Download 1 Exam Study Guide Chapter 11 Net Ionic Equations Precipitate

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Exam Study Guide
Chapter 11
Net Ionic Equations
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Precipitate – products falling out of a solution
A complete ionic equation shows dissolved ionic compounds as dissociated free ions
A spectator ion is an ion that appears on both sides of an equation as isn’t directly involved in
the reaction
The net ionic equation is an equation that shows only those particles involved in the reaction
and is balanced with respect to both mass and charge
For example:
Find the net ionic equation when Na2S (aq) is added to Cd(NO3)2
1) Write the double displacement reaction
Na2S (aq) + Cd(NO3)2 (aq)  CdS (s)+ NaNO3 (aq)
-Remember: you switch the 2 cations: Na and Cd
-Then you balance the charges for the new compounds:
-For example: Cd +2 S –2 becomes CdS
2) Write the complete ionic equation
2Na+(aq) + S–2 (aq) + Cd+2(aq) + 2NO3 –1(aq)  CdS (s) + 2Na+(aq) + 2NO3 –1(aq)
-You basically break each ion up and write their charge next to it
-CdS doesn’t ionize because it is a solid
3) Write the net ionic equation
2Na+(aq) + S–2 (aq) + Cd+2(aq) + 2NO3 –1(aq)  CdS (s) + 2Na+(aq) + 2NO3 –1(aq)
-First you have to cancel all the spectator ions
-Then you rewrite the equation
So the net ionic equation is:
S–2 (aq) + Cd+2(aq)  CdS (s)
Predicting the Formation of a Precipitate
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You can use solubility rules for ionic compounds to predict the formation of a precipitate
The chart on page 344 (table 11.3) shows all the different solubility rules
For example: Is AgCl a solid?
-If you look on the chart, it says that chloride salts are soluble unless they are with Ag,
Pb, or Hg. Because AgCl is a combination of a chloride salt and Ag, this compound is NOT
soluble. This means it can’t dissolve so it isn’t aqueous – it’s a solid.
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Chapter 12: Stoichiometry
Stoichiometry= the calculation of amounts of substances involved in chemical reactions
In chemical reactions:
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The number of atoms never changes
The mass never changes
Molecules can change, but not always
Moles can change, but not always
Ratios stay the same
 See Figure 12.3 for an example
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Remember:
o Representative particles = atoms, molecules, and formula units
o Ionic compound – ionic bond
o Molecules – covalent bonds
Mole Ratios
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Mole ratios are ratios between coefficients in balanced chemical equations
Used to convert between moles of reactants and products, just reactants, or just products
o For example: 1N2 + 3H2  2NH3
6 possible mole ratios:
Between reactants:
1N2/3H2
3H2/1N2
Between product and reactant:
1N2/2NH3
2NH3/1N2
Between product and reactant:
3H2/2NH3
2NH3/3H2
Mole to Mole Calculations
aG  bW
a = coefficient
G = given
B= coefficient
W = unknown amount
x mol G * b mol W = mol W
a mol G
-Basically, multiply moles of G (the amount of moles of the given substance) by the mole ratio to get to
moles of W (the amount of moles of the substance that the question asked for)
 See Sample Problem 12.2 for an example (page 360)
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Mass to Mass Calculations
1) First, go from grams  moles
*To do this, use molar mass
2) Then go from moles of the given  moles of the other substance
*To do this, use mole ratios
3) Then go from moles  mass
*To do this, use molar mass
For example:
-With 5.40 grams of H2, how many grams of NH3 are produced
1N2 + 3H2  2NH3
First: grams  moles
5.40 grams H2 * I mole H2 = 2.70 mol H2
2 g H2 (that’s the molar mass of H2)
This is the mole ratio
Second: moles  moles
2. 70 mol H2 * 2 mole NH3 = 1.80 mol NH3
3 mole H2
This is the molar mass of NH3
Third: moles  mass
1.80 mol NH3 * 17 grams NH3 = 30.60 g NH3
Basically:
aG
bW
(given quantity)
(wanted quantity)
a = coefficient
G = given
B= coefficient
W = unknown amount
Representative
particles of G
X
6.02 x 1023
1 mol W
1 mol G
6.02 x 1023
mol G
Mass of G X
1 mol G
molar mass
G
X
b mol W
a mol G
=
Representative
particles of W
mol W
molar mass W
1 mol W
= mass of W
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Limiting and Excess Reagents
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The limiting reagent is the reagent that determines the amount of product that can be
formed by a reaction. It’s all used up.
o For example, if you have 2 tabletops, but you need 3  it’s the limiting reagent
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The excess reagent is the reagent that isn’t completely used up.
o For example, if you have 12 legs, but you only need 8  it’s the excess reagent
For example: What is the limiting reagent when 80.0 g Cu reacts with 25.0 g S?
2Cu + S  Cu2S
First: grams  moles
80.0 g Cu X 1 mole Cu = 1.26 mol Cu (remember this is what you have)
63.5 g Cu
25.0 g S X 1 mole S = 0.779 mol S (this is what you have)
32.1 g S
molar mass
Second: moles  moles
1.26 mol Cu X 1 mol S
= 0.630 mol S (this is what you need)
2 mol Cu
mole ratio
You have more moles of Sulfur than you need  so sulfur is the excess reagent
This means copper is the limiting reagent
Theoretical yield- the amount of product that should be formed
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The maximum amount of product that could be formed from given amounts of reactants
Derived from the amount of limiting reagent
This is found by the calculations you do
Actual yield- the amount of product that actually forms when the reaction is carried out in the lab
o This is given to you in the problem
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Chapter 13
Kinetic Theory and a Model for Gases
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Kinetic energy is the energy that an object has because of its motion
According to kinetic theory, all matter consists of tiny particles that are in constant motion
-The gas particles are usually molecules or atoms
 General characteristics of gas particles:
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Gas particles are small, hard spheres with an insignificant volume
Gas particles are very far apart. In between particles, there is empty space.
Gas particles move rapidly
Gas particles move independently
All collisions between particles in a gas are elastic.
-This means that no energy is lost
Particles travel in straight-line paths between collisions
-Particles only change direction when they collide with other particle
A gas fills all the available space container, regardless of volume and shape
Gas Pressure
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Gas pressure results from the force exerted by a gas per unit surface area of an object
Gas pressure is the result of simultaneous collisions of billions of rapidly moving particles
A vacuum is an empty space with no particles and no pressure
Atmospheric pressure results from the collisions of atoms and molecules in air with objects
-Atmospheric pressure decreases as you climb a mountain because the density of
earth’s atmosphere decreases as the elevation increases
The SI unit of pressure is the pascal (Pa)
 1 atm = 760 mm Hg = 101.3 kPa
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Standard temperature and pressure (STP): at 0º Celsius and 101.3 kPa, or 1 atm
Kinetic Energy and Temperature
**Heating a substance raises temperature, kinetic energy, and speed**
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Particles of the same substance have the same average kinetic energy
In hot water, particles move faster
Kelvin temperature is directly proportional to the average kinetic energy of a particle
-If you double the Kelvin temperature, you double the average kinetic energy
Absolute zero Kelvin has absolutely no vibrations, and therefore no kinetic energy
Liquids
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Liquids can flow and conform to the shape of a container
-Substances that can flow are called fluids
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According to kinetic theory, both particles in gases and liquids have kinetic energy
-There are no attractions between the particles in a gas
-BUT there are attractions between the particles of a liquid
These significant interactions between particles hold the particles in liquids close together.
-This is why liquids:
-have a definite volume (lower than gas)
