Transcript Chapter 4 Notes: Types of Reactions & Solution

Types of Chemical Reactions
and Solution Stoichiometry
Classification
Solutions
are
homogeneous
mixtures
of Matter
Solute
A solute is the dissolved substance in a
solution.
Salt in salt water
Sugar in soda drinks
Carbon dioxide in soda drinks
Solvent
A solvent is the dissolving medium in a
solution.
Water in salt water
Water in soda
Saturation of Solutions
 A solution that contains the maximum amount of solute that
may be dissolved under existing conditions is saturated.
 A solution that contains less solute than a saturated solution
under existing conditions is unsaturated.
 A solution that contains more dissolved solute than a
saturated solution under the same conditions is
supersaturated.
Electrolytes vs. Nonelectrolytes
The ammeter measures the flow of electrons (current)
through the circuit.
 If the ammeter measures a current, and the bulb
glows, then the solution conducts.
 If the ammeter fails to measure a current, and the
bulb does not glow, the solution is non-conducting.
Definition of Electrolytes and
Nonelectrolytes
An electrolyte is:
 A substance whose aqueous solution conducts
an electric current.
A nonelectrolyte is:
 A substance whose aqueous solution does not
conduct an electric current.
Try to classify the following substances as
electrolytes or nonelectrolytes…
Electrolytes?
1.Pure water
2.Tap water
3.Sugar solution
4.Sodium chloride solution
5.Hydrochloric acid solution
6.Lactic acid solution
7.Ethyl alcohol solution
8.Pure, solid sodium chloride
Answers…
ELECTROLYTES:
NONELECTROLYTES:
Tap water (weak)
Pure water
NaCl solution
Sugar solution
HCl solution
Ethanol solution
Lactate solution (weak)
Pure, solid NaCl
But why do some compounds conduct electricity in
solution while others do not…?
We must understand the nature of water
The Nature of Water
 Water molecules are BENT (105⁰<)
 H & O share electrons UNEVENLY
 O pulls harder on e-s, so the O end of the molecule is
slightly negative
 H pulls less hard on e-s, so the H end of the molecule
is slightly positive
 This makes H2O a POLAR molecule (oppositely
charged ends), like a little magnet
 AKA “a dipole”
The Nature of Water
 The polarity of water gives it great ability to dissolve
compounds
 Imany ionic compounds
 Some acids: HCl, NO3,
1. Ionic Compounds Ionize in Solution
• + ions associate with the - (oxygen) end
of the water dipole.
• - ions associate with the + (hydrogen)
end of the water dipole.
•
Ions tend to stay in solution where they
can conduct a current rather than
reforming a solid.
• IONIC CPDS ARE ELECTROLYTES!
• Dissociation of
sodium chloride
Many Ionic Compounds Dissociate
(break apart)
H2O
NaCl(s)  Na+(aq) + Cl-(aq)
H2O
AgNO3(s)  Ag+(aq) + NO3-(aq)
H2O
MgCl2(s)  Mg2+(aq) + 2 Cl-(aq)
H2O
Na2SO4(s)  2 Na+(aq) + SO42-(aq)
H2O
AlCl3(s)  Al3+(aq) + 3 Cl-(aq)
Note: write “H2O” above each arrow to
show that the ionic compound is placed in
water, NOT reacting with water.
Practice Dissociation Equations
 Study Guide p 106 #1, 2
2. Covalent (molecular) acids IONIZE in
solution
•Again, b/c water is polar…and also b/c
covalent acids are also polar!
•For instance, hydrogen chloride molecules, which
are polar, give up their hydrogens to water,
•forming chloride ions (Cl-)
•and hydronium ions (H3O+).
Strong acids are completely ionized in
solution
(Strong Electrolytes)
Examples of strong acids include:
 Hydrochloric acid, HCl
Sulfuric acid, H2SO4
 Nitric acid, HNO3
 Hydroiodic acid, HI
 Perchloric acid, HClO4
In general, we can assume nearly all
other acids are weak
Weak acids ionize
only slightly
(Weak Electrolytes)
 HC2H3O2 (acetic acid or
vinegar)is a weak acid
Many of these weaker acids
are “organic” acids
that contain a “carboxyl”
group.
The carboxyl group does not easily give up its
hydrogen.
FYI: Because of the carboxyl group, organic acids are
sometimes called “carboxylic acids”.
Other organic acids and their sources include:
o
o
o
o
o
o
Citric acid – citrus fruit
Malic acid – apples
Butyric acid – rancid butter
Amino acids – protein
Nucleic acids – DNA and RNA
Ascorbic acid – Vitamin C
This is an enormous group of compounds; these
are only a few examples.
3.Some Other Polar Covalent
Compounds are Weak Electrolytes
 They ionize very slightly in
water
 Ex: Ammonia, NH3
 Only about 1% of the
molecules dissociate!
4. MOST covalent compounds do not
ionize at all in solution. (Nonelectrolyes)
•Sugar (sucrose – C12H22O11),
•and ethanol (ethyl alcohol – C2H5OH)
NOTE: These molecular compounds
DISSOLVE (the water pulls the molecules
away from each other, but do not IONIZE (the
molecules are not charged.)
Practice Identifying Electrolytes
 Study Guide, p 106 # 4 & 5
Molarity
The concentration of a solution measured
in moles of solute per liter of solution.
mol = M
L
Preparation of Molar Solutions
Problem: How many grams of sodium chloride are needed
to prepare 1.50 liters of 0.500 M NaCl solution?
 Step #1: Ask “How Much?” (What volume to prepare?)
 Step #2: Ask “How Strong?” (What molarity?)
 Step #3: Ask “What does it weigh?” (Molar mass is?)
1.500 L
0.500 mol
58.44 g
1 L
1 mol
= 43.8 g
Practice calculating mass needed to
make a solution
 How many grams of NaOH are needed to make 3.5 L of a 2.5
molar solution?
Given
3.5 L
x
2.5 mol NaoH
1 L of soln
x
39.9g NaoH
1 mol NaOH
Unknown
=?g NaOH
Serial Dilution
It’s not practical to keep solutions of many
different concentrations on hand, so chemists
prepare more dilute solutions from a more
concentrated “stock” solution.
Problem: What volume of stock (11.6 M) hydrochloric
acid is needed to prepare 250. mL of 3.0 M HCl
solution? (Note: must convert volumes to L)
MstockVstock = MdiluteVdilute
M1V1=M2V2
OR
(11.6 M)(x Liters) = (3.0 M)(0.250 Liters)
x Liters = (3.0 M)(0.250 Liters)
11.6 M
= 0.065 L
Molarity of Ions in Solution
 When an ionic compound ionizes in solution, the number of
ions formed may be different than the moles of compound.
H2O(l)
 Ex: CrCl3 (s)  Cr3+(aq) + 3Cl-(aq)
 Say “1 mole of CrCl3 ionizes to form 1 mole of Cr3+ ions
and 3 moles Cl- ions”
Calculating Molarity of Ions in Solution
0.25 M CrCl3 What is the molarity of Cr3+ ions? Cl- ions?
CrCl3 (s)  Cr3+(aq) + 3Cl-(aq)
 ANSWER Molarity of Cr3+ = molarity of CrCl3
= 0.25 M
Molarity of Cl- = 3 x molarity of CrCl3
= 3 x 0.25M = 0.75M
Practice Dilutions & Molarity Calcs
 Study Guide, p 106 # 7, 9, 11, 13,
Types of Reactions
 5 basic
 Precipitation
 Net ionic
 Oxidation-reduction (redox)
 Acid-Base
1. Single Replacement Reactions
A + BX  AX + B
BX + Y  BY + X
Replacement of:




