated with lack of chemical stewardship. Animals, build up toxic levels of chemicals by living in an environment rich in chemical waste. Although many of us would not worry about a dead fish or two or even a dead bird of prey, they serve as a warning to us that we are harming our Earth in which we live. Ultimately, we will harm ourselves.
It may be that the increase in the level of cancers is due, in part, to food preservatives and the many chemicals that are used in modern society. The facts are not all known.
“Excerpted in the New Yorker three months before it was published as a book, biologist Rachel Carson’s eloquent, rigorous attack on the overuse of DDT and other pesticides – she called them ‘elixirs of death’-had already upset the chemical industry. Velsicol, maker of two top bug killers, threatened to sue the book’s publisher, Houghton Mifflin, which stood firm but asked a toxicologist to recheck Carson’s facts before it shipped Silent Spring to bookstores.
Carson spent publication day in her home in Silver Springs, Md., preparing for speeches and a book tour, according to biographer Linda Lear. In a letter to a friend, Carson called Silent Spring “something I believed in so deeply that there was no other course; nothing that ever happened made me even consider turning back.” When the book appeared, industry critics assailed “the hysterical woman.” but it became an instant best seller with lasting impact. It spurred the banning off DDT in the U.S., the passage of major environmental laws and eventually a global treaty to phase out 12 pesticides known as “the dirty dozen.” Carson dies, at 56, of cancer less than two years after the book’s publication, but if she were alive today, she would undoubtedly warn about hundreds of other chemicals still released recklessly into nature.”
-By Charles Alexander TIME magazine, March 31, 2002 Pg A36.
Frog with three legs – it has
mutated from chemical exposure.

3. Solutions What the solute and the solvent are determines
whether a substance will dissolve.
how much will dissolve.
A substance dissolves faster if it is stirred or
shaken.
The particles are made smaller.
The temperature is increased.
Why?
Stirring
In order to dissolve the solvent molecules must touch the solute.
Solvent molecules hold on to and surround the solute
Stirring moves fresh solvent next to the solute.
Dissolves faster
Particle Size
The solvent touches the surface of the solute.
Smaller pieces increase the amount of surface of the solute.
Solvent and solute touch each other more often
Smaller particles dissolve faster
Temperature
Higher temperature makes the molecules of the solvent move around faster and contact the solute harder and more often.
More pieces are broken off
Speeds up dissolving.
Usually increases the amount of solid that will dissolve.

4. Solution = Solute + Solvent
Solute - gets dissolved
Solvent - does the dissolving
Aqueous (water)
Tincture (alcohol)
Amalgam (mercury)
Organic
Polar
Non-polar
Dental filling
Solutions are always homogeneous – evenly mixed.
Solutions
– In all solutions, whether gaseous, liquid, or solid, the substance present in the greatest amount is the solvent, and the substance or substances present in lesser amounts are the solute(s).
– Solute does not have to be in the same physical state as the solvent but the physical state of the solvent determines the state of the solution.
– If solute and solvent combine to give a homogeneous solution, solute is said to be soluble in the solvent.
The difference between hydrophilic and hydrophobic substances has substantial consequences in biological systems.
– Vitamins can be classified as either fat soluble or water soluble.
1. Fat-soluble vitamins (Vitamin A) are nonpolar, hydrophobic molecules and tend to be absorbed into fatty tissues and stored there.
2. Water-soluble vitamins (Vitamin C) are polar, hydrophilic molecules that circulate in the blood and intracellular fluids and are excreted from the body and must be replenished in the daily diet.
Amalgam
Amalgam is a commonly used dental filling that has been used for over 150 years. It is a mixture of mercury with at least one other metal. Currently dental amalgams are composed of about 40% mercury, and 60% powder where the powder is made up of Silver (~65%), Tin (~29%), copper (~10%) and zinc (~2%). Amalgam has many advantages over other restorative material, such as low cost, strength, durability and bacteriostatic effects. Its main disadvantage is poor esthetics on anterior teeth. There have been some concerns over the years about the detrimental health effects from the low levels of mercury released from amalgam, however there is no scientific evidence to support any detrimental effect.
Nightmare on White Street
Chem Matters, December 1996

5. Solution Definitions solution: a homogeneous mixture
-- evenly mixed at the particle level
-- e.g., salt water
alloy:
a solid solution of metals
-- e.g., bronze = Cu + Sn; brass = Cu + Zn
solvent: the substance that dissolves the solute
water
salt
Solutions
– A homogeneous mixture in which substances present in lesser amounts, called solutes, are dispersed uniformly throughout the substance in the greater amount, the solvent
– Aqueous solution — a solution in which the solvent is water
– Nonaqueous solution — any substance other than water is the solvent
– Water is essential for life and makes up about 70% of the mass of the human body.
– Many of the chemical reactions that are essential for life depend on the interaction of water molecules with dissolved compounds.
soluble:
“will dissolve in”
miscible:
refers to two gases or two liquids that form
a solution; more specific than “soluble”
-- e.g., food coloring and water

6. Types of Solutions Solute Solvent Solution Gaseous Solutions
liquid
air (nitrogen, oxygen, argon gases)
humid air (water vapor in air)
Liquid Solutions
solid
carbonated drinks (CO2 in water)
vinegar (CH3COOH in water)
salt water (NaCl in water)
Solid Solutions
dental amalgam (Hg in Ag)
sterling silver (Cu in Ag)
Solutions are not limited to gases and liquids; solid solutions also exist.
• Amalgams, which are usually solids, are solutions of metals in liquid mercury.
• Network solids are insoluble in all solvents with which they do not react chemically; covalent bonds that hold the network together are too strong to be broken and are much stronger than any combination of intermolecular interactions that might occur in solution.
• Most metals are insoluble in all solvents but do react with solutions such as aqueous acid or base to produce a solution; in these cases the metal undergoes a chemical transformation that cannot be reversed by removing the solvent.
Charles H.Corwin, Introductory Chemistry 2005, page 369

7. Factors Affecting the Rate of Dissolution
As To , rate
1. temperature
As size , rate
2. particle size
More mixing, rate
Formation of a solution from a solute and a solvent is a physical process, not a chemical one.
Both solute and solvent can be recovered in chemically unchanged form using appropriate separation methods.
Dissolution of a solute in a solvent to form a solution does not involve a chemical transformation.
Substances that form a single homogeneous phase in all proportions are said to be completely miscible in one another.
If two substances are essentially insoluble in each other, they are immiscible.
3. mixing
4. nature of solvent or solute

8. Classes of Solutions aqueous solution: solvent = water
water = “the universal solvent”
amalgam:
solvent = Hg
e.g., dental amalgam
tincture:
solvent = alcohol
e.g., tincture of iodine (for cuts)
organic solution:
solvent contains carbon
e.g., gasoline, benzene, toluene, hexane

9. Solubility Experiment 1: Add 1 drop of red food coloring A B A B
Before
Water
HOT
AFTER
COLD
A B
Miscible – “mixable”
two gases or two liquids
that mix evenly
Water
COLD
Water
HOT
You should observe that temperature effects the rate of solution. As the temperature of the liquid solvent increases, the molecules move faster, and the food coloring dissolves more quickly.
A B

10. Solubility Experiment 2: Add oil to water and shake T0 sec T30 sec
AFTER
Before
Immiscible – “does not mix”
two liquids or two gases
that DO NOT MIX
Oil
You should observe that for solutions to mix they must be chemically similar. Polar and polar molecules will mix, non-polar and non-polar molecules will mix, but polar and non-polar molecules will not mix. The reasons for this will be explained later.
Remember, ‘like dissolves like’. polar dissolves polar
non-polar dissolves non-polar
Water
Water
T0 sec
T30 sec

11. Muddy Water: Dissolved Solids
Experiment 3:
Add soil to water, shake well, and allow to settle
AFTER
Before
Muddy
Water
Dissolved solids can be calculated
as a percentage:
v/v (volume/volume)
w/v (weight/volume)
w/w (weight/weight)
Water
5% v/v soil in water
5 mL solid / 95 mL water
T1 min T5 min
5 mL / 100 mL = 5%

