Download + H 2 O

Chapter 4
Types of Chemical Reactions
and Solution Chemistry
Topics
Water as a solvent
 Electrolytes and nonelectrolytes
 Calculations involving molarity of
solutions
 Precipitation reactions
 Acid base reactions
 Oxidation reduction reactions

4.1 Water, the common solvent
The Water Molecule, Polarity
δ+ water
meanshas
a a
Thus,
δ- means a
partial
partial
negative
partial
endpositive
(0xygen) and
negative
charge
a partial
positive
δcharge
end (Hydrogen) –
O
and it is called
H
H
“polar” because of
+
+
δ
δ
the unequal
charge
bond angle of water = 105o
distribution
The Water Molecule
bent shape and ability to
hydrogen bond gives it many
special properties.
 Water molecules are attracted to one
another by dipole interactions
 This hydrogen bonding gives water:
a) its high surface tension, b) its
 Water’s
low vapor pressure.
Dissolving ionic salts in water
and Hydration
Ions have charges and attract the opposite
charges on the water molecules.
 The process of breaking the ions of salts
apart is called hydration

NH4NO3(s)
H2O (l)
NH4+(aq) + NO3-(aq)
Designates hydration of ions
How Ionic solids dissolve in water
These ions have been pulled away from the
main crystal structure by water’s polarity.
H
H
H
H
H
These ions have been
surrounded by water,
and are now dissolved!
Hydration
Solubility in water and
Aqueous Solutions
Water dissolves ionic compounds
(NaCl) and polar covalent
molecules (ethanol C2H5OH)
 The rule is: “like dissolves like”
 Polar dissolves polar.
 Nonpolar dissolves nonpolar.
 Oil is nonpolar.
– Oil and water don’t mix.
 Many Salts are ionic- sea water

The Solution Process
Called “solvation”.
 Water breaks the + and - charged
pieces apart and surrounds them.
 Solubility in water depends on the
relative attractions of ions for each
other and attraction of ions for water
molecules
 In some ionic compounds, the attraction
between ions is greater than the
attraction exerted by water (slightly
soluble slats)
– Barium sulfate and calcium carbonate






Solids will dissolve if the attractive force
of the water molecules is stronger than
the attractive force of the crystal.
If not, the solids are insoluble.
Water doesn’t dissolve nonpolar
molecules (like oil) because the water
molecules can’t hold onto them.
The water molecules hold onto other
water molecules, and separate from the
nonpolar molecules.
Nonpolars? No repulsion between them
How does ethanol dissolve in
water?
Ethanol Molecule
Contains a Polar
O-H Bond Similar
to Those in the
Water Molecule
The polar water molecule interacts strongly
with the polar-O-H bond in ethanol
4.2 The nature of aqueous solutions:
strong and weak electrolytes
A solution is a homogenous mixture of 2 or
more substances
The solute is(are) the substance(s) present in the
smaller amount(s)
The solvent is the substance present in the
larger amount
Solution
Solvent
Solute(s)
Soft drink (l)
H2O
Sugar, CO2
Air (g)
N2
O2, Ar, CH4
Soft Solder (s)
Pb
Sn
Electrolytes and Nonelectrolytes
Electrolytes- compounds that conduct
an electric current in aqueous solution,
or in the molten state
– all ionic compounds are electrolytes
because they dissociate into ions
(they are also called “salts”)
 barium sulfate- will conduct when
molten, but is insoluble in water!

 Nonelectrolytes-
Do not conduct
an electric current
– Most are molecular materials,
because they do not have ions
 Not
all electrolytes conduct to
the same degree
– there are strong electrolytes, and
weak electrolytes
– Conductivity depends on: degree
of dissociation or ionization
Electric conductivity of aqueous solutions
Nonelectrolyte
______________
Weak electrolyte
______________
Strong electrolyte
______________
4.1
Dissociation of acids and bases:
Strong and weak acids and bases
Acids- form H+ ions when dissolved in water
(According to Arrhenius)
 Strong acids dissociate completely into H+
and anions
 Strong acids- H2SO4 HNO3
HCl HBr HI
HClO4

Bases - form OH- ions when dissolved in
water
 Strong bases- KOH, NaOH

Weak acids- dissociate partially
 Acetic acid: HC2H3O2 has 1%
dissociation in aqueous solutions
 The most common weak base is
ammonia, NH3

