Transcript Slide 1

C H E M I S T R Y
Chapter 4
Reactions in Aqueous Solution
Acids, Bases, and Neutralization
Reactions
Acid (Arrhenius): A substance that dissociates in water to produce hydrogen ions,
H1+:
HA(aq)
H1+(aq) + A1-(aq)
HCl(aq)
H1+(aq) + Cl1-(aq)
In water, acids produce hydronium ions, H3O1+:
HCl(aq) + H2O(aq)
H3O1+(aq) + Cl1-(aq)
Acids, Bases, and Neutralization
Reactions
Base (Arrhenius): A substance that dissociates in water to produce hydroxide
ions, OH1-:
MOH(aq)
M1+(aq) + OH1-(aq)
NaOH(aq)
Na1+(aq) + OH1-(aq)
Ammonia, commonly called “ammonium hydroxide” is a base:
NH3(aq) + H2O(aq)
NH41+(aq) + OH1-(aq)
Acids, Bases, and Neutralization
Reactions
Strong acids and strong bases are strong electrolytes.
Weak acids and weak bases are weak electrolytes.
Acids, Bases, and Neutralization
Reactions
These acid-base neutralization reactions are double-replacement reactions just like
the precipitation reactions:
HA + MOH
MA + HOH
or
HA + MOH
Acid
Base
MA + H2O
Salt
Water
Acids, Bases, and Neutralization
Reactions
Write the molecular, ionic, and net ionic equations for the reaction of aqueous
HBr and aqueous Ba(OH)2.
1.
Write the chemical formulas of the products (use proper ionic rules for the salt).
HBr(aq) +
Acid
Ba(OH)2(aq)
Base
H2O
Water
+ BaBr2
Salt
Acids, Bases, and Neutralization
Reactions
Write the molecular, ionic, and net ionic equations for the reaction of aqueous
NaOH and aqueous HF.
1.
Write the chemical formulas of the products (use proper ionic rules for the salt).
HF(aq) + NaOH(aq)
Acid
Base
H2O
+ NaF
Water
Salt
Examples
 Predict the product and write a molecular equation, ionic
equation and net ionic equation for the following reactions
 K2CO3(aq) + NiCl2(aq) 
 HNO3(aq) + LiOH(aq) 
 HCN(aq) + Mg(OH)2
Oxidation-Reduction (Redox)
Reactions
4Fe(s) + 3O2(g)
2Fe2O3(s) + 3C(s)
2Fe2O3(s)
4Fe(s) + 3CO2(g)
Rusting of iron:
an oxidation of Fe
Manufacture of iron: a
reduction of Fe
Oxidation-Reduction (Redox)
Reactions
Oxidation Number (State): A value which indicates whether an atom is neutral,
electron-rich, or electron-poor.
Rules for Assigning Oxidation Numbers
1.
An atom in its elemental state has an oxidation number of 0.
Na
H2
Br2
Oxidation number 0
S
Ne
Oxidation-Reduction (Redox)
Reactions
2.
A monatomic ion has an oxidation number identical to its charge.
Na1+
Ca2+
Al3+
Cl1-
O2-
+1
+2
+3
-1
-2
Oxidation-Reduction (Redox)
Reactions
3.
An atom in a polyatomic ion or in a molecular compound usually has the
same oxidation number it would have if it were a monatomic ion.
a) Hydrogen can be either +1 or -1.
H
+1
b)
O
1-
H
Ca
-1
-2
H
+2
-1
Oxygen usually has an oxidation number of -2.
H
+1
O
-2
H
+1
H
+1
O
-1
O
-1
H
+1
Oxidation-Reduction (Redox)
Reactions
3.
c)
Halogens usually have an oxidation number of -1.
H
Cl
+1
-1
Cl
+1
O
-2
Cl
+1
Oxidation-Reduction (Redox)
Reactions
4.
The sum of the oxidation numbers is 0 for a neutral compound and is equal
to the net charge for a polyatomic ion.
