Chemistry 100(02) Fall 2010
Instructor: Dr. Upali Siriwardane
e-mail: upali@chem.latech.edu
Office: CTH 311 Phone 257-4941
Office Hours: M,W, 8:00-9:00 & 11:00-12:00 a.m Tu,Th,F 9:00 10:00 a.m.
Test Dates: March 25, April 26, and May 18; Comprehensive Fina
Exam: 9:30-10:45 am, CTH 328.
October 4,
October 27,
November 17,
November 18,
2010 (Test 1): Chapter 1 & 2
2010 (Test 2): Chapter 3 & 4
2010 (Test 3): (Chapter 5 & 6)
2010 (Make-up test) comprehensive: Chapters 1-6
9:30-10:45:15 AM, CTH 328
CHEM 100, Fall 2010, LA TECH
5-1
Chapter 5. Chemical Reactions
5.1 Exchange Reactions: Precipitation and Net Ionic
Equations Page164
5.2 Acids, Bases, and Acid-Base Exchange Reactions
Page171
5.3 Oxidation-Reduction Reactions Page179
5.4 Oxidation Numbers and Redox Reactions Page184
5.5 Displacement Reactions, Redox, and the Activity Series
Page187
5.6 Solution Concentration Page191
5.7 Molarity and Reactions in Aqueous Solutions Page198
5.8 Aqueous Solution Titrations Page201
CHEMISTRY IN THE NEWS: The Breathalyzer Page189
CHEMISTRY YOU CAN DO: Pennies, Redox, and the Activity
Series of Metals Page192
5-2
CHEM 100, Fall 2010, LA TECH
Chapter 5. KEY CONCEPTS
Chemical Reactions
Solution Chemistry
Solubilities of chemical compounds
Types of chemical equations
Write molecular equation
Write complete ionic equation
Write a net ionic equation
Common types of chemical reactions
Precipitation
Acid/base
Neutralization reactions
Common acids and bases
CHEM 100, Fall 2010, LA TECH
Gas-forming
Oxidation-reduction reactions
Redox reactions
Oxidizing and reducing agents
Assigning oxidation numbers
Activity series
Single displacement
Redox reactions
Molarity
Preparing solutions
Dilutions
Titrations
Solution stoichiometry
5-3
Solution Chemistry
Collisions of reactants decides the rates of
reactions
Why Solution Reactions?
Why not gaseous Reactions?
In Gases: Very fast Reactions
Why not solid Reactions?
Solids: Very slow Reactions
Solution reactions are manageble!
Liquids:
Fast Reactions
CHEM 100, Fall 2010, LA TECH
5-4
What is a Solution?
A solution: A homogeneous mixture of two or
more components.
Sugar in water
Oxygen in water
Air
Dental fillings
Saline
CHEM 100, Fall 2010, LA TECH
5-5
Solute
substance that is present in smallest
quantity
dissolved substance(s)
can be either a gas, a liquid, or a solid
one or more present in a solution
Solvent
substance present in largest quantity
only one per solution
water in aqueous solutions
CHEM 100, Fall 2010, LA TECH
5-6
Cola Drinks
Solvent
• water
Solutes
• carbon dioxide (gas)
• sweetner (solid)
• phosphoric acid (liquid)
• caramel color (solid)
CHEM 100, Fall 2010, LA TECH
5-7
Reactions in Solutions
There are many solvents However,
water is most abundant dissolving
many chemicals.
Water can interact with both cations
and anions making it the best
solvent for ionic compounds.
CHEM 100, Fall 2010, LA TECH
5-8
Dissolution of (a) Ionic and
(b) Molecular Compounds
CHEM 100, Fall 2010, LA TECH
5-9
Types of Chemical Equations
Molecular equation:
Equation with formula, correct stoichiometric coefficients
and physical form written within parenthesis.
Ionic equation:
All the ionic compounds soluble in water are separated into
ions written with their ionic charge and (aq).
Net Ionic equation:
Ionic equation with all spectator ions removed from both
sides.