Metals by another metal
Hydrogen in water by a metal
Hydrogen in an acid by a metal
Halogens by more active halogens
The Activity Series of the Metals
Lithium
Potassium
Calcium
Sodium*
Magnesium
Aluminum
Zinc
Chromium
Iron
Nickel
Lead
Hydrogen
Bismuth
Copper
Mercury
Silver
Platinum
Gold
Metals can replace other metals
provided that they are above the
metal that they are trying to
replace.
Metals above hydrogen can
replace hydrogen in acids.
*Metals from sodium upward can
replace hydrogen in water
The Activity Series of the Halogens
Fluorine
Chlorine
Bromine
Iodine
Halogens can replace other
halogens in compounds, provided
that they are above the halogen
that they are trying to replace.
2NaCl(s) + F2(g) 
??? 2NaF(s) + Cl2(g)
MgCl2(s) + Br2(g) 
???
No Reaction
2.Double Replacement Reactions
The ions of two compounds exchange places in an
aqueous solution to form two new compounds.
AX + BY  AY + BX
One of the compounds formed is usually
• a precipitate (an insoluble solid),
• an insoluble gas that bubbles out of solution,
• or a molecular compound, usually water.
Formation of a Precipitate
Highly Soluble Ionic Compounds (& their
exceptions) MEMORIZE THESE RULES!
Ion
Solubility
Exceptions
NO3-
Soluble
None
ClO4-
Soluble
None
Na+
Soluble
None
K+
Soluble
None
NH4+
Soluble
None
Cl-,I-, Br-
Soluble
Pb2+, Ag+, Hg22+
SO42-
Soluble
Ca2+, Ba2+, Sr2+, Pb2+, Ag+, Hg2+
Slightly Soluble (usually considered insoluble)
Ionic Compounds MEMORIZE THESE RULES!
Ion
CO32-
Solubility
Slightly
soluble
Exceptions
Group 1 (IA) and NH4+
PO43-
“
Group 1 (IA) and NH4+
OH-
“
Group 1 (IA) and Ca2+, Ba2+, Sr2+
S2-
“
Groups 1 (IA), 2 (IIA), and NH4+
CrO4 2-
“
none
Completing a Double replacement Rxn:
Ex: Pb(NO3)2(aq) + 2KI(aq) PbI2 (s) + 2KNO3(aq)
1. Determine the products formed when the ions are exchanged, & balance
2. Decide if the products are soluble (aq) or insoluble (s)
PbI2
KNO3
Insoluble, so it will be a solid
Soluble, so it will be in aqueous solution
Complete Ionic Equation
 Shows all soluble compounds as aqueous ions
Pb(NO3)2(aq)
+
2KI
 PbI2(s)
+ 2KNO3 (aq)
Pb2+(aq) + 2 NO3-(aq) + 2 K+(aq) +2 I-(aq)  PbI2(s) + 2K+(aq) + 2 NO3-(aq)
Shows all insoluble compounds as a unit, use the
symbol (s) to show it is a solid.
Net Ionic Equation
 Eliminates all “spectator” ions (ions that appear identically on both sides of equation)
Pb2+(aq) + 2NO3-(aq) + 2K+(aq) +2I-(aq) 
PbI2(s) + 2K+(aq) + 2NO3-(aq)
Pb2+(aq) + 2 I-(aq)  PbI2(s)
Practice Complete Ionic Equation & Net
Ionic Equation
CuSO4 + Na2S  CuS (insoluble) + Na2SO4 (soluble)
Complete Ionic
Cu2+(aq) +SO42-(aq) +2Na+(aq) +S2-(aq)  CuS(s) + 2Na+(aq)+SO42- (aq)
Net Ionic
Cu2+(aq) + S2-(aq)  CuS(s)
Practice Problems
 Study Guide: Please Note: show all work in your notebook,
not your study guide!
 Pp 108 #28
 Net Ionic Equation Worksheet Q#1-10
Stoichiometry of Precipitation Rxns
 It is helpful to be able to predict the amount of precipitant
formed b/c it is often collected & used.
 Ex: Calculate the mass of solid NaCl that must be added to 1.50L
of a 0.100 M AgNO3 solution to precipitate all the Ag+ ions in
the form of AgCl.
 First, we need a balanced equation:
NaCl(s) + AgNO3(aq)  AgCl(s) + NaNO3(aq)
Given
1.5 L AgNO3 x 0.100 mol AgNO3 x
1 L AgNO3
1 mol NaCl x
1 mol AgNO3
58.45 g NaCl =8.77
___g NaCl
1 mol NaCl
Stoichiometry of Precipitation Rxns, cont.
 Please note: you will also encounter limiting reactant problems
with precip. Rxns!
 Remember, set up 2 separate equations to see which reactant
forms the LEAST product. This is your limiting reactant. The
equations will be identical to those on the prior page.
Stoichiometry of Precipitation Rxns, cont.
 HINT: label all equations with your givens & unknown.
Example: 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.
Na2SO4 (aq) + Pb(NO3)2 (aq)  PbSO4 (s) + 2NaNO3 (aq)
V=2.00 L
M=0.0250mol/L
V=1.25 L
M=0.0500mol/L
m=? g
NOTE: set up 2 separate
equations & solve for limiting
reactant. See prior slide.
Equations are set up same way.
Practice Problems:
 Study Guide, pp __________ Q # _________
 Please show all your work in your notebook, not your study
guide.
Oxidation and Reduction Reactions (Redox)
Def: Rxns in which electrons are transferred
Ex: 2Na(s) + Cl2(g) 2 NaCl(s)
An electron transfers from the Na atom to the Cl atom.