-are denser than gases
-The same is true of solids. This is why liquids and solids are known as condensed states of matter
Evaporation
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Vaporization is a conversion of a liquid into a gas or vapor
Evaporation is when vaporization occurs at surface of a liquid that isn’t boiling
To vaporize, particles need a minimum kinetic energy
-Most particles don’t have enough kinetic energy to overcome the attractive forces and
escape into the gaseous state
Liquid evaporates faster when heated
-This is because heating a liquid increases average kinetic energy, and more kinetic
energy allows more particles to escape
As evaporation occurs, the particles with the highest kinetic energy tend to escape first
Evaporation cools a liquid
-Why?? – When the particles with the highest kinetic energy escape, the particles with lowest
kinetic energy are left
-Example: sweating cools you off when it evaporates
Liquid
evaporation
condensation
Vapor (gas)
Vapor Pressure
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Vapor pressure is a measure of the pressure of a molecule after it evaporates in a container
Evaporation of a liquid in a closed contain is different from evaporation in an open container.
-Closed container: no particles can escape so the particles at the surface vaporize and
become vapor particles. As the number of vapor particles increase, some condense and
become a liquid.
Equilibrium is when the ratio of vaporization equals the rate of condensation, so that vapor
pressure remains constant
-The number of particles condensing = the number of particles vaporizing
-At equilibrium, particles continue to evaporate and condense, but the number of
particles doesn’t change  there’s a constant amount of vapor
Easily evaporated molecules have high vapor pressures
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-These molecules are “lone wolfs” because they have very little intermolecular
interactions
-For example: ethanol will evaporate before water because it has a higher vapor
pressure
Vapor Pressure and Temperature Change
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An increase in temperature increases vapor pressure.
This is because the kinetic energy increases too, so more particles escape the liquid, and
collide with the walls more
The number of particles in the air depend on temperature
-Particles with higher temperature have higher kinetic energy, and therefore escape into
the air, so there are more particles in the air
Boiling Point
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Increase in evaporation rate of a liquid  increase in kinetic energy  increase in temperature
A liquid boils when: the liquid is heated o a temperature where the particles have enough
kinetic energy to vaporize
*this temperature is called the boiling point
The boiling point is the temperature at which the vapor pressure of the liquid is just equal to the
external pressure on the liquid
o This occurs when the pressure of the liquid = the external pressure
Bubbles of vapor form throughout the liquid, rise to the surface, and escape into the air
Boiling Point and Pressure Changes
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Because a liquid boils when its vapor pressure is equal to external pressure, liquids don’t always
boil at the same temperature
 At low external pressure, boiling point decreases
At higher external pressure, the boiling point increases
 Boiling points decrease at higher altitudes
o This is because higher elevation = less particles = less pressure
 Boiling is a cooling process similar to evaporation
During boiling, the particles with the highest kinetic energy escape first, like with
evaporation
 The temperature of a boiling liquid never rises above its boiling point
-if a liquid is heated after it reaches it boiling point, it just boils faster
 The vapor produced is the same temperature of the boiling liquid
**Atmospheric pressure goes down  vapor pressure goes down  boiling point goes down**
Normal Boiling Point
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The normal boiling point is at a pressure of 101.3 kPa
The normal boiling point of water is 100 degrees Celsius
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Chapter 14
Compressibility
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Gases are compressible
-They can be squeezed together more tightly and decrease their volume
Compressibility is a measure of how much the volume of matter decreases under pressure
Gases are easily compressible because of the empty space between particles
-So the particles can be forced closer together
-This is not possible with liquids and solids
Factors Effecting Gas Pressure
Number of Particles
More particles  more collisions  higher pressure
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If you double the particles, you double the pressure
Unit = number of moles
Once the pressure exceeds the strength of the container, the container will burst
Gas particles tend to move from areas of higher pressure to areas of lower pressure
Volume
Higher volume  lower concentration  less collisions  lower pressure
o
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If you double the volume, you will get ½ the pressure
Unit = liters
Temperature
Higher temperature  higher average kinetic energy  higher speed  higher force of impact
 higher pressure
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If you double the temperature, you double the pressure
*Unit = must always be in Kelvin!!
Boyle’s Law: Pressure and Volume
As pressure increases, volume decreases (if temperature is constant)
 They are inversely proportional
Equation: P1 * V1 = P2 * V2
Charles’s Law: Temperature and Volume
As temperature increases, volume increases (if pressure is constant)
 They are directly proportional
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Equation: V1 = V2
T1
T2
Gay Lussac’s Law: Temperature and Pressure
As temperature increases, pressure increases (if volume is constant)
 They are directly proportional
Equation: P1 = P2
T1
T2
For example: If you heat a tire (with no volume change), the pressure increases
-during the summer: the pressure will be higher
-during the winter: pressure will be lower
The Combined Gas Law
Allows for calculations when 2 or 3 variables are changing
Equation: P1V1 = P2V2
T1
T2
*number of gas particles (or number of moles) must be constant
As you go higher:
 Pressure goes down, temperature goes down, but volume goes up
o This is because:
-Pressure and volume are inversely proportional
-Temperature and volume are directly proportional
Ideal Gas Laws (Mr. Kub said that there won’t be any calculations on this)
Used to calculate the number of moles (n) in a contained gas
R = P1V1 = P2V2
T1n1
T 2n 2
Ideal gas constant = R = 8.31 (L*kPa)/(K*mol)
Ideal gas law: PV = nRT
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Real gases differ most from an ideal gas at low temperatures and high pressures
-Ideal gases: perfect conditions
-Real gases: in the real world, so they’re not perfect
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Dalton’s Law of Partial Pressure
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Gas pressure results from collisions of particles
Pressure depends on:
1) Number of moles
2) Volume
3) Temperature (average kinetic energy)
Partial pressure is the contribution that each gas in a mixture makes to the total pressure
In a mixture of gases, the total pressure in the sum of the partial pressures of the gases
 Dalton’s law = Ptotal = P1 + P2 + P3 + P4 + …
Graham’s Law
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Diffusion is the tendency of molecules to move toward areas of lower concentration until the
concentration is uniform throughout
During effusion, a gas escapes through a tiny hole in its container
Gases of lower molar mass diffuse and effuse faster than gases of higher molar mass
o Lighter = faster
o Heavier = slower
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Chapter 15
Water in the Liquid State
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Water is a simple triatomic molecule called H2O
The oxygen atom forms a covalent bond with each of the hydrogen atoms
Oxygen has a greater electronegativity and attracts the electrons more than hydrogen  the
oxygen atom has a partial negative charge (δ—) and the hydrogen atom has a partial positive
charge (δ+)
Water is highly polar and has a bent shape
Many unique and important properties of water result from hydrogen bonding