CHEM 100, Fall 2010, LA TECH
5-10
NaCl(aq)+AgNO3(aq)
-->AgCl(s)+ NaNO3(aq)
Molecular equation:
NaCl (aq) + AgNO3 (aq) --> AgCl (s) + NaNO3 (aq)
Ionic Equation:
Na+ (aq) + Cl-(aq) + Ag+ (aq) + NO3-(aq)
--> AgCl(s) + Na+(aq) + NO3- (aq)
Spectator Ions:
Na+ (aq) and NO-3 (aq)
Net Ionic Equation:
Cl- (aq) + Ag+ (aq) --> AgCl (s)
CHEM 100, Fall 2010, LA TECH
5-11
Spectator Ions
Ions appearing on both side of an
ionic equation.
Ionic Equation:
Na+ (aq) + Cl-(aq) + Ag+ (aq) + NO3-(aq)
--> AgCl(s) + Na+(aq) + NO3- (aq)
CHEM 100, Fall 2010, LA TECH
5-12
More Examples
HCl(aq) + NaOH(aq) ----> NaCl(aq) + H2O(l)
NaOH(aq) + HC2H3O2(aq) ---> NaC2H3O2(aq) + H2O(l)
CHEM 100, Fall 2010, LA TECH
5-13
Types of Chemical Reactions
Based on driving force
a) Precipitation
b)
Reactions
Acid-base Reactions
c) Gas-forming
Reactions
d)Oxidation-reduction
CHEM 100, Fall 2010, LA TECH
(REDOX)Reactions
5-14
Precipitation reactions
They are double displacement reactions
of ionic compounds where an insoluble
salt is formed when two aqueous salt
solutions are mixed.
CHEM 100, Fall 2010, LA TECH
5-15
Solubility rules for ionic compounds
All acids are soluble.
All Na+, K+ and NH4+ salts are soluble.
All nitrate and acetate salts are soluble.
All chlorides except AgCl and Hg2Cl2 PbCl2 are soluble.
All sulfates are soluble except PbSO4, Hg2SO4, SrSO4 and
BaSO4.
All sulfides are insoluble except those of the Group IA (1), IIA
(2) and ammonium sulfide.
All hydroxides are insoluble except those of the group IA(1)
and IIA Ba(OH)2. Sr(OH)2 and Ca(OH)2
CHEM 100, Fall 2010, LA TECH
5-16
Illustration of
Some
Solubility
Rules
CHEM 100, Fall 2010, LA TECH
5-17
Precipitation of Silver Chloride
AgNO3 + NaCl  AgCl + NaNO3
precipitate
CHEM 100, Fall 2010, LA TECH
5-18
Ionic equations
When ionic substances dissolve in water, they
dissociate into ions.
H2O
AgNO3
Ag+ + NO3KCl
H2O
K+ + Cl-
When a reaction occurs, only some of the ions are
actually involved in the reaction.
Ag+ + NO3- + K+ + Cl-
CHEM 100, Fall 2010, LA TECH
AgCl(s) + K+ + NO3-
5-19
Ionic equations
To help make the reaction easier to see, we
commonly list only the species actually
involved in the reaction.
Full ionic equation
Ag+ + NO3- + K+ + ClNet ionic equation
Ag+ + Cl-
AgCl(s) + K+ + NO3-
AgCl(s)
NO3- and K+ are referred to as spectator ions.