12. Centrifugation
Spin sample very rapidly: denser materials go to bottom (outside)
Separate blood into serum and plasma
Serum (clear)
Plasma (contains red blood cells ‘RBCs’)
Check for anemia (lack of iron)
AFTER
Before
RBC’s
Serum
Blood
When you donate blood, the phlebotomist will check for anemia before drawing your blood. A small amount of blood will be taken from your fingertip into a thin glass tube. The blood may be placed in a centrifuge as described above to check for sufficient amounts of red blood cells. An alternate method is to drop the blood in a solution of copper sulfate. If the blood contains enough RBC’s the blood drop will sink to the bottom of the container, if it is not dense enough (lacks adequate number of RBC’s), the blood will float and you will not be allowed to donate blood. A
B C

13. Making solutions
In order to dissolve - the solvent molecules must come in contact with the solute.
Stirring moves fresh solvent next to the solute.
The solvent touches the surface of the solute.
Smaller pieces increase the amount of surface of the solute.
When a solute dissolves, its individual atoms, molecules, or ions interact with the solvent, become solvated, and are able to diffuse independently throughout the solution.
• If the molecule or ion collides with the surface of a particle of the undissolved solute, it may adhere to the particle in a process called crystallization (formation of a solid with a well-defined crystalline structure).
• Dissolution and crystallization continue as long as excess solid is present, resulting in a dynamic equilibrium.
dissolution
solute + solvent crystallization solution

14. Water Molecule H2O d+ d+ d- d- Water is a POLAR molecule O2- H+ H+
Draw the Lewis dot structure of a water molecule.
Recall, oxygen has 6 valence electrons and hydrogen has one valence electron.
The ‘electron cloud’ is more dense around the oxygen than the hydrogen. Therefore, the oxygen side of the molecule is more negatively charged and has a partial negative charge.
The hydrogen have a lower electron density around them and receive a partial positive charge. This molecule does not have a formal (+) and (-) charge – as in an ionic compound.

15. create surface tension
Water molecules
“stick” together to
create surface tension
to support
light weight objects.
Copyright © 2007 Pearson Benjamin Cummings. All rights reserved.

16. Water Molecule What is a polar molecule?d-
Hydrogen
bond d+ O H
How does the polarity of water effect this molecule?
An individual water molecule consists of two hydrogen atoms bonded to an oxygen atom in a bent (V-shaped) structure.
The oxygen atom in each O–H covalent bond attracts the electrons more strongly than the hydrogen atom.
O and H nuclei do not share the electrons equally.
– Hydrogen atoms are electron-poor compared with a neutral hydrogen atom and have a partial positive charge, indicated by the symbol δ+.
– The oxygen atom is more electron-rich than a neutral oxygen atom and has a partial negative charge, indicated by the symbol 2δ-.
Unequal distribution of charge creates a polar bond.

17. Hydrogen bonds occur between two polar molecules, or between different polar regions of one large macro-molecule.
One “relatively” negative region is attracted to a second “relatively” positive region. O H
Electronegative
atoms
Hydrogen
bond H N

18. Interstitial Spaces Oil Non-polar "immiscible" Polar dissolved solid
Layer
Water
Water
Water
Water
Water
Water
Water
Water
Polar
This is an overly simplified model of spaces between water molecules. However, it is in these interstitial spaces that gases (e.g. oxygen and carbon dioxide) dissolve and solids also dissolve.
red food
coloring

19. Dissolving of solid NaCl
salt
Na+
NaCl solid
NaCl (aq)
= Na+
= Cl-
Animation by Raymond Chang
All rights reserved.

20. Dissolving of NaCl - - - O + + + + Cl- Na+ hydrated ions H
Timberlake, Chemistry 7th Edition, page 287

21. Dissolving of Salt in Water
Na+
ions
Water molecules
Cl-
ions
When sodium chloride crystals are dissolved in water, the polar water molecules exert attracting forces which weaken the ionic bonds. The process of solution occurs the ions of sodium and chloride become hydrated.
NaCl(s) + H2O  Na+(aq) + Cl-(aq)

22. Solubility vs. Temperature
200
180
160
140
120
100 80 60 40 20 KI
KNO3
Solubility (g solute / 100 g H2O)
NaNO3
The general rule of thumb is that solubility of solids increases with increases in temperature.
Maximum amount of a solute that can dissolve in a solvent at a specified temperature and pressure is its solubility.
– Solubility is expressed as the mass of solute per volume (g/L) or mass of solute per mass of solvent (g/g) or as the moles of solute per volume (mol/L).
– Solubility of a substance depends on energetic factors and on the temperature and, for gases, the pressure.
• A solution that contains the maximum possible amount of solute is saturated.
• If a solution contains less than the maximum amount of solute, it is unsaturated.
When a solution is saturated and excess solute is present, the rate of dissolution is equal to the rate of crystallization.
• Solubility increases with increasing temperature — a saturated solution that was prepared at a higher temperature contains more dissolved solute than it would contain at a lower temperature, when the solution is cooled, it can become supersaturated.
Solubility of a substance generally increases with increasing temperature
No relationship between the structure of a substance and the temperature dependence of its solubility
Solubility may increase or decrease with temperature; the magnitude of this temperature dependence varies widely among compounds
This variation of solubility with temperature is used to separate the components of a mixture by fractional crystallization, the separation of compounds based on their solubilities in a given solvent
Fractional crystallization is a common technique for purifying compounds; the compound of interest must be more soluble at high temperature than at low temperature, so that lowering the temperature causes it to crystallize out of solution
Solubility of gases in liquids decreases with increasing temperature
Attractive intermolecular interactions in the gas phase are essentially zero for most substances
When a gas dissolves, its molecules interact with solvent molecules and heat is released when these new attractive interactions form, therefore, dissolving most gases in liquids is an exothermic process (Hsoln < 0)
Adding heat to the solution provides thermal energy that overcomes the attractive forces between the gas and the solvent molecules, thereby decreasing the solubility of the gas
Na3PO4
NaCl 20 40 60 80
100
Temperature (oC)
Timberlake, Chemistry 7th Edition, page 297

23. Gas Solubility Higher Temperature …Gas is LESS Soluble CH4 O2
2.0 O2
Higher Temperature
…Gas is LESS Soluble CO
Solubility (mM)
1.0
The general rule of thumb is that the solubility of gases decreases when temperature increases.
Solubility of a substance generally increases with increasing temperature
No relationship between the structure of a substance and the temperature dependence of its solubility
Solubility may increase or decrease with temperature; the magnitude of this temperature dependence varies widely among compounds
This variation of solubility with temperature is used to separate the components of a mixture by fractional crystallization, the separation of compounds based on their solubilities in a given solvent
Fractional crystallization is a common technique for purifying compounds; the compound of interest must be more soluble at high temperature than at low temperature, so that lowering the temperature causes it to crystallize out of solution
Solubility of gases in liquids decreases with increasing temperature
Attractive intermolecular interactions in the gas phase are essentially zero for most substances
When a gas dissolves, its molecules interact with solvent molecules and heat is released when these new attractive interactions form, therefore, dissolving most gases in liquids is an exothermic process (Hsoln < 0)
Adding heat to the solution provides thermal energy that overcomes the attractive forces between the gas and the solvent molecules, thereby decreasing the solubility of the gas He 10 20 30 40 50
Temperature (oC)

24. Electrolytes Electrolytes - solutions that carry an electric current
Electrolyte — any compound that can form ions when it dissolves in water
– When strong electrolytes dissolve, constituent ions dissociate completely, producing aqueous solutions that conduct electricity very well.
– When weak electrolytes dissolve, they produce relatively few ions in solution, and aqueous solutions, of weak electrolytes do not conduct electricity as well as solutions of strong electrolytes.
– Nonelectrolytes dissolve in water as neutral molecules and have no effect on conductivity.
strong electrolyte
weak electrolyte
nonelectrolyte
NaCl(aq) Na+ + Cl-
HF(aq) H+ + F-
Timberlake, Chemistry 7th Edition, page 290

25. Effect of Salinity on Cells
isotonic solution
no change
hypotonic solution
hemolysis
Salt can be sprinkled on snails to cause crenation. The snail will automatically secrete water from inside its cells to dilute the high salt content. This ends up killing the snail.
Many older adults must be careful not to consume too much salt. The excess salt in their cells causes excess water weight. Their ankles swell and possible their heart.
For this reasons, a prescription diuretic (a substance to help you remove water from your body) is prescribed.
Salt peter helps male dancers too…shrinkage!
hypertonic solution
crenation
Timberlake, Chemistry 7th Edition, page 312

26. Isotonic Hypotonic Hypertonic (a) Cells in dilute salt solution
(b) Cells in distilled water
(c) Cells in concentrated salt solution
Copyright © 2007 Pearson Benjamin Cummings. All rights reserved.