What about molecular compounds?

Non-ionic compounds can also be formed from
redox reactions. Even though e-s aren’t FULLY
transferred, they can be assumed to involve a
transfer…let’s see how!
Oxidation and Reduction Reactions (Redox)
Ex:CH4(g) + 2O2(g)  CO2(g) + 2H2O + energy
Carbon is less EN than oxygen, so we assume there is a
transfer of e-s from C to O.
NOTE: We will study electronegativity (EN) in Chapter 8.
•EN is the strength with which an atom in a bond pulls on electrons.
(Check Figure 8.3 on p 353.)
•This shows us the value for EN of O is 3.5 and the EN value for C is 2.5.
Oxidation and Reduction Rxns, cont.
Oxidation States (Oxidation #s) provide a
way to track e-s in redox reactions,
especially in molecular substances.
Rules for Assigning Oxidation Numbers
Rules 1 & 2
1. The oxidation number of any uncombined
element is zero
2. The oxidation number of a monatomic ion
equals its charge
0
0
1
1
2 Na  Cl 2  2 Na Cl
Rules for Assigning Oxidation Numbers
Rules 3 & 4
3. The oxidation number of oxygen in
compounds is -2 (except in peroxides, in
which the oxidation number is -1)
1 2
H2 O
1
1
H 2 O2
4. The oxidation number of hydrogen in
compounds is +1
Rules for Assigning Oxidation Number
Rule 5
5. The sum of the oxidation numbers
in the formula of a compound is 0
1
2
H2O
2(+1) + (-2) = 0
H
O
2
2 1
Ca(O H ) 2
(+2) + 2(-2) + 2(+1) = 0
Ca
O
H
Rules for Assigning Oxidation Numbers
Rule 6
6. The sum of the oxidation numbers in the
formula of a polyatomic ion is equal to
its charge
? 2
N O3
? 2

