Types of Chemical Reactions
and Solution Stoichiometry
Chapter 4- Types of Chemical
Reactions and Solution
Stoichiometry
Pg. 133-187 in text
The Bottom Line:Chapter 4
• Aqueous reactions account for virtually
all chemistry that takes place in living
systems.
Water, the Common Solvent
• Shape- bent ~1050
• Electrons not
distributed evenly
• Polar molecule
• Likes dissolve likes
– Polar and ionic
– Polar and polar
– Nonpolar and
nonpolar
Likes Dissolve Likes
State whether each pair of substances will
mix. State Why or why not.
a. NaNO3 and H2O
b. C6H14 and H2O
c. I2 and C6H14
d. I2 and H2O
Polar Water Molecules Interact with the
Positive and Negative Ions of a Salt
Water Also Dissolves Many
Nonionic Substances
Why is Ethanol
(C2H5OH) very
soluble in water?
• Ethanol contains a
polar -OH bond like
those in water
Polar Bond
Water Does Not Dissolve all
Nonionic Substances
Water = Polar
Fats= Nonpolar
Likes dissolve likes
Hydrophilic /
Hydrophobic
Solute / Solvent
• Solute:
1. If it and the solvent are present in the same
phase, it is the one in lesser amount.
2. If it and the solvent are present in different
phases, it is the one that changes phase.
3. The one that dissolves into the solvent.
Aqueous solution means that water is the
solvent.
Strong, Weak, Non-Electrolytes
Electrolyte
Conductivity Degree of
Examples
Dissociation
Strong
Strong acids, many
salts, strong bases
Weak
Weak organic
acids, weak bases
Non
Highly nonpolar
organic compounds
like sugar, insoluble
salts like AgCl
Acids
Strong acids hydrochloric and sulfuric
acid
Weak acids –
acetic and formic acid
Bases
Strong bases –
sodium hydroxide
Weak bases –
ammonia
Strong AcidsVirtually every molecule
dissociates into given
ions
Strong Bases- soluble
compounds containing
the hydroxide (-OH-) ion
Group IA metal
hydroxides = LiOH,
NaOH, KOH, RbOH,
CsOH
Heavy group 2 metal
hydroxides =
Ca(OH)2, Sr(OH)2,
Ba(OH)2
List whether each of the following is a
strong, weak, or nonelectrolyte
a. HClO4
b. C6H12
c. LiOH
d. NH3
e. CaCl2
f. HC2H3O2
Common Terms of Solution
Concentration
Stock - routinely used solutions prepared
in concentrated form.
Concentrated - relatively large ratio of
solute to solvent. (5.0 M NaCl)
Dilute - relatively small ratio of solute to
solvent. (0.01 M NaCl)
The Composition of Solutions
• Molarity = moles of solute / liters of solution.
Example 4.1 pg. 96:
Calculate the molarity of a solution prepared by bubbling
1.56 g of gaseous HCl into enough water to make 26.8
mL of solution.
Standard Solutions- a solution whose
concentration is accurately known.
• Example 4.4 pg. 97:
To analyze the alcohol content of a certain wine,
a chemist needs 1.00 L of an aqueous 0.200
M K2Cr2O7 (potassium dichromate) solution.
Describe how to prepare this solution.
Ion Concentration in Solutions
Determine the molarity of Fe3+ ions and
SO42- ions in a solution prepared by
dissolving 48.05 g of Fe2(SO4)3 in
enough water to make 800. mL of
solution.
Dilution- water is added to a stock
concentrated stock solution to achieve
desired molarity.
Concept:
Moles of solute after dilution = moles of solute before dilution
M1V1 = M2V2
Example:
What volume of 12 M hydrochloric acid
must be used to prepare 600. mL of a
0.30 M HCl solution?
Types of Chemical Reactions
o
Precipitation reactions
AgNO3(aq) + NaCl(aq)  AgCl(s) +NaNO3(aq)
o
Acid-base reactions
NaOH(aq) + HCl(aq)  NaCl(aq) + H2O(l)
o
Oxidation-reduction reactions
Fe2O3(s) + Al(s)  Fe(l) + Al2O3(s)
Precipitation Reactions
• Remember: In virtually every case,
when a solid containing ions dissolves
in water, the ions separate and move
around independently
MgSO4 (aq) + Na2CO3 (aq)
Predicting whether or not a precipitation
reaction will occur depends on the solubility
of the potential end products.