CHEM 100, Fall 2010, LA TECH
5-20
Precipitation of Barium Sulfate
Double Displacement:
BaCl2(aq) + Na2SO4(aq) 
2NaCl(aq) + BaSO4(s)
precipitate
CHEM 100, Fall 2010, LA TECH
5-21
Ionic Equations
Molecular Equation:
BaCl2(aq) + Na2SO4(aq)  2NaCl(aq) + BaSO4(s)
precipitate
Total Ionic Equation:
Ba+2 + 2Cl-1 + 2Na+ + SO4-2  2Na+ +Cl-1 + BaSO4(s)
Net Ionic Equation:
Ba+2 + SO4-2  BaSO4
CHEM 100, Fall 2010, LA TECH
5-22
Precipitation or Not
MgI2 + NaNO3= 2 NaI
+ Mg(NO3)2
Ba(NO3)2+Na2SO4= BaSO4 + 2 NaNO3
AgCl +NaNO3
CHEM 100, Fall 2010, LA TECH
= AgNO3 + NaCl
5-23
Acid/base Reactions
Acid
substance that donates H+ ions to solution
sour-tasting substances
substances whose aqueous solutions are capable of
turning blue litmus indicators red
dissolves certain metals to form salts
react with bases or alkalis to form salts
Base
substance that donates a OH-1 ion to solution
hydroxides and oxides of metals
bitter tasting, slippery solutions
turn litmus blue
react with acids to form salts
CHEM 100, Fall 2010, LA TECH
5-24
Neutralization Reactions
Formation of water is the diving
force
acid + base  “salt” + water
HCl + NaOH  NaCl + H2O
Salt
H2SO4 + 2KOH  K2SO4 + 2H2O
•Substances produced by the reaction of an acid with a base
•Characterized by ionic bonds and high melting points
•Electrical conductivity when melted or when in solution
•Has a crystalline structure when in the solid state
CHEM 100, Fall 2010, LA TECH
5-25
How do find precursor
Acid and base of a Salt
Acid (A) + Base(B) = Salt + water (H2O)
HA + BOH = BA + H2O
E.g. LiNO3
B (Li) A (NO3)
OH
H
BOH(LiOH) HA(HNO3)
CHEM 100, Fall 2010, LA TECH
5-26
Some acids, bases and their salts
Acid
Name
Formula
Acetic acid
Hydrogen chloride
Nitric acid
Phosphoric acid
Sulfuric acid
HC2H3O2
HCl
HNO3
H3PO4
H2SO4
Base
Chloride salt
Name
Formula
Sodium hydroxide
NaOH
Barium oxide
BaO
Sodium oxide
Na2O
Ammonia
NH3
CHEM 100, Fall 2010, LA TECH
Sodium salt
Name
Formula
Sodium acetate
NaC2H3O2
Sodium chloride
NaCl
Sodium nitrate
NaNO3
Sodium phosphate
Na3PO4
Sodium sulfateNa2SO4
Name
Sodium chloride
Barium chloride
Sodium chloride
Ammonium chloride
Formula
NaCl
BaCl2
NaCl
NH4Cl
5-27
Ionization of Acids in Water
CHEM 100, Fall 2010, LA TECH
5-28
Common Acids and Bases
CHEM 100, Fall 2010, LA TECH
5-29
Ionic Equations Strong
Acid/base
Molecular Equation:
HCl(aq) + NaOH(aq) ----> NaCl(aq) + H2O(l)
Total Ionic Equation:
H+ + Cl-1 + Na+ + OH-1  Na+ +Cl-1 + H2O
Net Ionic Equation:
H+ + OH-1  H2O
NaOH(aq) + HC2H3O2(aq) -----> NaC2H3O2(aq) + H2O(l)
Na+(aq) + OH -(aq) + HC2H3O2(aq) -----> Na+(aq) +C2H3O2-(aq) + H2O(l)
HC2H3O2(aq) + OH-1  C2H3O2-(aq) + H2O
CHEM 100, Fall 2010, LA TECH
5-30
Ionic Equations Weak
Acid/base
Molecular Equation:
NaOH(aq) + HC2H3O2(aq) -----> NaC2H3O2 (aq) + H2O(l)
Total Ionic Equation:
Na+(aq) + OH - (aq) + HC2H3O2aq) -----> Na+(aq) +C2H3O2 -(aq) + H2O(l)
Net Ionic Equation:
HC2H3O2 (aq) + OH-  C2H3O2 -(aq) + H2O
CHEM 100, Fall 2010, LA TECH
5-31
Acid/base Ionic Equations
H2SO4 + 2KOH  K2SO4 + 2H2O
Total Ionic Equation:
2H+ + SO4-2 + 2Na+ + 2OH-1  2Na+ +2Cl-1 + 2H2O
Net Ionic Equation:
2H+ + 2OH-1  2H2O
CHEM 100, Fall 2010, LA TECH
5-32
Strong Acids
strong – completely ionized
weak – partially ionized
Strong Acids(strong electrolytes)
HCl, HBr, HI, H2SO4, HNO3, HClO4 (all others are
weak)
Weak acids( weak electrolytes)
CH3COOH, HF, HCN, H3PO4, HCOOH, H2CO3
Table 5.2 page 171.