X + 3(-2) = -1
N
O
 X = +5
S O4
2
X + 4(-2) = -2
S
O
 X = +6
 X = +6
Practice Together
Assign oxidation states to
all atoms in the
following:
1. CO2
2. SF6
3. NO2-
1. C+4
O-2
2. S+6
F-1
3. N+3
O-2
Check the math on #3 (NO2-)
+3 + 2(-2) = -1
Practice Assigning Oxidation Numbers
 Study Guide, p 110
 Q# 50-52,
 Please Note: show all work in your notebook, not your study
guide!
Oxidation and Reduction
An old memory device for oxidation
and reduction goes like this…
LEO says GER
Lose Electrons = Oxidation
Gain Electrons = Reduction
Using Half Reactions to understand Redox Reactions
0
1
0
1
2 Na  Cl 2  2 Na Cl
Each sodium atom loses one electron:
1
0
Na  Na  e

Each chlorine atom gains one electron:
0

1
Cl  e  Cl
LEO says GER :
Lose Electrons = Oxidation
0
1
Na  Na  e

Sodium is oxidized
Gain Electrons = Reduction
0

1
Cl  e  Cl
Chlorine is reduced
Reducing Agents and Oxidizing Agents
The substance reduced is the oxidizing agent
The substance oxidized is the reducing agent
1
0
Na  Na  e