Memorize:
1.
2.
3.
4.
5.
6.
Most nitrate (NO3) salts are soluble.
Most alkali (group 1A) salts and NH4+ are soluble.
Most Cl, Br, and I salts are soluble (NOT Ag+,
Pb2+, Hg22+)
Most sulfate salts are soluble (NOT BaSO4, PbSO4,
HgSO4, CaSO4)
Most OH salts are only slightly soluble (NaOH,
KOH are soluble, Ba(OH)2, Ca(OH)2 are marginally
soluble)
Most S2, CO32, CrO42, PO43 salts are only
slightly soluble.
The Reaction of K2CrO4(aq) and
Ba(NO3)2(aq)
Describing Reactions in Solution
The
Reaction of
KCI(aq) and
AgNO3(aq)
Complete and balance the following
reactions, determining, in each case, if
a precipitate is formed.
a.
b.
c.
d.
KCl (aq) + Pb(NO3)2 (aq) -->
AgNO3 (aq) + MgBr2 (aq) -->
Ca(OH)2 (aq) + FeCl3 (aq) -->
NaOH (aq) + HCl (aq) -->
Describing Reactions in
Solution
1.
Molecular equation (reactants and
products as compounds)
AgNO3(aq) + NaCl(aq)  AgCl(s) + NaNO3(aq)
2.
Complete ionic equation (all strong
electrolytes shown as ions)
Describing Reactions in
Solution (continued)
3.
Net ionic equation (show only
components that actually react)
Write the molecular, complete
ionic, and net ionic forms
Aqueous nickel (II) chloride reacts with
aqueous sodium hydroxide
Selective Precipitation
Precipitation reactions allow us to target
specific substances, and separate and
recover them from a solution.
Example:
A solution contains Ca2+, Cu2+, and Pb2+.
What anions can we add, and in what
order , to separate and recover each
cation?
Stoichiometry of Precipitation Reactions
• Solving problems involving precipitates from solutions
makes use of molarity, solubility rules, balancing
equations, and limiting reactant calculations.
• Take a systematic approach!
Example:
What mass of precipitate is produced when 35.mL of a
0.250 M Fe(NO3)3 solution is mixed with 55 mL of a
0.180 M KOH solution?
Performing Calculations for
Acid-Base Reactions
1.
2.
3.
4.
5.
List initial species and predict reaction.
Write balanced net ionic reaction.
Calculate moles of reactants.
Determine limiting reactant.
Calculate moles of required
reactant/product.
6. Convert to grams or volume, as required.
Key Titration Terms
Titrant - solution of known concentration
used in titration
Analyte - substance being analyzed
Equivalence point –
Endpoint -
Acid-Base Reactions
• Acid =
• Base =
Neutralization of a Strong Acid- products will
always be water and a salt.
Example 1: How many mL of a 0.800M NaOH
solution is needed to just neutralize 40.00 mL
of a 0.600M HCl solution?
Example 2:
You wish to determine the molarity of a solution
of sodium hydroxide. To do this, you titrate a
25.00 mL aliquot of your sample, which has
had 3 drops of phenolphthalein indicator
added so that it is pink, with 0.1067 M HCl.
The sample turns clear (indicating that the
NaOH (aq) has been precisely neutralized by
the HCl solution) after the addition of 42.95
mL of the HCl. Calculate the molarity of your
NaOH solution.
Oxidation-Reduction Reactions
Redox reactions involve a transfer of
electrons. In order to determine if an
electron has been transferred, one must
be able to assign oxidation states.
Assigning oxidation states to an element
in a molecule requires knowledge of a
set of rules. These rules are outlined on
page 120 of your text.
Rules for Assigning Oxidation States
1.
OS of an atom in an element is 0.
2.
OS of a monatomic ion is the same as its charge.
3.
In its covalent compounds with nonmetals,
hydrogen is assigned an OS of +1.
4.
Oxygen is assigned an OS of -2 in its covalent
compounds.
The exception to this rules occurs in peroxides
(compounds contains the O22- group), where each
oxygen is assigned an OS of -1.
5. In binary compounds the element with
the greater attraction for the electrons in
the bond is assigned a negative OS
equal to its charge in its ionic
compounds.