HCl
H 2O
HNO3
H 2O
HC2H3O2(aq)
H+ + ClH+ + NO3H+ + C2H3CO2-
Strong electrolyte, strong acid
Strong electrolyte, strong acid
Weak electrolyte, weak acid
H2SO4
H+ + HSO4-
Strong electrolyte, strong acid
HSO4-
H+ + SO42-
Weak electrolyte, weak acid
H3PO4
H2PO4HPO42-
H+ + H2PO4H+ + HPO42H+ + PO43-
Weak electrolyte, weak acid
Weak electrolyte, weak acid
Weak electrolyte, weak acid
4.3
H2 O

NaOH(s)

NH3(aq) + H2O
Na+(aq) + OH-(aq)
NH4+(aq) + OH-(aq)
4.3 The composition of solutions
Molarity (M)
Molarity: A concentration that expresses
the moles of solute in 1 L of solution
Molarity (M) =
moles of solute
liters solution
Molarity Calculation
If 4.0 g NaOH are used to make 500. mL
of NaOH solution, what is the molarity
(M) of the solution?
Calculating Molarity
4.0 g NaOH x 1 mole NaOH = 0.10 mole NaOH
40.0 g NaOH
500. mL x
1L_
1000 mL
= 0.500 L
0.10 mole NaOH = 0.20 mole NaOH
0.500 L
1L
= 0.20 M NaOH
An acid solution is a 0.10 M HCl. How
many moles of HCl are in 1500 mL of this
acid solution?
1500 mL x
1 L
=
1000 mL
1.5 L
1.5 L x 0.10 mole HCl = 0.15 mole HCl
1L
How many grams of KCl are present
in 2.5 L of 0.50 M KCl?
2.5 L x 0.50 mole x 74.6 g KCl = 93
g KCl
1L
1 mole KCl
How many milliliters of an acid solution,
which is 0.10 M HCl, contain 0.15 mole
HCl?
0.15 mole HCl x 1 L soln
x 1000 mL
0.10 mole HCl
1L
= 1500 mL HCl
How many grams of NaOH are required to prepare 400.
mL of 3.0 M NaOH solution?
400. mL x 1 L
= 0.400 L
1000 mL
0.400 L x 3.0 mole NaOH x 40.0 g NaOH
1L
1 mole NaOH
=
48 g NaOH
4.5
A sample of 0.14 M NaCl. What volume of
sample contains 1.0 mg NaCl?
# mol = M X V (L)
V ( L) 
1g
# mol
M

# mol
V ( L) 
M
1L
1 mol NaCl
x
1.0mg NaCl x
= 1.2X10-4 L
x
1000 mg 58.5g NaCl 0.14mol NaCl
Dilution

Adding more solvent to a known solution.
The moles of solute stay the same.
#moles = M x volume (L)

# moles before dilution (1) = # moles after dilution


 M1 V1

= M2 V2
Stock solution is a solution of known concentration
used to make more dilute solutions
Preparing a less concentrated solution
from a more concentrated solution by dilution
Dilution
Add Solvent
Moles of solute
before dilution (i)
=
Moles of solute
after dilution (f)
MiVi
=
MfVf
Example
How would you prepare 60.0 mL of 0.200 M HNO3
from a stock solution of 4.00 M HNO3?
M1 = 4.00
M2 = 0.200
V2 = 60.0 m L V1 = ?mL
M1V1 = M2V2
V1 =
M2V2
M1
0.200 M X 60.0 mL
=
4.00 M
= 3.00 mL
Dilution process
Wash bottle
Funnel
pipits
Volumetric
flask
4.4 Types of Chemical Reactions
Precipitation reactions
 Acid-base reactions
 Oxidation-reduction reactions

4.5 Precipitation Reactions
When aqueous solutions of ionic
compounds are mixed together a
solid forms.
 A solid that forms from mixed
solutions is called precipitate
 If the substance is not part of the
solution, it is a precipitate