Sodium is oxidized – it is the reducing agent
0

1
Cl  e  Cl
Chlorine is reduced – it is the oxidizing agent
Using Half-reactions to Balance Redox Equations
 Sometimes it is helpful to separate the rxn into two half-reactions:
Oxidation
2. Reduction
1.
Example: Ce4+ (aq) + Sn2+ (aq)  Ce3+ (aq) + Sn4+ (aq)
1.
Oxidation: Sn2+ (aq)  Sn4+ (aq) + 2 e-
2.
Reduction: Ce4+ (aq) + e- Ce3+ (aq)
NOTE: If you combine these equations, the e-s and charges don’t balance.
So, multiply the 2nd equation by 2.
Using Half-reactions to Balance Redox Equations
1.
Oxidation:
Sn2+ (aq)  Sn4+ (aq) + 2 e-
2.
Reduction:
2(Ce4+ (aq) + e-  Ce3+ (aq))
2. Reduction:
=
2Ce4+ (aq) + 2e- 2Ce3+ (aq)
Sn2+ (aq) + 2Ce4+ (aq) + 2 e-  + Sn4+(aq) +2Ce3+ (aq) + 2 e-
Now, combine (add) the two equations, cancelling items that appear identically
on both sides of the arrow.
EXTRA: Which is the oxidizing agent?
Reducing agent?
cesium
tin
Practice Problems
 text book: p 183-184, practice problems #57, 61, 62, 64
 Show all your work in your notebook.
Let’s do # 64a together
Zn(s) + HCl(aq)  Zn2+(aq) + H2(g) + Cl- (aq)
Let’s try 53 c
 Cu + H+ + NO3-  Cu2+ + NO2 + H2O
 Cu  Cu2+ + 2e 2 (NO3- + 1e-  NO2 )
 Cu + 2 NO3- + 2e- 2NO2 + 2e-
Trends in Oxidation and Reduction
Active metals:
Lose electrons easily
Are easily oxidized
Are strong reducing agents
Active nonmetals:
Gain electrons easily
Are easily reduced
Are strong oxidizing agents
Redox Reaction Prediction #1
Important Oxidizers
Formed in reaction
MnO4- (acid solution)
MnO4- (basic solution)
MnO2 (acid solution)
Cr2O72- (acid)
CrO42HNO3, concentrated
HNO3, dilute
H2SO4, hot conc
Metallic Ions
Free Halogens
HClO4
Na2O2
H2O2
Mn(II)
MnO2
Mn(II)
Cr(III)
Cr(III)
NO2
NO
SO2
Metallous Ions
Halide ions
ClOHO2
Redox Reaction Prediction #2
Important Reducers
Formed in reaction
Halide Ions
Free Metals
Metalous Ions
Nitrite Ions
Sulfite Ions
Free Halogens (dil, basic sol)
Free Halogens (conc, basic sol)
C2O42-
Halogens
Metal Ions
Metallic ions
Nitrate Ions
SO42Hypohalite ions
Halate ions
CO2
Not All Reactions are Redox Reactions
Reactions in which there has been no change
in oxidation number are not redox rxns.
Examples:
1 5 2
1
1
1
1
1 5 2
Ag N O3 (aq)  Na Cl (aq)  Ag Cl ( s)  Na N O3 (aq)
1 2 1
1
6 2
1
6 2
1
2
2 Na O H (aq)  H 2 S O 4 (aq)   Na 2 S O 4 (aq)  H 2 O(l )
Acid-Base Reactions
 Arrhenius definition:
 Acid is a proton (hydrogen ion) donor
 Base is a hydroxide (OH-) donor
 Bronsted-Lowry Definition (applies to more compounds)
 Acid is a proton (hydrogen ion) donor
 Base is a proton (hydrogen ion) acceptor
Strong Acids: there are 7
 Some oxyacids: HClO4, HClO3, H2SO4, HNO3
 Most halide acids are strong:HI, HCl, HBr
 Except HF, which is weak
Strong Bases: there are 8
 Hydroxides of Group 1 metals are all strong bases:
 LiOH, NaOH, KOH, RbOH, CsOH
 3 Hydroxides of Group 2 metals are strong bases
 Ca(OH)2 Sr(OH)2 Ba(OH)2
Acid-Base Reactions:
Strong Acid/Strong Base
 To predict results of rxn, focus on the species present in
mixed solution.