27. Concentration = 1 “fishar”
# of fish
volume (L)
1 fish
Concentration =
1 (L)
Concentration = 1 “fishar”
V = 1000 mL
V = 1000 mL
V = 5000 mL
n = 2 fish
n = 4 fish
n = 20 fish
Concentration = 2 “fishar”
[ ] = 4 “fishar”
[ ] = 4 “fishar”

28. Concentration = # of moles volume (L) V = 1000 mL V = 1000 mL
n = 8 moles
[ ] = 32 molar
V = 1000 mL
V = 1000 mL
V = 5000 mL
n = 2 moles
n = 4 moles
n = 20 moles
Concentration = 2 molar
[ ] = 4 molar
[ ] = 4 molar

29. Making Molar Solutions
…from liquids
(More accurately, from stock solutions)

30. Concentration…a measure of solute-to-solvent ratio
concentrated vs dilute
“lots of solute” “not much solute”
“watery”
Concentration of a solution describes the quantity of a solute that is contained in a particular quantity of solvent or solution
Knowing the concentration of solutes is important in controlling the stoichiometry of reactant for reactions that occur in solution
A concentrated solution contains a large amount of solute in a given amount of solution. A 10 mol/L solution would be called concentrated.
A dilute solution contains a small amount of solute in a given amount of solution. A 0.01 mol/L solution would be called dilute.
Add water to dilute a solution; boil water off to concentrate it.

31. Making a Dilute Solution
remove
sample
moles of
solute
initial solution
same number of
moles of solute
in a larger volume
mix
Making a Dilute Solution
diluted solution
Timberlake, Chemistry 7th Edition, page 344

32. Concentration “The amount of solute in a solution”
A. mass % = mass of solute
mass of sol’n
B. parts per million (ppm)  also, ppb and ppt
– commonly used for minerals or
contaminants in water supplies
C. molarity (M) = moles of solute
L of sol’n
– used most often in this class
% by mass – medicated creams
% by volume – rubbing alcohol
mol L M
MOLARITY - Most common unit of concentration
Most useful for calculations involving the stoichiometry of reactions in solution
Molarity of a solution is the number of moles of solute present in exactly 1 L of solution:
moles of solute
molarity = liters of solution
Units of molarity — moles per liter of solution (mol/L),
abbreviated as M
Relationship among volume, molarity, and moles is expressed as
VL M Mol/L = L (mol) = moles
(L)
There are several different ways to quantitatively describe the concentration of a solution, which is the amount of solute in a given quantity of solution.
1. Molarity
– Useful way to describe solution concentrations for reactions that are carried out in solution or for titrations
– Molarity is the number of moles of solute divided by the olume of the solution
Molarity = moles of solute = mol/L
liter of solution
– Volume of a solution depends on its density, which is a function of temperature
2. Molality
– Concentration of a solution can also be described by its molality (m), the number of moles of solute per kilogram of solvent
– Molality = moles of solute
kilogram solvent
– Depends on the masses of the solute and solvent, which are independent of temperature
– Used in determining how colligative properties vary with solute concentrations
3. Mole fraction
– Used to describe gas concentrations and to determine the vapor pressures of mixtures of similar liquids
– Mole fraction () = moles of component
total moles in the solution
– Depends on only the masses of the solute and solvent and is temperature independent
4. Mass percentage (%)
– The ratio of the mass of the solute to the total mass of the solution
– Result can be expressed as mass percentage, parts per million (ppm), or parts per billion (ppb)
mass percentage = mass of solute  100%
mass of solution
parts per million (ppm) = mass of solute  106
parts per billion (ppb) = mass of solute  109
– Parts per million (ppm) and parts per billion (ppb) are used to describe concentrations of highly dilute solutions, and these measurements correspond to milligrams (mg) and micrograms (g) of solute per kilogram of solution, respectively
– Mass percentage and parts per million or billion can express the concentrations of substances even if their molecular mass is unknown because these are simply different ways of expressing the ratios of the mass of a solute to the mass of the solution
M =
mol L
D. molality (m) = moles of solute
kg of solvent

33. Glassware – Precision and Cost
beaker vs. volumetric flask
When filled to 1000 mL line, how much liquid is present?
beaker
5% of 1000 mL =
volumetric flask
50 mL
1000 mL mL
Range:
950 mL – 1050 mL
Range:
mL– mL
imprecise; cheap
precise; expensive

34. Markings on Glassware Beaker 500 mL + 5% Range = 500 mL + 25 mL
Graduated Cylinder
500 mL mL
Range = 500 mL + 5 mL
495 – 505 mL
Volumetric Flask
500 mL mL
Range = – mL
TC 20oC “to contain at a temperature of 20 oC” 22
TD “to deliver” T s
“time in seconds”

35. How to mix solid chemicals
Lets mix chemicals for the upcoming soap lab.
We will need 1000 mL of 3 M NaOH per class.
How much sodium hydroxide will I need, for five classes, for this lab?
M =
mol L
3 M =
? mol
1 L
? = 3 mol NaOH/class
x 5 classes
15 mol NaOH
How much will this weigh?
1 23g/mol + 16g/mol g/mol
MMNaOH = 40g/mol
To prepare a solution that contains a specified concentration of a substance, it is necessary to dissolve the desired number of moles of solute in enough solvent to give the desired final volume of solution.
Solute occupies space in the solution so the volume of the solvent that is needed is less than the desired volume of solution.
To prepare a particular volume of a solution that contains a specified concentration of a solute, calculate the number of moles of solute in the desired volume of solution and then covert the number of moles of solute to the corresponding mass of solute needed.
40.0 g NaOH
X g NaOH = mol NaOH =
600 g NaOH
1 mol NaOH
FOR EACH CLASS:
To mix this, add 120 g NaOH into 1L volumetric flask with
~750 mL cold H2O.
Mix, allow to return to room temperature – bring volume to 1 L.

