Download Ch 8 Lecture Notes

Ch 8: Aqueous Solutions:
Chemistry of the Hydrosphere
H2S + Cu2+ → CuS(s) + 2H+
(Fe, Ni, Mn also)
HS- + 2 O2  HSO4- + energy
(supports life)
1
2
Figure taken from “Principles of Biochemistry”, 2nd Ed. By Lehninger, Nelson, and Cox
Chapter Outline
 8.1 Solutions and Their Concentrations
 8.2 Dilutions
 8.3 Electrolytes and Nonelectrolytes
 8.4 Acids, Bases, and Neutralization Reactions
 8.5 Precipitation Reactions
 8.6 Oxidation-Reduction Reactions
 8.7 Titrations
 8.8 Ion Exchange
3
Concentration of Solutions - Molarity
The concentration of a solution is the amount of solute
present in a given quantity of solvent or solution.
Mass-to-mass ratios:
mg/kg solvent
Mass-to-volume ratios: mg/L solvent
1 g = mg = mg
103 g
L
106 g
Parts per million:
dH2O = 1.00
g/mL @ 25 oC
Molarity (M) =
moles of solute
liters of solution
Sample Exercise 8.1: Converting Mass-perVolume Concentrations into Molarity
Vinyl chloride (C2H3Cl, MW = 62.49) is one of the most widely used
industrial chemicals. It is also one of the most toxic, and it’s a known
carcinogen. The maximum concentration of vinyl chloride allowed in
drinking water in the U.S. is 0.002 mg/L. What is that concentration in
moles per liter?
Plan: mg/L  g/L  mol/L = Molarity
1 mol
0.002 mg x 1 g
x 62.49 g = 3.2 x 10-8
1000
mg
L
mol/L
4
Sample Exercise 8.2: Converting Mass-per-Mass
Concentrations into Molarity
A water sample from the Great Salt Lake in Utah contains 83.6 mg Na+ per
gram of lake water (soln). What is the molar concentration of Na+ ions if the
density of the lake water is 1.160 g/mL? AW Na = 22.99
mg Na+  g  mol
Plan:
mol = M
L
1 g soln  mL  L
dsoln = 1.160 g/mL
1 mol Na+
1g
83.6 mg Na+ x 1000 mg x 22.99 g = 3.636 x 10-3 mol Na+
1 g soln
mol
L
M =
1 mL soln
x 1.160 g
=
1L
x 1000 mL
= 8.6221 x 10-4 L
3.636 x 10-3 mol Na+
= 4.22 M
8.6221 x 10-4 L
Sample Exercise 8.3: Calculating the Quantity of
Solute Needed to Prepare a Solution
An aqueous solution called phosphate-buffered saline (PBS) is used in biology
research to wash and store living cells. It contains ionic solutes including 10.0
mM Na2HPO42H2O (MW = 178.0). How many grams of this solute would you
need to prepare 10.0 L of PBS?
MM
Plan:
mM  M  mol  g
use the volume
Useful equation: M x V = moles
178.0 g
1 mol
10.0 mmol x
x 10.0 L x
mol
1000 mmol
L
= 17.8 g
M x V = moles
5
Chapter Outline
 8.1 Solutions and Their Concentrations
 8.2 Dilutions
 8.3 Electrolytes and Nonelectrolytes
 8.4 Acids, Bases, and Neutralization Reactions
 8.5 Precipitation Reactions
 8.6 Oxidation-Reduction Reactions
 8.7 Titrations
 8.8 Ion Exchange
Dilution is the procedure for preparing a less concentrated
solution from a more concentrated solution.
Dilution
Add Solvent
Moles of solute
before dilution (i)
=
Moles of solute
after dilution (f)
MiVi
=
MfVf
6
Dilution Equation: Mi Vi = Mf Vf
Sample Exercise 8.4: Diluting Solutions
The solution used in hospitals for intravenous infusion—called
physiological or normal saline - is 0.155 M NaCl. It may be prepared by
diluting a commercially available standard solution that is 1.76 M NaCl.
What volume of standard solution (Vi) is required to prepare 10.0 L of
physiological saline?
Collect data:
Mi = 1.76 M
Vi = ?