-Hydrogen bonding = when the positive end of one water molecule is attracted to the
negative end of another water molecule
**Remember: Covalent bonding = within water molecules
Hydrogen bonding = between different water molecules
Covalent bond
-between hydrogen
and oxygen of the
same atom
δ+
δ—
δ—
Hydrogen bond
δ+
-between hydrogen
and oxygen of
different atoms
δ+
= oxygen
δ+
= hydrogen
High Surface Tension
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Water has a very high surface tension
Surface tension is the inward pull that minimizes surface area of a liquid
-Water wants to minimize its interactions with anything but itself
Water becomes spherical to minimize its surface area
A surfactant decreases the surface tension of water and is any substance that interferes with
the hydrogen bonding between water molecules
-For example: soaps and detergents are surfactants
Low Vapor Pressure
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Water has a very low vapor pressure
Remember: vapor pressure is the result of molecules escaping from the surface of a liquid and
entering the gas phase
Because hydrogen bonds hold water molecules to one another, the tendency of these molecules
to escape is low and evaporation is slow
Solvents and Solutes
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An aqueous solution is water that contains dissolved substances
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Solvent = the liquid
Solute = the dissolved particles
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A solvent dissolves the solute. The solute is dispersed in the solvent
A solution can’t be separated by a filtration
-the dissolved particles are small enough to go through a filter
The Solution Process
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As individual solute ions break away from the crystal, the negatively and positively charged
ions become surrounded by the solvent molecules and the ionic crystal dissolves
-For example: salt molecules break down into ions, Na+ and Cl—  water molecules
surround these ions  the ionic crystal dissolves
-The negative oxygen atoms of water surround the positive Na ions
-The positive hydrogen atoms of water surround the negative Cl ions
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Solvation is the process by which the positive and negative ions become surrounded by solvent
molecules
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“Like dissolves like”
Polar dissolves polar and some ionic compounds
Nonpolar dissolves nonpolar
-Water is polar, so other polar molecules and ionic compounds dissolve in it
-for example: water dissolves salt and ethanol
-Nonpolar molecules don’t dissolve in water
-for example: oil, gasoline, oil, and methane don’t dissolve in water, but they can
dissolve themselves
Electrolytes and Nonelectrolytes
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Electrolytes are compounds that conduct electric current when in an aqueous solution or
molten state
-requires mobile ions to conduct an electric current
All soluble ionic compounds are electrolytes because they dissociate into ions
Nonelectrolytes do not conduct electricity
-are often non-ionic compounds because they aren’t composed of ions
-for example: sugar and rubbing alcohol
Strong electrolyte = NaCl and inorganic acids and bases completely breaks into ions
Weak electrolyte = NH3 and organic acids and bases  only breaks into a few ions
Non-electrolyte = ethanol and glucose  doesn’t break into ions
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Chapter 16
Solution Formation
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The nature of the solvent and the solute determine if a substance will dissolve
Stirring (agitation), temperature, and the surface area of the dissolving particles all determine
how fast the substance will dissolve
Stirring
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The more you stir the solution  the faster the dissolving rate
Why?
-while you stir, the fresh solvent is continually brought into contact with the surface of the
solute
Agitation (stirring or shaking) affects only the rate of dissolving, not the amount that will
dissolve
An insoluble substance will not dissolve, no matter how much you shake or stir it
Temperature
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The higher the temperature  the faster the dissolving rate
Why?
-Increase in temperature  increase in kinetic energy  increase in speed of particles 
increase in the force of collisions
Particle Size
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The smaller the particles  the more surface area you have  the faster the dissolving rate
-The more surface of the solute that is exposed, the faster the rate of dissolving
For example: a spoonful of granulated sugar will dissolve faster than a sugar cube
Solubility
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A saturated solution contains the maximum amount of solute for a given amount of solvent
In a saturated solution, a state of dynamic equilibrium exists between the solution and the
excess solute.
-The rate of the salvation (dissolving) = the rate of crystallization
-The total amount of dissolved solute remains constant
-if more solute is added, it will not dissolve
Solubility is the amount of solute that dissolves to produce a saturated solution
o Solubility is measured in grams of solute per 100 g of solvent
An unsaturated solution is a solution that contains less solute that a saturated solution
-It is not full
-If more solute is added, the solute will dissolve until the solution is saturated
If two liquids are miscible, they dissolve in each other in all proportions
-For example: ethanol and water are infinitely soluble in each other
If two liquids are immiscible, they are not able to be dissolved at all
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Factors Affecting Solubility
Temperature
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Solubility increases as temperature of the solvent increases
A supersaturated solution holds more solute than theoretically possible at a given temperature
-It’s superfull
-the crystallization of a supersaturated solution can be started if a very small crystal, called a
seed crystal, is added
-the rate of crystallization is usually very rapid
The solubility of a gas increases as temperature decreases
Pressure
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The solubility of a gas increases as the pressure of the gas above the solution increases
Changes in pressure have little effect on the solubility of solids and liquids, but strongly affect
the solubility of gases
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Henry’s law states that the solubility of a gas is directly proportional to the pressure of the gas
above the liquid
S1 = P1
S2 = P2
Concentrations of Solutions
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Concentration is the measure of how much solute is dissolved in a solute
A diluted solution contains a small amount of solute
A concentrated solution contains a large amount of solute
Molality = moles of solute
kg of solute
Molarity (M) = moles of solute
liters of solute
Sample Problem 16.2
o
o
One saline solution contains 0.90 g NaCl in exactly 100 mL of solution. What is the
molarity of the solution?
1st convert grams to moles and 2nd convert milliliters to liters
0.90 g NaCl X 1 mol NaCl X 1000 mL =
0.15 M
100 mL
58.5 g NaCl
1L
molar mass of NaCl
Sample Problem 16.3
o
How many moles are in 1.5 L of 0.70M NaClO?
1.5 L X 0.70M NaClO = 1.1 moles of solute
1L
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Chapter 17
Energy Transformations
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Thermochemistry: the study of energy changes during chemical reactions and changes in state
Chemical potential energy: energy contained in bonds
Heat (q) – energy that is transferred
-not energy itself – it’s transferred energy due to temperature change
Heat flows from warmer objects to cooler objects, until both objects are at the same
temperature
Exothermic and Endothermic Processes
system = what
we are focused
on)
surroundings = everything else
(like water)
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Together, the system and its surroundings make up the universe
Law of conservation of energy- energy is neither created nor destroyed
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Exothermic process – system’s energy goes down; surrounding’s energy goes up