36. How to mix a Standard Solution
Wash bottle
Volume marker
(calibration mark)
Weighed
amount
of solute
Use a VOLUMETRIC FLASK to make a standard solution of known concentration
Step 1> add the weighed amount of solute in the volumetric flask
Step 2> add distilled water (about half of final volume)
Step 3> cap volumetric flask, and shake to dissolve solute completely
Step 4> add distilled water to volume marker (calibration mark)
The solution process may be exothermic (release heat). This may cause the liquid to show a larger volume than is real. Allow the solution to return to ambient (room) temperature and check volume again.
Zumdahl, Zumdahl, DeCoste, World of Chemistry 2002, page 480

37. Process of Making a Standard Solution from Liquids
Solutions can be made using liquids or solids (or gases).
To make a 5% solution v/v (volume to volume)
This means to add 5 mL of solute in 95 mL of solvent. The total is 5 mL / 100 mL or 5%.
For the diagram add 25 mL of liquid solute and add water to bring volume to 500 mL (about 475 mL water).
SAFETY NOTE: Always add acid concentrate to water…never add water to concentrated acid.
If you’ve seen what happens when water or ice crystals hit hot oil…a similar phenomenon occurs when water is added to concentrated acid.
The addition of water to concentrated dissipates a large amount of heat. This heat rapidly boils the acid and causes it to spatter.
If however, you start with a large volume of water and slowly add acid, the same amount of heat is generated.
This time, the large volume of water is capable of absorbing the heat. The solution will not splatter.
Zumdahl, Zumdahl, DeCoste, World of Chemistry 2002, page 483

38. Reading a pipette 4.48 - 4.50 mL 4.86 - 4.87 mL 5.00 mL
Identify each volume to two decimal places
(values tell you how much you have expelled) mL mL
5.00 mL

39. MConc.VConc. = MDiluteVDilute
Dilution of Solutions
Solution Guide
Formula
Weight
Specific Gravity
Molarity
Reagent
Percent
To Prepare 1
Liter of one molar
Solution
Acetic Acid Glacial (CH3COOH)
60.05
1.05
17.45
99.8%
57.3 mL
Ammonium Hydroxide (NH4OH)
35.05
0.90
14.53
56.6%
69.0 mL
Formic Acid (HCOOH)
46.03
1.20
23.6
90.5%
42.5 mL
Hydrochloric Acid (HCl)
36.46
1.19
12.1
37.2%
82.5 mL
Hydrofluoric Acid (HF)
20.0
1.18
28.9
49.0%
34.5 mL
Nitric Acid (HNO3)
63.01
1.42
15.9
70.0%
63.0 mL
Perchloric Acid 60% (HClO4)
100.47
1.54
9.1
60.0%
110 mL
Perchloric Acid 70% (HClO4)
1.67
11.7
70.5%
85.5 mL
Phosphoric Acid (H3PO4)
97.1
1.70
14.8
85.5%
67.5 mL
Potassium Hydroxide (KOH)
Sodium Hydroxide (NaOH)
40.0
19.4
45.0%
Sulfuric Acid (H2SO4)
98.08
1.84
18.0
50.5%
51.5 mL
This chart quickly shows you the amount of concentrated acid needed to make 1 liter of a 1 M solution. If you need a 5 M solution, add 5x the amount of acid in the same volume.
MConc.VConc. = MDiluteVDilute

40. **Safety Tip: When diluting, add acid or base to water.**
Acids (and sometimes bases) are
purchased in concentrated form (“concentrate”) and are easily
diluted to any desired concentration.
Dilutions of Solutions 
**Safety Tip: When diluting, add acid or base to water.**
C = concentrate
D = dilute
Dilution Equation:
Concentrated H3PO4 is 14.8 M. What volume of concentrate
is required to make L of M H3PO4?
VC = L = 845 mL

41. How would you mix the above solution?
1. Measure out L of concentrated H3PO4 .
2. In separate container, obtain ~20 L of cold H2O.
3. In fume hood, slowly pour [H3PO4] into cold H2O.
4. Add enough H2O until L of solution is obtained.
Be sure to wear your safety glasses!

42. You have 75 mL of conc. HF (28.9 M); you need 15.0 L of
0.100 M HF. Do you have enough to do the experiment?
MCVC = MDVD
28.9 M (0.075 L) = M (15.0 L)
Yes;
we’re OK.
mol HAVE
>
1.50 mol NEED

43. Dilution
Preparation of a desired solution by adding water to a concentrate.
Moles of solute remain the same.

44. Dilution
What volume of 15.8M HNO3 is required to make 250 mL of a 6.0M solution?
GIVEN:
M1 = 15.8M
V1 = ?
M2 = 6.0M
V2 = 250 mL
WORK:
M1 V1 = M2 V2
(15.8M) V1 = (6.0M)(250mL)
V1 = 95 mL of 15.8M HNO3

45. Preparing Solutions How to prepare 500 mL of 1.54 M NaCl solution
mass 45.0 g of NaCl
add water until total volume is 500 mL
Shake to dissolve salt
500 mL
volumetric
flask
500 mL
mark
45.0 g NaCl
solute

46. Preparing Solutions 500 mL of 1.54M NaCl
molality
molarity
1.54m NaCl in kg of water
500 mL of 1.54M NaCl
mass 45.0 g of NaCl
add kg of water
mass 45.0 g of NaCl
add water until total volume is 500 mL
500 mL water
500 mL
volumetric
flask
45.0 g NaCl
500 mL
mark

47. Spec-20 Instrument Wavelength Light control Amplifier Sample holder
control knob
Sample holder
cover
Amplifier
control knob
Light control
knob

48. Spec-20 Absorbance Absorbance l (wavelength) Insert Photograph Spec-20
Sample-holder
Meter
Zero adjust
and power control
Wavelength scale
Wavelength control
Light control
100 % Absorbance
0 % Transmittance
Absorbance
0 % Absorbance
100 % Transmittance
l (wavelength)

49. Schematic representation of a spectrophotometer
Monochromator
Detector
Light
Source
(1)
Sample
cell
(3)
Meter
(2)
(4)
(5)
A spectrophotometer is an instrument that measures the fraction (I/Io) of an incident
beam of light (Io) which is transmitted (I) by a sample at a particular wavelength.
A light source that produces light with a wavelength range from about 275 to 650 nm;
A monochromator that selects a particular wavelength of light and sends it to the sample cell with an intensity of Io;
The sample cell, which contains the solution being analyzed (NOTE: It the intensity of the incident light is Io and the solution absorbs light, the intensity of the transmitted light, I , is less than Io);
A detector that measures the intensity of the light is transmitted, I, from the sample cell
Sample-holder
Meter
Zero adjust
and power control
Wavelength scale
Wavelength control
Light control
For a given substance,
the amount of light
absorbed depends on:
a) the concentration;
b) the cell or path length;
c) the wavelength of light; and
d) the solvent.

50. Absorbance of Chlorophyll
1.5
1026
1024
1022
1020
1018
1016
1014
1010
108
106
104
102 1
10-8
10-6
10-4
10-2
1012
Frequency (Hz)
Wavelength (nm)
cosmic
rays
gamma
x-rays
ultra-
violet
infra-
red
radio
(microwave)
radar
tele-
vision
power
transmission
Violet
Blue
Green
Yellow
Orange
Red UV
Near
Infrared
400 nm
500 nm
600 nm
700 nm
1.2
0.9
Absorbance
0.6
1. The two peak absorbances occur at 440 nm and 660 nanometers with absorbances at 1.27 and 0.30 respectively.
nanometers is in the red region of the visible spectrum nanometers is in the blue region of the spectrum.
3. Chlorophyll is green because it does not absorb green light. It transmits the green light so that is what we see when we look at green plants.
Flinn Spec-20 instruction manual & Kelter, Carr, Scott, , Chemistry: A World of Choices 1999, page 480
0.3
0.0
300
400
500
600
700
800
Wavelength (nm)

51. Calibration Curve (fixed wavelength) ? Absorbance Concentration
0 % Transmittance
out of linear range
(too concentrated) ?
Absorbance
0 % Absorbance
100 % Transmittance
x 2
Concentration
Dilute sample with water 50:50. Run sample, read concentration.