Mf = 0.155 M
Vf = 10.0 L
add
H2O
Add 881 mL of
1.76 M NaCl,
and then “dilute
to the mark”
(10.0 L)
Mi Vi = Mf Vf
Vi = Mf Vf
Mi
Vi = (0.155 M )(10.0 L)
(1.76 M)
= 0.881 L
= 881 mL
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Chapter Outline
 8.1 Solutions and Their Concentrations
 8.2 Dilutions
 8.3 Electrolytes and Nonelectrolytes
 8.4 Acids, Bases, and Neutralization Reactions
 8.5 Precipitation Reactions
 8.6 Oxidation-Reduction Reactions
 8.7 Titrations
 8.8 Ion Exchange
8
Strong Electrolytes:
• Nearly 100% dissociated
into ions
• Conduct current efficiently
 Examples: ionic compounds such as
NaCl, strong acids like HCl
• NaCl(s) → Na+(aq) + Cl−(aq)
• HCl(aq) + H2O(l)  H3O+ + Cl-(aq)
Figure from John J. Fortman, J. Chem. Ed. Vol. 71, No. 1, 1994, p. 27-28
9
Weak Electrolytes:
• Only partially dissociate into
ions
• Slightly conductive
 Example: weak acids such as
acetic acid
Hydronium Ion: H3O+
The higher the concentration of H3O+ in solution, the stronger the acid.
+
10
Non-electrolytes
 Substances in which no ionization
occurs. There is no flow of electrical
current through the solution
 Examples: molecular or covalent
compounds such as ethanol
11
Chapter Outline
 8.1 Solutions and Their Concentrations
 8.2 Dilutions
 8.3 Electrolytes and Nonelectrolytes
 8.4 Acids, Bases, and Neutralization Reactions
 8.5 Precipitation Reactions
 8.6 Oxidation-Reduction Reactions
 8.7 Titrations
 8.8 Ion Exchange
12
Acids
Have a sour taste. Vinegar owes its taste to
acetic acid. Citrus fruits contain citric acid.
Cause color changes in plant dyes (litmus paper).
React with certain metals to produce hydrogen gas.
2HCl (aq) + Mg (s)
MgCl2 (aq) + H2 (g)
React with carbonates and bicarbonates to produce carbon
dioxide gas
2HCl (aq) + CaCO3 (s)
CaCl2 (aq) + CO2 (g) + H2O (l)
Aqueous acid solutions conduct electricity.
Limestone
dissolving in acid =
cave
Most metals
dissolve in acids to
produce H2
13
Strong Acids and Bases
 Strong Acids/Bases:
• Completely ionized
in aqueous solution
(i.e., strong electrolytes)
• Memorize the 6 strong
acids 
• All other acids are weak
KNOW!
14
Bases
Have a bitter taste.
Feel slippery. Many soaps contain bases.
Cause color changes in plant dyes (litmus paper).
Aqueous base solutions conduct electricity.
A Brønsted acid is a proton donor
A Brønsted base is a proton acceptor
base
acid
conjugate
acid
conjugate
base
For every acid there is a conjugate base, and for every
base there is a conjugate acid (conj. acid-base pairs).
15
Water: Acid or Base?
 Water as Base:
HCl(aq) + H2O(l) → H3O+(aq) + Cl−(aq)
 Water as Acid:
NH3(aq) + H2O(l) ⇌ NH4+(aq) + OH−(aq)
 Amphiprotic: Acts as an Acid or Base.
Preface: Double-Displacement Reactions
Many of the reactions in this chapter follow this pattern:
(+) (-)
(+) (-)
(+) (-)
(+) (-)
AB + CD → AD + CB
16
Acid-Base (Neutralization) Reactions
acid + base
HCl(aq) + NaOH(aq)
H2SO4(aq) + 2 KOH(aq)
salt + water
HOH(aq) + NaCl(aq)
2 HOH(aq) + K2SO4(aq)
Net Ionic Equations
1. Write the balanced molecular equation.
2. Write the ionic equation showing the strong
electrolytes completely dissociated into cations and
anions.
3. Cancel the spectator ions on both sides of the ionic
equation
Tips:
• Do not break apart solids, liquids (such as H2O), and gases (e.g. CO2)
• Do not break apart weak electrolytes (such as weak acids, e.g.