-in the same order of magnitude
-q is negative
-heat is transferred from the system to its surroundings  heat is released by the
system
-example: A hot metal in cold water – the water’s temperature increases
Endothermic process – system’s energy goes up; surrounding’s energy goes down
-in same order of magnitude
-q is positive
-heat is transferred from the surroundings to the system  heat is absorbed by the
system
-examples: A cold metal in hot water – the water’s temperature decreases
A person (system) next to a fire gains heat from the fire (its surroundings)
Units for Measuring Heat Flow
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Heat flow is measured in 2 units: the calorie and the joule
1 Calorie = 1 kilo calorie = 1000 calories
1 calorie is the amount of heat needed to raise the temperature of water by 1ºC
The Joule is the SI unit of energy
 1 joule = 0.2390 calories
 1 cal = 4.184 Joules
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**1 calorie (or 4.184 J) is the specific heat of water** (the amount of heat needed to
raise the temperature of 1 gram of H2O by 1ºC)
Heat Capacity and Specific Heat
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Heat capacity – the amount of heat needed to raise the temperature of an entire object by 1ºC
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Specific heat – the amount of heat needed to raise the temperature of 1 gram of an object by
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Heat capacity depends on both mass and chemical composition
-Mass: It takes more heat to raise the temperature of a whole bowl of water than to
raise the temperature of a drop of water
-Chemical composition: different substances with the same mass may have different
heat capacities
-a puddle of water might be cool, but a puddle of steel might be burning hot
1ºC
Water has a very high heat capacity
-It takes a lot of heat to raise its temperature
-It releases a lot of heat when it is lowering in temperature
Units for specific heat: J/(g · ºC) and cal/(g · ºC)
Formula for Specific Heat
C =
q
m x ΔT
Formula for Heat
q = m x C x ΔT
C = specific heat
q = heat
m = mass
Δ T = change in temperature = T final – T initial
Sample Problem 17.1
o
The temperature of a 95.4 g piece of copper increases from 25ºC to 48ºC when the copper
absorbs 849 J of heat. What is the specific heat of copper?
-Use the formula for specific heat, plugging in:
m = 95.4
ΔT = 48ºC – 25ºC = 23ºC
q = 849 J
C =
q
=
849
f = 0.387 J/(g x ºC)
m x ΔT
95.4 g x 23ºC
Calorimetry
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Calorimetry – the measure of heat flow into or out of a system
In calorimetry, the heat released by a system = the heat absorbed by the surroundings
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Calorimeter – the well-insulated device used to measure the heat that is released or absorbed

