1
6.4
6.5
6.6
6.7
Lesson Topic
6.1 Properties of Water and Intermolecular Forces / Lab: Don’t Flip Your Lid
6.2
6.3
Introduction to Solutions; Factors Affecting Solubility and Rates of Dissolution
Lab: Sugar Dissolving Race
7
Developing and Using Solubility Rules; Writing Equations using Solubility Rules 9
Page
2
Lab: Eight Solution Problem
Types of Solutions/ Lab: Unsaturated, Saturated, Supersaturated Solutions
Molarity and Dilution
Lab: Preparing Solutions
11
14
6.8
6.9
6.10
6.11
Introduction to Acids, Bases and pH; Acid-Base Theories; Acid-Base Strength pH and pOH Calculations for Strong Acids and Bases
Neutralization and Solution Stoichiometry
Lab: % Acetic Acid in Mustard
19
22
25
6.12: Unit 6 Exam
Part I Objectives:
Properties of Water

Describe the structure of a water molecule

Explain how the physical properties of water are determined by the structure of water.

Relate the structure and properties of water to its unique role in chemical and biological systems.
Intermolecular Forces (IMFs)

Describe and compare hydrogen bonding, London Dispersion Forces, and Dipole-Dipole Forces

Identify the type of IMF based on the substance (solid, liquid, gas, or solution)

Relate the IMF type and strength to the physical properties of the substance (solid, liquid, gas, or solution)
Solutions

Investigate, describe, and explain how physical and chemical factors affect solubility and rates of dissolution for solid and gaseous solutes

Describe the dissociation of ionic compounds and the ionization of some molecular compounds when they dissolve in water.

Distinguish solutions from suspensions and colloids.

Define electrolyte, and distinguish between strong and weak electrolytes

Distinguish solutions based on electrolytic properties and on degree of saturation.

Interpret and draw conclusions based on descriptions, observations, and solubility curves.

Define and perform molarity and percent by volume calculations, and dilution calculations.

Develop and use solubility rules to determine if an ionic compound is soluble or insoluble in water.

Use a solubility chart to determine if a solution is unsaturated, saturated, or supersaturated.

Write molecular and net ionic equations based on correct application of the solubility rules.

2
Surface tension
Polarizability
Intramolecular force
Intermolecular Force van der Waals forces
Hydrogen Bonding
London Dispersion Forces
Dipole-Dipole Forces
Van der Waals Forces
Vapor pressure
Soluble
Insoluble
Miscible
Immiscible
Hydrophobic / Hydrophillic
Aqueous (solution)
Solution
Solute
Solvent
(Rate of) Dissolution
Solvation
Hydration
Agitation
Tyndall effect
Suspension
Colloid
Electrolyte
Nonelectrolyte
Strong electrolyte
Weak electrolyte
Henry’s Law
Agitation
Saturated
Unsaturated
Supersaturated
Crystallization
Seed Crystal
Concentration
Molarity
Dilution
Standard solution
Dissociation
Ionization
Molecular equation
Net ionic equation
Precipitate / Precipitation
Spectator ion
A.
Properties of Water
W ATER EXISTS IN THREE FORMS
Water can exist on our planet in three physical states.
Water can be a liquid (water), a gas (clouds), or a solid
(ice). If the ambient conditions of Earth were much cooler
(or at higher pressure), we would all be frozen.
Alternatively, if the Earth was hotter than Hades, we would be bathed in perpetual clouds. (like Venus).
High Specific Heat:
CE
LOATS
The solid state of most things are much denser than the liquid state and therefore sink. Usually what happens when a solid is formed is that the molecules become more tightly packed together. But water is weird - the solid state is less dense than the liquid.
The tetrahedral bent molecular geometry of water
(as ice) allows it to form a hexagonal arrangement with other water molecules, increasing the volume and decreasing the density of ice.

3
ATER AS A
OLVENT
The partial charge that develops across the water molecule helps make it an excellent solvent
Water’s polarity is due to its shape. It is essential to life, because most biological molecules (nutrients, waste, gases, etc) are at least partly soluble in water. Water dissolves by surrounding charged particles and 'pulling' them into solution. Any polar substance carries a net electrical charge, including ionic compounds and polar covalent molecules. Polar molecules can thus dissolve in water. This idea also explains why some substances do NOT dissolve in water.
Oil, for example, is a non-polar molecule.
Because there is no net electrical charge across an oil molecule, it is not attracted to water molecules and does not dissolve in water.
URFACE
ENSION
Hydrogen bonding causes neighboring water molecules to be attracted to one another. Molecules at the surface of liquid water have fewer neighbors and, as a result, have a greater attraction to the few water molecules that are nearby. It makes the surface of the liquid slightly more difficult to break.
When a small object is placed carefully on the surface, it can remain suspended on the surface due to surface tension.
Surface tension is related to the cohesive properties of water.
APILLARY
CTION
Surface tension is related to the cohesive properties of water. Capillary
action is related to the adhesive properties of water. You can see capillary action 'in action' by using a drinking straw. The water molecules are attracted to the straw molecules. When one water molecule moves closer to the straw molecules, the other water molecules (which are cohesively attracted to that water molecule) also move up into the straw. Plants take advantage of capillary action to pull water from the roots into and up their stems.
Review: Use the reading and your notes to answer the following questions.
1.
Draw two water molecules. Label each atom, each atom’s partial charge, bond angle, and molecular shape. Label the hydrogen bonds between them.
2.
How does the structure and hydrogen bonding of water affect its properties?
3.
Explain water’s unique role in biological systems.
4.
Explain water’s unique role in chemical systems, such as in climate regulation.

4
B.
Comparing Intermolecular Forces
IntrAmolecular forces are the bonds (ionic or covalent) that hold atoms together within a molecule. These forces arise from unequal distributions of the electrons in the molecule and the electrostatic attraction between oppositely charged portions of molecules. The forces that hold one molecule to another molecule are referred to as intermolecular forces (IMFs). Another term for IMFs is van der Waals forces.
Understanding intermolecular forces is important because they determine the properties of a substance.
Substances, in general, can be classified as “molecular” or “ionic”. The strength and predominant type of force present determines properties such as state at room temperature, hardness, solubility in water, melting point, boiling point, and conductivity. We will start first with molecular compounds; compounds consisting of molecules, held together by covalent bonds.
Let us assume that the molecule involved is nonpolar. A good example would be O
2
. Pretend that the molecule is alone in the universe. If that were the case, the electrons in the molecule would be perfectly symmetrical. However, the molecule is not really alone. It is surrounded by other molecules that are constantly colliding with it. When these collisions occur, the electron cloud around the molecule is distorted.
This produces a momentary induced dipole within the molecule. The amount of distortion of the electron could is referred to as polarizability. Since the molecule now has a positive side and a negative side, it can be attracted to the other molecules. This attractive force is called a London dispersion force. Since all molecules have electrons, all molecules have London forces. These forces range from 5 – 40 kJ/mol. An example is the force between two methane (CH
4
) molecules. These compounds tend to be soft, waxy, insoluble in water, and have low melting/boiling points. They also tend to be gases or liquids at room temperature.
Some molecules are naturally polar; therefore, in addition to dispersion forces, they can also have a permanent dipole which attracts other polar molecules
(either induced or permanent). This is called a dipole-dipole force. An example is the force between two hydrogen chloride molecules. On the scale of things, dipole-dipole force is stronger than London dispersion force; that is, it takes more energy per mole to disrupt dipole forces than it does to disrupt
London dispersion force. The properties of molecular substances held by dipole-dipole forces change, compared to those held by London dispersion force. Melting and boiling points are slightly higher as more energy per mole is required to interrupt the forces between these molecules, and their solubility in water is higher since they can now form attractive forces with the polar water molecules. Conductivity is increased since each end of the molecule has a permanent charge.

