SOL REMEDIATION
2016
Date & Time
SESSIONS
AND
TOPICS
Monday 4/18/16
8:00 – 9:00 Am
Wednesday 4/20/16
8:00- 9:00 Am
Friday 4/22/16
8:00 – 9:00 Am
Monday 4/25/16
8:00 – 9:00 Am
Wednesday 4/27/16
8:00- 9:00 Am
Friday 4/29/16
8:00 – 9:00 Am
Monday 5/2/16
8:00 – 9:00 Am
Wednesday 5/4/16
8:00 – 9:00 Pm
Friday 5/6/16
8:00 – 9:00 Am
Monday 5/9/16
8:00 – 9:00 Am
Wednesday 5/11/16
8:00 – 9:00 Am
Friday 5/13/16
8:00 – 9:00 Am
Monday 5/18/16
8:00 – 9:00 Am
Wednesday 5/20/16
8:00 – 9:00 Am
Friday 5/22/16
8:00 – 9:00 Am
Subject
Foundation
Matter
Atomic Theory
Modern Atomic Theory
Periodic Table/ Period Trends
Bonding
Naming
Reactions
Moles
Stoichiometry
Gas Laws
Energy
Solutions
Equilibrium
Acid and Base
SESSION 1: FOUNDATIONS
Topics
1. Metric conversion
2. Scientific Notation
3. Significant Figures
4. Percent Error
5. SI units
METRIC CONVERSIONS
When converting any unit
follow the steps below
1. Start with the information
given
2. Set up a conversion bar
 On the top place the new
unit on the top
 Place the old unit on the
bottom
 Smaller unit hold the
non- one factor
3. Do the math!
4. Write the answer WITH units
http://www.kentchemistry.com/links/Measurements/metricconversions.htm
SAMPLE PROBLEMS
1. How many grams are in 1000 mg
1000 mg = _____ g
1000 mg x
1 𝑔𝑟𝑎𝑚
1000 𝑚𝑔
= 1 gram
2. How many centimeters are in 8 meter
8 m = _____ cm
8mx
100 𝑐𝑚
1𝑚
= 800 cm
SCIENTIFIC NOTATION
• A system of writing big and small numbers
4.67 x 106
5.23 x10-3
4670000
.00523
Base Number (argument) – tells the value
Times and Exponent – tells the scale of the number
• If exponent is negative
• Number is smaller than one
• If exponent is positive
• Number is greater than one
SIGNIFICANT FIGURES
• A system of rounding to account for the
precision of the measurement
• Measurements that do not apply to Significant
Figures
• Counted numbers
• Conversion factors
SIGNIFICANT DIGITS
1.All non-zero numbers are
significant
2.Zeros that are sandwiched between
non-zero numbers are significant
3.The base of a number in scientific
notation
NON-SIGNIFICANT NUMBERS
• Place holders
• Zeros to the right of a number WITHOUT decimal
• 100000000 ONLY 1 sig fig
• Zeros to the left of decimal point
• .0000025 ONLY 2 sig figs
SAMPLE PROBLEMS
1) 2804 m
2) 2.84
km
3) 5.029 m
4) 0.003068 m
5) 4.6 x 105 m
6) 4.06 x 10-5 m
1) 4
2) 3
3) 4
4) 4
5) 2
6) 3
PERCENT ERROR
⃓𝐸𝑥𝑝𝑒𝑟𝑖𝑚𝑒𝑛𝑡𝑎𝑙 −𝐴𝑐𝑐𝑒𝑝𝑡𝑒𝑑⃓⃓
𝐴𝑐𝑐𝑒𝑝𝑡𝑒𝑑⃓
x 100
• Experimental is found in lab or given in the
problem
• Accepted is the known or calculated value
SCIENTIFIC INTERNATIONAL UNITS
• For every measurable amount there is a standard unit in which it is to be
reported.
http://www.learnalberta.ca/content/memg/division03/International%20System%20of%20Units/index.html
MATTER- SESSION 2
• Topics
1. Basic Atomic Structure
A. Isotope
B. Ion
2.
3.
4.
5.
Nuclear Decay
Half life
Average Atomic Mass
Physical and Chemical
A. Properties
B. Changes
ATOMIC STRUCTURE
Outside
nucleus are
electrons
N
P
E
N
P
E
Nucleus is the
center of an atom
ISOTOPE AND ION
• Isotope: The same atom with different number of neutrons
• Ion: The same atom with different numbers of electrons
• Atoms from ion to become more stable  achieve a full
shell
• Metals: lose electron(s) to form cation (positive)
• Non-Metals gain electron(s) to for anion (negative)
http://mrbloch516.edublogs.org/2015/01/page/2/
NUCLEAR DECAY
• Stability
• When the number of protons and neutrons are one to one the nucleus is stable
• When it is more than one to one nucleus is unstable and decay/transmutation
can occur
Particles
Symbols
Alpha
4
2𝐻𝑒
0
−1β
0
+1β
Beta
Beta (positron)
Neutron
Gamma
1
0𝑛
0
0γ
TRANSMUTATION EXAMPLES
210
84𝑃𝑜
241
94𝑃𝑢
239
93𝑁𝑝
http://www.chemteam.info/ChemTeamIndex.html
HALF LIFE
• The amount of time it takes for half the
nuclear material within a substance to
transmute/ decay
• Calculated using a time and mass chart
• Always start with time zero
HALF LIFE
Sample Problem
The half-life of chromium-51 is 28 days. If the sample contained 510 grams,
how much chromium would remain after 56 days?
Divide
by 2
Divide
by 2
Mass
Time
510 g
0 days
255 g
28 days
127.5 g
56 days
Add ½
life
Add ½
life
AVERAGE ATOMIC MASS
• The mass of an atom factors the abundance of the all naturally
occurring isotopes
• Weighted average
• How to solve
1. Make a table of values that include
A. Isotope
B. Mass
C. Percent
2. Covert the percent to decimals (divide by 100)
3. Multiple by the mass of the isotope
4. Add the answers together
SAMPLE PROBLEM
The element copper has naturally occurring isotopes with mass numbers of 63
and 65. The relative abundance and atomic masses are 69.2% for a mass of
62.93amu and 30.8% for a mass of 64.93amu. Calculate the average atomic
mass of copper.
Isotope
Cu - 63
Cu – 65
Percent
69.2 % 0.692
30.8 % 0.308
(0.692) (62.93) = 43.55
(0.308) (64.93) = 24.67
68.22 amu
Mass
62.93 amu
64.93 amu
NOW YOU TRY!
Magnesium has three naturally occurring isotopes. 78.70% of
Magnesium atoms exist as Magnesium-24 (23.9850 amu), 10.03% exist
as Magnesium-25 (24.9858 amu) and 11.17% exist as Magnesium-26
(25.9826 amu). What is the average atomic mass of Magnesium?