HC2H3O2)
• Carbonic acid, H2CO3, decomposes into CO2 and H2O
• An acid such as H2SO4 breaks apart into 2H+(aq) + SO42-, not H22+ +
SO42-
17
1. Write the balanced molecular equation.
H2SO4(aq) + 2 KOH(aq)
strong
2 H2O(aq) + K2SO4(aq)
strong
Not a strong
electrolyte!!
strong
2. Write the ionic equation showing the strong electrolytes
completely dissociated into cations and anions.
2 H+
SO42-
2 K+ 2OH-
2 H2O
2 K+
SO42-
3. Cancel the spectator ions on both sides of the ionic equation
Net ionic equation:
2 H+ + 2 OH-
2 H2O
H+ + OH-
H2O
Sample Exercise 8.5: Writing the Net Ionic
Equation for a Neutralization Reaction
Write the net ionic equation describing the reaction that takes place when
acid rain containing sulfuric acid reacts with a marble statue (calcium
carbonate) 1.
Write the balanced molecular equation.
H2SO4(aq) + CaCO3(s)
strong
solid
H2O(l) + CO2(g) + CaSO4(aq)
Not strong
electrolytes!!
strong*
2. Write the ionic equation showing the strong electrolytes completely dissociated
into cations and anions.
2H+(aq) + SO42-(aq) + CaCO3(s)
H2O(l) + CO2(g) + Ca2+(aq) + SO42-(aq)
*generally speaking, CaSO4 is insoluble in water, but not in strong acid. More on
this later when we study solubility rules.
18
Sample Exercise 8.5: Writing the Net Ionic
Equation for a Neutralization Reaction
3. Cancel the spectator ions on both sides of the ionic equation
2H+(aq) + SO42-(aq) + CaCO3(s)
H2O(l) + CO2(g) + Ca2+(aq) + SO42-(aq)
Net ionic equation:
2H+(aq) + CaCO3(s)
H2O(l) + CO2(g) + Ca2+(aq)
Chapter Outline
 8.1 Solutions and Their Concentrations
 8.2 Dilutions
 8.3 Electrolytes and Nonelectrolytes
 8.4 Acids, Bases, and Neutralization Reactions
 8.5 Precipitation Reactions
 8.6 Oxidation-Reduction Reactions
 8.7 Titrations
 8.8 Ion Exchange
19
Double-Displacement Reactions Again
Precipitation reactions in this section follow the same
pattern as acid-base reactions:
(+) (-)
(+) (-)
(+) (-)
(+) (-)
AB + CD → AD + CB
Types of Solutions
 Saturated solution:
• Contains the maximum amount of solute that can
dissolve in a given volume at a given temperature.
• Solubility = g solute/100 mL solvent.
 Supersaturated solution:
• Contains more dissolved solute than predicted at a
given temperature.
20
A Saturated Solution Example
Supersaturated Solution
Sodium acetate precipitates from a supersaturated solution.
21
22
 Precipitate:
• Solid product formed from reactants in
solution.
• AgNO3(aq) + NaCl(aq) → NaNO3(aq) + AgCl(s)
 Can predict formation of precipitates
based on solubility “rules.”
We learned in Ch 4, p. 134, that the electrostatic energy (strength) of an
ionic bond depended on the product of the charges, divided by the
internuclear distance. The solubility of ionic compounds generally
follows the trend that the greater the charge, and the shorter the
internuclear distance, the more insoluble the compound is.