 Ex#1 : HCl(aq) + NaOH(aq)  ???
Species Present before rxn occurs:
H+ (aq) + Cl- (aq) + Na+ (aq) + OH- (aq)
(b/c HCl is a strong acid and NaOH is a strong base=complete
dissociation of both acid & base)
 NaCl is soluble in water (check your table) so Na+ & Cl- are
spectator ions.
Acid-Base Reactions:
Strong Acid/Strong Base
Ex#1(continued): HCl(aq) + NaOH(aq)  ???
 H+ (aq) + OH- (aq) are the only other species present.
 Water is a nonelectrolyte, so H+ (aq) + OH- will NOT exist in
solution together. Therefore, they must combine to form H2O.
 Our net ionic equation for this reaction is therefore:
H+ (aq) + OH- (aq)  H2O (l)
Acid-Base Reactions:
Weak Acid/Strong Base
 Again, to predict results of rxn, focus on the species present in
mixed solution.
 Ex#2 : HC2H3O2(aq) + KOH(aq)  ???
Species Present before rxn occurs:
HC2H3O2(aq) + K+ (aq) + OH- (aq)
 b/c HC2H3O2(aq) is a weak acid (almost no dissociation = 1%) &
 b/c KOH is a strong base (complete dissociation)
Acid-Base Reactions:
Weak Acid/Strong Base
 Note: OH- is such a strong base that it will strip H+ ions from
the HC2H3O2(acetic acid)
• we can assume it reacts completely with any weak
acid it will be combined with
•But the net ionic equation is quite different than for a strong
acid/strong base rxn:
Remember, acetic acid is NOT dissociated
before the reaction!
HC2H3O2(aq) + K+ (aq)
+ OH- (aq)  HOH + C2H3O2-(aq) + K+(aq)
Net Ionic: HC2H3O2(aq) + OH- (aq)  HOH + C2H3O2-(aq)
Acid-Base Reactions
Arrhenius Acids & Bases
 Often called “neutralization reactions”
 Complete equation shows formation of
 a salt
 and water
 Ex: HCl(aq) + NaOH(aq)  NaCl(aq) + H2O(l)
Acids that Undergo Multi-step Rxn
 H2CO3 (aq) breaks into water and carbon dioxide
 H2CO3  H2O + CO2
Stoichiometry of Neutralization
Reactions
List the species present in the combined solution before
any reaction occurs & decide what will occur.
2. Write balanced net ionic equation.
3. Set up your calculation, identifying Given, Unknown, and
conversion factors that cancel the necessary units. Or
identify the proper formula and solve for unknown.
1.
Stoichiometry of Neutralization
Reactions: Example 1 (p159 of text)
Example: What volume of a 0.100 M HCl solution is needed to
neutralize 25.0 mL of 0.350 M NaOH?
1. List the species present in the combined solution before any
reaction occurs & decide what will occur.
 H+ (aq) + Cl- (aq)
+ Na+ (aq) + OH- (aq)
(HCl is a strong acid and NaOH is a strong base, so both completely
ionize in aqueous solution)
• NaCl is soluble, so the Na+ and Cl- ions are spectators. (This
reaction :Na+ (aq) + Cl- (aq)  NaCl will not occur.)
• H+ and OH- do not coexist, so H2O will form
Stoichiometry of Acid-Base Rxns
2. Write the balanced net ionic equation for this reaction.
 H+ (aq) + OH- (aq)  H2O (l)
Stoichiometry of Acid-Base Rxns
3. Set up your calculation, identifying Given, Unknown, and
conversion factors that cancel the necessary units. Or identify
the proper formula and solve for the unknown.
H+ (aq)
+
OH- (aq)

