Chapter 4
Chemical Reactions and Solution
Stoichiometry
Water:
The Common Solvent
• Essential for most
biological reactions
• Polar covalent molecule
• Partial charges create
poles (dipole moments)
• Polarity gives water the
ability to dissolve
compounds
Water
Hydration of a Solute
Hydration-the interaction between solute particles and water molecules
Water
Hydration of a Solute
• When ions dissolve,
they break up into
individual ions
• NH4NO3(s)
NH4+(aq) + NO3-(aq)
• HOWEVER, solubility varies
greatly among ionic compounds
– Solubility depends on relative
attractions of ions for each other (to
be covered in CHpt 11)
• Take home message:
– An ionic solid dissolves into dispersed
ions in water upon hydration
Non-ionic solubility
• Non ionic compounds
can be soluble if they
have a polar bond.
• Examples:
– ethanol C2H5OH
– Methanol CH3OH
– Acetone (CH3)2CO
• Rule of Thumb
– Polar molecules are
soluble, “like dissolves
like”
Non soluble Compounds/ions
• Carbonate (CO32- )
• Phosphate (PO43- )
• ETC………………………………
……………………………………
……………………………………
……………………………………
•
Non-polar molecules (animal fat)
Characterization of Solutions
• Electrical Conductivity
– Strong electrolytes: dissolve and completely ionize
in water
• Strong conductors of electricity
• Soluble Salts, Strong Acids, Strong Bases
– Weak electrolytes: small degree of ionization
• Weak conductors of electricity (but STILL CONDUCT!!)
• Weak acids (Acetic Acid), weak bases (NH3)
– Non-electrolytes: dissolve but no ionization
• Ie; Sucrose (C12H22O11), Ethanol (C2H5OH)
Composition of Solutions
• Molarity=
Moles of solute
liters of solution
• What is the molarity of a
0.30 liter solution
containing 0.50 moles of
NaCl?
• How many liters of solution
are needed to make a 1.66
M solution containing 2.11
moles of KMnO4?
Composition of Solutions
• What is the molarity in 650. ml of solution
containing 63 grams of NaCl?
• How many grams of Ca(OH)2 are needed to
produce 500. ml of 1.66 M Ca(OH)2 solution?
• What volume of a 0.88 M solution can be
made using 130. grams of FeCl2?
Composition of Solutions
• Dilutions
•
Moles solute before = moles after
M1V1=M2V2
• How do you prepare a
250.-ml of a 2.35 M HF
dilution from a 15.0 M
stock solution?
• If 65.5 ml of HCl stock
solution is used to make
450.-ml of a 0.675 M
HCl dilution, what is the
molarity of the stock
solution?
Tonight’s Homework
• Worksheet I just handed you 
• Pg 180-#’s 15,16,21-25
• Read Chapter 4, Sections 1-5 (20 pgs)
– DO IT OR YOU WILL BE BEHIND TOMORROW!!!!
Types of Solutions
• Precipitation Reactions- aqueous solutions
form an insoluble precipitate (duh…)
– Involves using solubility rules (pg 150 Table 4.1) to
determine the solid precipitate.
FOR EXAMPLE
Pb(NO3)2(aq) + KCl(aq)
Pb(NO3)2(aq) + KCl(aq
)
• We know that:
– Most nitrates salts are soluble.
– Most Group I salts are soluble.
– Most Chloride salts are soluble with the exception
of Pb+2 and others.
– Therefore, we determine the products to be:
Pb(NO3)2(aq) + KCl(aq)  PbCl2(s) + KNO3(aq)
Describing reactions/
Writing equations
• Molecular Equation:
• Pb(NO3)2(aq) + 2KCl(aq)  PbCl2(s) + 2KNO3(aq)
• Complete ionic Equation
• Pb+(aq) + 2NO3-(aq) + 2K+(aq) + 2Cl-(aq)  Pb+2(s) + 2Cl-(s) + 2K+(aq) + 2NO3-(aq)
• Net Ionic Equation (get rid of the spectator ions)
– Spectator ions: present in the aqueous solution before and after
reaction. In this case, K+and NO3-
– Indicate ions that form precipitate.