α (alpha) means “is proportional to”
Q+ = charge on the cation
Q- = charge on the anion
d = internuclear distance
-
+
𝑄+ 𝑥 𝑄−
𝐸 α
𝑑
d
cations
+1
compounds
alkali metals
NH4+
+2
+3
Al3+
anions
compounds
solubility
-1
NO3-
Always soluble
CH3COO-2
SO42-
Soluble with exceptions
-2
S2-,
O2-
Insoluble with
exceptions
-3
PO43-
Insoluble with
exceptions
23
Always
soluble
Soluble
w/exceptions
and Ag+
Add to table
Insoluble
w/exceptions
Table 8.3 - Soluble Compounds
 Soluble Cations:
• Group I ions (alkali metals) and NH4+
 Soluble Anions:
• NO3− and CH3COO− (acetate)
• Halides (Group 17)
»
Exceptions: Cu+, Hg22+, Ag+, Pb2+
• Sulfates (SO42−)
»
Exceptions: Hg22+, Ag+, Pb2+, Ca2+, Ba2+, Sr2+
Add to table
24
Table 8.3 - Insoluble Compounds
 All hydroxides (OH−) except:
• Group IA = Li, Na etc and NH4+
• Slightly soluble: Group IIA = Ca(OH)2, Sr(OH)2, and
Ba(OH)2
 All sulfides (S2−)
 All carbonates (CO32−)
except IA, NH4+
 All phosphates (PO43−)
 All chromates (CrO42-)
Mnemonic device to help remember the
exceptions to the soluble compounds -
Cute HAPpy halides
HAPpy sulfates "2"
Cl-, Br-, I-
Cu+
Sr2+
Hg22+
Ag+
Ca2+
Pb2+
Hg22+
Ag+
Pb2+
Ba2+
from
Group 2
(slightly)
25
Mnemonic device to help remember the insoluble
compounds -
Shop for a chrome car
u
l
f
i
d
e
s
S2-
y
d
r
o
x
i
d
e
s
OH-
h
o
s
p
h
a
t
e
s
h
r
o
m
a
t
e
s
a
r
b
o
n
a
t
e
s
CrO42-
CO32-
PO43-
Applying the Solubility Rules – Will a Precipitate Form?
Does a precipitate form when sodium chloride is mixed with silver nitrate?
If so, write the net ionic equation for the formation of the precipitate:
Cute HAPpy halides
Cl-, Br-, I-
Cu+ Hg22+ Ag+ Pb2+
NaCl(aq) + AgNO3(aq)  AgCl(s) + NaNO3(aq)
Na+(aq) + Ag+(aq) + Cl-(aq) + NO3-(aq)  AgCl(s) + Na+(aq) + NO3-(aq)
Ag+(aq) + Cl-(aq)  AgCl(s)
26
Formation of an Insoluble Product
NaI(aq) + Pb(NO3)2(aq) →
The soluble compounds break up
into ions and recombine -
K+(aq)
I-(aq)
Pb2+(aq)
NO3-(aq)
Sample Exercise 8.6
A precipitate forms when aqueous solutions of ammonium sulfate and
barium chloride are mixed. Write the net ionic equation for the reaction.
(NH4)2SO4(aq) + BaCl2(aq)  2 NH4Cl(aq) + BaSO4(s)
2NH4+(aq) + SO42-(aq) + Ba2+(aq) + 2Cl-(aq)  BaSO4(s) + 2NH4+(aq) + 2Cl-(aq)
Ba2+(aq) + SO42-(aq)  BaSO4(s)
27
Sample Exercise 8.7 – Prediction the Mass of a Precipitate
Barium sulfate is used to enhance Xray imaging of the upper and lower GI tract. In
upper GI imaging, patients drink a suspension of solid barium sulfate in H 2O. The
compound is not toxic because of its limited solubility. To make pure barium sulfate,
a precipitation reaction is employed: aqueous solutions of soluble barium nitrate
and sodium sulfate are mixed together, and solid barium sulfate is separated by
filtration. How many grams of barium sulfate (MW = 233.40) will be produced if
exactly 1.00 L of 1.55 M barium nitrate is reacted with excess sodium sulfate?
Ba(NO3)2(aq) + Na2SO4(aq)  BaSO4(s) + 2 NaNO3(aq)
V = 1.00 L
1.55 M
Plan:
g BaSO4 =
Excess
no LR
?g
MW = 233.40
ratio
M x V = mol Ba(NO3)2  mol BaSO4  g BaSO4
233.40 g BaSO4
1.55 mol
1 mol BaSO4
x 1.00 L x
x
= 362 g BaSO4
1 mol BaSO4
1
mol
Ba(NO
)
L
3 2
Sample Exercise 8.8 – Calculating Solute Concentration
from Mass of Precipitate
To determine the concentration of chloride ion in a 100.0 mL sample of
groundwater, a chemist adds a large enough volume of a solution of AgNO 3
to precipitate all the Cl- ions as silver chloride. The mass of the resulting
precipitate is 71.7 mg. What is the Cl- concentration in the sample in mg/L?
