Transcript Lecture 2 - Chemistry at Winthrop University

Course Logistics
• Textbook
– An online option is available for the Campbell and Farrell text
• Problem Sets
– The problem sets will consist of problems very similar to
those assigned on the class schedule
– They are due on the date posted on the class schedule
• CHEM108
– The lab course is a corequisite and you must pass the lab
class to receive a grade for CHEM106
• Chemistry tutors
– They should start next week
• Amino Acids
– Start learning them now.
– By the end of next week, you should know the chemical
structure of: A, V, S, D, E, R, K, C, H, F
Law of Conservation of Matter
• “Matter can neither be created nor destroyed”
– Antoine Lavoisier, 1774
If a complete chemical reaction has occurred, all of the
reactant atoms must be present in the product(s)
Law of Conservation of Matter
a)
b)
•Stoichiometric coefficients are necessary to balance the
equation so that the Law of Conservation of Matter is not
violated
•6 molecules of Cl2 react with 1 molecule of P4
•3 molecules of Cl2 react with 2 molecules of Fe
Example of Using Stoichiometric
Coefficients
Balancing Chemical Reactions
• Let’s look at Oxide Formation
• Metals/Nonmetals may react with
oxygen to form an oxide with the
formula MxOy
• Example 1: Iron reacts with oxygen to
give Iron (III) Oxide
Fe (s) + O2 (g) → Fe2O3 (s)
How do we solve it?
Fe (s) + O2 (g) → Fe2O3 (s)
• Step 1: Look at the product. There are 3
atoms of oxygen in the product, but we
start with an even number of oxygen
atoms.
– Let’s convert the # of oxygens in the product
to an even number
Result:
Fe (s) + O2 (g) → 2Fe2O3 (s)
How do we Solve It?
Fe (s) + O2 (g) → 2Fe2O3 (s)
• Then, balance the reactant side and
make sure the number/type of atoms on
each side balance.
Balanced Equation: 4Fe (s) + 3O2 (g) → 2Fe2O3 (s)
How do we Solve It?
• Example 2: Sulfur and oxygen react to
form sulfur dioxide.
S (s) + O2 (g) → SO2 (g)
• Step 1: Look at the reaction. We lucked
out!
Balanced Equation: S (s) + O2 (g) → SO2 (g)
How do we Solve It?
• Example 3: Phosphorus (P4) reacts with oxygen to
give tetraphosphorus decaoxide.
P4 (s) + O2 (g) → P4O10 (s)
• Step 1: Look at the reaction. The
phosphorus atoms are balanced, so let’s
balance the oxygens.
Balanced Equation: P4 (s) + 5O2 (g) → P4O10 (g)
How do we Solve It?
• Example 4: Combustion of Octane
(C8H18).
C8H18 (l) + O2 (g) → CO2 (g) + H2O (g)
• Step 1: Look at the reaction. Then:
– Balance the Carbons
C8H18 (l) + O2 (g) → 8CO2 (g) + H2O (g)
How do we Solve It?
C8H18 (l) + O2 (g) → 8CO2 (g) + H2O (g)
• Step 2: Balance the Hydrogens
C8H18 (l) + O2 (g) → 8CO2 (g) + 9H2O (g)
• Step 3: Balance the Oxygens
– Problem! Odd number of oxygen atoms
• 12.5 Oxygens on reactant side
– Solution: Double EVERY coefficient (even
those with a value of ‘1’)
How do we Solve It?
C8H18 (l) + 12.5O2 (g) → 8CO2 (g) + 9H2O (g)
• Step 3 (cont’d): Balance the Oxygens
2C8H18 (l) + 25O2 (g) → 16CO2 (g) + 18H2O (g)
• Step 4: Make sure everything checks out
Review of Balancing Equations
Aqueous Solutions and
Precipitation Reactions
Terminology:
1. Soluble substance: A
substance that dissolves to
a significant extent in a
specific solvent
•
Electrolytes: Strong acid,
strong base, soluble ionic
compounds
2. Insoluble substance: A
substance that does not
dissolve significantly in a
specific solvent
Soluble K2CrO4 (yellow)
Soluble AgNO3 (clear)
Insoluble Ag2CrO4 (red)
Electrolytes
• Remember that when dealing with strong
acids and strong bases, the two will react to
make a salt and water
– Strong acids:
• Any hydrohalogen acid
• Nitric, Chloric and sulfuric acids
– Strong bases:
• Group I hydroxides
• Group 2, Period 3 or higher hydroxides
• Group 1/Group 2 Oxides
• To determine how much strong acid or base
to add necessary to neutralize a strong base
or acid, you’ll need to know what?

Precipitation Reactions
• In a Precipitation Reaction,
an insoluble product forms
when we mix two
electrolyte solutions
• What determines if
something precipitates
when the solutions are
mixed?
– Intermolecular forces
between the solute and
solvent
– The entropy change that
occurs as a result of
solvation
If we mix the soluble solutions
NaCl (aq) and AgNO3 (aq)
How do we determine if a precipitate forms?
Step 1: Write the balanced chemical equation for
the double displacement reaction
AB + CD --> AD + BC
• Remember the charges on the ions
Step 2: Using the Solubility rules, determine if
either product is insoluble
– If all products are insoluble, then no reaction occurs
Step 3: Write the Complete and Net Ionic
equations for the reaction
Complete and Net Ionic Equations
• A Complete Ionic Equation shows all
chemical species present in the reaction
• A Net Ionic Equation shows the net
change taking place in the reaction
– The Net Ionic Equation is made by taking
the Spectator Ions out of the complete
ionic equation

Solubility Equilibria
• When we put an ionic compound in a
solvent, even if it is considered to be
insoluble, some small number of ions
dissolve into the bulk phase
• For an ionic solid of formula AB, this can
be expressed as the equilibrium
reaction:
AB (s)  A+ (aq) + B- (aq)
Solubility Product
• The equilibrium constant for the solubility
equilibrium of an ionic solid is called the
solubility product, Ksp
AB(s)  mA+ (aq) + nB- (aq)
Ksp = [A+]m [B-]n
We don’t put [AB] in the denominator, why?
Solubility Product
•The smaller the value of Ksp, the less soluble an ionic solid is
•We calculate Ksp using Molar Solubility
•Molar concentration of the compound in a saturated solution

Solubility Product
• We can use the solubility products and
molar solubilities to predict how
precipitation will proceed
• Remember:
Low Ksp, Low Solubility
The Common Ion Effect
We can use Le Chatelier’s Principle to find a
way to remove ions from solution
– Application: Heavy metal cleanup
CdCO3  Cd2+ + CO32If we add potassium carbonate to the solution,
what will happen?
The decrease in CdCO3 soubility caused by
addition of another carbonate salt is an
example of The Common Ion Effect