-For example: a coffee cup
Enthalpy (H)– the heat (energy) of a system

Changes in enthalpy = ΔH = q = the heat released or absorbed
*The heat absorbed by the surroundings is equal to, but has the opposite sign of, the
heat released by the system
q system = ΔH = – q surroundings = – (m x C x ΔT)
q surroundings = – ΔH = – q system = m x C x ΔT
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ΔH is negative for exothermic reactions
ΔH is positive for endothermic reactions
Sample Problem 17.2
o 25 mL of water containing HCl is added to another 25 mL of water containing NaOH.
Both containers of water started out at 25ºC and once they were combined, the final
temperature was 32ºC. Find the ΔH.
1) FInd the pieces you need to plug into the equation: m, C, and ΔT
m = 50 mL x 1.00 g = 50 g
1 mL
C(water) is always 4.184 J
ΔT = 32ºC – 25ºC = 7.0ºC
2) Plug in the numbers and solve the equation
ΔH = – (m x C x ΔT)
= – (50 g x 4.184 J/gºC x 7ºC)
= –1463 J
= – 1.5 kJ
*if it’s negative then this means that it’s an exothermic reaction  in exothermic
reactions, heat is released  so, this means that 1.5 kJ was released
Thermochemical Equations
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Thermochemical equation – a chemical equation that includes a change in enthalpy
In a chemical equation, the enthalpy change can be written as a product or a reactant
*As a product = exothermic reaction
*As a reactant = endothermic reaction
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Exothermic reactions
CaO + H2O  Ca(OH)2 + 65.2 kJ
*Because energy is on the right energy is a product  energy is released from the reaction 
it’s an exothermic reaction
So you can rewrite the equation as:
CaO + H2O  Ca(OH)2
ΔH = – 65.2 kJ
*The ΔH is negative because it’s an exothermic reaction
CaO + H2O has a HIGHER energy than Ca(OH)2
So Ca(OH)2 is stable, unreactive, and happy
And CaO + H2O is unstable, reactive, and unhappy
Endothermic reactions
2NaHCO3 + 129 kJ  Na2CO3 + H2O + CO2
*Because energy is on the left  energy is a reactant  heat is absorbed  It’s an endothermic
reaction
So you can rewrite the equation as:
2NaHCO3  Na2CO3 + H2O + CO2 ΔH = 129 kJ
*The ΔH is positive because it’s an endothermic reaction
-This equation shows that for every 2 mol of NaHCO3 the ΔH of decomposition is 129 kJ
-SO, for 4 mol of NaHCO3 the ΔH of decomposition is double the energy, or 254 kJ
2NaHCO3 has a LOWER energy than Na2CO3 + H2O + CO2
So 2NaHCO3 is stable, unreactive, and happy
And Na2CO3 + H2O + CO2is unstable, reactive, and unhappy
Sample Problem 17.3
o
Find the amount of heat (in kJ) needed to decompose 2.24 mol NaHCO3 . Use the chemical
equation 2NaHCO3  Na2CO3 + H2O + CO2 ΔH = 129 kJ
2.24 mol NaHCO3 x
129 kJ
= 144 kJ
2 mol NaHCO3
Hess’s Law