AgNO3(aq) + Cl-(aq)
excess
no LR

AgCl(s) + NO3-(aq)
V = 100.0 mL 71.7 mg
= 0.100 L
= 0.0717 g
? mg/L
ratio
Plan: mg AgCl  g AgCl  mol AgCl  mol Cl-  g Cl-
Then mg/L Cl- = g Cl-  mg Cl-  mg/L
Divide by liters of sample
28
Chapter Outline
 8.1 Solutions and Their Concentrations
 8.2 Dilutions
 8.3 Electrolytes and Nonelectrolytes
 8.4 Acids, Bases, and Neutralization Reactions
 8.5 Precipitation Reactions
 8.6 Oxidation-Reduction Reactions
 8.7 Titrations
 8.8 Ion Exchange
29
(electron transfer reactions)
Section 8.6 - Oxidation-Reduction Reactions
Oxidation-Reduction Reactions (Redox):
Characterized by gain or loss of electrons by atoms
involved in the reaction.
Oxidation:
Historical definition = gain of oxygen (sometimes with
a simultaneous loss of hydrogen)
e.g. C2H6 + O2  CO2 + H2O
Modern definition = loss of electrons
Reduction:
Historical definition = loss of oxygen (sometimes with
a simultaneous gain in hydrogen)
e.g. Fe2O3 + 3 CO  2 Fe + 3 CO2
“oil rig”
Modern definition = gain of electrons
Table 8.4 - Oxidation Number Rules
The charge the atom would have in a molecule (or an ionic
compound) if electrons were completely transferred, i.e.
applies to both ionic and covalent compounds.
1. The O.N. before a reaction for pure elements = 0
(including diatomics).
 Mg 


HH
O O
2. O.N = the charge on monovalent ions


Cl



Cl  + e 


 Mg   Mg2+ + 2 e
3. O.N. of fluorine = -1 for all of its compounds
30
4. O.N. of oxygen = -2 in nearly all of its compounds


O


+2e
2-



O O
Exceptions are peroxides
(O.N. = -1) and superoxides
(O.N. = -1/2), e.g. H2O2, KO2
5. O.N. of hydrogen = +1 in nearly all of its compounds
HH
 2 H+ + 2 e
Exception = hydrides, e.g. CaH2
6. O.N. values of the atoms in a neutral molecule sum up
to zero
CaCl2
+2 + 2(Cl) = 0
so Cl = -1
7. O.N. values of the atoms in a polyatomic ion sum up to
the charge on the ion
3-
PO4
P + 4(-2) = -3
so P = +5
Sample Exercise 8.9
What are the oxidation states for sulfur in (a) S8, (b) SO2, (c)
Na2S, and (d) CaSO4 ?
31
Electron Transfer in Redox Reactions
What follows is a method that keeps track of the flow of electrons
within the atoms and molecules taking part in a chemical reaction.
O.N. = final O.N.
- initial O.N.
oxidation
lose n electrons
O.N. = +n
reduction
gain n electrons
O.N. = -n
Pb0  Pb2+ + 2 e
O.N. = (2 – 0) = +2
Cu2+ + 2 e  Cu0
O.N. = (0 – 2) = -2
32
What is the net ionic equation for the reaction of zinc metal plus cupric sulfate?
Oxidizing Agent = oxidizes something else while being reduced (Cu 2+  Cu0)
Reducing Agent = reduces something else while being oxidized (Zn  Zn2+)
33
Sample Exercise 8.10:
Identifying Oxidizing and Reducing Agents and Determining
Number of Electrons Transferred
Energy released by the reaction of hydrazine and
dinitrogen tetroxide is used to orient and maneuver
spacecraft and to propel rockets into space. Identify
the elements that are oxidized and reduced, the
oxidizing agent, and the reducing agent, and
determine the number of electrons transferred in
the balanced chemical equation.
