Types of Chemical Reactions and
Solution Stoichiometry
Aqueous Solutions : Water
Most reactions take place in aqueous solutions. Water is the
dissolving medium.
Water
 Polar molecule- gives the ability to dissolve many
compounds
 Hydration- process of separating and surrounding particles of
an ionic solid by water
 The O-H bonds are covalent
 (aq) means surrounded by water molecules
 NH4NO3  NH4+ + NO3-
Solutions
 Solution- Homogeneous substance
 Solute- dissolving substance; in lesser quantity;
changes phase
 Solvent- dissolving substance; in greater quantity
 Solubility of ionic substances in water depends
upon the relative attractions of the ions for each
other and the attractions of the ions for water.
Generally if the substance is polar, most likely it will
dissolve in water. “Like dissolves like.”
Electrolytes
 Electrolytes- ions in water
 Strong electrolyte-
conducts electricity
 Soluble salts- ionic soluble solids break down to form ions
 Strong acids- completely dissociates its ions
 Binary acids- HCl, HBr, HI; all others are weak
 Ternary acids- If the number of oxygen out number the number of
hydrogen by two or more the acid is strong; H2SO4, HClO4, HClO3,
HNO3
 Strong Bases- soluble hydroxides; Group IA, Group IIA
from calcium down
Weak and Non-electrolytes
 Weak electrolyte- weakly conducts electricity
o Weak acids
o Weak bases
 Nonelectrolyte- does not conduct electricity
o Highly insoluble salts
Arrhenius Definition
 Arrhenius
 Acid- substance containing ionizable H+
 Base- contains OH Ex. Write an equation for the dissociation of sodium
hydroxide.
 Ex. Write and equation for the dissociation of acetic
acid.
Examples
 Ex. Write the net ionic equation for:
 Aqueous potassium chloride is added to aqueous
silver nitrate to form silver chloride plus aqueous
potassium nitrate
 Aqueous potassium hydroxide is mixed with aqueous
iron (III) nitrate to form a precipitate of iron (III)
hydroxide and aqueous potassium nitrate
Concentrations
 Common Terms of Solution Concentration
 Stock- routinely used solutions prepared in
concentrated form
 Concentrated- relatively large ratio of solute to solvent
 Dilute- relatively small ratio of solute to solvent
Concentrations
 Molarity- unit of concentration
 M = moles of solute/ L of solution
(given before dissociation)
 NaCl  Na+ + Cl1M
1M
1M
 Co(NO3)2  Co2+ + 2 NO3 0.50 M
0.50 M 1. 0 M
Ex. 1 M NaCl
Example
 Ex. What are the concentrations of each ion in a
solution of 0.32 M Ba(NO3)2?
Dilution
 M = mols/L
 Mol = M x L or MV
 Dilution
 Moles before dilution = moles after dilution
 M1V1= M2V2
Precipitation Reactions
 Precipitation reaction- precipitate forms- insoluble
substance formed from a reaction
 Solubility rules
Concentration
 Weight percent- concentration unit; used in
medical field
 # of grams of solute/ 100 g of water
 Ex. How would you prepare a 5% glucose
solution?
Net Ionic Equations
 Molecular Equation
 Ionic Equation
 Net Ionic Equation
 Spectator Ions
Stoichiometry of Chemical Reactions
 Ex. Calculate the mass of solid KCl necessary to
precipitate all of the Ag+ ions from a solution of 0.75
L of a 0.15 M AgNO3 solution.
Stoichiometry
 Limiting Reactant- Must take into consideration
which one is limiting.
 Ex. When aqueous solutions of Na2SO4 and
Pb(NO3)2 are mixed, PbSO4 precipitates. Calculate
the mass of PbSO4 formed when 1.25 L of 0.0500 M
Pb(NO3)2 and 2.00 L of 0.0250 M Na2SO4 are
mixed.
 Bronsted-Lowry Acid-Base Theory

 KOH + HC2H3O2  KC2H3O2 + H2O
 The hydroxide ion is a very strong base and can strip
the proton from a weak electrolyte like HC2H3O2
 OH- is base and HC2H3O2 is acid.
 Acid-Base Reaction= Neutralization Reaction
 When exactly the same amount of acid and base has
been added (according to the coefficients in the
balanced equation) the acid has been neutralized
Example
 Ex. What volume of 0.100 M HCl solution is
needed to neutralize 25.0 mL of a 0.350 M NaOH
solution?
Example
 Ex. In a certain experiment 28.0 mL of 0.250 M
HNO3 and 53.0 mL of 0.125 M KOH are mixed.