PHYSICAL AND CHEMICAL PROPERTIES
Physical
• Intrinsic
• Is dependent on the amount of
matter
• Extrinsic
• Does not depend on the amount
of matter
• Density
• Describes the sample
• Think 5 senses
Chemical
• A description of a change
• Examples
• Flammability
• Reactivity
PHYSICAL AND CHEMICAL CHANGES
Physical
• Does not change the
identity of the substance
• State changes
Chemical
• Changes the identity of the
substance
• Color change
• Precipitate
• Production of a gas
• Change in energy
ATOMIC THEORY –SESSION THREE
Topics
1. Dalton
2. Thompson
A. Cathode Ray
experiment
B. Electrons
3. Rutherford
A. Gold Sheet
experiment
B. Nucleus
C. Atomic structure
4. Bohr
A. Energy level
B. Emission spectra
DALTON
• Five principles
• All matter was made up of atoms
• Atoms were the smallest particle
• Atoms of the same elements had the same properties
• Atoms of different elements had different properties
• Atoms can rearrange and combine to form other
substances
• Two Principles where incorrect
• atoms can only combine with other elements
• Diatomic molecules
• Atoms are NOT the smallest particle
THOMPSON
• Plum Pudding model
• Atoms were positive sphere
with negative specs in them to
be neutral in charge
• Cathode Ray experiment
• Beam was run through a vacuum
and when introduced to magnetic
fields the beam bent away from
negative and toward positive
• Proved that there was a negative
subatomic particle
RUTHERFORD
• Fired an alpha particle through
a thin sheet of gold
• Found three things
1. Most went through foil
A. Atoms are mostly empty space
2. Some reflected at an high angle
A. There are positive particles in an
atom
3. Some bounced back
A. There is a dense mass in the center
of an atom
B. Nucleus
Neutrons
Protons
http://www.daviddarling.info/encyclopedia/R/Rutherfords_experiment_and_atomic_model.html
• Previously scientists could not
explain why electrons did not
crash into the positively charged
nucleus
• Bohr determined that electron
could only exist in specific
energy level
• Quanta: specific amount of energy
• Each element has its own set of
energy levels
• Each element produces an light
emission spectra that corresponds to its
energy level
BOHR
EMISSION SPECTRA
• An electron absorbs a photon of
energy and jumps to a higher
energy level.
• Becomes excited
• moves to an excited state
• An electron releases a photon in
the form of wavelength
• De-excites back down to it normal
energy state
• Ground state
• The wavelength is seen as color
http://imagine.gsfc.nasa.gov/educators/lessons/xray_spectra/background-atoms.html
MODERN ATOMIC THEORY-SESSION 4
1. Modern Model
2. Electron Configurations
MODERN MODEL
• Atoms is made up of three main subatomic particles
• Proton (positively charged)
• Neutron (no charge)
• Electron (negatively charged)
• Proton and neutrons make up the nucleus in the center of an
atom
• Electrons exist in the electron cloud around the nucleus
• Atoms do not have definite edges
• Electron only exist within energy levels NEVER between
ELECTRON CONFIGURATIONS
• A system that details where the electrons can most likely be found
• Principle number (n) corresponds to the row of the periodic table
• Orbital type (s,p,d,f) corresponds to the region of the periodic table
• Aufbau
• Energy levels are filled lowest to highest energy
• Exception are transition metal
• Hund’s rule
• Every orbital must be occupied with one electron before electrons
can double up
• Pauli Exclusion
• Electrons of opposite spins can exist within a single orbital
ORBITAL TYPE
• 4 types of orbitals
• Orbital can hold a total of two electrons
Orbital type
Where on Periodic
Table
Number of orbital
Total number of
electrons
s
Groups 1 – 2 and
Helium
1
2
d
Groups 3-12
5
10
p
Groups 13-18
3
6
f
Rows under the main
periodic table
7
14
HOW TO WRITE AN ELECTRON
CONFIGURATION
1. Find the element on the
periodic table
2. Start at the first row on
the periodic table
3. Write the row number
4. Write the orbital block
(s,p,d,f)
5. Write the number of
electrons
SAMPLE PROBLEMS
• Li
1s2
2s1
•V
1s2
2s2
2p6
3s2
3p6
4s2
3d3
• Al
1s2
2s2
2p6
3s2
3p1
• Br
1s2
2s2
2p6
3s2
3p6
4s2
3d10
4p5
AUFBAU ORDER
slideplayer.com
chemistry.tutorcircle.com
USING HUND’S RULE AND AUFBAU ORDER
3d
Carbon
4s
3p
2p
3s
2s
2p
1s
2s
1s
Iron
PERIODIC TABLE AND TRENDS –SESSION 5
1. Layout
2. Groups
3. Valence
A. Ion formation
4. Trends
A. Atomic radii
B. Ionization energy
C. Electronegativity
LAYOUT
Non- Metals
H
Actinide Series
Metals
Noble Gases
Lanthanide Series
Halogens
Alkaline Earth Metals
Alkali Metals
Transition Metals
DEFINITIONS
• Period: Row/ energy level
• Groups: Column / Valence
• Valence: Electrons in highest energy level
GROUPS/ REGIONS
Regions
Groups
• Metals make up most of the periodic
table. Left of the stair step.
• Alkali Metals: Group 1, most reactive
metals.
• Alkaline Earth Metals: Group 2, reactive
metals.
• Transition Metals: Charge and valence
change depending on what they are
bonded with.
• Halogens: Group 17, most reactive nonmetals
• Noble Gases: Group 18, inert gases.
Non-reactive due to their full outer
shell.
• Metal lose electron to have a full
outer shell
• Non-metals are to the right of the
stair step.
• Non-Metal gain electron to have a
full outer shell
• Metalloids: can act as either metal
or non metal depending on
conditions
VALENCE
Group
Valence number
Charge
Gains or loses
electrons
Alkali Metals
1
1+
Loses 1
Alkaline Earth
Metals
2
2+
Loses 2
Transition Metals
Changes
Dependent on
bonding pair
Loses
Halogen
7
1-
Gains 1
Noble Gas
8
0
N/A
TRENDS
• Atomic Radii: The distance between two
bonded nuclei. The size of a atom.
• Ionization Energy: The energy it takes to
remove a valence electron.
• Electronegativity: The amount of relative
attraction of electrons by the nucleus.
• Increases:
Down a group
because of
shielding and
the addition of
energy levels.
• Decreases:
Across a period
due to
increased
nuclear pull
ATOMIC RADII
www.shodor.org
www.dayasriojm.top
• Decreases: Down
a groups because
inner electrons
shield the pull of
the nucleus which
increases as
energy levels are
added.
• Increases: Across
a period due to
increase nuclear
charge of the
same energy
level.