CHEM 100, Fall 2010, LA TECH
5-33
Strong Bases
Strong bases
Metal Hydroxides
(Group IA):Li, Na K. (Group IIA): Ca,
Sr, Ba (all others metal hydroxides
are weak)
Weak bases
NH3, amines -CH3NH2
Table 5.2 page 171.
CHEM 100, Fall 2010, LA TECH
5-34
Gas-Forming
Exchange
Reaction:
CO2, SO2,
H2S
CHEM 100, Fall 2010, LA TECH
5-35
Reaction of Metal Carbonates with Acids
Molecular Equation:
CaCO3(s) + 2HCl(aq)  CaCl2(aq) + H2CO3(aq)
H2CO3(aq)  H2O + CO2(g)
Total Ionic Equation:
CaCO3(s) + 2H+ + 2Cl-1  Ca+2 + 2Cl-1 + H2O + CO2(g)
Net Ionic Equation:
CaCO3(s) + 2H+  Ca+2 + H2O + CO2(g)
CHEM 100, Fall 2010, LA TECH
5-36
Reaction of Metal Carbonates with Acids
Alka-Seltzer
NaHCO3(aq) + HCl(aq)  NaCl(aq) + H2O + CO2(g)
Net Ionic Equation:
HCO3-1 + 2H+  H2O + CO2(g)
Tums
CaCO3(s) + 2HCl(aq)  CaCl2(aq) + H2O + CO2(g)
Net Ionic Equation:
CO3-2 + 2H+  H2O + CO2(g)
CHEM 100, Fall 2010, LA TECH
5-37
Reaction of Metal Sulfites with Acids
Molecular Equation:
CaSO3(s) + 2HCl(aq)  CaCl2(aq) + H2SO3(aq)
H2SO3(aq)  H2O + SO2(g)
Total Ionic Equation:
CaSO3(s) + 2H+ + 2Cl-1  Ca+2 + 2Cl-1 + H2O + SO2(g)
Net Ionic Equation:
CaSO3(s) + 2H+  Ca+2 + H2O + SO2(g)
CHEM 100, Fall 2010, LA TECH
5-38
Reaction of Metal Sulfides with Acids
Molecular Equation:
Na2S(aq) + 2HCl(aq)  2NaCl(aq) + H2S(g)
Total Ionic Equation:
2Na+ + S-2 + 2H+ + 2Cl-1  2Na+ + 2Cl-1 + H2S(g)
Net Ionic Equation:
S-2 + 2H+  H2S(g)
CHEM 100, Fall 2010, LA TECH
5-39
Oxidation-Reduction Reactions
CHEM 100, Fall 2010, LA TECH
5-40
Recognizing Redox Reactions
CHEM 100, Fall 2010, LA TECH
5-41
Oxidation-Reduction
(Redox) Reactions
Electrons are transferred from one compound to the other
resulting in a chemical change.
E.g. Zn(s) + 2HCl(aq) --->ZnCl2(aq) + H2(g)
includes Single Replacement Reactions
Oxidation –loss of electrons
Reduction – gain of electrons
oxidizing agent – substance that causes oxidation
reducing agent – substance that cause reduction
CHEM 100, Fall 2010, LA TECH
5-42
Oxidation number or State
A number assigned to a atom in
compounds, ions and
polyatomic ions to show the
number of electrons relative to
an atom in the element.
CHEM 100, Fall 2010, LA TECH
5-43
Rules for Assigning Ox #
a) Oxidation number of atoms in an element
is zero (0). e.g. O2
b) Monoatomic ions: Ox # equal to charge.
E.g. Na+, Ox # = +1
c) Sum of the oxidation numbers in an element,
compound is equal to zero. Sum of the oxidation
numbers in an ion, cation or anion is equal to the ionic
charge
d) As a rule ONs of H =+1, and O=-2 almost most of
the time.