Calculate the amount of water formed in the
resulting reaction. What is the concentration of H+
and OH- ions in excess after the reaction goes to
completion?
Monoprotic and Diprotic Acids
 Monoprotic acid- contains only one ionizable
hydrogen Ex. HCl, HNO3, HC2H3O2
 Diprotic acid- contains two ionizable hydrogens
Ex. H2SO4
Titrations
 Titration- delivery of a measured volume of a solution of
known concentration (titrant) into a solution containing the
substance being analyzed (analyte).
 Point where enough titrant has been added is the point
where the indicator actually changes color is called
endpoint.
 Use an indictor to determine where the point is
 Acid/Base Titration
 Common indicator for strong acids and bases is
phenolphthalein
o Turns from colorless (acid) to pink (base)
 Before beginning the titration, the titrant is measured for
exact concentration, called standardizing the solution. Use a
solid to get exact number of moles of acid or base. Does not
matter how much water is used.
Example
 Ex. A sample of an analgesic drug was analyzed
for aspirin, a monoprotic acid, HC9H7O4, by titration
with a base. In a titration, a 0.500 g sample of the
drug required 21.50 mL of 0.100 M NaOH for
complete neutralization. What percent by mass of
the drug was aspirin?
Example
 Ex. A student carries out an experiment to standardize
a sodium hydroxide solution. To do this the student
weighs out a 1.3009 g sample of potassium hydrogen
phthalate (KHP, MM=204.22), a compound with the
formula KHC8H4O4 which has one acidic hydrogen. The
student dissolves the KHP in distilled water and titrates it
with the sodium hydroxide solution. The difference
between the final and initial buret readings indicates that
41.20 mL of sodium hydroxide solution is required to
react with the 1.3009 g of KHP. Calculate the
concentration of the sodium hydroxide solution.
Example
 Ex. In a reaction involving Ba(OH)2 and HCl,
determine the number of mL of a 0.50 M solution of
Ba(OH)2 necessary to neutralize 25.0 mL of a 0.35
M solution of HCl.
Redox
 Electrons are transferred- called oxidation-
reduction reactions or redox.
 Use oxidation numbers (states) to keep track of
electron transfer.
Oxidation numbers
 Rules for assigning oxidation states:
1. The oxidation state of an atom in an element is 0.
2. The oxidation state of a monatomic ion is the same as its
3.
4.
5.
6.
charge.
In its compounds, fluorine is always assigned an oxidation
state of –1.
Oxygen is usually assigned an oxidation state of –2 in its
covalent compounds. Exceptions are peroxides O22- and in
OF2 where oxygen is +2.
In its covalent compounds with nonmetals, hydrogen is
assigned an oxidation state of +1.
The sum of the oxidation states must be 0 for an
electrically neutral compound. For an ion, the sum of the
charges must equal the charge on the ion.
Examples
 Ex. Assign oxidation numbers for:
 CH4, SO3, SO42-, Fe3O4, C12H22O11
Oxidation Reductions Reactions
 Oxidation – increase in oxidation state ( a loss of
electrons)
 Reduction- decrease in oxidation state ( a gain of
electrons)
 Oxidizing agent- electron acceptor- substance
being reduced
 Reducing agent- electron donor- substance being
oxidized.
Redox
 When identifying what is oxidized or reduced use
specific element.
 When identifying oxidizing or reducing agent
identify whole compound.
Example
 Ex. 2Al + 3 I2  2 AlI3
 Identify the substance oxidized, reduced, oxidizing
agent and reducing agent.

Balancing Redox Equations- Acid
Solutions
 Balancing Redox Reactions
 In acidic solutions, there are excess hydrogen ions and
water molecules:
1. Break into two half reactions: one reduction, one
oxidation
2. Write half reactions.
3. Balance all elements other that H and O.
4. Balance oxygen using H2O.
5. Balance hydrogen by using H+.
6. Balance charge using electrons.
7. If needed multiply to make sure all electrons lost by
one substance are gained by another. Same number of
electrons lost and gained.
Example
 Ex. MnO4- + Fe2+  Fe3+ + Mn2+
Balancing Redox Equations- Basic
Solutions
 In basic solutions, there are excess hydroxide ions
and water molecules:
1. Balance according to acid rules.
2. For every H+ add OH- to both sides of the
equation.
3. Form water on side containing H+ and OH-.
 Add together and eliminate species on both sides
Example
 Ex. Ag + CN- + O2  Ag(CN)2-