5
A third type of intermolecular force is hydrogen bonding. When hydrogen is covalently bonded to a small electronegative atom like nitrogen, oxygen or fluorine, the electron cloud on the hydrogen is very distorted and pulled toward the electronegative atom. Since hydrogen has no inner core electrons, the positive nuclear charge is somewhat exposed. This sets up the potential for a reasonably strong attraction between this hydrogen and an electronegative atom in another molecule. Hydrogen bonds are significant in determining such factors as the high boiling point of water, solubility of acetone in water, and the shape and structure of proteins and DNA.
Ionic bonds are formed between charged particles. The smallest unit of an ionic bond is a formula unit, and formula units are held together in a crystal by electrostatic attraction. These electrostatic attractions are strong and are omnidirectional. Although you have been taught that ionic bonds are intramolecular forces (that is, they are chemical bonds), we include ionic bonds here in a discussion of IMFs because they are the major factor that determines the strength and type of properties exerted by ionic compounds. An ionic bond is extremely strong and consists of ions, so expect ionic compounds to have high melting and boiling points (when they are liquid), to conduct electricity when molten or dissolved in water
(but not solid), and to dissolve in water but not in nonpolar substances.
So, basically:
Molecular compounds Ionic compounds
Each molecule is held All formula units are held
by covalent bonds by ionic bonds
Molecules held Molecules held to Molecules held to each to each other by each other by other by
London dispersion forces dipole-dipole forces hydrogen bonds
Lowest BP/MP Higher BP/MP Highest BP/MP
No/lowest conductivity Highest conductivity Conductive when molten
Soft, waxy, slippery Soluble in water (or dissolved)
Gases or liquids at room temp Hard and brittle
Insoluble in water Solid at room temp
As an example of how properties are determined by IMF type and strength, let’s consider vapor pressure as compared between different molecular compounds in the liquid state. Vapor pressure is the pressure due to particles of a substance in the vapor phase above its liquid in a closed container at a given temperature. The weaker the forces holding the liquid together, the higher the vapor pressure of the liquid will be. So, it should make sense that liquids held by London dispersion forces will vaporize more easily, so their vapor pressure will be higher. (We call liquids with high vapor pressure “volatile”; think gasoline.)

800
6
(C
2
H
5
)
2
O C
2
H
5
OH H
2
O
600
400
200
0
0 20 40 60 80
Temperature (ÞC)
100 120
Boiling point is the temperature at which the vapor pressure of a liquid equals the atmospheric pressure. The normal boiling point is the temperature at which the vapor pressure of the liquid equals 1 atmosphere. Liquids held by London dispersion forces have the highest vapor pressures, because they have little or few attraction existing between its molecules. So it does not take as much energy to cause that liquid to boil; there are already quite a few molecules existing as vapor above the liquid, so its boiling point is lower than that of a substance with a lower vapor pressure. Water has a relatively high boiling point because its vapor pressure is lower; water molecules have high attraction for each other, and do not vaporize as readily.
The molar heat of vaporization (ΔH vap
) is the amount of heat required to change one mole of a pure liquid into one mole of a pure gas. The molar heat of fusion (ΔH fus
) is the amount of heat required to change one mole of a pure substance from a solid to a liquid.
Review:
1.
Identify the predominant IMF in samples of each of the following pure substances:
a) HF (l) b) H
2
S (l) c) CH
2
Cl
2
(l) d) Na
2
CO
3
(s)
2.
Compare a), b) and c) from question 1. Which compound should have the highest boiling point? Explain in terms of IMF type, strength, and the energy required.
3.
What generalization can be made between IMF strength and type and the heat of vaporization or heat of fusion? Explain in terms of IMF type, strength, and the energy required.

7
A.
Introduction to Solutions
Solutions are…
 homogenous mixtures of two or more substances (solutes) dissolved in a single phase. The solvent is the part of the solution that does the dissolving, and has the highest concentration in the solution.
 gases, liquids or solids (think: fog, shaving cream, cheese, soda, wood)
When a liquid solute successfully dissolves in a solvent, they are miscible with each other (they “mix”)
When a liquid solute doesn’t dissolve in a solvent, they are immiscible. (they “don’t mix”)
Most solutions we will deal with are those in a liquid state, where the solvent is H
2
O (i.e. aqueous solutions)
Solvents can be hydrophobic- water fearing (nonpolar) or hydrophilic - water loving (polar)
Dissolution is a general term for dissolving. Solvation is when solute-solvent interactions are strong enough to separate, surround and disperse a solute. If the solvent is H
2
O, then solvation is referred to as hydration. The energy changes associated with solvation is known as the enthalpy of solvation (ΔH solv
).
Review: Draw the hydration of a sodium chloride formula unit by several water molecules. Then, explain how the structure and properties of water allows for the easy dissolution of sodium chloride in water.
The solubility of a substance is the maximum mass of material that will dissolve in a given amount of solvent at a given temperature to produce a stable solution. Various factors can increase or decrease the solubility of a solute in a solvent.
B.
Factors Affecting Solubility and Rates of Dissolution a) Solubility and Nature of Solute and Solvent
In general… “Like dissolves like”


Polar or ionic solutes dissolve in polar solvents. Examples of polar solvents: water, alcohols, any molecule containing –OH groups if the hydrocarbon chain attached to it is long)
Non-polar solutes dissolve in non-polar solvents Examples of nonpolar solvents: hydrocarbons, especially those with long carbon chains; fats and oils; gasoline) b) Solubility and Pressure for Gaseous Solutes
 The solubility of a gas is higher with increased pressure. Pressure has very little effect on the solubility of
liquids and solids. (Carbonated beverages must be bottled at high pressures to ensure a high
concentration of carbon dioxide in the liquid.)