IONIZATION ENERGY
www.shodor.org
http://2012books.lardbucket.org/
ELECTRONEGATIVITY
• Decreases: Down
a groups because
inner electrons
shield the
attractive force of
the nucleus which
increase as energy
levels are added.
• Increases: Across
a period due to
increase nuclear
charge of the
same energy level.
www.webelements.com
www.shodor.org
Electronegativity
High and tight to the right
Low, Left, and Loose
chemistry.about.com
www.dayasriojf.top
BONDING – SESSION 6
• Topics
1. Ionic bonding
A. Polyatoms ions
2. Covalent bonding
A.Lewis dot structures
B. Geometric shapes
IONIC BONDING
• Metal and a non-metal
• Ions form
• attract to oppositely charged ions
• Crystalline structures
• Metals form cations
• Lose electrons
• Positively charged
• Non-metals form anions
• Gain electrons
• Negatively charged
www.geo.arizona.edu
PROPERTIES OF IONIC SALTS
• All salts have a unique bonding ratio
• Dependent on the size of the ions
• Causes different crystalline patterns
• High melting points and boiling points
• Rigid structures
• Fracture: the amount of pressure needed to break a salt
along a line
• Can conduct electricity when dissolved
• Not when in their solid state
www.science.uwaterloo.ca
POLYATOMIC IONS
• When two or more elements are
covalently bonded together and carry a
charge
COVALENT COMPOUNDS
• When electrons are shared
• Can share up to 6
electrons
• Single bond (2 electrons)
• Double bond (4 electrons)
• Triple Bonds (6 electrons
www.chemistryforkids.net
• Between two non-metals
www.bbc.co.uk
LEWIS DOT STRUCTURES
• Steps to draw a LDS
1. Calculate valence electrons for
molecule
A. That is the maximum number
2. Determine the central atom
A. Most electro-positive
3. Place the electrons to make octet for
EACH atom
A. Use multiple bonds if needed
4. Double check the electron number
and octet
A. Duet rule for H and He
SAMPLE PROBLEMS
H2O
CO2
Valence calculation
Valence calculation
H: 1 x 2 = 2
O: 6 x 1 = 6
Total: 8
C: 4 x1 = 4
O: 6 x 2 = 12
Total: 16
chemwiki.ucdavis.edu
MOLECULAR GEOMETRY
• How to predict the shape
1. Draw the Lewis Dot
Structure
2. Count the number of lone
pairs and bonded pairs
around the CENTRAL
atom
A. Predict the shape
www.studyblue.com
PROPERTIES OF COVALENT COMPOUNDS
• Low melting and boiling points
• Have different bond natures
• Polar
• Non-polar
• Exhibit intermolecular forces
• London Dispersion/Van Der Walls
• Dipole- dipole
• Hydrogen bonding
SHAPE DETERMINES PROPERTY
• Polar
• Regions of partial positive (electron deficient) and partial negative
(electron rich) at the poles of a molecule
• Due to differences in electronegativity – uneven sharing of
electrons
• Intermolecular forces
• Dipole-dipole
• Hydrogen bonding
• N,F,O and H
• Non-Polar
• Electronegativity differences at the poles of the molecules are small
or zero
• Intermolecular forces
• London Dispersion
NAMING- SESSION 7
1. Ionic Naming
A. Monatomic
B. Polyatomic
C. Transition metals
2. Covalent Naming
A. Number prefixes
IONIC NAMING
Main group metals
• Cation (metal)
keeps it’s name
• Anion (nonmetal)
• Take root of name
and end it in “ide”
Transition metal
• Cation (metal)
keeps it’s name
• Followed by a roman
numeral
• Indicates charge on
the metal
• Is determined by
the anion
• Anion (non-metal)
• Take root of name and
end it in “ide”
Polyatomic
• Polyatomic ion
keeps its name no
matter if it is the
anion or cation
EXAMPLES/ NOW YOU TRY!
1. Na2CO3
2. MgBr2
3. FeCl2
4. FeCl3
5. Zn(OH)2
6. Al2S3
• Sodium Carbonate
• Magnesium Bromide
• Iron (II) Chloride
• Iron (III) Chloride
• Zinc Hydroxide
• Aluminum Sulfide
FORMULAS TO NAME
• Ionic compounds have an
overall charge of zero
• Cations and anions
charges are equal and
opposite
• If they are not then
balance for charge by
using subscript
sodium phosphate
PO43-
Na1+
1+
3+
?
3-
?
3-
Na3(PO4)1
iron (II) bromide
Br1-
Fe2+
2+
?
Fe1Br2
2+
1-
?
2-
NOW YOU TRY!
1.
2.
3.
4.
5.
vanadium (V) phosphate
calcium oxide
magnesium acetate
aluminum sulfate
copper (I) carbonate
• V3(PO4)5
• CaO
• Mg(C2H3O2)2
• Al2(SO4)3
• Cu2CO3
COVALENT NAMING
• Binary Covalent compounds
• First element in formula
keeps it’s name
• If there is more than one of
the first element must include
number prefix
• Second element
• Take the root and end it in
“ide”
• Always include number prefix
www.uzinggo.com
EXAMPLE/ NOW YOU TRY
1. BrO3
2. BN
3. N2O3
4. NI3
5. SF6
6. XeF4
7. PCl3
1.
2.
3.
4.
5.
6.
7.
bromine trioxide
boron mononitride
dinitrogen trioxide
nitrogen triiodide
sulfur hexafluoride
xenon tetrafluoride
phosphorous trichloride
REACTIONS- SESSION 8
• Types of Reactions
• Balancing equations
REACTION TYPES
• Synthesis
• Decomposition
• Single Replacement
• Double Replacement
• Neutralization
• Combustion
• Incomplete
• Complete
SYNTHESIS/ DECOMPOSITION
Synthesis
• Two reactants form one product
X +Y  XY
2H2 + O2  2H2O
Decomposition
• One reactant break into two or more
products
• Binary- two elements
• Breaks into the elements
XY  X + Y
2H2O  2H2 + O2
• Ternary- three or more elements
• Breaks in to simpler compounds
CaCO3  CaO + CO2
SINGLE/ DOUBLE REPLACEMENT
Single
Double
• When an element replaces an
another element within a
compound
• Determined by an activity
series
• Like replaces like
• Occurs between ions in aqueous
solution. A reaction will occur
when pair of ions come together
to produce at least one of the
following
• A precipitate (solid)
• A gas
• Water or some other nonionized substance
• Metal replaced metal
• Non-metal replaces nonmetal
A + BX AX + B
AX + BY  AY + BX
Incomplete
COMBUSTION
Complete
• When a hydrocarbon burns in
oxygen gas and produces Carbon
Monoxide, Carbon Dioxide, and
Water
• When a hydrocarbon burns in
oxygen gas and produces Carbon
Dioxide and Water
4CH4 + 7O2 2CO + 8H2O
CH4 + 2O2 CO2 + 2H2O
BALANCING EQUATIONS
• Law of Conservation of Mass
• Mass of Reactants must equal mass of the products
• Same number and type of atoms must be on both sides of the
equation
3 KOH  ____
3 NaOH + ____ K3PO4
____ Na3PO4 + ____
3 Na 1 3
1
3 1
PO4 1
OH 1
1
K 3
3
NOW YOU TRY!