52. What mass of CaF2 must be added to 1,000 L of water
so that fluoride atoms are present at a conc. of 1.5 ppm?
1000 mL
1 g
1 mol
6.02 x1023 m’cule
X m’cule H2O = 1000 L
= 3.34 x 1028 m’cules H2O
1 L
1 mL
18 g
1 mol
1.5 atom F X atoms F
1,000,000 m’cule H2O x 1028 m’cule H2O =
1 molecule CaF2
X = 5.01 x 1022 atoms F
times =
2.505 x 1022 molecules CaF2
2 atoms F
1 mol CaF2
78.1 g CaF2
X g CaF2 = x 1022 molecules =
3.25 g CaF2
6.02 x 1023 molecules
1 mol CaF2

53. mol L M
How many moles solute are required to make 1.35 L of 2.50 M solution?
mol = M L
= 2.50 M (1.35 L) =
3.38 mol
A. What mass sodium hydroxide is this?
40.0 g NaOH
X g NaOH = mol NaOH =
135 g NaOH
1 mol NaOH
B. What mass magnesium phosphate is this?
262.9 g Mg3(PO4)2
X g Mg3(PO4)2 = mol Mg3(PO4)2 =
889 g Mg3(PO4)2
1 mol Mg3(PO4)2

54. Find molarity if 58.6 g barium hydroxide are in 5.65 L solution.
Step 1). How many moles barium hydroxide is this?
1 mol Ba(OH)2
X mol Ba(OH)2 = g Ba(OH)2 =
0.342 mol Ba(OH)2
171.3 g Ba(OH)2
Step 2). What is the molarity of a 5.65 L solution containing mol solute?
M =
mol L
0.342 mol
M = =
0.061 M Ba(OH)2
5.65 L

55. You have 10.8 g potassium nitrate.
How many mL of solution will make this a 0.14 M solution?
convert to mL

56. occurs when neutral combinations of particles separate into ions while in aqueous solution.
Dissociation
sodium chloride
NaCl  Na Cl1–
sodium hydroxide
NaOH  Na OH1–
hydrochloric acid
HCl  H Cl1–
sulfuric acid
H2SO4  2 H SO42–
acetic acid
CH3COOH  CH3COO1– H1+
In general, acids yield hydrogen ions
(H1+) ?
in aqueous solution; bases yield hydroxide ions.
(OH1–) ?

57. Strong electrolytes exhibit nearly 100% dissociation.
NaCl Na Cl1–
NOT in water:
in aq. solution:
Weak electrolytes exhibit little dissociation.
CH3COOH CH3COO1– H1+
NOT in water:
in aq. solution:
“Strong” or “weak” is a property of the substance.
We can’t change one into the other.

58. electrolytes: solutes that dissociate in solution
-- conduct electric current because of free-moving ions
e.g., acids, bases, most ionic compounds
-- are crucial for many cellular processes
-- obtained in a healthy diet
-- For sustained exercise or a bout of the flu, sports drinks
ensure adequate electrolytes.
nonelectrolytes: solutes that DO NOT dissociate
-- DO NOT conduct electric current (not enough ions)
e.g., any type of sugar

59. Colligative Properties  depend on concentration of a solution
Compared to solvent’s… a solution w/that solvent has a…
…normal freezing point (NFP) …lower FP
FREEZING PT.
DEPRESSION
…normal boiling point (NBP) …higher BP
BOILING PT.
ELEVATION

60. Applications (NOTE: Data are fictitious.)
1. salting roads in winter FP BP
water
0oC (NFP)
100oC (NBP)
water + a little salt
–11oC
103oC
water + more salt
–18oC
105oC
2. antifreeze (AF) /coolant FP BP
water
0oC (NFP)
100oC (NBP)
water + a little AF
–10oC
110oC
50% water + 50% AF
–35oC
130oC
Why do you think some towns use calcium chloride on roads in the winter versus sodium chloride?
CaCl2 yields three ions while NaCl yields only two ions. Calcium chloride will work in colder weather. Calcium chloride will work better than sodium chloride. Secondly, calcium chloride doesn’t kill grass like sodium chloride.

61. 3. law enforcement white powder starts melting at… finishes
penalty, if
convicted A
109oC
175oC
comm. service B
150oC
180oC
2 years C
194oC
196oC
20 years
Drugs, especially pharmaceutical drugs, are tested for purity. A simple test is to measure the melting point range of the substance. A pure sample will have a very narrow range for its melting point. As impurities are introduced into the sample, an increase in the melting point range is observed.

62. Effect of Pressure on Boiling Point
Boiling Point of Water at Various Locations
Location
Feet above sea level
Patm
(kPa)
Boiling Point
(C)
Top of Mt. Everest, Tibet
29,028 32 70
Top of Mt. Denali, Alaska
20,320
45.3 79
Top of Mt. Whitney, California
14,494
57.3 85
Leadville, Colorado
10,150 68 89
Top of Mt. Washington, N.H.
6,293
78.6 93
Boulder, Colorado
5,430
81.3 94
Madison, Wisconsin
900
97.3 99
New York City, New York 10
101.3
100
Death Valley, California
-282
102.6
100.3

63. Calculations for Colligative Properties
The change in FP or BP is found using… DTx = Kx m i
DTx = change in To (below NFP or above NBP)
Kx = constant depending on… (A) solvent
(B) freezing or boiling
m = molality of solute = mol solute / kg solvent
i = integer that accounts for any solute dissociation
any sugar (all nonelectrolytes)……………...i = 1
table salt, NaCl  Na1+ + Cl1–………………i = 2
barium bromide, BaBr2  Ba Br1–……i = 3

64. Freezing Point Depression Boiling Point Elevation
DTf = Kf m i DTb = Kb m i
Then use these in conjunction with the NFP and NBP to
find the FP and BP of the mixture.
The normal boiling point of a substance is the temperature at which the vapor pressure equals 1 atm.
Because the vapor pressure of the solution at a given temperature is lower than the vapor pressure of the pure solvent, achieving a vapor pressure of 1 atm for the solution requires a higher temperature than the normal boiling point of the solvent.
The boiling point of a solution is always higher than that of the pure solvent.
The magnitude of the increase in the boiling point is related to the magnitude of the decrease in the vapor pressure; the decrease in the vapor pressure is proportional to the concentration of the solute in solution.
The boiling point elevation (Tb) is defined as the difference between the boiling points of the solution and the pure solvent:
Tb = Tb – T0b
where Tb is the boiling point of the solution and T0b is the boiling point of the pure solvent.
The relationship between Tb and concentration can be expressed as
Tb = mKb
where m is the concentration of the solute expressed in molality, and Kb is the molal boiling-point elevation constant of the solvent, which has units of ºC/m.
Concentration of the solute is expressed as molality rather than mole fraction or molarity for two reasons:
1. Because the density of a solution changes with temperature, the value of molarity also varies with temperature, and if the boiling point depends on the solute concentration, then the system is not maintained at a constant temperature.
2. Molality and mole fraction are proportional for relatively dilute solutions, but molality has a larger numerical value and eliminates nonsignificant zeros.
• Boiling-point elevation depends on the total number of dissolved nonvolatile solute particles.
Dissolving a nonvolatile solute in water not only raises the boiling point of the water but also lowers its freezing point.
The freezing-point depression (Tf) is defined as the difference between the freezing point of the pure solvent and the freezing point of the solution:
Tf = T0f – Tf
where T0f is the freezing point of the pure solvent and Tf is the freezing point of the solution:
The relationship between Tf and the solute concentration is
Tf = mKf
where m is the molality of the solution and Kf is the molal freezing-point depression constant for the solvent in units of ºC/m.
Freezing-point depression depends on the total number of dissolved nonvolatile solute particles.
The molar mass of an unknown compound can be determined by measuring the freezing point of a solution that contains a known mass of solute, which is accurate for dilute solutions.
Changes in the boiling point are smaller than changes in the freezing point for a given solvent, so boiling point elevations are difficult to measure and are not used to determine molar mass.
(Kf = cryoscopic constant, which is 1.86 K kg/mol for the freezing point of water)
(Kb = ebullioscopic constant, which is 0.51 K kg/mol for the boiling point of water)

65. 168 g glucose (C6H12O6) are mixed w/2.50 kg H2O.
Find BP and FP of mixture. For H2O, Kb = 0.512, Kf = –1.86.
i = 1
(NONELECTROLYTE)
DTb = Kb m i = (0.373) (1) = 0.19oC
BP = ( )oC = oC
DTf = Kf m i = –1.86 (0.373) (1) = –0.69oC
FP = (0 + –0.69)oC = –0.69oC

66. 168 g cesium bromide are mixed w/2. 50 kg H2O
168 g cesium bromide are mixed w/2.50 kg H2O. Find BP and FP of mixture.
For H2O, Kb = 0.512, Kf = –1.86.
Cs1+ Br1– i = 2
CsBr  Cs Br1–
DTb = Kb m i = (0.316) (2) = 0.32oC
BP = ( )oC = oC
DTf = Kf m i = –1.86 (0.316) (2) = –1.18oC
FP = (0 + –1.18)oC = –1.18oC

67. Molarity and StoichiometryV P
mol
M =
mol L
mol = M L
M L
M L
The coefficients in the balanced chemical equation indicate the number of moles of each reactant that is needed and the number of moles of each product that can be produced.
It doesn’t matter if you are dealing with volumes of solutions of reactants or masses of reactants. __ 1
Pb(NO3)2(aq) KI (aq)  PbI2(s) KNO3(aq) 2 1 2
What volume of 4.0 M KI solution is required to yield 89 g PbI2?