8
Henry’s Law- the amount of a gas dissolved in a solution is directly proportional to the pressure of the gas above the solution.
P= kC
P = partial pressure of the gaseous solute above the solution
k= constant (depends on the solution)
C=concentration of dissolved gas
Henry’s Law is obeyed best for dilute solutions of gases that don’t dissociate or react with the solvent.
 Why does a bottle of soda “fizz up” when you pour the soda out? There is less pressure outside
the bottle than inside; the solubility of the carbon dioxide decreases when you pour the soda into the
glass; thus, the carbon dioxide moves out of solution and into the atmosphere as a gas.
 Pressure does not affect the solubility of a solid or liquid solute. Pressure affects solubility of gases
because there are large spaces between each molecule; gases are compressible, so increasing the
pressure of a gas exerts huge consequences on its properties – including solubility. Solids and liquids are
not, since the space between each molecule is negligible. c) Solubility and Temperature:
 The amount of solid or liquid solute that will dissolve usually increases with increasing temperature.
An increased temperature means more kinetic energy is available to help separate ions from each other and overcome their attractive forces, and dissolve; thus, solubility of a solid usually increases (more solid solute dissolves) when the solvent is warmer.
This is always true for solutes that consume energy as they dissolve – i.e. the heat of solution is endothermic. This is obviously much easier if the solute and solvent have similar (polar-polar/ionic or nonpolar-nonpolar) intermolecular forces. IF the solute and solvent have dissimilar IMFs, it will take too much energy to form a solution, so no dissolving takes place.
Some solid solutes decrease in solubility as temperature is increased (i.e. ammonium nitrate). This is
because these solutes, when they dissolve in water, give off heat during the dissolving process. Adding
more heat, i.e. heating up the solvent, while this is going on, forces the formation of more solid,
not the dissolving of the original mass of solid.
 The solubility of a gas in water always decreases with increasing temperature.
Most gases are nonpolar. Their intermolecular forces are weak, and don’t form attractive forces well with the solvent (usually water), so the kinetic energy of the gas is sizeable enough that they escape the solvent.
The greater the temperature, the greater their kinetic energy, so higher temperatures mean the gas escapes and thus its solubility decreases.
A real-life consequence of this is thermal pollution – water being returned to its natural source at a higher
ambient temperature kills aquatic wildlife, as less oxygen can dissolve in warmer water.

9
d) Rates of Dissolution and Surface Area
 Increasing surface area of a solid solute increases the rate of dissolution.
This is accomplished by crushing the solute. A larger surface area allows more solvent molecules to
surround the solute, allowing more intermolecular forces to form per unit of time. Note: this only
speeds up the rate of dissolution, not the solubility. 5 g of sugar in 1.00 L of water will dissolve faster
when the sugar is fine (obviously) but 5 g of sugar cubes in 1.00 L of water will still dissolve completely –
it will just take much longer.
e) Rates of Dissolution and Agitation
 Increasing agitation increases the rate of dissolution.
This is accomplished by vigorous stirring. Stirring will bring fresh solvent molecules into contact with
solute so they can form more intermolecular forces with each other per unit of time. It will also help
speed up the movement of both solute and solvent particles. Again, this affects the rate of dissolution,
not the solubility.
Recall: Electrolytes are substances that form ions in aqueous solutions (dissolved in water). Positive and negative ions carry current (conduct electricity) in an aqueous solution. Usually these are soluble ionic salts, strong acids (completely dissociate), and strong bases (completely dissociate). Substances that will not dissociate in solution are insoluble and nonelectrolytes (cannot conduct electricity). The following table is a guide to whether a solution is soluble or not.
1. Most alkali metal salts AND NH
4
+ salts ARE soluble
2. Cl , Br , I are soluble, *except for Ag + , Hg
2
+2 , Pb +2
3. F are soluble, *except for IIA metals
4. NO
3
, ClO
3
, ClO
4
, and CH
3
COO are soluble
5. SO
4
-2 are soluble, except for Ca 2+ , Sr +2 , Ba +2 , Ag + , Pb +2 , Hg
2
2+
6. CO
3
-2 , PO
4
-3 , C
2
O
4
-2 , CrO
4
-2 , S -2 , OH , and O -2 are INSOLUBLE (rule 1 takes priority!)*
Heavy metal
BAD GUYS!
It can be assumed that ionic cmpds. that dissolve in water are strong electrolytes and are therefore soluble.
*hydroxides of Ca 2+ , Sr +2 , Ba +2 are soluble
Review: Identify the following compounds as either electrolytes or nonelectrolytes. If the solution is electrolytic, write a dissociation equation to show how the ions form. Assume the solvent is water.
1) LiCl
2) (NH
4
)
2
S
3) Ammonium chlorate
4) Ca(OH)
2
5) Mg(OH)
2

10
6) Silver chloride
Precipitates: These are insoluble solids that emerge from an aqueous solution. The emergence of the insoluble solid from solution is called precipitation.
Precipitates can form when two soluble salts react in solution to form one or more insoluble products. The insoluble product separates from the liquid and is called a precipitate. Precipitates can also form when the temperature of a solution is lowered. The lower temperature lower temperature reduces the solubility of the salt resulting in the formation of a precipitate. This process is called crystallization.
Any ions that do not participate in the formation of precipitate are called spectator Ions.
Writing Equations Using Solubility Rules
Typically, these are double replacement/displacement reaction types that, if a reaction occurs, will form one solid precipitate. Net ionic equations are used to show which chemical species are actively reacting and eliminates spectator ions.
Below is an example of how these are done.
EXAMPLE: KCl
(aq)
+ Pb(NO
3
)
2(aq)

1. a. Take only one of the first cation(s) and match it with one of the second anion(s). (Write the cation first) b. Take only one of the second cation(s) and match it with one of the first anion(s). (Write the cation first)
KCl
(aq)
+ Pb(NO
3
)
2(aq)
 KNO
3
+PbCl
2. Correct the formulas of the products based on the charges of the ions. (Note the subscript change on
Cl)
KCl
(aq)
+ Pb(NO
3
)
2(aq)
 KNO
3
+PbCl
2
3. Balance the equation.
2 KCl
(aq)
+ Pb(NO
3
)
2(aq)
 2 KNO
3
+PbCl
2
4.
Consult the solubility rules and assign the correct state symbol. This should agree with any observations concerning
the formation of a precipitate which gets the symbol (s). If water is formed, water is a molecule; it does not ionize to any significant extent. It is given (l). Note the state variables on the product side. Why does KNO
3 get (aq) and PbCl
2 get (s).
2 KCl (aq) + Pb(NO
3
)
2
(aq)  2 KNO
3
(aq) + PbCl
2
(s)
5. Write the total ionic equation. All compounds that are aqueous (aq) break up into individual cations and anions
2 K + (aq) + 2Cl (aq) + Pb 2+ (aq) + 2NO
3
(aq)  2 K + (aq) +2 NO
3
(aq) +PbCl
2
(s)
6. Eliminate spectator ions. Spectator ions are in the same form on each side of the equation arrow.
2 K +
(aq)
+ 2Cl -
(aq)
+ Pb 2+
(aq)
+ 2NO
3
-
(aq)
 2 K +
(aq)
+2 NO
3
-
(aq)
+PbCl
2(s)