____ MgF2 + ____ Li2CO3  ____ MgCO3 + ____ LiF
____ P4 + ____ O2  ____ P2O3
____ RbNO3 + ____ BeF2  ____ Be(NO3)2 + ____ RbF
MOLES- SESSION 9
1. Percent composition
2. Mole
A. Avogadro's number
Simple conversions
3. Empirical Formula
4. Molecular Formula
PERCENT COMPOSITION
• The percent of different elements/ions within a compound
• How to find percent composition
1. Find the molar mass of the compound
A. Add up all the elements masses from the periodic table
• If there are subscripts multiply mass by the subscripts
2. Take the mass of the element an divide by the total
mass of the compound
3. Multiple by 100
SAMPLE PROBLEMS
Sample
• What is the percent of Carbon in
C2H6.
• C : 12.01(2) = 24.02 g/mol
• H: 1.009 (6) = 6.054 g/mol
30.074 g/mol
24.02 g/mol
30.074 g/mol
100 = 79.86 %
Now you try!
• NaHSO4
THE MOLE
• S.I. Unit for amount
• In one mole of ANY substance there are 6.022 x 10
• Particles can be
•
•
•
•
23
particles
Atoms
Ions
Molecules
Formula units
• The amount of mass in one mole is equal to the molar mass
• Calculated from periodic table
• The subscripts of a compound are the relative number of mole
• C6H12
• Carbon has 6 moles and Hydrogen has 12 moles in that compound
THE MOLE CONVERSION
Particles
Mass
(grams)
Ions
Atoms
Molecules
Formula units
x by 22.4 L
÷ by 22.4 L
Mole
Volume
(Liters)
Gases at STP
PARTICLE TO MOLE/ MOLES TO PARTICLES
1. How many moles are 1.20 x 1025 atoms of phosphorous?
1.20 x 1025 atoms X
1 mole
6.022 x1023atoms
= 20.0 moles
2. How many atoms are in 0.750 moles of zinc?
0.750 moles X 6.022 x1023atoms = 4.52 x 10
1 mole
23
atoms
MASS TO MOLE/ MOLES TO MASS
1. Find the number of moles of argon in 452 g of argon.
452 g Ar x 1 mol = 11.3 moles
39.94 g
2. Find the mass in 2.6 mol of lithium bromide.
2.6 moles x [(6.941 g Li) +(79.904 g Br)]
= 225.7 gram
1 mole
= 230 grams (sig fig)
NOW YOU TRY
1. How many molecules are in 0.400 moles of N2O5
2. Find the grams in 1.26 x 10-4 mol of HC2H3O2
EMPIRICAL FORMULA
• Lowest ratio of atoms in a compound
• How to solve
1. Take the percent's and “convert” to grams
• Just switch the sign because we assume there are 100 grams
2. Divide each mass amount by the elements molar mass
3. Re-divide by the lowest answer
4. Write the compound
• Your answers become the subscripts of the elements within the
compound
SAMPLE PROBLEM
Find the empirical formula for a compound which contains 32.8%
chromium and 67.2% chlorine.
Element
Percent Divide by
molar Mass
Moles
Cr
32.8%
0.631mol
32.8 g
51.99 g/mol
Cl
67.2 %
67.2 g
35.45 g/mol
Re-divide
0.631 mol
subscripts
1
CrCl3
0.631
1.90 mol
1.90 mol
0.631
3
NOW YOU TRY!
What is the empirical formula for a compound
which contains 67.1% zinc and 32.9 % oxygen?
MOLECULAR FORMULA
• This is a multiple of the empirical formula
• There is a common multiple
• How to solve
1. Find the mass of the empirical formula
2. Divide the molecular formula mass by the empirical
formula
3. Multiple the subscripts of the empirical formula by the
multiple
SAMPLE PROBLEM
• You and your lab partner determined that the mass of the sample
was 475.02 grams using the empirical formula from the previous
sample. What is the molecular formula of the compound.
Empirical formula
CrCl3
Cr: 51.99
Cl: 35.45 x 3 = 106.35
158.34 g/mol
475.02
158.34
=3
Cr3Cl9
STOICHIOMETRY- SESSION 10
1. Topics
A. Mole ratio
How to determine
B. Mass to mass problems
DEFINITION
• Stoichiometry
• using molar relationships between reactants and/or
products in a chemical reaction to determine desired
quantitative data
MOLE RATIO
• Comes from the balance equation
• The coefficients are the relative number of moles
in a balanced equation
2 Na3PO4 + 3 CaCl2  6 NaCl + 1 Ca3(PO4)2
2:1
Mole ratio between Na3PO4 and Ca3(PO4)2 is 2:1
MASS TO MASS
How to solve:
1. Identify the knowns and unknowns
2. Find the mole ratio
3. Setup the problem
A. Write what you are given
B. Convert to Moles
C. Switch substances using mole ratio
D. Convert to the unit the problem calls for
4. Solve!
SAMPLE PROBLEM
How many grams of potassium chloride, KCl, are produced if 25.0g of
potassium chlorate, KClO3, decompose?
25.O g
Xg
2KClO3 → 2KCl + 3O2
2:2
1 mole
25.O g KClO3 X
= .204 mole KClO3
122.518 g
2 mole KCl
.204 mole KClO3 x
= .204 mole KCl
2 mole KClO3
74.548 g KCl
.204 mole KCl x
= 15.208 grams KCl
1 mole KCl
NOW YOU TRY!
How many grams of AgCl, silver chloride, are produced from
5.0 g of AgNO3, silver nitrate?
2AgNO3 + BaCl2 → 2AgCl + Ba(NO3) 2
GAS LAWS- SESSION 11
Topics
1.
2.
3.
4.
5.
6.
7.