The group number in the periodic table could be used
for main group elements (p and s blocks). d and f block
elements show variable ONs E.g. Fe shows either +3 or
+2.
CHEM 100, Fall 2010, LA TECH
5-44
Oxidation State
What is the oxidation state of Cl in HClO4?
H  +1
O  -2
neutral compound, thus sum equals zero
4O  4  -2 = -8
H  1  +1 = +1
0 = +1 + (y) + (-8)
y = +7
CHEM 100, Fall 2010, LA TECH
5-45
Oxidation State
What is the oxidation state of S in H2SO4?
H  +1
O  -2
neutral compound, thus sum equals zero
4O  4  -2 = -8
2H  2  +1 = +2
0 = +2 + (x) + (-8)
x = +6
CHEM 100, Fall 2010, LA TECH
5-46
Assigning the Oxidation State
Assign the oxidation states to each atom in
NaCl
O2
CBr4
S8
MnO2
KMnO4
K2Cr2O7
CHEM 100, Fall 2010, LA TECH
5-47
Which of the following reactions
are redox?
a) NaCl + AgNO3 ----> AgCl + NaNO3
b) NaOH + HCl
----> NaCl + H2O
c) Zn + 2HCl
----> ZnCl2 + H2
d) 2Cr + 6HCl
----> 2CrCl3 + 3H2
e) MnO2 + 4HBr ----> Br2 + MnBr2 + 2H2O
CHEM 100, Fall 2010, LA TECH
5-48
Half Redox Reactions
two half-reactions in a redox
reaction: one where the oxidation
is talking place and the other
where reduction is taking place.
CHEM 100, Fall 2010, LA TECH
5-49
Half Reactions
E.g.
2 Na + Cl2 ----> 2NaCl
ON
0
0
+1 -1
Oxidation Na ----> Na+ + e- ; Na
increase ON, 0 ----> +1
Reduction
Cl2 + 2e- ----> 2Cl- ;
Cl decrease ON, 0
---->
-1
CHEM 100, Fall 2010, LA TECH
5-50
Separating Half Reactions
a) Zn + 2HCl
----> ZnCl2 + H2
b) MnO2+4HBr----> Br2 + MnBr2 + 2H2O
c) 10K + 2KNO3
CHEM 100, Fall 2010, LA TECH
----> N2 + 6K2O
5-51
Copper and
Nitric Acid
CHEM 100, Fall 2010, LA TECH
5-52
Copper Oxide and Hydrogen Gas
CHEM 100, Fall 2010, LA TECH
5-53
Single replacement reactions
If various metals are in water, we observe that some
are more reactive than others.
2Na (s) + 2H2O (l)
Ca (s) + 2H2O (l)
Mg (s) + H2O (l)
2NaOH (aq) + H2 (g) (fast)
Ca(OH)2 (s) + H2 (g) (slow)
no reaction
This indicates that the order of reactivity of these
metals towards water is
Na > Ca > Mg
We can show the reactivity of metals towards water
and acids using an activity series.
CHEM 100, Fall 2010, LA TECH
5-54
Activation
Series of
Metals
CHEM 100, Fall 2010, LA TECH
5-55
Activation Series of Metals
metals higher in series react with compounds of
those below
metals become less reactive to water top to bottom
metals become less able to displace H2 from acids
top to bottom
CHEM 100, Fall 2010, LA TECH
5-56
Potassium +
Water
CHEM 100, Fall 2010, LA TECH
5-57
Activity series of metals various metals in HCl
Iron
CHEM 100, Fall 2010, LA TECH
Zinc
Magnesium
5-58
Metal + Metal Salt Displacement
CHEM 100, Fall 2010, LA TECH
5-59
Activation Series of Metals
Zn(s) + CuSO4(aq)  ZnSO4(aq) + Cu(s)
Cu(s) + 2AgNO3(aq)  Cu(NO3)2(aq) + Ag(s)
Fe(s) + 2HCl(aq)  FeCl2(aq) + H2(g)
Zn(s) + 2HBr(aq)  ZnBr2(aq) + H2(g)
CHEM 100, Fall 2010, LA TECH
5-60
Common Oxidizing and Reducing
Agents
CHEM 100, Fall 2010, LA TECH
5-61
Metal + Acid Displacement
Single Displacement
CHEM 100, Fall 2010, LA TECH
5-62
Concentration Units
a) Molarity (M)
b) Molality (m)
c) Mole fraction (Ca)
d) Mass percent (% weight)
e) Volume percent (% volume)
f) "Proof"
g) ppm and ppb
CHEM 100, Fall 2010, LA TECH
5-63
Molarity
The number of moles of solute per liter of solution.