11
5.
Write the Net Ionic Equation. The convention is to write the cation first followed by the anion on the “reactants”
side. Don’t forget that chemical equations are written using the lowest common coefficients (including net ionic equations). If all ions cancel each other out then it is NR, and no precipitates will form.
Pb 2+
(aq)
+ 2Cl -
(aq)
 +PbCl
2(s)
Review: Write and balance the net ionic equation for the following reactions. All states must be present.
1.. K
3
PO
4
(aq) + Al(NO
3
)
3
(aq)

2. BeI
2
(aq) + Cu
2
SO
4
(aq)

3. Ni(NO
3
)
3
(aq) + KBr (aq)

4. cobalt(III) bromide + potassium carbonate

5. barium nitrate + ammonium phosphate

There are many ways to classify solutions: a) By particle size and ability to scatter light – solution, colloid or suspension
These can be distinguished from each other by passing a bright light (like a flashlight) through a sample.
Depending on the size of the solute particle, light may either pass through the sample straight through or will be scattered by larger solute particles. This effect is called the Tyndall effect.
 Solutions have dissolved solute particles that measure less than 1 nm in diameter. A solution is a
homogeneous mixture; it cannot be filtered to separate the solute from the solvent. You must use
distillation to separate the parts. The particles are invisible to your eye. The particles will not scatter
light. Solutions appear transparent. Example: tap water, clear drinks

12
 Colloids are heterogeneous mixtures with solute particles measuring between 1 – 1000 nm in diameter.
Solute cannot be seen with the naked eye and pass through filter paper, but could be blocked by
other kinds of paper. Colloids appear cloudy but uniform and homogeneous. They show the Tyndall
effect. The solute particles do not settle to the bottom. It appears translucent.
Examples: fog, smoky air, “colloidal” silver
 Suspensions are heterogeneous mixtures with solute particles measuring larger than 1000 nm in
diameter. They appear cloudy and heterogeneous. The solute particles are large enough to be filtered.
Eventually, the solute particles may settle to the bottom of their container. Tyndall effects are variable.
Examples: oil in water, milk of magnesia b) By its ability to conduct electricity – electrolyte or nonelectrolyte, strong or weak electrolyte
 Electrolytes: solutions containing dissolved ions in aqueous solution. Positive and negative ions carry
current (conduct electricity) in an aqueous solution. Usually these are soluble ionic salts, strong acids
(completely dissociate), and strong bases (completely dissociate).
Strong electrolytes contain high ion concentrations and conduct electricity very well. This is detected by
a conductivity tester or by a simple circuit – the lightbulb will glow very brightly. Soluble ionic salts,
strong acids (ex. HCl or H2SO4) and strong bases (ex. NaOH) are all strong electrolytes because close to
100% of the solute dissociates (separates) into ions.
Weak electrolytes contain moderately high ion concentration and conduct electricity well. The reading
on the tester will be lower, and a lightbulb in the circuit will glow brightly. The best examples of weak
electrolytes are weak acids (i.e. HC
2
H
3
O
2
, vinegar) and weak bases (i.e. NH
3
, ammonia). Weak
electrolytes cause the formation of some ions, but far more solute particles remain undissociated.
 Nonelectrolytes: solutions containing undissociated solute in aqueous solution. The solute remains
completely undissociated. Alcohol is an example. Alcohols are polar and are usually miscible, but
they do not dissociate or form ions with water. c) By the mass of solute dissolved in solvent – saturated, unsaturated or supersaturated
 Saturated solution - a solution containing the maximum amount of solute that will dissolve at a specific
temperature. At a microscopic level, the solute particles dissolve and recrystallize at equal
rates.
Saturated solutions will not dissolve any more solute at that temperature. If you were to add more
solute to a saturated solution while keeping the temperature the same, it will sink to the bottom,
undissolved. Saturated solutions often have solid settled at the bottom, but since it is a solution and
not a suspension, the liquid above the solid will appear clear, not cloudy. (If you want to dissolve that
solid, you will have to increase the temperature – which means it will now longer be saturated.)

13
 Unsaturated solution- a solution containing less than the maximum amount of solute that will dissolve
at a specific temperature. Unsaturated solutions will dissolve more solute, but only up to a certain limit.
 Supersaturated solution- a solution that has been prepared at an elevated temperature and then slowly
cooled. It contains more than the usual maximum amount of solution dissolved. A supersaturated
solution is very unstable and the addition of a “seed crystal’ will cause all excess solute to crystallize out
of solution to form a saturated solution (with crystal at the bottom). (Rock candy is made this way.)
One way to distinguish between saturated, supersaturated and unsaturated solutions is by the interpretation of a
solubility curve.
 The solubility of most solids increases (that is, more mass of solid dissolves successfully) as t
temperature increases. This is true if the enthalpy of solvation is positive – that is, if dissolving
in water is an endothermic process. Their solubility curve will increase with temperature.
 Some solids and most gases will have decreasing solubility curves with temperature.
Based on the solubility curve, decide whether each of the following (1-4) is A: unsaturated, B: saturated, C: supersaturated, or whether D: not enough information is given. * assume itʼs dissolved *
1) 50 g KCl in 100 g of water at 90°C. ____ 3) 50 g KNO
3
in 25 g of water at 60°C. ____
2) 50 g KCl in 100 g of water at 60°C. ____ 4) 65 g KNO
3
in 400 g of water at 70°C. ____

14
5) If a saturated sodium nitrate solution at 30 ℃ is evaporated to dryness, how many grams will crystallize out?
6) Which solid salt has the greatest increase in solubility from 20 ℃ to 70 ℃ ?
7) You have a potassium chlorate solution containing 10 g of solute in 100 g of water at 50 ℃ . What mass of solute do
you have to add to make this solution saturated?
8) You have 80 g ammonium chloride in 100 g of water at 80 ℃ . How many grams of solute must crystallize out of the
solution for it to become saturated at the same temperature?
9) You have 100 g ammonium chloride in 100 g of water at 10 ℃ . What temperature do you have to heat this to in order
for the solution to become saturated?
9) How much ammonia gas escapes out as a saturated solution originally held at 40 ℃ is warmed to 90 ℃ ?
10) What are the steps needed to make a saturated solution of potassium chloride in 300 g of water at 50 ℃?
The concentration of a solution is called its molarity. It is, simply, the number of solute particles dissolved (expressed as moles) per liter of solution. The solution includes both the solute and the solvent.
Practically speaking, you can increase the molarity of a solution (or make it more concentrated) by:
 Adding more solute to the solution and dissolving it all – if the temperature is high enough
 Decreasing the volume of the solution – not by pouring out some of the sample (that will do nothing to
the concentration) but by removing some of the solvent. This is best done by evaporating it.
To make a solution less concentrated (or dilute), it is easiest to just add more solvent to the sample in order to increase the volume of the solution.
Relevant formulas:
Percent by Mass %