68. Stoichiometry for Reactions in Solution
Step 1)
Identify the species present in the combined solution,
and determine what reaction occurs.
Step 2)
Write the balanced net ionic equation for the reaction.
Step 3)
Calculate the moles of reactants.
Step 4)
Determine which reactant is limiting.
Step 5)
Calculate the moles of product or products, as required.
Step 6)
Convert to grams or other units, as required.
Stoichiometry steps for reactions in solution

69. 1 Pb(NO3)2(aq) + 2 KI (aq)  1 PbI2(s) + 2 KNO3(aq)
? L 4.0 M
89 g
What volume of 4.0 M KI solution is required to yield 89 g PbI2?
Strategy:
(1) Find mol KI needed to yield 89 g PbI2.
(2) Based on (1), find volume of 4.0 M KI solution.
1 mol PbI2
2 mol KI
X mol KI = 89 g PbI2
= mol KI
461 g PbI2
1 mol PbI2
M =
mol L
L =
mol M =
0.39 mol KI
4.0 M KI =
0.098 L of 4.0 M KI

70. How many mL of a 0.500 M CuSO4 solution will react
w/excess Al to produce 11.0 g Cu?
Al3+ SO42–
__CuSO4(aq) + __Al (s) 
__Cu(s) +
__Al2(SO4)3(aq) 3
CuSO4(aq) Al (s)  Cu(s) + Al2(SO4)3(aq) 2 3 1
x mol
11 g
1 mol Cu
3 mol CuSO4
X mol CuSO4 = 11 g Cu
= mol CuSO4
63.5 g Cu
3 mol Cu
M =
mol L
L =
mol M
0.173 mol CuSO4
0.500 M CuSO4
= L
1000 mL
0.346 L
= 346 mL
1 L

71. Stoichiometry Problems
How many grams of Cu are required to react with 1.5 L of 0.10M AgNO3?
Cu AgNO3  2Ag Cu(NO3)2
? g
1.5L
0.10M
1.5 L
.10 mol
AgNO3
1 L
1 mol Cu
2 mol
AgNO3
63.55
g Cu
1 mol Cu
= 4.8 g Cu
Courtesy Christy Johannesson

72. Limiting Reactants Zn + 2HCl  ZnCl2 + H2 79.1 g ? L 0.90 L 2.5M
79.1 g of zinc react with 0.90 L of 2.5M HCl. Identify the limiting and excess reactants. How many liters of hydrogen are formed at STP?
The concept of limiting reactants applies to reactions that are carried out in solution and reactions that involve pure substances.
If all the reactants but one are present in excess, then the amount of the limiting reactant can be calculated.
When the limiting reactant is not known, one can determine which reactant is limiting by comparing the molar amounts of the reactants with their coefficients in the balanced chemical equation.
Use volumes and concentrations of solutions of reactants to calculate the number of moles of reactants.
Zn HCl  ZnCl H2
79.1 g
0.90 L
2.5M
? L
Courtesy Christy Johannesson

73. Limiting Reactants Zn + 2HCl  ZnCl2 + H2 79.1 g ? L 0.90 L 2.5M 79.1
g Zn
1 mol Zn
65.39
g Zn
1 mol H2 Zn
22.4 L H2
1 mol
= 27.1 L H2
Courtesy Christy Johannesson

74. Limiting Reactants Zn + 2HCl  ZnCl2 + H2 79.1 g 0.90 L 2.5M ? L 0.90
2.5 mol
HCl
1 L
1 mol H2
2 mol
HCl
22.4
L H2
1 mol H2
= 25 L H2
Courtesy Christy Johannesson

75. Limiting Reactants Zn: 27.1 L H2 HCl: 25 L H2 Limiting reactant: HCl
Excess reactant: Zn
Product Formed: 25 L H2
left over zinc
Courtesy Christy Johannesson

76. A Hydrocarbon Typical petroleum product Non-polar C18H38 CH2 CH2 CH2

77. Oil and Water Don’t Mix Oil is nonpolar Water is polar
“Like dissolves like”
Lycopodium Powder demonstration.
Lycopodium powder is a nonpolar hydrocarbon that doesn’t mix with water. It is dry to the touch and not oily feeling. Sprinkle lycopodium powder on top of a beaker of water. Slowly, push your finger through the surface of the lycopodium powder into the water. Slowly withdraw your finger, it remains dry the entire time. Why?
Zumdahl, Zumdahl, DeCoste, World of Chemistry 2002, page 470

78. Hydrogenation + H2 shortening stick margarine tub (soft) margarine
vegetable oils
+ H2
FOOD SUBSTITUTES
“Chemistry is widely used in the food industry to manufacture new convenient foodstuffs and to check for impurities. Butter is known as a saturated fat, whereas most vegetable oils, which are normally liquid, are polyunsaturated fats. They can be changed to solids, such as margarine, by a chemical process called hydrogenation – the addition of hydrogen to the oils using a catalyst. Hydrogenation makes the oil more like butter and easier to spread. Margarine was first developed in the 1860’s in France as a cheap butter substitute.”
Eyewitness Science “Chemistry” , Dr. Ann Newmark, DK Publishing, Inc., 1993, pg 7
heat, nickel
catalyst
Timberlake, Chemistry 7th Edition, page 570

79. Molecular Polarity H nonpolar molecules: -- e– are shared equally
H–C–H H
nonpolar molecules: -- e– are shared equally
-- tend to be symmetric
e.g., fats and oils H O
polar molecules: -- e– NOT shared equally
e.g., water
“Like dissolves like.”
polar + polar = solution
nonpolar + nonpolar = solution
polar + nonpolar = suspension (won’t mix evenly)

80. Using Solubility Principles
Chemicals used by body obey solubility principles.
-- water-soluble vitamins: e.g., vit. C
-- fat-soluble vitamins: e.g., vits. A, D
Dry cleaning employs nonpolar liquids.
-- polar liquids damage wool, silk
-- also, dry clean for stubborn stains (ink, rust, grease)
-- (tetra) perchloroethylene is in common use
C=C Cl

81. MODEL OF A SOAP MOLECULE
emulsifying agent (emulsifier):
-- molecules w/both a polar AND a nonpolar end
-- allows polar and nonpolar substances to mix
e.g., soap
detergent
lecithin
eggs
MODEL OF A SOAP MOLECULE
Na1+
NONPOLAR
HYDROCARBON
TAIL
POLAR
HEAD

82. glyceryl tripalmitate
Saponification
Process of making soap from animal fat or vegetable oil using a base.
CH2 – O – C – (CH2)14CH3
CH – O – C – (CH2)14CH3 O
CH2 – OH
CH – OH O
3 Na+ -OC – (CH2)14CH3 +
3 NaOH +
SOAP MAKING
Saponification is the process of making soap from fat. The chemical reaction for this is shown in the slide.
3 C19H37O NaOH > C18H35O21- Na C3H8O3
glyceryl tristearate (fat) sodium hydroxide sodium stearate (soap) glycerin
Read Chem Matters “Soap Making”
glyceryl tripalmitate
(tripalmitin)
sodium
hydroxide
glycerol
3 sodium palmitate
(soap)