Hess’s Law of Summation – add the reactions together to get another reaction of interest and

add their ΔHs together to find the ΔH of the reaction of interest
Hess’s law allows you to determine the heat of reaction indirectly
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For example:
C (s, diamond)  C (s, graphite) is the reaction of interest  What is it’s ΔH?
The two given reactions are:
C (s, diamond) + O2  CO2 ΔH = -395.4 kJ
C (s, graphite) + O2  CO2 ΔH = -393.5 kJ
First: you have to switch the 2nd reaction, which changes the sign of the ΔH
CO2  C (s, graphite) + O2  CO2 ΔH = 393.5 kJ
Then add them up and cancel out:
C (s, diamond) + O2 + CO2  C (s, graphite) + O2 + CO2
So you get:
C (s, diamond)  C (s, graphite)
This is the answer: the ΔH of the reaction of interest
Then you add up the ΔHs
–395.4 kJ + 393.5 kJ = -1.9 kJ
Standard Heats of Formation

ΔHf = standard heat of formation = change in energy during formation of a compound from


its free elements
-free elements – any element by itself
You can calculate the heat of reaction by using standard heats of formation.
Free elements = H2, O2, N2, F2, Br2, Cl2, I2, C(graphite)
*For these free elements: Hf0 = 0
£ ΔHf0 (products) – £ ΔHf0 (reactants)
Sample Problem 17.7
o 2CO + O2  2CO2. What is the ΔHf0 for this reaction?
ΔHf0 O2 = 0 kJ/mol
ΔHf0 CO = -110.5 kJ/mol
ΔHf0 CO2 = -393.5 kJ/mol
First: add up all the ΔHf0 of the reactants and the products
ΔHf0 (reactants) = (2 x -110.5) + 0 = -221.0 kJ
ΔHf0 (products) = 2 x -393.5 = -787.0 kJ
Then: solve the equation
ΔHf0 = ΔHf0 (products) – ΔHf0 (reactants)
= (-787 kJ) – (-221 kJ)
= –566 kJ
The ΔHf0 is negative, so the reaction is exothermic
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Chapter 18
Rates of Reaction







Rate is a measure of speed of any change that occurs within an interval of time
In chemistry, the reaction rate (rate of a chemical change) = amount of reactant per unit time
o For example: 0.2 mol/1 month
Collision theory- atoms, molecules, and ions can react to form products when they collide with
one another ONLY IF they have enough kinetic energy
o If 2 particles collide and they do NOT have enough chemical energy, then they bounce
off each other and remain unchanged
Activation energy = the minimum amount of energy that colliding particles must have in order
to react
o activation energy = barrier
Activated complex = an unstable arrangement of atoms that forms momentarily at the peak of
the activation energy barrier
o When 2 reactants combine with the necessary activation energy, a new product is
formed, called the activated complex
o Only forms if there is enough activation energy
o Transition state = activated complex
 It’s the top of the curve on the chart on page 543
Factors affecting Reaction Rates
 Temperature increases, reaction rate increases
-more particles  greater force in collisions  faster rate
 Concentration increases, reaction rate increases
-more particles  greater number of collisions  faster rate
 Small particles, reaction rate increases
 More surface area, reaction rate increases
Catalysts increase reaction rates
o Catalysts speed up both the forward and reverse reactions equally because the reverse
reaction is exactly the opposite of the forward reaction
o The catalyst lowers the activation energy of the reaction
o Catalysts remain unchanged and don’t affect the amount of reactants at products, only
the rate that it takes them to achieve equilibrium
o Inhibitors interfere with catalysts
Reversible Reactions and Equilibrium

Reversible reaction- reaction in which conversion of reactants to products and products to
reactants occur simultaneously
o In principle all reactions are reversible
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o