Precipitation Reactions
precipitate
Pb(NO3)2 (aq) + 2NaI (aq)
PbI2 (s) + 2NaNO3 (aq)
________________________
Pb2+ + 2NO3- + 2Na+ + 2I-
PbI2 (s) + 2Na+ + 2NO3-
Anions and cations switch partners
________________________
PbI2
Pb2+ + 2I-
PbI2 (s)
________________________
Na+ and NO3- are Spectator ions
Solubility rules for common ionic compounds in water at 250 C
Soluble Compounds
Exceptions
Compounds containing alkali
metal ions and NH4+
NO3-, HCO3-, ClO3Cl-, Br-, I-
Halides of Ag+, Hg22+, Pb2+
SO42-
Sulfates of Ag+, Ca2+, Sr2+,
Ba2+, Hg2+, Pb2+
Slightly soluble Compounds
Exceptions
CO3
OH-
2-,
PO4
3-,
CrO4
2-,
S2-
Compounds containing alkali
metal ions and NH4+
Compounds containing alkali
metal ions and Ba2+, Ca2+,
Mg2+ are marginally soluble
Solubility Rules
Predicting reaction’s product
All nitrates are soluble
Alkali metals ions and NH4+ ions are
soluble
Halides are soluble except Ag+,
Pb+2, and Hg2+2
Most sulfates are soluble, except
Pb+2, Ba+2, Hg+2,and Ca+2
Solubility Rules
Most hydroxides are slightly soluble
(insoluble) except NaOH and KOH
Sulfides, carbonates, chromates, and
phosphates are insoluble
Lower number rules supersede so
Na2S is soluble
4.6 Describing reactions in solution

Types of equations used to represent
chemical reactions:
– Formula equation
– Complete ionic equation
– Net ionic equation
Writing Net Ionic Equations
1. Write the balanced formula equation.
2. Write the net ionic equation showing the strong
electrolytes
3. Determine precipitate from solubility rules
4. Cancel the spectator ions on both sides of the ionic
equation
Write the net ionic equation for the reaction of
silver nitrate with sodium chloride.
AgNO3 (aq) + NaCl (aq)
AgCl (s) + NaNO3 (aq)
4.7 Stoichiometry of precipitation reactions
Exactly the same, except you may have
to figure out what the pieces are.
 What mass of solid is formed when
100.00 mL of 0.100 M Barium chloride is
mixed with 100.00 mL of 0.100 M sodium
hydroxide?
 What volume of 0.204 M HCl is needed to
precipitate the silver from 50.ml of
0.0500 M silver nitrate solution ?

Example
Calculate mass of solid NaCl that must be to
1.50 L 0f 0.100M AgNO3 solution to precipitate
all of the Ag+ ions in the form of AgCl
Ag+(aq) + Cl- (aq)
AgCl(s)
#moles Ag+ = #moles AgNO3

= 1.50L X
0.100 mol Ag+
1L
= 0.15 mol Ag+
(1:1 mole ratio)
#mole Cl- required =#mole NaCl = #moles Ag+
58.45 g NaCl
= 0.15 mol NaCl X
1mol NaCl
= 8.77 g NaCl
4.8 Acid-Base Reactions
For the purpose of this chapter:
An acid is a proton donor
 a base is a proton acceptor usually

(donates OH- ions to the solution)
base
acid
acid
base
Describing acid-base reactions
Strong acid
HCl (aq) + NaOH (aq)
H+ + Cl- + Na+ + OHH+(aq) + OH- (aq)
NaCl (aq) + H2O
Na+ + Cl- + H2O
H2O (l)
Weak acid
HC2H3O2(aq)+NaOH (aq)
HC2H3O2(aq) +Na+ + OHHC2H3O2(aq) + OH-
NaC2H3O2(aq) +H2O
C2H3O2- +Na+ +H2O
C2H3O2- +H2O
Neutralization Reactions and Salts

•
•
•
•
When acid and bases with equal amounts of
hydrogen ion H+ and hydroxide ions OH- are
mixed, the resulting solution is neutral.
 This reaction is called Neutralization
reaction
• HCl(aq) + NaOH(aq)  H2O(l) + NaCl(aq)
H2SO4(aq) + 2KOH (aq)  K2SO4(aq) + 2H2O(l)
Notice salt (NaCl) and water are the products
Salt = ionic compound whose cation comes from
a base and anion from an acid.
Neutralization between acid and metal
hydroxide produces water and a salt.
Example
How many mL of 2.00 M H2SO4 are required to
neutralize 50.0 mL of 1.00 M KOH?
H2SO4 + 2KOH
K2SO4 + 2H2O
0.0500 L x 1.00 mole KOH x 1 mole H2SO4 x
1L
1L
2 mole H2SO4
x
2 mole KOH
1000 mL =
1L
12.5 mL
Another solution
H2SO4 + 2KOH
K2SO4 + 2H2O
#mole KOH = (M) KOH X (V) KOH
#mole KOH = 0.05 L X1M = 0.05 mol KOH
1 mol H2SO4
#mole H2SO4 = # mole KOH X