(+4) (-2)
(-2) (+1)
2N + 4(+1) = 0
So N = -2
2N + 4(-2) = 0
So N = +4
(0)
Element in its
natural state
(+1) (-2)
O.N. rules
H = +1
O = -2
O.N. = 0 – (-2) = +2
lost electrons
oxidation
(-2)
(+4)
(0)
O.N. = (0 – 4) = -4
gained electrons
reduction
34
Net number of electrons transferred during the reaction:
= (#molecules) x (#atoms in molecule) x O.N.
O.N. from
previous slide
2 x 2 x +2 = +8 (oxidation)
1 x 2 x -4 = -8 (reduction)
O.N. from
previous slide
So the net number of electrons flowing is 8
Balancing Redox Reactions in Acid and Base
(We’re going to use a simpler method than the book
called the Method of Half Reactions)
Examples of Half Reactions:
copper wire immersed in a silver nitrate solution:
(+1) (+5) (-2)
(+2) (+5) (-2)
(0)
(0)
Cu(s) + 2 AgNO3(aq)  Cu(NO3)2(aq) + 2 Ag(s)
O.N. Cu = +2 – (0) = +2
oxidation
O.N. Ag = 0 – (1) = -1
reduction
The half-reactions are:
ox: Cu(s)  Cu2+(aq) + 2 e
red: Ag+(aq) + e  Ag(s)
Cu(s)
Ag(s)
AgNO3(aq)
(clear soln)
Cu(NO3)2(aq)
(blue soln)
35
Examples of Half Reactions (cont’d):
copper wire immersed in a silver nitrate solution:
The total reaction is obtained by adding the half reactions together, after
balancing electrons ox:
Cu(s)  Cu2+(aq) + 2 e
red: Ag+(aq) + e  Ag(s)
ox:
x2
Cu(s)  Cu2+(aq) + 2 e
red: 2 Ag+(aq) + 2 e  2 Ag(s)
Net: Cu(s) + 2 Ag+(aq)  Cu2+(aq) + 2 Ag(s)
The next topic is how to balance a redox reaction that occurs in acid or base.
Procedure for balancing a redox reaction in acid or base.
1.
2.
3.
4.
5.
Determine the oxidation numbers for the reactants and compare to the products.
Write the oxidation and reduction half-reactions without electrons (yet)
Balance everything but oxygen and hydrogen
Balance oxygen by adding water
Balance hydrogen by adding (a) H+ in acidic solutions, (b) in basic solutions,
continue as if in acidic solution, but at the end each H + ion will be neutralized by
adding OH- ions
6. Balance charge by adding electrons; for the oxidation half-reaction, the electrons
will be on the right, for the reduction half-reaction, the electrons will appear on the
left
7. Make sure the number of electrons in each half-reaction are the same. Then add
the half reactions together
8. Make sure that the equation is balanced for mass and for charge
NOTE: sometimes you have to cancel H2O’s that are on each side, as well
as H+ or OH-
36
Sample Exercise 8.11
Balancing Redox Equations I: Acidic Solutions
Microorganisms called denitrifying bacteria that grow in waterlogged soil convert
NO3- ions into N2O gas as they feed on dead plant tissue (empirical formula =
CH2O), converting it into CO2 and H2O. Write a balanced net ionic equation
describing this conversion of dissolved nitrates to N 2O gas. Assume the reaction
occurs in slightly acidic water.