Properties of Gases
Boyle’s Law
Charles’ Law
Combine Gas Law
Dalton’s Law of Partial Pressure
Avogadro's relationship
Ideal Gas Law
A. Density
B. Molecular mass
PROPERTIES OF GASES
• No definite volume
• Must be enclosed in a container
• No definite shape
• Move
• Independently of each other
• At high speeds
• Behave ideally under most conditions
• No appreciable volume
• No attractive forces
• Pressure
• The amount of force, per unit area, in which the molecules hit
the container
STANDARD TEMPERATURE AND PRESSURE
Pressure
• 3 main units of pressure
1. Atmospheres (atm)
2. Millimeters of Mercury (mm Hg)
3. KiloPascals (kPa)
• Atmospheric pressure is standard
pressure (sea level)
1 atm = 760 mm Hg = 101.325 kPa
Temperature
• 2 temperature units
1. Celsius (oC)
2. Kelvin (K)
• For gas law problem MUST convert
to Kelvin
• TK = TC + 273
• Cannot divide by zero
• Standard temperature
• 0 oC
• 273 K
BOYLE’S LAW
• Relationship between pressure and volume
• Indirect relationship
• Pressure increases Volume decreases
• Pressure decrease Volume increases
P1V1 = P2V2
www.kentchemistry.com
BOYLE’S LAW PROBLEM
A gas occupies 12.3 liters at a pressure of 40.0 mm Hg. What
is the volume when the pressure is increased to 60.0 mm Hg?
P1V1 = P2V2
P1 = 40.0 mm Hg
V1 = 12.3 L
P2 = 60.0 mm Hg
(40.0 mm Hg)(12.3 L) = (60.0 mm Hg) (V2)
(40.0 mm Hg)(12.3 L)
= V2
(60.0 mm Hg)
V2 = X
V2 = 8.2 L
NOW YOU TRY!
If a gas at 25.0 °C occupies 3.60 liters at a pressure of 1.00
atm, what will be its volume at a pressure of 2.50 atm?
CHARLES LAW
• Relationship between temperature and volume
• Direct relationship
• Volume increases Temperature increase
• Volume decreases Temperature decreases
V1
T1
=
V2
T2
www.kentchemistry.com
CHARLES' LAW PROBLEM
A sample of oxygen occupies a volume of 350 mL at 35 oC.
What volume will it occupy at 85 oC?
V1: 350 mL
T1: 35 oC + 273 = 308 K
V2: X
T2: 85 oC + 273 = 358 K
𝐕𝟏
𝐓𝟏
=
𝐕𝟐
𝐓𝟐
𝟑𝟓𝟎 𝐦𝐋
𝐕𝟐
=
𝟑𝟎𝟖 𝐊
𝟑𝟓𝟖 𝑲
V2 = 406.82 mL
NOW YOU TRY!
Neon gas was heated from 50 oC to 150 oC. Its new
volume is 175 mL. What was the original volume?
COMBINED GAS LAW
• Relates Pressure, Temperature, and Volume together
V1P1
T1
=
V2P2
T2
SAMPLE PROBLEM
400 mL of a gas is contained at 300 mm Hg and 0oC. What will its
volume be in mL at 140 mm Hg and 10oC?
V1 = 400 mL
V1P1
V2P2
=
P1 = 300 mm Hg
T1
T2
T1 = OoC + 273 = 273K
V2 = ?
(400mL)(300 mmHg)
V2( 140mm Hg)
P2 = 140 mm Hg
=
o
273 K
283 K
T2 = 10 C + 273 = 283K
V2 = 888.54 mL
NOW YOU TRY!
A gas has a volume of 39 liters at STP. What will its
volume be at 4 atm and 25oC?
DALTON’S PARTIAL PRESSURE
• Each gas in a system exerts it’s
own pressure
• All the partial pressures of
the gas can be added up to
find the total pressure.
A metal tank contains three gases:
oxygen, helium, and nitrogen. If the
partial pressures of the three gases in
the tank are 35 atm of O2, 5 atm of
N2, and 25 atm of He, what is the total
pressure exerted inside the tank?
PN2 + PO2 +PHe = Ptotal
P1 + P2 +…..Pn = Ptotal
5 atm + 35 atm + 25 atm = Ptotal
65 atm = Ptotal
NOW YOU TRY!
Blast furnaces give off many unpleasant and unhealthy gases. If the
total air pressure is 0.99 atm, the partial pressure of carbon dioxide is
0.05 atm, and the partial pressure of hydrogen sulfide is 0.02 atm, what
is the partial pressure of the remaining air?
AVOGADRO'S RELATIONSHIP
• ONLY AT Standard
Temperature and
Pressure (STP) does this
relationship apply
• Standard Pressure: 1
atm
• Standard Temperature:
0o C
• A mole of a gas is
equal to 22.4 L
• Used for conversions
50 g of nitrogen (N2) has a volume
of ___ liters at STP. (40 L).
1 mole
50 g N2 x
= 1.786 mole N2
28.0 g
22.4 L
1.786 mole N2 x
= 40 L N2
1 mole
NOW YOU TRY!
100 g of oxygen(O2) is added to the gas in Question 16. What
is the volume of the combined gases at STP.
IDEAL GAS LAW
• Relates the measurable quantities of an ideal gas together
using a gas constant (R)
• R : 8.314
kPa· L
mol·K
or 0.0821
PV = nRT
P: pressure (must match R)
V: Volume (must be in L)
n: number of moles
R: gas constant
T: Temperature (must be in K)
atm· L
mol·K
ALTERNATE FORM
To Find Density
• PM = DRT
P: Pressure
M: Molecular Mass (periodic table)
D: Density
R: Gas Constant
T: Temperature (K)
To Find Molecular Mass
• PV= nRT
• Solve for number of moles (n)
• Then divide the mass given in the
problem by the number of mole
calculated.
Grams
• Molecular mass =
Moles
SAMPLE PROBLEM
How many moles of gas would be present in a gas trapped
within a 100.0 mL vessel at 25.0 °C at a pressure of 2.50
atmospheres?
P: 2.50 atm
V: 100.0 x
1L
1000 mL
PV=nRT
= 0.1 L
(2.5atm)(0.1L) =
n: ?
R:
atm· L
0.0821 mol·K
T: 25.0 °C + 273 = 298K
n=
atm· L
n(0.0821
)
mol·K
(2.5 atm)(0.1L)
atm· L
0.0821
mol·K
(298K)
(298K)
n = 0.01 moles
NOW YOU TRY!
At what pressure would 0.150 mole of nitrogen gas at
23.0 °C occupy 8.90 L?