H2O (l)
MA =0.100 mol/L
VA= ??? L
MB=0.350 mol/L
VB = 0.025 L
• Analysis: MAVA = MBVB
Solution:
Given
0.025L OH-
VA
=(0.350M) (0.025L)
(0.100 M)
VA = 8.75 x 10-3L H+
NOTE: Our usual
analysis will also
work!
x
0.350 mol OH1 L OH-
x 1 mol H+
1 mol OH-
x
1L H+
0.100 mol H+
Unknown
= 8.75 x 10-3L H+
Stoichiometry of Acid-Base Rxns:
Example 2 (p160 of text)
Example2 : 28.0 mL of 0.250M HNO3 and 53.0 mL of 0.320 M
KOH are mixed. Calculate the amount of water formed in the
resulting reaction. What is the concentration of H+ or OHions in excess after the rxn goes to completion?
1. List the species present in the combined solution before any
reaction occurs & decide what will occur.
 H+ (aq) + NO3- (aq)
+ K+ (aq) + OH- (aq)
(HNO3 is a strong acid and KOH is a strong base, so both completely
ionize in aqueous solution)
• KNO3 is soluble, so the K+ and NO3- ions are spectators. (This
reaction :K+ (aq) + NO3- (aq)  KNO3 will not occur.)
• H+ and OH- do not coexist, so H2O will form
Stoichiometry Acid-Base Rxns:
Example 2 (p160 of text)
2. Write the balanced net ionic equation for this reaction.
 H+ (aq) + OH- (aq)  H2O (l)
Practice Acid-Base Stoichiometry
Problems
 Text Book, p 183, Q # 45, 47, 49, 50 , 51, 53
 Show all your work in your notebook
Stoichiometry of Neutralization
Reactions
3. Set up your calculation, identifying Given, Unknown, and
conversion factors that cancel the necessary units. Or identify
the proper formula and solve for the unknown.
H+ (aq)
+
MA =0.250 mol/L
VA= 0.0280 L
• Analysis:
OH- (aq)
MB=0.320 mol/L
VB = 0.0530 L

H2O (l)
Unknown 1: L H2O
Unknown 2: M H+ or M OH- in excess
MAVA = MBVB Will NOT work for this problem. All
your variables have values.
•This is a limiting reactant problem
Stoichiometry of Neutralization
Reactions
 Solution
H+ (aq)
+
OH- (aq)

H2O (l)
Unknown 1: L H2O
Unknown 2: M H+ or M OH- in excess
Given
0.0280L H+ x
Given
0.0530L OH- x
0.250 mol H+
1 L H+
0.350 mol OH1 L OH-
Unknown
= 0.00700
__molH+
Unknown
= 0.0170 mol OH-
Titration