83. A Phospholipid (a) chemical structure of a phospholipid
polar head
nonpolar tails
(a) chemical structure of a phospholipid
(b) simplified way to draw a phospholipid
Phospholipids
– are a large class of biological molecules that consist of detergent-like molecules that contain a hydrophilic head and two hydrophobic tails; additional tail results in a cylindrical shape that prevents phospholipids from forming a spherical micelle
– form bilayers, extended sheets consisting of a double layer of molecules; the hydrophobic tails are in the center of the bilayer, where they are not in contact with water, and the hydrophilic heads are on the two surfaces, in contact with the surrounding aqueous solution
• Cells
– are collections of molecules that are surrounded by a phospholipid bilayer called a cell membrane and are able to reproduce themselves
Timberlake, Chemistry 7th Edition, page 576

84. A Model of a Cell Membrane
Polar
Nonpolar
Cholesterol
Proteins
Phospholipid
bilayer
Timberlake, Chemistry 7th Edition, page 587

85. Cleaning Action of Soap
Micelle
Timberlake, Chemistry 7th Edition, page 573

86. SOAP vs. DETERGENT
-- made from animal and -- made from petroleum
vegetable fats -- works better in hard water
Hard water contains minerals w/ions like Ca2+, Mg2+, and Fe3+ that
replace Na1+ at polar end of soap molecule. Soap is changed into
an insoluble precipitate (i.e., soap scum).
micelle: a liquid droplet covered
w/soap or detergent molecules

87. Solvation
“Like Dissolves Like”
NONPOLAR
POLAR

88. Solvation Soap / Detergent polar “head” with long nonpolar “tail”
dissolves nonpolar grease in polar water
micelle

89. Lava Lamp It is… a philosophy the primordial ooze once ruled our world
Polar mixture H
H-O-C-C-O-C-C-O-C-C-O-C-C-O-H
Water
the primordial ooze
Polyethylene glycol
once ruled our world
the moment
Nonpolar mixture Cl H
H-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-H
an art form
Chlorinated paraffin
Paraffin from kerosene
a classic
Heat transfer coil
progressive H
H-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-H
prehistoric
Bulb gives heat and light
post-modern
here to stay

90. Dialysis A semi-permeable membrane allows small particles to pass
through while blocking larger
particles.
Dialysis is used to clean blood
when people suffer kidney
failure.
graphic (upper right corner)
(woman receiving dialysis)
Dialysis uses a semipermeable membrane with pores that allow only small solute molecules and solvent molecules to pass through but not large molecules.
The same principle used to preserve fruits and a similar process using salt prevents bacteria from growing.
A process called reverse osmosis can be used to produce pure water from seawater.

91. Solution, Suspension, Colloid
The main difference between a solution, colloid and suspension is in PARTICLE SIZE. With a solution, the particles are very small (too small to be seen with the naked eye) and are dissolved between the molecules of the solvent (in the interstitial spaces). With a colloid, the particles are still small enough to fit in the interstitial spaces, but are large enough to be seen with the naked eye. Both a solution and a colloid are homogeneous mixtures. In a colloid, the particles are not dissolved in the solution, they are too large to fit in the interstitial spaces. Instead, they are stirred up and suspended in the solvent for short periods of time. Suspensions settle out when allowed to stand undisturbed.
Two common types of mixtures whose properties are intermediate between those of true solutions and heterogeneous mixtures are:
1. Suspension — a heterogeneous mixture of particles with diameters of about 1 m (1000 nm) that are distributed throughout a second phase, if allowed to stand, the two phases will separate
2. Colloid — a heterogeneous mixture with particles smaller than those of a suspension (2 – 500 nm in diameter) that do not separate into two phases on standing
Mixture of gases is the only combination of substances that cannot produce a suspension or colloid because their particles are small and form true solutions
Graham characterized colloids in 1860 and found that they diffuse very slowly or not at all.
• A colloid can be classified as
1. a sol or gel, a dispersion of solid particles in a liquid or solid in which all the solvent has been absorbed by the solid particles, thus preventing the mixture from flowing readily;
2. An aerosol, a dispersion of solid or liquid particles in a gas;
3. An emulsion, a dispersion of one liquid phase in another liquid with which it is immiscible.
• Colloids share many properties with solutions — the particles in both are invisible without a microscope, do not settle on standing, and pass through most filters.
• A colloid can be distinguished from a true solution by its ability to scatter a beam of light, known as the Tyndall effect.
Colloids and suspensions can have particles similar in size, but the two differ in stability.
– The particles of a colloid remain dispersed indefinitely unless the temperature or chemical composition of the dispersing medium is changed.
– Chemical explanation for the stability of colloids depends on whether the colloidal particles are hydrophilic or hydrophobic.
– Hydrophilic colloids contain an outer shell of groups that interact favorably with water, whereas hydrophobic colloids have an outer surface with little affinity for water.
– A stable colloid can be transformed to an aggregated suspension by a minor chemical modification.
– Aggregation and precipitation can result when the outer, charged layer of a particle is neutralized by ions with the opposite charge.
Emulsions are colloids formed by the dispersion of a hydrophobic liquid in water, thereby bringing two mutually insoluble liquids in close contact.
• Various agents have been developed to stabilize emulsions, the most successful being molecules that combine a relatively long hydrophobic “tail” with a hydrophilic “head.”
• Examples of emulsifying agents include soaps and detergents.
Timberlake, Chemistry 7th Edition, page 309

92. SUPERSATURATED SOLUTION
Solubility
UNSATURATED SOLUTION
more solute dissolves
SATURATED SOLUTION
no more solute dissolves
SUPERSATURATED SOLUTION
becomes unstable, crystals form
Maximum amount of a solute that can dissolve in a solvent at a specified temperature and pressure is its solubility.
– Solubility is expressed as the mass of solute per volume (g/L) or mass of solute per mass of solvent (g/g) or as the moles of solute per volume (mol/L).
– Solubility of a substance depends on energetic factors and on the temperature and, for gases, the pressure.
• A solution that contains the maximum possible amount of solute is saturated.
• If a solution contains less than the maximum amount of solute, it is unsaturated.
When a solution is saturated and excess solute is present, the rate of dissolution is equal to the rate of crystallization.
• Solubility increases with increasing temperature — a saturated solution that was prepared at a higher temperature contains more dissolved solute than it would contain at a lower temperature, when the solution is cooled, it can become supersaturated.
increasing concentration

93. of solubility on temperature
Solubility vs. Temperature for Solids
140 KI
130
120
gases
solids
NaNO3
110
Solubility Table
100
KNO3 90 80
HCl
NH4Cl
shows the dependence
of solubility on temperature 70
Solubility (grams of solute/100 g H2O) 60
NH3
KCl 50
“Solubility Curves for Selected Solutes”
Description: This slide is a graph of solubility curves for 10 solutes. It shows the number of grams of solute that will dissolve in 100 grams of water over a temperature range of 0cC to 10 cC.
Basic Concepts
The maximum amount of solute that will dissolve at a given temperature in 100 grams of water is given by the solubility curve for that substance.
When the temperature of a saturated solution decreases, a precipitate forms.
Most solids become more soluble in water as temperature increases, whereas gases become less soluble as temperature increases.
Teaching Suggestions
Use this slide to teach students how to use solubility curves to determine the solubilities of various substances at different temperatures. Direct their attention to the dashed lines; these can be used to find the solubility of KClO3 at 50 cC (about 21 g per 100 g of H2O). Make sure students understand that a point on a solubility curve represents the maximum quantity of a particular solute that can be dissolved in a specified quantity of solvent or solution at a particular temperature. Point out that the solubility curve for a particular solute does not depend on whether other solutes also are present in the solution (unless there is a common-ion effect; this subject usually is covered at a later stage in a chemistry course).
Questions
Determine the solubilities (in water) of the following substance at the indicated temperatures: NH3 at 50 oC; KCl at 90 oC; and NaNO3 at 0 oC.
Which of the substances shown on the graph is most soluble in water at 20 oC? Which is lease soluble at that temperature? For which substance is the solubility lease affected by changes in temperature?
Why do you think solubilities are only shown between 0 oC and 100 oC?
In a flask, you heat a mixture of 120 grams of KClO3 and 300 grams of water until all of the KClO3 has just been dissolved. At what temperature does this occur? You then allow the flask to cool. When you examine it later, the temperature is 64 oC and you notice a white powder in the solution. What has happened? What is the mass of the white powder?
Compare the solubility curves for the gases HCl, NH3, and SO2) with the solubility curves for the solid solutes. What generalizations(s) can you make about the relationship between solubility and temperature?
According to an article in an engineering journal, there is a salt whose solubility in water increases as the water temperature increases from 0 oC to 65 oC. The salt’s solubility then decreases at temperatures above 65 oC, the article states. In your opinion, is such a salt likely to exist? Explain your answer. What could you do to verify the claims of the article? 40 30
NaCl
KClO3 20 10
SO2
LeMay Jr, Beall, Robblee, Brower, Chemistry Connections to Our Changing World , 1996, page 517