100
Molarity

** changes w/ temp
Dilution M
1
V
1
= M
2
V
2
# moles solute # moles = M × V in L
Density of solution If given the volume and density of the solution, mass of solute = volume × density
Review: Calculate the molarity of each solution. a) 3.0 mol sugar dissolved in 2.0 L of solution. ___

15 b) 0.030 moles KNO
3
dissolved in 50.0 mL of soln. _____ c) 6.45 g of sodium sulfate dissolved in 250 mL of solution. ___ d) 465 mg potassium fluoride of dissolved in 0.054 L of soln. _____
Review: Calculate the number of moles, mass of solute, or volume of solution. a) How many moles of NaBr are needed to make 150 mL of 3.0 M NaBr solution? Ans: ______ b) How many grams of NaNO
2
are needed to make 3.5 L of 0.50 M NaNO
2
solution? Ans: ______ c) How many grams of K
2
CO
3
are needed to make 300.0 mL of 1.25 M K
2
CO
3
solution? Ans: ______ d) How many mL of 2.50 M Na
3
PO
4
solution can be made using 1.8 g of Na
3
PO
4
? Ans: ______
Review: Determine the concentrations for each of the following mixtures. a) equal volumes of 3.0 M KCl & water: b) equal volumes of 3.0 M KCl & 7.0 M KCl:
Review: Use the dilution equation to determine the concentrations of the following mixtures. a) 45 L of 3.6 M KCl and 71 L of water: b) 215 mL of 2.8 M KCl and 47 mL water:
Ans: ______ Ans: ______

16 c) 45 mL of 3.6 M KCl and 71 mL of 6.2 M KCl: d) 38 mL of 6.0 M KCl diluted to a total volume of 100 mL:
Ans: ______ Ans: ______
To what total volume must 26.0 mL of 4.80 M KCl be diluted to reduce its concentration to... e) ... 2.10 M f) ... 0.480 M
Ans: ______ Ans: ______
What volume of water must be added to 35 mL of 2.6 M KCl to reduce its concentration to... a) ... 1.2 M b) ... 0.26 M
Ans: ______ Ans: ______
What volume of 2.5 M KCl must be added to 37 mL of 6.0 M KCl to make the total concentration of: a) ... 1.5 M b) ... 4.2 M
Ans: ______ Ans: ______
Steps in making a solution 1) calculate and weigh out mass of solute required
(given molarity, calculate # moles solute # moles = M × V in L)
2) add small amount of distilled H
3) add solute to H
2
O in flask
2
O to appropriate volumetric flask
4) swirl/swish to dissolve, or use a glass stirring rod
5) add distilled water to just before the mark/line
6) use dropper to add drops of water up to the mark; add stopper

17
Review: Describe each step of the preparation of a standard solution of 500. mL of 0.150 M KMnO
4
, and include the necessary calculations in detail. (A “standard solution” is a solution that you precisely know the concentration of.)
Diluting a solution 1) calculate V
1
, given M
2
(required molarity), V
2
(required volume), and M
1
(Pipets/burets are accurate, original molarity). V
1
is the volume of original solution you need to use. but take more time to use) 2) measure out V
1
using volumetric pipet, buret or graduated cylinder
3) add V
1
to beaker or volumetric flask that matches V
2
4) add distilled water to beaker or flask containing V
1
to just before the mark
5) use dropper to add drops of water up to the mark; add stopper
** Caution: if the solute is a strong acid, switch steps 3 and 4. Always add acid to water, not in reverse!
Review: What volume of 12M hydrochloric acid must be used to prepare 600 mL of a 0.30 M HCl solution?
Describe the steps and equipment necessary to make this solution.

18
Part II Objectives:
Acids/Bases and their reactions
 explain and identify the essential qualities of acids and bases
 recognize characteristics of Arrhenius and Bronsted-Lowry definitions and differentiate between them based on these characteristics
 predict the products of acid-base reactions that form water. pH/pOH
 state the meaning of and explain pH
 use the hydrogen ion concentration to calculate pH of a solution.
 use the ionization constant of water and the hydroxide ion concentration of a solution to determine the hydrogen ion concentration of the solution.
 use the ionization constant of water and the hydrogen ion concentration of a solution to determine the hydroxide ion concentration of the solution.
 use pH to calculate the hydrogen ion concentration of a solution.
Acid/Base Strength
 recognize that degree of dissociation of a compound is independent of its concentration in solution.
 classify acids as strong or weak based on degree of dissociation
 classify bases as strong or weak based on degree of dissociation
Titrations
 perform titrations and calculations to predict the unknown concentration of a solution
Part II Vocabulary:
Acid / Acidic
Base / Basic/ Alkaline pH
Dissociation
Strong electrolyte
Weak electrolyte
Autoionization of water
Arrhenius Acid/Base
Bronsted-Lowry Acid/Base
Hydronium (H
3
O +
Equilibrium
)
Conjugate Acid/Conjugate Base (Pair)
Degree of dissociation
Neutral/Neutralize
Salt
Ionization
Strong Acid/Base
Weak Acid/Base
Monoprotic
Diprotic
Polyprotic
Amphoteric
K w pOH
Indicator
Phenolphthalein
Bromothymol blue
Equivalence point
Endpoint
Titration
Titrant

19
A.
Introduction to Acids, Bases, and pH
You most likely learned the properties of acids and bases in previous science classes. Acids taste sour, turn red litmus paper blue, react well with metals, and have low pH values; bases taste bitter, feel slippery, turn blue litmus paper red, and have high pH values. “pH”, or “potency of hydrogen”, reflects the concentration of hydrogen ion in solution; the acid or base forms predominantly hydrogen or hydroxide ion in solution as that solute dissociates. A neutral solution has equal concentrations of hydrogen and hydroxide ion in solution. Acids and bases neutralize with each other and form salts (ionic compounds) as a result.
Both acids and bases are electrolytic; strong acids and bases are strong electrolytes, and weak acids and bases are weak
electrolytes.
 Strong acids and bases completely dissociate in water; that is, 100% of the molecules dissociate completely, so a very high concentration of ions results, and the solution conducts electricity strongly.
 Another way to say this is that strong acids and bases have high degrees of dissociation.
 In weak acids and bases, perhaps 1 out of 100 molecules dissociate, and the rest stay together as molecules, so they have low degrees of dissociation. (Acids ionize; bases dissociate.)
The vast majority of acids and bases are weak.