1. Determine the oxidation numbers for the reactants and compare to the products.
(+5) (-2)
(0) (+1) (-2)
(+1) (-2) (+4) (-2)
(+1) (-2)
NO3-(aq) + CH2O(s)  N2O(g) + CO2(g) + H2O(l)
Given:
O.N. C = +4 – (0) = +4
oxidation
O.N. N = 1 – (5) = -4
reduction
2. Write the oxidation and reduction half-reactions without electrons (yet)
ox:
CH2O(s)  CO2(g)
red: NO3-(aq)  N2O(g)
3. Balance everything but oxygen and hydrogen
ox:
CH2O(s)  CO2(g)
red: 2 NO3-(aq)  N2O(g)
4. Balance oxygen by adding water
ox:
H2O + CH2O(s)  CO2(g)
red: 2 NO3-(aq)  N2O(g) + 5 H2O
37
5. Balance hydrogen by adding (a) H+ in acidic solutions, (b) in basic solutions,
continue as if in acidic solution, but at the end each H + ion will be neutralized by
adding OH- ions
ox:
H2O + CH2O(s)  CO2(g) + 4 H+(aq)
red: 10 H+(aq) + 2 NO3-(aq) 
N2O(g) + 5 H2O
6. Balance charge by adding electrons; for the oxidation half-reaction, the electrons
will be on the right, for the reduction half-reaction, the electrons will appear on the
left
ox:
H2O + CH2O(s)  CO2(g) + 4 H+(aq) + 4 e
red: 8 e + 10 H+(aq) + 2 NO3-(aq)  N2O(g) + 5 H2O
7. Make sure the number of electrons in each half-reaction are the same. Then add
the half reactions together
ox:
H2O + CH2O(s)  CO2(g) + 4 H+(aq) + 4 e
x2
red: 8 e + 10 H+(aq) + 2 NO3-(aq)  N2O(g) + 5 H2O
Net: 2H2O + 2CH2O(s) + 10H+(aq) + 2NO3-(aq)  2CO2(g) + 8H+(aq) + N2O(g) + 5H2O
NOTE: sometimes you have to cancel H2O’s that are on each side, as well
as H+ or OH2H2O + 2CH2O(s) + 10H+(aq) + 2NO3-(aq)  2CO2(g) + 8H+(aq) + N2O(g) + 5H2O
2CH2O(s) + 10H+(aq) + 2NO3-(aq)  2CO2(g) + 8H+(aq) + N2O(g) + 3H2O
2CH2O(s) + 2H+(aq) + 2NO3-(aq)  2CO2(g) + N2O(g) + 3H2O
38
8. Make sure that the equation is balanced for mass and for charge
2CH2O(s) + 2H+(aq) + 2NO3-(aq)  2CO2(g) + N2O(g) + 3H2O
C=2
C=2
H=6
O=8
H=6
O=8
N=2
N=2
Mass:
Charge:
2(+1) + 2(-1) = 0
=0
Homework Problem 98(c)
Balancing Redox Equations II: Basic Solutions
Permanganate ion is used in water purification to remove oxidizable substances.
Complete and balance the reaction for the removal of sulfite ion.
1. Determine the oxidation numbers for the reactants and compare to the products.
(+7) (-2)
(+4) (-2)
(+4) (-2)
(+6) (-2)
MnO4-(aq) + SO32-(aq)  MnO2(s) + SO42-(aq)
O.N. S = +6 – (+4) = +2
oxidation
O.N. Mn = +4 – (+7) = -3
reduction
2. Write the oxidation and reduction half-reactions without electrons (yet)
ox:
SO32-(aq) 
red: MnO4-(aq) 
SO42-(aq)
MnO2(s)
39
3. Balance everything but oxygen and hydrogen
ox:
SO32-(aq) 
red:
MnO4-(aq)

SO42-(aq)
MnO2(s)
S and Mn are
already balanced
4. Balance oxygen by adding water
ox:
H2O + SO32-(aq)  SO42-(aq)
red: MnO4-(aq) 
MnO2(s) + 2H2O
5. Balance hydrogen by adding (a) H+ in acidic solutions, (b) in basic solutions,
continue as if in acidic solution, but at the end each H+ ion will be neutralized
by adding OH- ions
ox:
H2O + SO32-(aq)  SO42-(aq) + 2H+(aq)
red: 4H+(aq) + MnO4-(aq)  MnO2(s) + 2H2O
6. Balance charge by adding electrons; for the oxidation half-reaction, the electrons
will be on the right, for the reduction half-reaction, the electrons will appear on the