6. The sum of the oxidation states must
be zero for an electrically neutral
compound and must be equal to the
overall charge for an ionic species.
Determine the Oxidation States
Fe2O3 + 2Al --> Al2O3 + 2Fe
LEO the Lion Says GER
OIL RIG
A
Summary
of an
OxidationReduction
Process
LEO the lion says GER or OIL RIG
Fe2O3 + 2Al --> Al2O3 + 2Fe
Which Atoms Undergo Redox?
1. 2H2 (g) + O2 (g) --> 2H2O (g)
2. Zn (s) + Cu2+(aq) --> Zn2+ (aq) + Cu(s)
3.
2AgCl (s) + H2 (g) --> 2H+ (aq) + 2Ag(s) + 2Cl- (aq)
4.
2MnO4- (aq) + 16H+ (aq) + 5C2O42- (aq) --> 2Mn2+(aq) + 10 CO2 (g) + 8 H2O (l)
Balancing by Half-Reaction
Method
1.
Write separate reduction, oxidation
reactions.
2.
For each half-reaction:
 Balance elements (except H, O)
 Balance O using H2O
 Balance H using H+
 Balance charge using electrons
Balancing by Half-Reaction
Method (continued)
3.
If necessary, multiply by integer to
equalize electron count.
4.
Add half-reactions.
5.
Check that elements and charges
are balanced.
The Half-Reaction Method for
Balancing Redox Reactions
Half Reactions = the two parts of an
oxidation reduction reaction, one
representing oxidation, the other
reduction.
Balance the following equation in acid
solution using the half reaction method.
Cu(s) + HNO3 (aq) --> Cu2+ (aq) + NO(g)
Cu(s) + HNO3 (aq) --> Cu2+ (aq) + NO(g)
1. Identify and write equations for the half
reactions.
Copper is being oxidized: Cu --> Cu2+
Nitrogen is being reduced: HNO3 --> NO
2. Balance each half reaction.
(oxidation) Cu --> Cu2+ + 2e(reduction) HNO3 --> NO
a. Balance all atoms that are neither oxygen nor
hydrogen.
b. Balance oxygens by adding water to the side that
needs oxygen.
c. Balance hydrogens by adding H+ to the side that
needs hydrogen.
d. Balance charges by adding electrons to the side that
is more positive
3. Equalize the electron transfer. The same number of
electrons must be gained as are lost in the reaction
4. Add the half reactions and cancel appropriately to get
a complete redox reaction.
Cancel electrons on both sides and double check. Do
we have the correct number of electrons on both
sides?
Cr2O72- (aq) + NO (g) --> Cr3+ (aq) + NO3- (aq)
Balancing Redox Equations in Basic Solutions
Balance the following equation (it is already
balanced in acid) assuming it is now in basic
solution.
Cr2O72-(aq) + 2NO (g) + 6H+ (aq) --> 2Cr3+ (aq) + 2NO3- (aq) + 3H2O (l)
We need to get rid of excess H+, because OH- is the dominate acid base
related species.
Solution: add 6 OH- to both sides of the equation.
Cr2O72-(aq) + 2NO (g) + 6H+ (aq) + 6OH- --> 2Cr3+ (aq) + 2NO3- (aq) + 3H2O (l) + 6OHCr2O72-(aq) + 2NO (g) + 6 H2O --> 2Cr3+ (aq) + 2NO3- (aq) + 3H2O (l) + 6OH-
Cancel waters on both sides
Cr2O72-(aq) + 2NO (g) + 3 H2O --> 2Cr3+ (aq) + 2NO3- (aq) + 6OH- (aq)
Simple Oxidation-Reduction Titrations
1. Balance the redox equation.
2. Determine the moles of titrant.
3. Use the balanced redox equation to
determine the number of moles of
unknown.
4. Convert from moles of unknown to
grams, percent, molarity, or whatever.
The use of potassium permanganate
(KMnO4) as an oxidizing agent is
described in your text. A 0.0483M
KMnO4 solution was used to titrate a
solution containing 0.8329 g of impure
calcium oxalate, CaC2O4. If 30.25 mL
of the KMnO4 solution was required to
reach the titration endpoint, calculate
the percent purity of the CaC2O4.
MnO4- (aq) + C2O42- (aq) --> Mn2+ (aq) +CO2 (g)