20
Strong Acids
HClO
4
HI
HBr
HCl
HNO
3
H
2
SO
4
Strong Bases perchloric acid hydroiodic acid hydrobromic acid hydrochloric acid nitric acid sulfuric acid
** All group I hydroxides and most group II hydroxides – Ca(OH)
2
, Sr(OH)
2
, Ba(OH)
2
Weak Acids (examples)
CH
3
COOH acetic acid
HCOOH
HF
HCN
HNO
2
HSO
4
formic acid hydrofluoric acid hydrocyanic acid nitrous acid hydrogen sulfate ion
Weak Bases (examples)
NH
3 ammonia
CH
3
NH
2
methylamine
C
5
H
5
N pyridine
Based on their formulas, acids can be defined as:
 Monoprotic: produce 1 H+ ion in solution (example: HCl (aq), HNO
3
(aq))
 or diprotic: produces 2 H+ ions in solution (example: H
2
SO
4
(aq))
 or polyprotic: produces several H+ ions in solution (example: H
3
PO
4
(aq))
 and/or amphoteric: acts as either an acid or a base (example: H
2
O (l) )
Polyprotic acids ionize stepwise; that is, one H + at a time.
Review:
1.
Write an equation for the dissociation (ionization) of the following acids in water: (follow example:) a) HClO
4
---> H + + ClO
4
c) H
2
S (complete ionization) b) HSO
4
d) HC
2
H
3
O
2
2.
Write an equation for the dissociation of the following bases in water: (follow example:) a) Mg(OH)
2
↔ Mg +2 + 2OH c) KOH

b) NaOH
21
d) Ba(OH)
2
There are two main theories in chemistry that attempt to define acid and base behavior. They attempt to explain their behavior in terms of ions that are either released or donated, or their ability to release or donate ions. Here is a summary:
Acid
Arrhenius Theory
Releases a H + ion in water
Example: HCl
Bronsted-Lowry Theory
Donates a H + ion in water to a Bronsted base, forming hydronium ion, H
3
O +
Example: HCl
HCl (aq)  H + (aq) + Cl (aq) HCl (aq) + H
2
O (l)   H
3
O + (aq) + Cl (aq)
Base Releases a OH- ion in water
Example: NaOH
NaOH (aq)  Na + (aq) + OH (aq)
Accepts a H + ion in water from a Bronsted acid, forming hydroxide ion, OH-
Example: NH
3
NH
3
(aq) + H
2
O (l)  NH
4
+ (aq) + OH (aq)
You should notice some interesting things about the two theories:
 An Arrhenius acid (or base) can also be classified as a Bronsted-Lowry acid (or base), BUT not all Bronsted acids
(or bases) are Arrhenius acids (or bases). The classic example is NH
3
. NH
3
does cause hydroxide to form in solution, but it doesn’t contain OH - in its formula.
 Water is amphoteric: it is acting as a Bronsted base (accepting H + ) in the first example and acting as a Bronsted acid (donating H + ) in the second. Generally any species that has more than one H + AND can accept at least one more H + can be amphoteric.
Water can either be amphoteric with itself in a process called the auto-ionization of water.
H
2
O(l) + H
2
O(l)  H
3
O + (aq) + OH - (aq)
When Bronsted acids and bases react, they form conjugate acids and bases:

22
Review:
1.
Using your knowledge of the Brønsted-Lowry theory of acids and bases, write equations for the following acid-base reactions and indicate each conjugate acid-base pair. a) HNO
3
+ OH  b) CH
3
NH
2
+ H
2
O  c) OH + HPO
4
-2 
Recall that water is amphoteric, meaning that it will behave like an acid or base, and will to a slight extent dissociate.
H
2
O(l) + H
2
O(l)  H
3
O + (aq) + OH - (aq)
The concentration of hydronium and hydroxide ions present from the dissociation of pure water at a given temperature always multiply to give you a constant. This constant is given the symbol K w
and is often called the ion product
constant. K w
does not have a given unit.
w
3
+
-
-14
This relationship till holds true even if additional hydronium or hydroxide ion is present from the ionization or dissociation of an acid or base.
If [H
3
O + ] > [OH ], the solution is acidic. If [OH-] > [H
3
O + ], the solution is basic. If they are equal (that is, both are
1.0 x 10 -7 M – the solution is neutral.
Review:
1.
Find [H+] for solutions having the following [OH-] value in molarity:
2. Calculate [OH ] of a solution when
its [H + ] has the following values in molarity: a) [OH-] = 1 x 10 -13 a) [H + ] = 1 x 10 -3 b) [OH-] = 2.7 x 10 - 4 b) [H + ] = 3.6 x 10 -5
c) [H + ] = 1 x 10 -2 c) [OH-] = 1 x 10 - 3 d) [OH-] = 6.3 x 10 -10 d) [H + ] = 7.8 x 10 -8

23
3. Go through each solution described in questions 1 and 2, and state if it is acidic, basic, or neutral.
The pH scale and pOH scale
K w
is the basis for the pH scale. The pH of a solution is calculated as the negative base 10 logarithm of the hydronium ion concentration [H
3
O + ] or [H + ].
+
+
– pH
Since K w
= [H + ] [OH - ] = 1.0 x 10 -14 and pure H
2
O has [H + ] = 1.0 x 10 -7 M. and [OH - ] = 1.0 x 10 -7 M. This is why the concentration of a neutral solution is 7 and the range of the pH scale is 0-14. If [H + ] > [OH - ] the pH is less than 7 and the solution is mostly H + and therefore acidic. If [OH - ] > [H + ] the pH is greater than 7 and the solution is mostly OH - and therefore basic (alkaline)
The pOH of a solution is calculated as the negative base 10 logarithm of the hydroxide ion concentration [OH ]
-
– pOH
Another helpful relationship is:
.
Note that the pH and pOH scales go up by powers of 10 so an increase of 1 unit means the concentration of H + increases tenfold and the concentration of OH- decreases tenfold. This means something with pH of 3 is
10 times more acidic than something that is pH 4, and something with a pH of 7 is 10 times less basic than something with a pH of 8.
***When solving pH and pOH problems if the acid or base is a strong acid or base assume it dissociates 100% and plug the concentration [mol / liter] values into the pH and pOH formula. ***
Review:
1. Calculate the pH and pOH for each of the following solutions, a) – f). All given values have the unit M.
a) [H + ] = 1 x 10 -3
b) [H + ] = 3.6 x 10 -5 c) [OH ] = 2.7 x 10 - 4

24
d) [OH ] = 6.3 x 10 -10
2.
For each of the following pH or pOH values, calculate the corresponding [H+] and [OH-]. a) pH = 3.092 b) pH = 8.319 c) pOH = 6.942 d) pOH = 2.574
3.
Be careful! If you place one drop of hydrochloric acid with a concentration of [H + ] = 1.0 x 10 -2 M into a full barrel of plain water, what will be the resulting pH?
Always write an ionization equation (for acids) or dissociation equation (for bases) if you are not directly given the hydronium or hydroxide ion concentration. Simple stoichiometry will help you figure out the ion concentration you need.
4.
What is the pH of a solution that contains 25 grams of hydrochloric acid (HCl) dissolved in 1.5 liters of water?
5.
What is the pH and pOH of a solution that contains 1.32 grams of sodium hydroxide dissolved in 750 mL of water?
6.
What is the molarity of a calcium hydroxide solution that has a pH of 10.07?