left
ox:
H2O + SO32-(aq)  SO42-(aq) + 2H+(aq) + 2 e
red: 3 e + 4H+(aq) + MnO4-(aq)  MnO2(s) + 2H2O
7. Make sure the number of electrons in each half-reaction are the same. Then add
the half reactions together
ox:
H2O + SO32-(aq)  SO42-(aq) + 2H+(aq) + 2 e
x3
red: 3 e + 4H+(aq) + MnO4-(aq)  MnO2(s) + 2H2O x 2
Net:
3H2O + 3SO32-(aq) + 8H+(aq) + 2MnO4-(aq) 
3SO42-(aq) + 6H+(aq) + 2MnO2(s) + 4H2O
40
NOTE: sometimes you have to cancel H2O’s that are on each side, as well
as H+ or OH3H2O + 3SO32-(aq) + 8H+(aq) + 2MnO4-(aq)  3SO42-(aq) + 6H+(aq) + 2MnO2(s) + 4H2O
3SO32-(aq) + 8H+(aq) + 2MnO4-(aq)  3SO42-(aq) + 6H+(aq) + 2MnO2(s) + H2O
3SO32-(aq) + 2H+(aq) + 2MnO4-(aq)  3SO42-(aq) + 2MnO2(s) + H2O + 2OH+ 2OH2H2O
3SO32-(aq) + H2O(l) + 2MnO4-(aq)  3SO42-(aq) + 2MnO2(s) + 2OH-(aq)
8. Make sure that the equation is balanced for mass and for charge
3SO32-(aq) + H2O(l) + 2MnO4-(aq)  3SO42-(aq) + 2MnO2(s) + 2OH-(aq)
Mass:
Charge:
S=3
S=3
O = 18
H=2
O = 18
H=2
Mn = 2
Mn = 2
3(-2) + 2(-1) = -8
3(-2) +2(-1) = -8
41
Chapter Outline
 8.1 Solutions and Their Concentrations
 8.2 Dilutions
 8.3 Electrolytes and Nonelectrolytes
 8.4 Acids, Bases, and Neutralization Reactions
 8.5 Precipitation Reactions
 8.6 Oxidation-Reduction Reactions
 8.7 Titrations
 8.8 Ion Exchange
Titrations
In a titration a solution of accurately known concentration is
added gradually added to another solution of unknown
concentration until the chemical reaction between the two
solutions is complete.
Equivalence point – the point at which the reaction is complete
Indicator – substance that changes color at (or near) the
equivalence point
Slowly add base
to unknown acid
UNTIL
the indicator
changes color
42
Titration Example, p. 344-345
H2SO4 + 2NaOH → Na2SO4 + 2H2O
(titrant)
End Point
H2SO4
(unknown)
Titration Calculations, p. 344
Suppose it takes 22.40 mL of 0.00100 M sodium hydroxide to react completely with
the sulfuric acid in a 100.0 mL sample of acidic mine drainage. What was the
molarity of the sulfuric acid?
2NaOH(aq) + H2SO4(aq)  2H2O + Na2SO4
0.00100 M
22.40 mL
= 0.02240 L
100.0 mL
= 0.1000 L
M=?
1 mol H2SO4
2 mol NaOH
Plan: M x V = mol NaOH

mol H2SO4
V
 M
mol H2SO4 = 0.00100 mol NaOH x 0.02240 L NaOH x 1 mol H2SO4
L
2 mol NaOH
= 1.120 x 10-5 mol
M = 1.120 x 10-5 mol
0.1000 L
= 1.120 x 10-4 M
43
Sample Exercise 8.13:
Apple cider vinegar is an aqueous solution of acetic acid (MW = 60.05) made from fermented
apple juice. Commercial vinegar must contain at least 4 grams of acetic acid per 100 mL of
vinegar. Suppose the titration of a 25.00 mL sample of vinegar requires 12.15 mL of 1.885 M
sodium hydroxide. What is the molarity of acetic acid in the vinegar sample? Does it meet the
4 g/100 mL standard?
HC2H3O2(aq) + NaOH(aq)  H2O + NaC2H3O2
25.00 mL
M=?
mg/L = ?
12.15 mL
= 0.01215 L
M = 1.885
1 mol HC2H3O2
1 mol NaOH
Plan: M x V = mol NaOH
M HC2H3O2 = 1.885 mol NaOH
L

mol HC2H3O2
MW
V
 M  g/mL
x 0.01215 L NaOH x 1 mol HC2H3O2 x
1
1 mol NaOH
0.02500 L
= 0.9161 M
g
L
= 0.9161 mol x 60.05 g
1mol
L
x
1L
1000 mL
= 0.05501 g/mL
= 5.501 g/100 mL
44