Pb+(aq) + 2Cl-(aq)  PbCl2(s)
Solution Stoichiometry
• Steps to solving Solution Stoichiometry
– 1. Identify Species, determine reaction
– 2. Write balanced net ionic equation
– 3. Calculate moles of reactants
– 4. Determine limiting reactant
– 5. Calculate moles of product
– 6. Convert to proper unit, sig figs, etc…
Calculate the mass of solid NaCl that must be added to 1.5L of a
.100M AgNO3 solution to precipitate all the Ag+ ions in the form
of AgCl
•
•
•
•
•
•
Step 1: NaCl(s) + AgNO3(aq)  AgCl(s) + NaNO3(aq)
Step 2: Ag+(aq) + Cl-(aq)  AgCl(s)
Step 3: 1.5L x .1M Ag+ = ?
Step 4?
Step 5?
Step 6?
Calculate the mass of solid NaCl that must be added to 1.5L of a
.100M AgNO3 solution to precipitate all the Ag+ ions in the form
of AgCl
•
•
•
•
•
•
1Ag+(aq) + 1Cl-(aq)  1AgCl(s)
Step 3: Using a 1:1 mole ratio from the balanced
equation
1.5L x .1M Ag+ = .150 mol Ag+
.150 mol NaCl x (58.45g NaCl/1mol)= 8.77g NaCl
Step 4?
Step 5?
Step 6?
More homework 
• Pg 181 #’s 29-36, 39-44
Acid-Base Reactions
• Acids-produce H+ ions when dissolved in
water. A proton donor.
• Bases-produce OH- ions when dissolved in
water. A proton acceptor
H+(aq) + OH-(aq)  H20(l)
Solution Stoichiometry for Acid-Base
Reactions
• Steps to solving Solution Stoichiometry
– 1. Identify Species, determine reaction
– 2. Write balanced net ionic equation
– 3. Calculate moles of reactants
– 4. Determine limiting reactant
– 5. Calculate moles of product
– 6. Convert to proper unit, sig figs, etc…
Solution Stoichiometry for Acid-Base
Reactions
FOR EXAMPLE
What Volume of a .100M HCl solution is needed to
neutralize 25.0 mL of .350M NaOH?
In reality, all you are trying to do is match the moles of
hydroxide ions with moles of hydrogen ions.
• 1. Identify Species, determine reaction
• 2. Write balanced net ionic equation
• Involves using solubility rules (pg 150 Table 4.1) to
determine a solid precipitate and spectator ions
What Volume of a .100M HCl solution is
needed to neutralize 25.0 mL of .350M NaOH?
• 1. Identify Species, determine reaction, mole
ratio
HCl(aq) + NaOH(aq)  NaCl(s) + H20(l)
– Mole ratio of 1:1:1:1
– Remember your solubility rules,
– Remember that you cannot dissolve water in
water, so water is always in the “liquid” state of
matter. (Sounds stupid but you would be
surprised).
What Volume of a .100M HCl solution is
needed to neutralize 25.0 mL of .350M NaOH?
• 2. Write balanced net ionic equation
• Involves using solubility rules (pg 150 Table 4.1) to
determine spectator ions.
– Complete ionic equation:
H+(aq) + Cl-(aq) + Na+ (aq) + OH- (aq)  H2O(l) + Na+(aq) + Cl-(aq)
– Na+ and Cl- are soluble, so they exist as ions in solution.
Therefore, the actual reaction does not take place. They are
spectator ions.
• H+(aq) + OH-(aq)  H2O(l)
What Volume of a .100M HCl solution is
needed to neutralize 25.0 mL of .350M NaOH?
• 3. Calculate moles of reactants (don’t forget your
mole ratios in this step!)
–M=n/V
– .350M HCL = n / 0.025L
– n= .00875 moles  account for sig figs 8.75 x 10-3
moles (preferred by AP graders)
What Volume of a .100M HCl solution is needed to
neutralize 25.0 mL of .350M NaOH?
• 4. Determine limiting reactant
– We are adding a reactant, therefore we do not
care about a limiting reactant. But in reality it is a
1:1 mole ratio so there isn’t one.
What Volume of a .100M HCl solution is
needed to neutralize 25.0 mL of .350M NaOH?
• 5. Calculate moles of reactant needed (HCl solution)
• We have a 1:1 mole ratio of hydrogen ions to hydroxide
ions.