25
7.
What is the pH of a solution that contains 10.0 g of nitric acid (HNO
3
) and 15.0 g moles of hydrochloric acid (HCl) dissolved in 1000 liters of water?
What happens when an acid is mixed with a base? NEUTRALIZATION!
Products of Neutralization: The products of acid-base neutralization are always a metallic salt and H
2
O.
The definition of a metallic salt is a class of compounds formed when the hydrogen ion of an acid is partly or wholly replaced by a metal. In order for neutralization to occur an equal number of moles of acid and base must combine, so that every 1 mole of H + combines with exactly 1 mole OH - to form 1 mole of H
2
O. Likewise, 1 mole of negative anion from the acid combines with 1 mole of positive cation from the base to form 1 mole of metallic salt.
It is a misconception that acid-base neutralization always yields a pH of 7. The resulting pH is 7 only if the number of moles of acid and base are exactly enough to react completely with each other. If there is any excess unreacted acid (or base) left over, it dictates the resulting pH of the solution. Excess unreacted hydronium ion, for example, will make the pH after neutralization be under 7. It is more correct to say that pH after neutralization will be closer to 7.
Examples of neutralizations:
HCl + NaOH  NaCl + H
2
O H
2
SO
4
+ Ca(OH)
2
 CaSO
4
+ 2 H
2
O 2 HNO
3
+ Mg(OH)
2
 Mg(NO
3
)
2
+ 2 H
2
O
Review: Write the balanced neutralization equation for each reaction. Assume that ionization is complete for all acids.
1.
Hydrobromic acid and sodium hydroxide
2.
Hydroiodic acid and calcium hydroxide
3.
Sulfuric acid and rubidium hydroxide
In solution stoichiometry (including acid-base stoichiometry), if you know the molarity and volume of a reactant or product in aqueous solution, you can calculate how many moles of solute are contained in it.
From there you can calculate the # of moles of any other reactant or product in the chemical reaction. If precipitation occurs, you can calculate and predict the mass of precipitate that should be recovered.

26
Example: 150. mL of 0.100 M calcium nitrate solution is mixed with excess sodium hydroxide solution. a) Write a balanced chemical equation for this process. (Hint: use your solubility rules.) b) How many moles of calcium nitrate will react? c) How many moles of precipitate will form? d) What is the mass of the precipitate?
Review: If 250. mL of 0.300 M sodium iodide react with excess lead (II) nitrate, how many grams of precipitate will form?
In acid-base neutralization, which is really just another type of solution stoichiometry, we are most often concerned with knowing about the other reactant as opposed to the yield of some product (although you could calculate that as well, if you really wanted to). In the food industry, which handles acids and bases quite often, a common measure of quality control is to identify the concentration of acid in a product, such as ketchup or vinegar. Ingredient quantities must be consistent, so samples of the same food are tested, over and over, to ensure that acid concentrations are always the same.
For neutralization: moles of acid = moles of base
M a
V a n b
= M b
V b n a
where M = molarity, n = mols of H + or OH - V = volume
Example: What volume of 2.00 M sulfuric acid would be required to neutralize 54.1 mL of 1.40 M calcium hydroxide?

27
Example: Calculate the molarity of 75.0 mL hydrochloric acid that could neutralize 49.1 mL of 1.40 M Ca(OH)
2
.
Review:
1) It was found that 33.8 mL of a triprotic acid was required to neutralize 43.1 mL of 1.10 M NaOH.
What is the molarity of the acid?
2) Industrially produced nitric acid is 8.0 M. What volume of nitric acid is needed to exactly neutralize 37.9 mL of
3.50
M calcium hydroxide spill?
Titration is a widely used quantitative method of analyzing an unknown solution to determine its molarity. Let’s say we want to determine the molarity of an acid.
 A known volume of that acid is placed in an Erlenmayer flask, along with a few drops of a chemical called an
indicator. The indicator changes color when the reaction between the acid and base has completed. This is only needed if the acid does not change color on its own. Two common indicators are: phenolphthalein, which is colorless in acids but turns pink when the solution is basic; and bromothymol blue, which is yellow in acids and blue in bases.
 The point at which the indicator changes color is called the endpoint. The endpoint occurs very close to the
equivalence point, which is the point when the reaction has completed.
 Prepare the buret. (See inset.)
 A known volume and concentration of a base are
placed in a buret. Take note of the initial volume of
liquid in the buret. Read one uncertain digit and read
the bottom of the meniscus. You turn the stopcock to
control the flow of buret solution into the Erlenmayer
flask.
 When the indicator in the flask has just barely changed color, stop adding base. Any hanging drops on the inside walls of the Erlenmayer flask and from the buret tip must be put in the Erlenmayer flask.
Take note of the final volume of liquid in the buret.
Technique: Preparing the Buret for Titration
1.
2.
3.
Rinse a clean buret several times with 5 mL portions of distilled water and then with portions of base.
Allow the titrant to drain through the stopcock so that the tip gets rinsed with titrant as well.
Discard the rinse solution in a waste beaker.
Clamp the buret into place, and fill it with the base. Remove air bubbles from the tip of the buret and the stopcock by draining several milliliters of titrant. Dispose of the drained titrant in a waste beaker .

28
Solving Titration Problems
1.
Subtract the initial buret reading and the final buret reading.
2.
You should know the buret solution molarity.
Calculate the # moles of solute delivered to the flask.
3.
Do solution stoichiometry to determine the # moles and molarity of the unknown solution in the flask.
The initial volume reading on a buret filled with 0.10 M sodium hydroxide is 1.52 mL, and when the
endpoint is reached, it reads 54.52 mL. What is the molarity of 125 mL HCl solution titrated by this
solution?
Example: 0.50 M calcium hydroxide solution is used to titrate 75.20 mL of perchloric acid. What is the molarity of the
acid?
Ca(OH)
2
Initial volume
Volume (mL)
17.40 mL
Final volume 49.32 mL
Review
1.
The concentration of a solution of potassium hydroxide is determined by titration with nitric acid. A 25.0 mL sample of potassium hydroxide reaches endpoint when 35.6 mL of 0.562 M nitric acid is added. What is the concentration of the potassium hydroxide solution?