94. Solubility Solubility maximum grams of solute that will dissolve
in 100 g of solvent at a given temperature
varies with temperature
based on a saturated solution

95. Boiling Points of Covalent Hydrides3 He 2 C 6 N 7 O 8 F 9 Ne 10 Na 11 B 5 Be 4 H 1 Al 13 Si 14 P 15 S 16 Cl 17 Ar 18 K 19 Ca 20 Sc 21 Ti 22 V 23 Cr 24 Mn 25 Fe 26 Co 27 Ni 28 Cu 29 Zn 30 Ga 31 Ge 32 As 33 Se 34 Br 35 Kr 36 Rb 37 Sr 38 Y 39 Zr 40 Nb 41 Mo 42 Tc 43 Ru 44 Rh 45 Pd 46 Ag 47 Cd 48 In 49 Sn 50 Sb 51 Te 52 I 53 Xe 54 Cs 55 Ba 56 Hf 72 Ta 73 W 74 Re 75 Os 76 Ir 77 Pt 78 Au 79 Hg 80 Tl 81 Pb 82 Bi 83 Po 84 At 85 Rn 86 Fr 87 Ra 88 Rf
104 Db
105 Sg
106 Bh
107 Hs
108 Mt
109 Mg 12 *
Boiling Points of Covalent Hydrides
H2O
100
Hydrogen Bonding
H2Te
Boiling point (oC)
H2Se
H2S
-100
-200 2 3 4 5
Period of X (H2X)
Zumdahl, Zumdahl, DeCoste, World of Chemistry 2002, page 443

96. Boiling Points of Covalent Hydrides
H2O
100
H2Te
Boiling point (oC)
H2Se
SnH4
H2S
-100
GeH4
SiH4
CH4
-200 50
100
150
Molecular mass

97. Group16 Hydrogen Compounds
Molar Mass
Melting Point (oC)
Boiling Point (oC)
H fusion
(cal/mol)
H vapor
H2O 18
0.0
100.0
1440
9720
H2S 34
-85.5
-60.7
568
4450
H2Se 81
-60.4
-41.5
899
4620
H2Te
130
-48.9
-2.2
1670
5570
Group 16 Hydrogen Compounds
Compound Molar Mass Melting Point Boiling Point H fusion H vapor
(oC) (oC) (cal/mol) (cal/mol)
H2O
H2S
H2Se
H2Te

98. AgNO3(aq) + KCl(aq)  KNO3 (aq) + AgCl(s)?
Precipitation Reaction Between
AgNO3 and KCl K+
NO3-
AgCl
AgNO3(aq) + KCl(aq)
unknown white solid
Ag+ + NO K+ + Cl-
in silver
nitrate
solution
in potassium
chloride
unknown white solid
Ag+
Cl-
Ag+ + NO K+ + Cl-
Ag+ + Cl K+ + NO3-
product
AgCl precipitate
AgNO3(aq) + KCl(aq) AgCl(s) KNO3(aq)
AgNO3(aq) + KCl(aq)  KNO3 (aq) + AgCl(s)

99. Clogged Pipes – Hard Water
Step 1: Acid rain is formed
H2O CO H2CO3
carbonic acid
Step 2: Acid rain dissolves limestone
H2CO CaCO Ca(HCO3)2
Water softener
H2CO MgCO Mg(HCO3)2
limestone ‘hard’ water

100. Pipes develop Scales Ca(HCO3)2 CaCO3(s) + H2O + CO2
Step 3: Hard water is heated and deposits scales
Ca(HCO3) CaCO3(s) H2O CO2
Mg(HCO3) MgCO3(s) H2O CO2 D
scales on pipes D
1 gram = 1/7000th pound
300 gallons / day with 10 grain water supply

101. No More Hard Water Scale
No More Ugly Hard Water Spotting
Protects Plumbing and Appliances
Saves Money on Cleaning Products

102. Water Purification Cation Exchanger Anion Exchanger Hard Water
Deionized
Water H+
OH- H+
OH- H+
Mg2+
OH-
Na+ H+
(a)
(b)
OH-
(c)
OH-
Fe3+
Ca2+ H+ H+
OH- H+
In some areas of the country, the minerals dissolved in water give it a high concentration of various ions that make the water suitable neither for drinking nor for agriculture. Such water is called hard water. Sometimes the mineral content in hard water is so great that it causes plumbing and corrosion problems. Hard water typically contains a high concentration of the following ions: Ca2+, Mg2+, Fe3+, Cl1-, CO32-, SO42-, and PO43-. Water is deionized (softened) by removing the minerals using an ion exchange system. Water is passed through a resin that has both cation and anion exchange components. First, cations such as Na+ in the water are exchanged for hydrogen ions on the resin. Second, anions such as Cl- in the water are exchanged for hydroxide ions. The net result is that the resin removes all ions from water passing through the deionizing system.
OH- H+
OH-
Hard water is softened by exchanging Na+ for Ca2+, Mg2+, and Fe3+.
The cations in hard water are exchanged for H+.
The anions in hard water are exchanged for OH-.
The H+ and OH- combine to give H2O.
Corwin, Introductory Chemistry 2005, page 361

103. Which ions are removed from hard water to produce soft water?
Na+(aq) H (resin) Na (resin) H+(aq)
Cl-(aq) (resin) OH (resin) Cl OH-(aq)
Notice that the ion exchange resin
produces both hydrogen ions and hydroxide ions
which can readily combine to give water.
H+(aq) OH- (aq) H2O (aq)
In some areas of the country, the minerals dissolved in water give it a high concentration of various ions that make the water suitable neither for drinking nor for agriculture. Such water is called hard water. Sometimes the mineral content in hard water is so great that it causes plumbing and corrosion problems. Hard water typically contains a high concentration of the following ions: Ca2+, Mg2+, Fe3+, Cl1-, CO32-, SO42-, and PO43-. Water is deionized (softened) by removing the minerals using an ion exchange system. Water is passed through a resin that has both cation and anion exchange components. First, cations such as Na+ in the water are exchanged for hydrogen ions on the resin. Second, anions such as Cl- in the water are exchanged for hydroxide ions. The net result is that the resin removes all ions from water passing through the deionizing system.
The net result is that the resin removes all ions from water passing through the deionizing system.
Hard water is softened by exchanging Na+ for Ca2+, Mg2+, and Fe3+.
Corwin, Introductory Chemistry 2005, page 361

104. Reverse Osmosis semipermeable membrane pressure fresh salt water water
A process by which a solvent such as water is purified of solutes by being forced through a semipermeable membrane through which the solvent, but not the solutes, may pass.
fresh
water
salt
water