Download Reactions in Aqueous Solution (Calculations): Chapter 11 Chapter

Reactions in Aqueous Solution
(Calculations): Chapter 11
Chem 101
Fall 2004
Chapter Outline
Aqueous Acid-Base Reactions
• Calculations Involving Molarity
• Titrations
• The Mole Method and Molarity
• Equivalent Weights and Normality
Oxidation-Reduction Reactions
• The Half-Reaction Method
• Adding in H+, OH- , or H2O to Balance Oxygen or
Hydrogen
• Stoichiometry of Redox Reactions
Chem 101
Fall 2004
Equivalent Weights and Normality
• Normality is another method of expressing
concentration.
• Normality is defined as the number of equivalent
weights of solute per liter of solution.
N =
# eq solute
# meq solute
or N =
L sol' n
mL sol' n
Chem 101
Fall 2004
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Equivalent Weights and Normality
• The equivalent weight of an acid is the mass in
grams of the acid necessary to furnish Avogadro’s
number of H+ ions.
• For monoprotic acids like HCl 1 mol = 1 eq
• For diprotic acids like H2SO4 1 mol = 2 eq
• For triprotic acids like H3PO4 1 mol = 3 eq
Chem 101
Fall 2004
Equivalent Weights and Normality
• Similarly for bases:
• 1 mol = 1 eq for NaOH
• 1 mol = 2 eq for Ba(OH)2
• 1 mol = 3 eq for Fe(OH)3
• The equivalent weights of some acids and bases are given
in Table 11-1 in your text.
• Any problem that can be done in normality can also be
done in molarity.
• The difference between the two is that in molarity you must
remember to use the reaction stoichiometry but in normality the
stoichiometry is included in the solution concentrations.
Chem 101
Fall 2004
Equivalent Weights and Normality
• Since M x L = moles then
• N x L = number of equivalents or
• N x mL = number of milliequivalents
• In reactions, including acid-base reactions, 1 equivalent
of an acid will react with 1 equivalent of a base.
Chem 101
Fall 2004
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Equivalent Weights and Normality
• What volume of 6.00 M phosphoric acid
solution is required to prepare 9.00 x 102 mL
of 0.200 N phosphoric acid solution?
Chem 101
Fall 2004
Oxidation-Reduction Reactions
• We have previously gone over the basic concepts
of oxidation & reduction in Chapter 4.
• Rules for assigning oxidation numbers were also
introduced in Chapter 4.
• Refresh your memory as necessary.
• We shall learn to balance redox reactions using the
half-reaction method.
Chem 101
Fall 2004
The Half-Reaction Method
Half reaction method rules:
• Write the unbalanced reaction.
• Break the reaction into 2 half reactions:
One oxidation half-reaction and
One reduction half-reaction
Each reaction must have complete formulas for molecules
and ions.
Chem 101
Fall 2004
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The Half-Reaction Method
Half reaction method rules:
• Mass balance each half reaction by adding appropriate
stoichiometric coefficients. To balance H and O we can
add:
H+ or H2O in acidic solutions.
OH- or H2O in basic solutions.
• Charge balance the half reactions by adding appropriate
numbers of electrons.
Electrons will be products in the oxidation half-reaction.
Electrons will be reactants in the reduction half-reaction.
Chem 101
Fall 2004
The Half-Reaction Method
Half reaction method rules:
• Multiply each half reaction by a number to make the
number of electrons in the oxidation half-reaction
equal to the number of electrons reduction halfreaction.
• Add the two half reactions.
• Eliminate any common terms and reduce coefficients
to smallest whole numbers.
Chem 101
Fall 2004
The Half-Reaction Method
• Tin (II) ions are oxidized to tin (IV) by bromine. Use
the half reaction method to write and balance the net
ionic equation.
Starting Reaction
Sn 2+ + Br2 → Sn 4+ + Br Mass balance the half - reaction.
Sn
2+
→ Sn
4+
Charge balance the half - reaction.
Sn
Chem 101
Fall 2004
2+
→ Sn
4+
+ 2e-
Electrons are products thus this
is the oxidation half - reaction
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The Half-Reaction Method
• Tin (II) ions are oxidized to tin (IV) by bromine. Use
the half reaction method to write and balance the net
ionic equation.
Mass balance the other half - reaction.
Br 2 → 2 Br Charge balance the other half - reaction.
Br 2 + 2 e - → 2 Br This is the reduction half reaction.
Chem 101
Fall 2004
The Half-Reaction Method
• Tin (II) ions are oxidized to tin (IV) by bromine. Use
the half reaction method to write and balance the net
ionic equation.
Sn 2+ + Br2 → Sn 4+ + Br- starting reaction
Add the two half reactions.
Sn 2+ → Sn 4+ + 2e- ox. half reaction
Br2 + 2e- → 2 Br- red. half reaction
Sn + Br2 → Sn 4+ + 2 Br- balanced reaction
2+
Chem 101
Fall 2004
Stoichiometry in Redox Reactions
• Just as we have done stoichiometry with acid-base
reactions, it can also be done with redox reactions.
• What volume of 0.200 M KMnO4 is required to
oxidize 35.0 mL of 0.150 M HCl?
2 KMnO 4 + 16 HCl → 2 KCl + 2 MnCl 2 + 5 Cl 2 + 8 H 2 O
Chem 101
Fall 2004
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Stoichiometry in Redox Reactions
2 KMnO 4 + 16 HCl → 2 KCl + 2 MnCl 2 + 5 Cl 2 + 8 H 2 O
.
M HCl) = 5.25 mmol HCl
(35 mL HCl)( 0150
(5.25 mmol HCl)
2 mmol KMnO 4 
 = 0.656 mmol KMnO 4
16 mmol HCl 


1 mL
 = 3.28 mL
 0.200 mmol KMnO 4 
(0.656 mmol KMnO 4 )
Chem 101
Fall 2004
Next Class: Gases and KineticMolecular Theory: Chapter 12
• Study for the Exam!
(Chapters 8, 10 and 11)
• Start to Read Chapter 12
Chem 101
Fall 2004
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