2mol KOH
= 0.025 mol H2SO4
#mole H2SO4 = M H2SO4 X vol (L) H2SO4
0.025 mol H2SO4 = 2.00 M X V
V = 0.0125 L = 12.5 mL
Acid - Base Titrations
Often called a neutralization titration Because
the acid neutralizes the base
 Often called volumetric analysis since titration
is made to determine concentrations
 It involves delivery of a measured volume of
solution of known concentration (titrant)
 Titrant is added to the unknown (analyte)


until the equivalence (stoichiometric) point is
reached where enough titrant has been added
to neutralize it.
Titration
Equivalence point is marked by using an
indicator
 Where the indicator changes color is the

endpoint
End point does not match always the
equivalence point.
 A successful titration requires:

– A rapid known exact reaction
– Endpoint is very close to the equivalence point
– Accurate Measurement of volume of titrant
Accurate determination of a solution
concnetration is called
Titration
Equivalence point is marked by using an
indicator
 Where the indicator changes color is the

endpoint
End point does not match always the
equivalence point.
 A 50.00 mL sample of aqueous Ca(OH)2
requires 34.66 mL of 0.0980 M Nitric acid
for neutralization. What is [Ca(OH)2 ]?


# of H+ x MA x VA = # of OH- x MB x VB
Titration
A solution of accurately known concentration is 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
Endpoint –the point at which the color of indicator changes
Slowly add base
to unknown acid
UNTIL
The indicator
changes color
(pink)
Example

In the titration of KHP (molar mass =
204.22g/mol), 34.67 mL of NaOH
solution was required to react with
0.1082 g KHP. Calculate the molarity
of NaOH solution.
Solution
NaOH(aq) + KHP(aq) → NaKP(aq) + H2O(l)
# mol NaOH = 0.1082 g KHP ×
1 mol KHP
1 mol NaOH

204.22 g KHP
mol KHP
= 5.298X10-4 mol NaOH
There are 5.298X10-4 mol of sodium hydroxide in
34.67 mL of solution.
#mole NaOH = MNaOH X VNaOH
MNaOH = # molNaOH
V
=
5.298  10 4 mol
34.67  10 3 L
NaOH
= 1.528
M NaOH
4.9 Oxidation - Reduction (Redox)
Reactions
0
1
0
1
2 Na ( s)  Cl ( g )  2 Na Cl ( s)
2
Each sodium atom loses one electron:
1
0
Na  Na  e

Each chlorine atom gains one electron:
0

1
Cl  e  Cl
Definitions
Loss of Electrons = Oxidation
1
0
Na  Na  e

Sodium is oxidized
Increase of the oxidation state
Gain of Electrons = Reduction
0

1
Cl  e  Cl
Chlorine is reduced
Decrease of the oxidation state
Definitions
- The substance that loses electrons is
called reducing agent (Electron donor)
- The substance that gains electrons is
called oxidizing agent (Electron acceptor)
+2
0
Mg is the
reducing
agent
-2
0
Mg(s) + S(s) → MgS(s)
Mg is oxidized – loses e-
S is reduced – gains eS is the oxidizing agent
Not All Reactions are Redox Reactions
- Reactions in which there has been
no change in oxidation number are
not redox reactions.
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 )
Assigning Oxidation States
• An “oxidation state” (oxidation
number) is a positive or negative
number assigned to an atom to
indicate its degree of oxidation or
reduction.
• Generally, a bonded atom’s
oxidation number is the charge it
would have if the electrons in the
bond were assigned to the atom of
the more electronegative element
Rules for Assigning Oxidation States
1) The oxidation State of any
uncombined element is zero.
2) The oxidation State of a
monatomic ion equals its charge.
0
0
1
1
2 Na  Cl 2  2 Na Cl
Rules for Assigning Oxidation States
3) The oxidation state of oxygen in
compounds is -2, except in peroxides,
such as H2O2 where it is -1.
4) The oxidation state of hydrogen in
compounds is +1, except in metal
hydrides, like NaH, where it is -1.
1
2
H2O
Rules for Assigning Oxidation states
5) The sum of the oxidation states of the
atoms in the compound must equal 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 States
6) The sum of the oxidation States in
the formula of a polyatomic ion is
equal to its ionic charge.
? 2
N O3