• Moles H+ = Moles OH- in order to reach neutralization
point
• Therefore, we need 8.75 x 10-3 mol H+ (present in HCl)
What Volume of a .100M HCl solution is needed to
neutralize 25.0 mL of .350M NaOH?
• 6. Convert to proper unit, sig figs, etc…
– The question asks us for the volume of .1M HCl
– So:
M=n/V  V=n/M
VHCl= 8.75 x 10-3 mol H+
.100 mol H+ / L
VHCl= 8.75 x 10-2
More homework…
• Pg 182 #’s 45-50
Acid-Base titrations
• Key vocabulary
– Volumetric Analysis-technique for determining the amount
of substance through titration
– Titration-delivery of a measured volume of a known
solution into a substance of unknown concentration
• Titrant
• Analyte
– Equivalence point-the point at which enough titrant has
been added to react exactly with the analyte
– Indicator-a substance that changes color at or near the
equivalence point
– Endpoint-point actual point where the indicator changes
color
Acid-Base titrations
• Requirements for sucessful titrations
– 1. exact reaction between titrant and analyte must
be known
– 2. Equivalence point must be marked accurately
(.15 mL of titrant added, etc…)
– 3. Volume of titrant required to reach equivalence
point must be known accurately (to calculate
percent error)
Acid-Base Titrations
• Basically, a titration is a neutralization
reaction, so we follow the same rules as
before.
– The goal is to react an equal hydrogen ions to
react with hydroxide ions.
Acid-Base Titrations
For Example:
A student dissolves 1.3009g KHC8H4O4 (KHP) in
distilled H20, adds Phenolphthaleine as an
indicator, and titrates with NaOH. It was found
that 41.20mL of NaOH is required to react
with the 1.3009g KHP. What is the
concentration of the sodium hydroxide?
A student dissolves 1.3009g KHC8H4O4 (KHP) in distilled H20, adds
Phenolphthaleine as an indicator, and titrates with NaOH. It was found that
41.20mL of NaOH is required to react with the 1.3009g KHP. What is the
concentration of the sodium hydroxide?
– 1. Identify Species, determine reaction, mole
ratios
• KHC8H4O4(aq) + NaOH(aq)  KOH + NaHC8H4O4(aq)
– 2. Write balanced net ionic equation
• K+ + HC8H4O4-(aq) + Na+ + OH-(aq)  K+ + OH- + NaHC8H4O4(aq)
Oxidation-Reduction Reactions
• Redox reaction-one or more electrons are
transferred
• Oxidation state (number)-the imaginary
charges the atoms would have if the shared
electrons were divided equally between
identical atoms bonded to each other (huh?)
Rules for assigning Oxidation States
• Pg 167, Table 4.2 of
your textbook
• Memorize it…
• More importantly,
memorize
exceptions to the
norm
Oxidation number Practice
• Assign Oxidation states to all atoms in the following
–
–
–
–
–
–
CO2
SF6
NO3CH3H2O2
H2O
–
–
–
–
–
–
KMnO4
NiO2
K4Fe(CN)6
N2H4
FeO4
Na2C2O4
– START WITH FIRST DRAWING THE MOLECULE, oxidation
states do not have to be whole numbers (for some atoms)
Oxidation-Reduction Reactions
• Oxidation-increase in oxidation state (loss of
electron). Ie, going from -1 to +1
• Reduction- decrease in oxidation state (gain of
an electron). Ie, going from +1 to -1
• Oxidizing agent-the electron acceptor
• Reducing agent-electron donor
Identification of
Oxidizers, reducers
• 2Al (s) + 3I2  2AlI3
– Step 1: balance equation, assign oxidation states
to reactants
– Step 2: assign oxidation states to products
– Step 3: Identify who gained electrons (reducing
agent
– Step 4: Identify who lost electrons (oxidizing
agent)
This was an easy problem
Identification of
Oxidizers, reducers
• Identify the oxidizing and reducing agents
– PbO(s) + CO2(g)  Pb(s) + CO2(g)
– CH4(g) + O2(g)  CO2(g) + H2O(l)
– Cr2O72-(aq) + HCl  ZnCl2(aq) + H2(g)
Homework 
• Pg 182-183, #s 57-62