Reactants go to products in the forward direction, and products go to reactants in the
reverse direction
Chemical equilibrium- state of balance in which forward and reverse reactions take place at the
same rate
o At chemical equilibrium, no net change occurs in the actual amounts of the components
of the system
o But chemical equilibrium is still a dynamic state and both the forward and reverse
reactions continue
Equilibrium position- relative concentrations of reactants and products of a reaction that has
reached equilibrium
o When rate of conversion of reactants = rate of conversion of products
o Even though the rates are equal, the concentrations don’t have to be the same
Le Chateliers Principle: when stress is applied to a system at equilibrium, the system changes to
relieve the stress
o This predicts the direction of change in the position of equilibrium
o Theses stresses include:
 Changes in concentration of reactants and products
-If you add something to the left, then the system shifts to the right
-Vice versa: If you add something to the right, then it shifts left
 Changes in temperature
-If you increase the heat on the left, then the system shifts right
-vice versa: if you add heat on the right, it shifts left
-If you decrease the heat on the left, then the system shifts left
-vice versa: if you lower heat on the right, it shifts right
 Changes in pressure
-If you increase pressure, it shifts in the direction that favors the
products
-if you decrease the pressure, it shifts in the direction that favors the
reactants
Equilibrium constant- It’s essentially the measure of the ratio of products to reactants at
equilibrium
o Keq
o
o
[C]c X [D]d
[A]a X [B]b
If Keq > 1, products favored at equilibrium
If Keq < 1, reactants favored at equilibrium
Entropy and Free Energy

Free energy- energy in a reaction that is available to do work
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



ΔH
–
+
–
+
Spontaneous reaction- reactions that favor formation of products under the specified
conditions
o Reactions that occur as written are called spontaneous reactions
o All spontaneous reactions release free energy
o All spontaneous reactions produce lots of product
o Spontaneous reactions go on and on
Nonspontaneous reactions- reactions that don’t give products under the specified conditions
o Equations for the reactions may be written, but the reactions are nonspontaneous
o All nonspontaneous reactions require free energy
o All nonspontaneous reactions produce lots of reactant
Entropy- measure of the disorder in the system
o Entropy increases if:
 Number of particles increases
 Shaking increases
 Temperature increases
 Change from a solid to liquid
 Change from a liquid to gas
 Things are divided into parts
 Things are mixed with dissolving
o Entropy increases if:
 Change from a liquid to solid
 Change from a gas to liquid
 Number of particles decreases
 Shaking decreases
During equilibrium, low temperature favors enthalpy and a high temperature favors entropy
ΔS
+
–
–
+

ΔG
–
+
– (if low temperature)
– (if high temperature)
Spontaneous?
Yes
No
Yes (if low temperature)
Yes (if high temperature)
Gibbs free-energy change: maximum amount of energy that can be coupled to another process
to do work
ΔG = ΔH – TΔS




ΔH = enthalpy
T = temperature in kelvins
ΔS = entropy
ΔG = gibbs free energy-change
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Chapter 19
Properties of Acids and Bases
Acids





Taste sour
Will change the color of an acid-base indicator
Can be strong or weak electrolytes
Will react with bases
Examples: vinegar = ethanoic acid = acetic acid, soda, and citrus foods
Bases






Taste bitter
Feel slippery, like soap
Will change the color of an acid-base indicator
Can be strong or weak electrolytes
Will react with acids
Examples: antacids and cleaning supplies
Arrhenius Acids and Bases
Arrhenius Acids




Arrhenius acids are hydrogen-containing compounds that ionize to give hydrogen ions (H+) in
aqueous solutions
Monoprotic = can donate 1 H+
Examples: HNO3, HCl, CH3COOH
Diprotic = can donate 2 H+
Examples: H2SO4, H2CO3
Triprotic = can donate 3 H+
Examples: H3PO4
Arrhenius Bases



Arrhenius bases are compounds that ionize to give hydroxide ions (OH–) in aqueous solutions
NaOH + H2O  Na+ + OH–
-dissociates into a sodium ion and a hydroxide ion
KOH + H2O  K+ + OH–
-dissociates into a potassium ion and a hydroxide ion
 Common bases
1)
2)
3)
4)
Sodium hydroxide = NaOH
Potassium hydroxide = KOH
Calcium hydroxide = Ca(OH)2
Magnesium hydroxide = Mg(OH)2
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Bronsted-Lowry Acids and Bases





Bronsted Lowry Acids – donate H+
Bronsted Lowry Bases – accept H+
Amphoteric: can act as both an acid and a base
-Example: water
A conjugate acid is the particle formed when a base gains a hydrogen ion and becomes an acid
A conjugate base is the particle formed when an acid loses a hydrogen ion and becomes a base
o A hydronium ion (H3O+) is a water molecule that gains a hydrogen ion and becomes
positively charged
Base to Acid
NH3
base
+
H2 O
acid
NH4
OH–
+
conjugate acid
conjugate base
o
To go from a base to an acid = add H+
o
Water loses an H+ and becomes OH–
Water loses an H+ and gives it to ammonia (NH3). Ammonia
becomes a conjugated acid, and water becomes a
conjugated base.
Acid to Base
CH3COO H + H2O
acid
o
o
base
CH3COO
conjugate base
To go from a acid to a base = lose H+
Water gains an H+ and becomes H3O
+
H3O
conjugate acid
Water gains an H+ and takes it from ethanoic
acid (CH3COOH). Ethanoic acid becomes a
conjugated base, and water becomes a
conjugated acid.
Lewis Acids and Bases