? 2
S O4
2
X + 3(-2) = -1
N
O
X + 4(-2) = -2
S
O
 X = +5
 X = +6
What is the oxidation states of
all atoms in the following ?
IF7
F=?
7x-1 + ? = 0
I=
NaIO3
Na =?
O=?
3x-2 + 1 + ? = 0
I=
K2Cr2O7
O=?
K=?
7x-2 + 2x+1 + 2x? = 0
Cr =??
4.10 Balancing oxidation-reduction Equations
Two systematic methods for balancing
redox equations are available, and are
based on the fact that the total electrons
gained in reduction equals the total lost
in oxidation.
The two methods:
1) Use oxidation state changes
2) Use half-reactions (the method to be
used her)
Balancing redox equations using half-reactions
1: write separate half-reaction equations
for oxidation and reduction
2. balance the atoms in the half reactions
3. add enough electrons to one side of each
half-reaction to balance the charges
4. multiply each half-reaction by a number to
make the electrons equal in both
5. add the balanced half-reactions to show an
overall equation and cancel identical
species
6. Check that elements and charges are balanced
Example
2Mg (s) + O2 (g)
2Mg
O2 + 4e-
2MgO (s)
2Mg2+ + 4e- Oxidation half-reaction (lose e-)
2O2-
Reduction half-reaction (gain e-)
2Mg + O2 + 4e2Mg + O2
2Mg2+ + 2O2- + 4e2MgO
Balancing redox equations
in an acidic solution
For reactions in acidic solution an 8
step procedure.
 Write separate half reactions
 For each half reaction balance all
reactants except H and O
 Balance O using H2O

 Balance H using H+
 Balance charge using e Multiply equations to make electrons
equal
 Add equations and cancel identical
species
 Check that charges and elements are
balanced.
Example: Balance the following equation
I- + OCI-

I-
Oxidation :
I3- + Cl- (acidic solution)


3I-
OC1-
I3-

3I-
Reduction : OC1-
I 3-


I3- + 2e-
C1C1- + H2O
OC1-

2H+ + OC1- 
1+
C1- + H2O
C1- + H2O
-1
2H+ + OC1- + 2e-  C1- + H2O (2)
3I-  I3- + 2e(1)
(1) + (2)
2H+ + OC1- + 3I-  C1- + I3- + H2O
Balancing equations in basic solution
• Do everything you would with acid, but add
one more step.
• Add enough OH- to both sides to neutralize
the H+
• For each H atom that is lacking, add one
molecule of H2O to the side that requires it.
At the same time add one unit of OH- in the
opposite side of the half-reaction
Example
Balance the following equation in basic solution
MnO4- + C2O42-  MnO2 + CO32First half reaction:
MnO4-  MnO2
(a) MnO4-  MnO2 + 2H2O
(b) 4 H2O + MnO4-  MnO2 + 2H2O + 4 OH(c) 4 H2O + MnO4- + 3 e-  MnO2 + 2H2O + 4 OH(d) 2 H2O + MnO4- + 3 e-  MnO2 + 4 OH-
(1)
Second half reaction:
C2O42-  CO32(a) C2O42-  2 CO32(b) 2 H2O + C2O42-  2 CO32(c) 4 OH- + 2 H2O + C2O42-  2 CO32- + 4 H2O
(d) 4 OH- + 2 H2O + C2O42-  2 CO32- + 4 H2O + 2 e(e) 4 OH- + C2O42-  2 CO32- + 2 H2O + 2 e-
2 H2O + MnO4- + 3 e-  MnO2 + 4 OH-
(2)
(1)
We eliminate electrons by multiplying (1) by 2 and (2) by 3.
(2)
X 3:
12 OH- + 3 C2O42-  6 CO32- + 6 H2O + 6 e-
(3)
(1) X 2:
4 H2O + 2 MnO4- + 6 e-  2 MnO2 + 8 OH-
(4)
(3) + (4) to eliminate the electrons and common species.
4 OH- + 3 C2O42- + 2 MnO4-  6 CO32- + 2 H2O + 2 MnO2