molarity  M
moles of solute
M=
liter of solution
units  molar = moles/liter = M
[CaCl2] - special symbol which means molar
(CaCl2 mol/L )
CHEM 100, Fall 2010, LA TECH
5-64
Examples
Calculate the molarity of a solution
prepared by dissolving 200.0 g of
K2SO4 in enough water to make 500.0
mL solution.
moles of solute
Molarity(M) = ---------------------Liters of solution
CHEM 100, Fall 2010, LA TECH
5-65
solute = K2SO4; F.W. = 174.27 g/mol; mass= 200g
moles of K2SO4 = ?
200 g /174.27 g K2SO4
= 1.148 mol K2SO4
500.0 mL = ?
Liters of solution = 0.5 L
Molarity?
1.148 mol K2SO4
Molarity of K2SO4 sol. = -----------------------0.5 Liters of solution
= 2.30 mole/Liter = 2.30 M (M = moles/liters)
CHEM 100, Fall 2010, LA TECH
5-66
Solution
Preparation from
Solid
CHEM 100, Fall 2010, LA TECH
5-67
How do you calculate moles of
substances in solutions
Use concentration of solution to convert L or
mL of solution in to moles
What is concentration of a solution?
The relative amounts of solute and solvent
There are so many ways to show amount: g,
mole, equivalents,volume
CHEM 100, Fall 2010, LA TECH
5-68
Ion Concentrations in Solutions
What is concntration of Cl- in 0.4 M MgCl2
aqeous solution if MgCl2 a strong electrolyte?
CHEM 100, Fall 2010, LA TECH
5-69
Examples
How many grams of KNO3 are contained in 500 mL
of a 0.500 M solution of potassium nitrate?
How many mL of 2.00 M solution of HNO3 are
required with water to make a 250 mL of 1.50 M
nitric acid solution?
CHEM 100, Fall 2010, LA TECH
5-70
c
Mole fraction ( a)
moles of solute (substance)
ca
= ------------------------------------moles of solute + moles of solvent
CHEM 100, Fall 2010, LA TECH
5-71
Calculate the mole fraction of
benzene in a benzene(C6H6)chloroform(CHCl3) solution which
contains 60 g of benzene and 30 g
of chloroform.
M.W. = 78.12 (C6H6)
M.W. = 119.37 (CHCl3)
CHEM 100, Fall 2010, LA TECH
5-72
Weight/Weight %
Weight/Weight %
=
Mass Solute x 100
Total Mass
Use the same units for both
If a ham contained 5 grams of fat in 200 g
of ham, what is the % wt/wt?
5 g / 200g * 100
=
2.5 wt/wt%
On the label, it would say 97.5 % fat free.
CHEM 100, Fall 2010, LA TECH
5-73
Volume/Volume %
Volume/Volume %
= Volume Solute x 100
Total Volume
Use the same units for both
If 10 ml of alcohol is dissolved in water to make
200 ml of solution, what is the concentration?
10 ml / 200 ml * 100
=
5 V/V%
Alcohol in wine is measured as a V/V%.
CHEM 100, Fall 2010, LA TECH
5-74
Weight/Volume %
Weight/Volume %
=
Mass solute x 100
Total Volume
use g and ml
If 5 grams of NaCl is dissolved in water to make
200 ml of solution, what is the concentration?
5 g / 200 ml * 100
=
2.5 wt/v%
Saline is a 0.9 wt/v% solution of NaCl in water.
CHEM 100, Fall 2010, LA TECH
5-75
Low concentrations in water
Mass percentages are used for water pollutants.
ppm =
mass solute
mass solution
x 106
ppb =
mass solute
mass solution
x 109
Example. One ppm of a toxin in water is the same
as 1 mg / liter since one liter of water has a mass
of approximately 106 mg.