29
2.
In a titration of a sample of vinegar with sodium hydroxide, your initial buret reading is 17.05 mL and your final buret reading is 28.27 mL. A 10.0 mL sample of vinegar (acetic acid [CH
3
COOH]) was titrated.
3.
What is the concentration of acid in rainwater when 100.0 mL is titrated with 25.12 mL of 0.00105 M NaOH? Since acid rain contains several acids, assume the acid is monoprotic and use the symbol HA to represent the acids that are present.
4.
A 0.250 M solution of barium hydroxide neutralizes 25.0 mL stomach acid. What is the molarity of the HCl in the solution? Assume stomach acid is the HCl solution.
Ba(OH)
2
Initial
Final
Volume (mL)
3.20 mL
47.02 mL
Unit 6 Homework
Lesson 6.1
1. The boiling points of the following liquids increase in the order in which they are listed below:
CH
4
< H
2
S < NH
3
Discuss the theoretical considerations involved and use them to account for this order.
2. Substance Melting Point, ºC
H
2
C
3
H
8
HF
CsI
-259
-190
-92
621
Discuss how the trend in the melting points of the substances tabulated above can be explained in terms of the types of attractive forces and/or bonds in these substances.
3. Alcohol dissolves in water to give a solution that boils at a lower temperature than pure water. Salt
dissolves in water to give a solution that boils at a higher temperature than pure water. Explain these
facts from the standpoint of vapor pressure and strength/type of intermolecular forces.

30
Lesson 6.2
1.
Summarize and explain the effects of various factors on the solubility of a solid solute.
2.
Summarize and explain the effects of various factors on the solubility of a gaseous solute.
3.
Summarize and explain the effects of various factors on the rate of dissolution of a solute.
Lesson 6.3
Write and balance the net ionic equation for the following reactions. All states must be present. If no precipitate forms, write “NR” at the end.
1.
Fe(NO
3
)
3
(aq) + NH
4
Cl (aq) 
2.
CaCl
2
(aq) + K
2
CO
3
(aq) 
3.
Sodium phosphate + copper (II) bromide 
4.
Silver nitrate + sodium sulfate 

31
Lesson 6.5
1.
Describe the properties of, and distinguish between: a) Solutions, colloids, and suspensions b) Electrolytes and nonelectrolytes c) Strong electrolytes and weak electrolytes
2.
Answer the questions using the given solubility curve .
Which of the salts is the least soluble in water at
10 o C ?
What two salts have the same solubility at 19 o C?
120. g of potassium nitrate is put in 200 mL of water at 60 o C. Is this solution saturated, unsaturated or supersaturated?
80. g of sodium nitrate is put in 50 mL of water at
10 o C. Is this solution saturated, unsaturated or supersaturated?
What mass of potassium chlorate must be added to 500. g water to produce a saturated solution at 50 o C?
What are the steps needed to make a saturated solution of ammonium chloride in 50 g water at
40ºC?
A saturated potassium nitrate solution is prepared at 60 o C using 100. mL of water. How many grams of solute will crystallize if the temperature is suddenly cooled to 30 o C?
30. grams of sodium nitrate are dissolved in 200. mL of water at 45 o C. How many grams should be added to this to make the solution saturated at 45 o C?

32
Lesson 6.6
1.
65.0 mL of a 1.30 M K
3
PO
4
solution is evaporated. How many grams of solid should be recovered after all the water boils away?
2.
Fill in the blank: To make orange juice from frozen concentrate, one usually mixes the can of concentrate with three cans of water. This dilutes the concentrate to _______ (what fraction?) its original concentration.
3.
Describe each step of the preparation of a standard solution of 750. mL of 0.250 M CuSO
4
, and include the necessary calculations in detail.
4.
Sketch a volumetric flask and explain precisely how you would use a 500.0 mL volumetric flask to make some 1.500
M NaNO
3
solution. (You have available some 2.000 M NaNO
3
solution and whatever other lab equipment you need)
How much 2.000 M solution is needed?
5.
You need to make up some 5.0 M KCl solution but all you have is 125 mL of 3.0 M KCl. Explain what to do to make up the 5.0 M solution. How much 5.0 M KCl will you get? Show calculations: (hint - calculate how much water to evaporate)
Lesson 6.8
1.
Write an equation for the ionization of the following acids in water. a) HNO
2
c) H
2
PO
4
(complete ionization) b) HCO
3
-

33
2.
Write an equation for the dissociation of each of the following bases in water.
- a) RbOH c) Sr(OH)
2 b) CsOH
3.
Using your knowledge of the Brønsted-Lowry theory of acids and bases, write equations for the following acidbase reactions and indicate each conjugate acid-base pair. a) CH
3
CH
2
NH
2
+ H
3
O +   b) NH
3
+ HSO
4
  c) HC
2
H
3
O
2
+ H
2
O   d) OH + HNO
3
 
Lesson 6.9
1) Complete the following table by calculating the desired quantities. All concentrations are in M.
[H + ] a 3.2 x 10 -3
[OH ] pH pOH acid, basic, neutral b 1.8 x 10 -14 c d
10.23
6.78 e 0.0050 f g h
0.000011
10.97
7.01
2. What would be the pH and pOH of each of the following solutions? a) 0.0010 M HCl ____ g) 1.0 M Ca(OH)
2
____ b) 0.0010 M HNO
3 c) 0.010 M HClO
4
____
____ h) 0.075 M KOH i) 0.000034 M Ba(OH)
2
____
____

34
Lesson 6.10
1.
Write the balanced equation for these neutralization reactions: (remember to balance charges for ionic compounds!) a) HCl + NaOH  b) HNO
3
+ KOH  c) Ca(OH)
2
+ H
2
SO
4

2.
Solve the following solution stoichiometry problems. a) How many grams of precipitate form after the complete reaction of 10.0 mL of 0.500 M barium nitrate and 20.0 mL of 0.250 M sodium sulfate? (Warning: limiting/excess) b) Calculate the molarity of phosphoric acid when 31.1 mL of it is neutralized by 21.3 mL of 0.357 M potassium hydroxide. c) What volume of 0.137 M rubidium hydroxide would be used to neutralize 15.3 mL of 0.0592 M sulfuric acid?
(Assume complete ionization of the acid.) d) 14.3 mL of calcium hydroxide solution is placed in an Erlenmeyer flask, along with a few drops of phenolphthalein indicator. The buret is filled with 0.400 M nitric acid. Figure 1 is the initial buret reading, and
Figure 2 is the final buret reading after the flask solution has turned pale pink. Calculate the molarity of calcium hydroxide in the flask.