2(+1) + x + 3(-2) = 0 (net charge)
H2SO3
+1
x
Cr2O72-
x = +4
-2
2(x) + 7(-2) = -2 (net charge)
x = +6
x
-2
Example
 Determine the oxidation number for each atom in the
following compounds/molecules
 CO2
 CCl4
 CoSO4
 K2O2
Identifying Redox Reactions
Reduction: gaining one or more electron
increasing in oxidation number
reduction
0
 Oxidizing agent
4Fe(s)
+
3O2(g)
0
-2
2Fe2 O3
+3
oxidation
Oxidation: losing one or more electrons
decreasing in oxidation number
 Reducing agent
(s)
Identifying Redox Reactions
Oxidizing Agent
•Causes oxidation
•Gains one or more electrons
•Undergoes reduction
•Oxidation number of atom decreases
Reducing Agent
Causes reduction
Loses one or more electrons
Undergoes oxidation
Oxidation number of atom increases
Example
Identify each of the following as
1) oxidation or 2) reduction.
__A. Sn(s)
Sn4+(aq) + 4e−
__B.
Fe3+(aq) + 1e−
Fe2+(aq)
__C.
Cl2(g) + 2e−
2Cl-(aq)
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Writing Oxidation and Reduction Reactions
Write the separate half oxidation and reduction
reactions for the following equation.
2Cs(s) + F2(g)
2CsF(s)
 3 Na(l) + AlCl3(l) 3  NaCl(l) + Al(l)
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The Activity Series of the Elements
Cu(s) + 2Ag1+(g)
Cu2+(aq) + 2Ag(s)
Which one of these reactions will occur?
2Ag(s) + Cu2+(g)
2Ag1+(aq) + Cu(s)
The Activity Series of the Elements
Fe(s) + Cu2+(aq)
Fe2+(aq) + Cu(s)
The Activity Series of the Elements
Elements that are
higher up in the table
are more likely to be
oxidized.
Thus, any element
higher in the activity
series will reduce the
ion of any element
lower in the activity
series.
The Activity Series of the Elements
Cu(s) + 2Ag1+(g)
Cu2+(aq) + 2Ag(s)
Which one of these reactions will occur?
2Ag(s) + Cu2+(g)
2Ag1+(aq) + Cu(s)
Example
 Predict whether the following redox reactions will occurred
or not. If so, predict the products
 Zn(s) + FeCl2(aq) 
 Ni(s) + Mg(NO3)2(aq) 
Redox Titrations
Titration: A procedure for determining the concentration of a solution by allowing a
carefully measured volume to react with a solution of another substance (the standard
solution) whose concentration is known.
5H2C2O4(aq) + 2MnO41-(aq) + 6H1+(aq)
10CO2(g) + 2Mn2+(aq) + 8H2O(l)
If the unknown concentration is the potassium permanganate solution, MnO41-, it can
be slowly added to a known amount of oxalic acid, H2C2O4, until a faint purple color
persists.
Redox Titrations
A solution is prepared with 0.2585 g of oxalic acid, H2C2O4. 22.35 mL of an unknown
solution of potassium permanganate are needed to titrate the solution. What is the
concentration of the potassium permanganate solution?
5H2C2O4(aq) + 2MnO41-(aq) + 6H1+(aq)
10CO2(g) + 2Mn2+(aq) + 8H2O(l)
Calculation Set up
Mass of
H2C2O4
Moles of
H2C2O4
Moles of
KMnO4
Mole Ratio
Molar Mass of
H2C2O4
Molarity of
KMnO4
Molarity of
KMnO4
Example
 A 0.0484M standard solution of potassium permanganate was
titrated against 25.00mL of an iron (II) sulfate solution. The
equivalence point, as indicated by a faint pink color, was reached
when 15.50mL of potassium permanganate solution had been
added. Calculate the concentration of the iron (II) sulfate
solution