ENERGY- SESSION 12
Topics
1. Energy
1. Heat
A. Temperature
2. Endothermic
3. Exothermic
4. Phase changes
A. Phase change diagram
5. Specific heat
A. Latent heat
B. Heating and cooling curves
6. Potential energy diagram
ENERGY
• SI unit for energy is Joules (J)
• Energy can be in different forms such as
• Heat
• Light
• Sound
• Chemical
• Mechanical
HEAT / TEMPERATURE
• Heat is a form of energy
• The transfer of energy between objects of different
temperatures
• Flows from Hot to cold
• More energy to less energy
• Temperature
• Average Kinetic Energy
• The average speed that the molecules are moving
ENDOTHERMIC / EXOTHERMIC
Endothermic
• Energy is added into the
system
• Heat of reaction (△H) is
negative
• Reactant have less energy
than the products
Exothermic
• Energy is released from the
system
• Heat of reaction (△H) is
positive
• Reactants have more energy
than the products
PHASE CHANGES
• Four phases of matter
1. Solid
2. Liquid
3. Gas
4. Plasma
• There are six main phase changes
1. Freezing (exo)
2. Melting (endo)
3. Evaporation/ Vaporization (endo)
4. Condensation (exo)
5. Sublimation (endo)
6. Deposition (exo)
www.elementalmatter.info
PHASE CHANGE DIAGRAM
Normal freezing point
1atm
Normal boiling point
www.course-notes.org
• Triple point: where all three
phases are at equilibrium
• Critical point: the point at
which the gas and liquid phase
can co-exist
• Normal Melting point: The
point where a substance melts
at normal atmospheric
pressure (standard pressure)
• Normal boiling point: The
temperature where a
substance will boil at standard
pressure
SPECIFIC HEAT/ LATENT HEAT
Specific Heat
• The amount of energy it takes to
raise one gram of a substance one
degree Celsius
Q = m X Cp X ΔT
Q: heat/energy (J)
M: mass (g)
Cp: specific heat J/gC
ΔT: change in temperature
Final temperature- initial temperature
Latent heat/ Heat of
Fusion/heat of Vaporization
• The amount of energy it take for a
substance to phase change.
Q = moles X ΔH
vap/fus
Q: heat/energy (J)
Moles
ΔHvap or ΔHfus: Heat of vaporization/
Heat of fusion
SAMPLE PROBLEMS
Specific heat
When a 500 gram piece of brass cools from
100°C to 60°C, it is found to have given up
7600 joules of energy. What is the specific
heat of the brass?
Q = m X Cp X ΔT
Q: -7600 J
m: 500 g
Cp: ?
ΔT: 60°C - 100°C = -40°C
7600 J = (500g)(Cp)(-40°C)
−7600 J
(500g)(−4o°C)
= Cp
Cp = 0.38 J/g°C
Latent Heat
How many joules of energy are needed to
boil 90 grams of water at its boiling point.
Assume that the molar heat of vaporization
of water is 41 kJ/mol.
Q = moles X ΔH
Q: ?
vap
Moles: 90g H2O x
ΔH
vap:
1 mole
18.02 g
= 4.99 moles
41 KJ/mol(4.99 mol)
Q = (41 KJ/mol)(4.99 mol)
Q = 204.77 KJ
HEAT AND COOLING CURVES
www.slideshare.net
www.teachinghighschoolchemistry.com
• The areas of the graph that
are sloped indicate and
temperature
• Increasing KE
• Use the specific heat
equations
• The flat segments of the
graph indicate a phase
change
• No temperature occurs
• All energy is being used
to over come
intermolecular forces
• Use latent heat equation
SAMPLE PROBLEM
A fluorescent light contains 0.1 g of mercury which needs to be vaporized to allow
the light to work. How much energy does it take to boil away the mercury if it starts
at 20 °C?
Boiling point
356.73 °C
Specific heat
0.14 J/g C
Heat of vaporization
59.11 KJ/mol
Part 1: 20 °C  356.73 °C
Q = m X Cp X ΔT
Q: ?
m: 0.1 g
Cp: 0.14 J/g C
ΔT: 356.73°C – 20 °C = 336.73°C
Q = (0.1g)(0.14 J/g C)(336.73°C)
Q = 4.714 J
Part 2: Liquid Hg  Gas Hg
Q = moles X ΔH
vap
Q: ?
1 mole Hg
Mole: 0.1 g Hg x
= 4.99 x 10
200.59 g
ΔH vap: 59.11 KJ/mol
Q = (4.99 x 10
-4
-4
moles
moles)(59.11 KJ/mol)
Q = 0.02946 KJ  29.46 J
Total Energy = 4.714 J + 29.46 J = 34.18 J
NOW YOU TRY
What is the total amount of energy needed to melt 54 grams
of ice starting at -5 oC? Assume that the molar heat of fusion
of ice is 6 kJ/mol. Specific Heat of Ice is 2.03 J/g oC.
POTENTIAL ENERGY DIAGRAMS
chemwiki.ucdavis.edu
1: Potential Energy (PE) of
reactants
2: Activation Energy (Ae) of
forward reaction
3: Potential Energy of
activation complex (transition
state)
4: Activation Energy of the
reverse reaction
5: Heat of Reaction (△H) Different of energy between
products and reactants
6: Potential Energy of products
SOLUTIONS – SESSION 13
Topics
1. Properties
2. Solubility
A. Rules
B. Curves
Calculations
C. Net ionic equations
3. Colligative properties
A. Boiling point elevation
B. Freezing point depression
PROPERTIES OF SOLUTIONS
• They are homogeneous mixture
• Light can pass through them
• The particles of the solution will pass through
a filter
• The particles will not separate on their own
SOLUBILITY
• The ability for a compound to dissolve
• Solute: the substance being dissolved
• Solvent: the medium of the solution
• The substance the solute is being dissolved into
FACTORS OF AFFECTING SOLUBILITY
AND RATE OF SOLUBILITY
Solubility
• Volume /pressure
• Gases ONLY
pressure
solubility
• Nature of Solute/Solvent
• Like dissolves like
• Temperature
•
temperature (solids/liquids)
solubility
• Gases
in temperature
in solubility
Rate
• Amount of solute already in
solution
• More of solute already in solution
rate
• Surface area
•
S.A. increase rate
• Stirring rate
• Temperature
•
temperature (solids/liquids)
rate
• Gases in temperature
rate
SOLUBILITY RULES
How to read:
1. Find the ion on
the chart
2. Check if it is
1. Soluble
2. Insoluble
3. Are there
any
exceptions
movies-in-theaters.net
IS THE COMPOUND SOLUBLE
1.
2.
3.
4.
5.
6.
7.
8.
9.
KBr
PbCO3
BaSO4
Mg3(PO4)2
KOH
NiCl2
NH4OH
Hg2SO4
PbI2
1.
2.
3.
4.
5.
6.
7.
8.
9.
Soluble
Insoluble
Insoluble
Insoluble
Soluble
Soluble
Soluble
Soluble
Insoluble
SOLUBILITY CURVE
• Graph showing the
relationship between amount
of solute and temperature.
• How to read:
• Find temperature
• Find the line for the
compound
• Read the amount of solute
www.sciencegeek.net
INFORMATION FROM THE CURVE
• Unsaturated: More solute can be added into the
solution (under the curve)
• Saturated: the maximum amount of solute is in
solution (on the curve)
• Supersaturated: there is more solute in solution
than at its saturation point (above the curve)
SAMPLE PROBLEMS
Amount change
Temperature Change
• At 10°C, 80 grams of NaNO3 are
dissolved in 100 grams of H2O.