CHEM 100, Fall 2010, LA TECH
5-76
Parts per Million
#g of solute
ppm =
 106 =
#g of solution
#mg of solute
#kg of solution
#micro-L solute
ppm =
#L of solution
CHEM 100, Fall 2010, LA TECH
5-77
Solution Preparation by Dilution
CHEM 100, Fall 2010, LA TECH
5-78
Dilution Problems
Why we dilute solutions?
Preparing solutions by adding water to
concentrated solutions
moles before = moles after
MiVi = MfVf
Mi
Vi
Mf
Vf
= initial molarity
= initial volume
=
final molarity
=
final molarity
CHEM 100, Fall 2010, LA TECH
5-79
Examples
How many mL of 2.00 M solution of HNO3 are
required with water to make a 250 mL of 1.50 M
nitric acid solution?
MiVi = MfVf
Mi = 2.00
Vi = ?
Mf = 1.50
Vf = 250 mL
MfVf
Vi = --------------- =
Mi
CHEM 100, Fall 2010, LA TECH
1.50 x 250
---------------- = 187.5 mL
2.00
5-80
Titrations
CHEM 100, Fall 2010, LA TECH
5-81
Titration
Method based on measurement of volume.
You must have a solution of known concentration standard solution.
It is added to an unknown solution while the
volume is measured.
The process is continued until the end point is
reached - a change that we can measure.
Acids and bases are commonly measured using
titrations.
CHEM 100, Fall 2010, LA TECH
5-82
Neutralization
The reaction of an acid with a base to produce a
salt and water.
HCl (aq) + NaOH (aq)
NaCl (aq) + H2O (l)
• If we prepare a standard solution of NaOH, we can then
use it to determine the concentration of HCl in a
sample.
• This is an example of Analytical Chemistry.
CHEM 100, Fall 2010, LA TECH
5-83
Stoichiometric Relationships
CHEM 100, Fall 2010, LA TECH
5-84
Stoichiometric calculations of
solutions reactions
Check whether chemical equation is balanced
get the moles from volume and “M” of solutions
find the limiting reactant
calculate moles of products from the limiting
reactant
convert moles of the products to grams
find the actual yield of the reaction
calculate % yield of the reaction
CHEM 100, Fall 2010, LA TECH
5-85
EXAMPLE: A sample of lye, sodium
hydroxide, is neutralized by sulfuric
acid. How many milliliters of 0.200 M
H2SO4 are needed to react completely
with 25.0 mL of 0.400 M NaOH?
2 NaOH(aq) + H2SO4(aq)  Na2SO4(aq) + 2 H2O
#mL H2SO4 =
(25.0 mL NaOH)
(0.400 mol NaOH)
(1 L NaOH)
(1 mol H2SO4)
(2 mol NaOH)
(1 L)
(1000 mL)
(1000 mL H2SO4)
(0.200 mol H2SO4)
= 25.0 mL H2SO4
CHEM 100, Fall 2010, LA TECH
5-86
Examples
How many mLs of 0.100 M BaCl2 are
required to react completely with 25 mL
of 0.200 M Fe2(SO4)3?
3 BaCl2(aq) + Fe2(SO4)3(aq)---> 3 BaSO4(s) + 2 Fe Cl3(aq)
3 BaCl2 = 1 Fe2(SO4)3
CHEM 100, Fall 2010, LA TECH
5-87
Calculate the moles of Fe2(SO4)3 :
moles = Molarity x
Liters of solution
0.200 M x 0.025 L = 0.005 mole Fe2(SO4)3
Then convert Fe2(SO4)3 to BaCl2 mole, BaCl2 moles
to liters and liters to mL.
0.005 mole Fe2(SO4)3 ----> BaCl2 moles
0.005 mol Fe2(SO4)3 x 3 = 0.015 BaCl2 moles
moles = Molarity x
Liters of solution
0.015 = 0.100 x Liters
Liters BaCl2 = 0.15L
= 150 mL of BaCl2
CHEM 100, Fall 2010, LA TECH
5-88