A Lewis Acid accepts a pair of electrons
-forms a covalent bond
A Lewis base donates a pair of electrons
-also forms a covalent bond
Hydrogen Ions from Water



A water molecule that loses a hydrogen ion becomes a negatively charged hydroxide ion (OH–)
A water molecule that gains a hydrogen ion becomes a positively charged hydronium ion (H3O+)
The self-ionization of water is the reaction in which water molecules produce ions
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H+
H2O
Hydrogen ion

+ OH–
Hydroxide ion

Hydrogen ions have many names
-H+
-H3O+
-protons
-hydronium ions
In pure water, the concentration of H+ and OH— both equal 1.00 X 10-7  so water is neutral

A neutral solution is any aqueous solution is which [H+] and [OH–] are equal
Ion Product Constant
Kw = [H+] X [OH—] = 1.0 X 10-14

The ion-product constant for water (Kw ) is the product of the concentrations of hydrogen ions
and hydroxide ions, and is always 1.0 X 10-14




So as concentration of hydrogen [H+] goes up, the concentration of hydroxide ions [OH–]
goes down
o And vice versa: as concentration of hydrogen [H+] goes down, the concentration
of hydroxide ions [OH–] goes up
A solution is acidic if:
- you have more [H+] than [OH--]
-the [H+] is more than 1.00 X 10-7
-the [OH--] is less than 1.00 X 10-7
A solution is basic or alkaline if:
- you have more [OH--] than [H+]
-the [H+] is less than 1.00 X 10-7
-the [OH--] is more than 1.00 X 10-7
A solution is neutral if:
-you have the same [H+] as [OH--]
The pH Concept
Calculating pH
pH = – log [H+]


If pH is less than 7:
-greater than 1.0 X 10 –7
-it’s acidic
If pH is equal to 7:
-equals 1.0 X 10 –7
-it’s neutral
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
If pH is greater than 7:
-less than 1.0 X 10 –7
-it’s basic/alkaline
Sample Problem 19.2
o
What is the pH of a solution with an [H+] of 4.2 X 10-10
pH = – log [H+]
pH = – log (4.2 X 10-10 )
pH = 9.38
Finding the [OH–] of a Solution
Sample Problem 19.1
o
If the [H+] of a solution is 1.0 X 10-5 M, is the solution acidic basic of neutral? What is the [OH--]
of this solution?
[H+] X [OH—] = 1.0 X 10-14
[OH—] = 1.0 X 10-14
[H+]
[OH—] = 1.0 X 10-14
1.0 X 10-5
[OH—] = 1.0 X 10-9
It’s acidic because this [OH–] is less than 1.0 X 10-7
Using pH to find [H+]
[H+] = 10 –pH
Sample Problem 19.3
o
The pH of a solution is 6.35. What is the hydrogen concentration [H+]?
[H+] = 10 –pH
[H+] = 10 –6.35
[H+] = 4.5 X 10–7
Calculating pOH
pOH = – log [OH–]
pH + pOH = 14
Sample Problem 19.4
o
What is the pH of a solution if [OH–] = 4.0 X 10-11?
pOH = – log [OH–]
pOH = – log (4.0 X 10-11)
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pOH = 10.4
pH + pOH = 14
pH + 10.4 = 14
pH = 14 – 10.4
pH = 3.60
Neutralization Reactions
Acid + Base  Salt + Water
-For example:
HCl + NaOH  NaCl + HOH
(Acid) (Base)



(Salt)
(Water)
Strong acid + strong base  neutral solution
Strong acid + weak base acidic solution
Weak acid + strong base  basic solution
Titrations- adding just enough acid or base to neutralize all bases or acids


Equivalence point = when all acid/base is neutralized
*moles of OH – = moles of H+
The point of neutralization is the end point of the titration
Sample Problem 19.6
o
How many moles of H2SO4 are required to neutralize 0.50 mol of NaOH?
1. Write the equation. It’s always the acid + the base, and then you switch the cations,
just like a double displacement reaction.
-the cations are H and Na, so you switch them
H2SO4 + NaOH  Na2SO4 + HOH
-Remember: subscripts don’t carry over. This is why H2 becomes H and it doesn’t
matter.
**Overall charge is important. You must balance the charges by switching them
Na +1 and SO4 2—  This becomes Na2SO4
2. Balance the equation. You must have equal numbers of OH – and H+ on both sides.
H2SO4 + 2NaOH  Na2SO4 + 2HOH
3. Use the mole ratio to multiply
0.50 mol NaOH X 1 mol H2SO4 = 0.25 mol H2SO4
2 mol NaOH
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