How much NaNO3 would be
dissolved in 75 grams of water at
the same temperature?
A saturated solution of KClO3 is
formed from one hundred grams of
water. If the saturated solution is
cooled from 90°C to 50°C, how
many grams of precipitate are
formed?
80 g NaNO3
100 g H2O
90°C: 47 g KClO3
50°C: 18 g KClO3
=
x g NaNO3
75 g H20
X = 60 g NaNO3
29 g KClO3
NET IONIC EQUATIONS
• Phase indicators
• Solids (s)
• Liquids (l)
• Gases (g)
• Aqueous (aq)
• Solutions which are dissolved in water determine if
the compound breaks into ions
PROCESS TO SOLVE
1. Write a balanced equation with phase indicators
A. Balance for charge (subscripts)
B. Balance for mass (coefficients)
2. Determine if there is an insoluble product
3. Separate all soluble compounds
A. Do not separate
A.
B.
C.
D.
Solids
Pure liquids
Gases
Insoluble compounds
4. Cross out spectator ions
A. Ions that do not directly contribute to the reaction
5. Re-write the equations with ions/cpmpound that contribute to the
reactions
SAMPLE PROBLEM
2
2
____Ni(NO3)2 (aq) + ____NaOH
(aq)  _____ Ni(OH)2 (s) + _____ NaNO3 (aq)
Ni
2+(aq) +2(NO )1-(aq)+2Na1+(aq)+2(OH 1-)(aq)
3
Ni
2+(aq)
+2(OH
 Ni(OH)2(s) + 2Na1+(aq)+ 2(NO3)1-(aq)
1-)(aq) 
Ni(OH)2(s)
NOW YOU TRY!
____ NaCl + ____ AgC2H3O2  ____ NaC2H3O2 + _____ AgCl
COLLIGATIVE PROPERTIES
• Properties dependent on concentration of solute/ions in solution, but
not on the identify of the solute.
• Boiling point elevation
• Freezing point depression
Might have
to do side
𝜟T = m (# of particles) Kb/Kf
calculations
ΔT: temperature change
moles of solute
m: molality (
)
kg of solvent
# of particles: the number of particles the compound breaks into
Kb/Kf: freezing/boiling point constant for solvent
SAMPLE PROBLEM
When 5.0 g of CaCl2 dissolves in 50.0 g of water, what is the
new boiling point of the solution? (Kb Water = 0.512 oC/m)
𝜟T = m (# of particles) Kb
𝟏 𝐦𝐨𝐥𝐞
5.0 g CaCl2 x
𝟏𝟏𝟎.𝟗𝟖 𝒈
Ca x 1: 40.08g
Cl X 2: 70.9
110.98 g/mol
50.0 g H2O x
= 0.045 moles
𝟏 𝐤𝐠
= 0.050 kg
𝟏𝟎𝟎𝟎 𝒈
STEP 2
𝛥T :?
m:
moles of solute
0.045 mole
=
kg of solvent
.050 Kg
= 0.90 mol/Kg
# of particles : 3
Kb: 0.512 oC/m
𝛥T = (0.90 mol/Kg)(3)(0.512 oC/m)
𝛥T =1.38 oC
100+1.38 = 101.38 oC
NOW YOU TRY!
Find the boiling point of a solution containing 6.0 g benzene,
C6H6, in 35 g of napthalene. (Kb of naphthalene = 5.65 oC/m)
EQUILIBRIUM/KINETICS- SESSION 13
• Topics
1. Kinetics
A. Collison theory
B. Reaction rates
2. Equilibrium
A. Constants
B. Le Chatelier Shifts
COLLISION THEORY
• In order to react molecules must
• Impact each other with the
• Sufficient amount of Kinetic energy
• Correct orientation
iverson.cm.utexas.edu
• Concentration
moles
• Molarity M =
liter
KINETICS
•
• Factors that affect
reaction rates
•
•
•
•
•
Concentration
Particle size
Pressure/volume
Temperature
Activation energy
•
•
•
• concentration reaction rate
Particle size
•
surface area reaction rate
Pressure/volume
• volume inc. reaction rate(solutions)
• pressure inc. reaction rate (gases)
Temperature – average K.E.
• temperature inc.
Lower activation energy(Ae)
• Catalyst
• reaction rate
WHAT WILL HAPPEN WHEN?
Scenario
Adding heat.
Removing heat
Adding a catalyst
Diluting a solution
Removing an enzyme
Lowering the temperature
Decreasing the surface area
Increasing concentration of a solution
Breaking reactants into smaller pieces
Increase or Decrease
Increases
Decreases
Increases
Decreases
Decreases
Decreases
Decreases
Increases
Increases
• Equilibrium: when the forward and reverse
reaction rate become constant
• In a closed system ONLY
• K or Keq – equilibrium constant
• Temperature specific
• Aqueous solutions and gases are included in
the Keq
• Concentration of products over concentration
of reactants
EQUILIBRIUM
• Coefficients become the exponents
Keq =
[𝐩𝐫𝐨𝐝𝐮𝐜𝐭𝐬]
[𝐫𝐞𝐚𝐜𝐭𝐚𝐧𝐭𝐬]
aA + bB  cC +dD
Keq =
AaB
CcD
b
d
jahschem.wikispaces.com
EQUILIBRIUM CONSTANTS
• Keq > 1
• Products are favored
• Keq < 1
• Reactants are favored
• Keq = 1
• Neither products nor reactants are favored
SAMPLE PROBLEM
A reaction vessel with a capacity of 1.0 L, in which the following reaction:
SO2(g) + NO2(g) ↔ SO3(g) + NO(g)
Had reached a state of equilibrium, was found to contain 0.40 mol of SO3, 0.30
mol of NO, 0.15 mol of NO2, and 0.20 mol of SO2. Write and calculate the Keq
for this reaction. Which side of the reaction is favored?
Keq =
NO 1 SO3 1
Keq =
SO2 1 NO2 1
[𝐩𝐫𝐨𝐝𝐮𝐜𝐭𝐬]
[𝐫𝐞𝐚𝐜𝐭𝐚𝐧𝐭𝐬]
0.30 1 0.40 1
Keq =
0.20 1 0.15 1
Keq= 4
Products are favored
NOW YOU TRY!
Consider the following equation:
H2(g) + I2(g) ↔ 2HI(g)
Write and calculate the Keq if at 300 K the concentrations are [H2] = 0.40
M, [I2] = 0.45 M and [HI] = 0.30 M. Which side of the reaction is favored?
LE CHATELIER’S PRINCIPLE
• How an equilibrium system reacts when a stress is
placed on the system
• The system will shift in a direction that relieves
the stress
• The equilibrium position changes but NOT the Keq
• Exception is when temperature changes
• Temperature is the ONLY variable that affects the Keq
FACTORS THAT AFFECT LE CHATELIER
• Temperature: treat as a
product/reactant
• Exothermic: temp shifts to left,
Keq increases
• Endothermic: temp. shifts right,
Keq decreases
• Concentration
• Reactants
• concentration
shifts Right
• When removed shifts left
• Products
• concentration shift right
• When removed shifts left
• Volume/Pressure
•
volume Dec. Pressure
• Shifts toward the side with more
moles
•
volume Inc. Pressure
• Shift towards the side with less
moles
WHAT WILL HAPPEN WHEN?
Reaction
Stressor
2 SO2 (g) + O2 (g)⇄ 2 SO3 (g) + energy
decrease temperature
Right
C
increase temperature
+ CO2 (g) + energy ⇄ 2 CO
Shift
N2O4 (g) ⇄ 2 NO2 (g)
increase total pressure
Left
Left
CO
decrease total pressure
No shift
add Fe(s)
No shift
add catalyst
[CO2] is decreased
No shift
Left
add H2O
Right
(s)
(g)
3 Fe
+ H2 O
(s)
(g)
+ 4 H2 O
(g)
⇄ CO2 (g) + H2 (g)
(g)
⇄ Fe3O4 (s) + 4 H2 (g)
2 SO2 (g) + O2 (g) ⇄ 2 SO3 (g)
CaCO3 (s) +170 KJ
⇄ CaO
4 NH3 (g) + 5 O2 (g) ⇄ 4 NO
(g)
(s)
+ CO2
+ 6 H2 O
(g)
(g)
N2(g) + 3H2(g) ↔ 2NH3(g)
Stress
Add N2
Add H2
Add NH3
Remove N2
Remove H2
Remove NH3
Increase
Pressure
Decrease
pressure
Equilibrium
shift
[N2]
[H2]
[NH3]
ACID AND BASE- SESSION 14
Topics
1. Properties
2. Categorizing
A. Arrhenius
B. Bronsted Lowry
C. Lewis
3. Conjugate pairs
4. Water constant (Kw)
5. Calculations
A. pH/pOH
B. Concentration
6. Titrations
PROPERTIES
Acid
• Are electrolytes
• Sour
• pH under 7
• Reacts with metal to produce
H2
• Colorless in phenolphthalein
• Turns blue litmus paper red
Base
• Are electrolytes
• Tastes bitter
• pH over 7
• Feels slippery to the skin
• Pink in phenolphthalein
• Turns red litmus paper
blue
CLASSIFICATION
Arrhenius Theory
Bronsted-Lowry Theory
Lewis Theory
Acid: any
substance which
releases H+ ion in
a water solution
Acid: Any
substance which
donates a proton
Acid: Any
substance which
accepts an
electron pair
Base: Any
substance which
releases OH- ions
in water solution
Base: Any
substance which
can accepts a
proton
Base: Any
substance which
can donate an
electron pair.
DEFINITIONS/CONCEPTS
• Amphoteric: any substance that can act as both an acid or a
base
• Dissociation: ability to dissolve
• Monoprotic: gives off one proton
• Polyprotic: can give off more than one proton
CONJUGATE ACID/BASE PAIRS
• In an equilibrium system acid/base conjugate pairs are
formed
• Acid forms conjugate base
• Base forms conjugate acid
Base
Conjugate Acid
H2SO4 + H2O  HSO4- + H3O+
Acid
Conjugate Base
IDENTIFY THE FOLLOWING
HClO4(aq) + H2O(l) ⇄ H3O+(aq) + ClO4–(aq)
NH3(g) + H2O(l) ⇄ NH4+(aq) + OH–(aq)
HC2H3O2(aq) + H2O(l) ⇄ H3O+(aq) + C2H3O2–(aq)
H2S(g) + H2O(l) ⇄ H3O+(aq) + HS–(aq)
WATER CONSTANT
• Water self ionized into OH- and H3O+
• For pure water the [OH-] = [H3O+]
• Kw is the water constant  1.0 x 10 -14
• [OH-] x [H3O+] = 1.0 x 10 -14
• [OH-] = 1.0 x 10 -7
• [H3O+] = 1.0 x 10 -7
• Basis of the pH scale
• Neutral = 7
• -log [1.0 x 10
-7]
• top of the pH scale is 14
• -log[1.0 x 10
-14]
CALCULATIONS
[H+]= 10
pH = -log [H+]
^ -pH
pH + pOH = 14
pOH = -log [OH-]
[OH-] = 10
^ -pOH
SAMPLE PROBLEM/NOW YOU TRY!
pH
3.78
- log [3.89 x 10–4]
= 3.41
[ H3O1+ ]
10^-3.78 = 1.65 x10
-4M
3.89 x 10–4 M
14-5.19 = 8.81
10^-8.81
14-5.31 = 8.68
10^-8.68
8.46
pOH
=
1.5 x 10-9M
= 2.04 x
10-9M
10^-8.46 = 3.47 x 10-9M
[ OH1– ]
14- 3.78 = 10.22 10^-10.22 =6.02 x 10-11 M
ACID or
BASE?
Acid
10^-10.59 = 2.57 x 10-11M
Acid
5.19
10^-5.19 = 6.4 x 10-6M
Base
-log [4.88 x 10–6]
= 5.31
4.88 x 10–6 M
Base
14-8.46= 5.54
10^-5.54 = 2.88 x 10-6 M
Base
14-3.41 = 10.59
TITRATIONS
• Purpose is to determine an unknown concentration of
an acid or base solution
• This is done by slowly adding a solution of a known
concentration (titrant) into a reaction vessel, with
solution of unknown concentration and indicator,
until the equivalence point is reached
• Equivalence point: moles of acid = moles of base
• Indicators: chemical change color according to the pH
range of the solution
TITRATION SETUP
intranet.tdmu.edu.ua
www.scimath.org
CALCULATIONS
If it takes 54 mL of 0.1 M NaOH to neutralize 125 mL
of an HCl solution, what is the concentration of the
HCl?
MAVA =MBVB
MA: ?
VA: 125 mL x
1𝐿
= 0.125L
1000 mL
MB: 0.1 M
• MAVA =MBVB
MA: molarity of acid
VA: Volume of acid
MB: molarity of base
VB: volume of base
1L
VB: 54 mL x
= 0.054 L
1000 mL
(X)(.125L HCl) = (0.1MNaOH)(0.054 L NaOH)
(0.1 M NaOH)(0.054 L NaOH)
X=
0.125 L HCl
X = 0.0432 M
NOW YOU TRY!
How many milliliters of 0.360 M H2SO4 are required to
neutralize 25.0 mL of 0.100 M Ba(OH)2?