CHE 106: General Chemistry
CHAPTER
FOUR
Copyright © James T. Spencer 1995 - 1999
All Rights Reserved
Chem 106, Prof. J.T. Spencer
1
CHAPTER 4
2
Aqueous Reactions
and
Solution Chemistry
Chapt. 4.1
Chem 106, Prof. J.T. Spencer
Aqueous Solution
Stoichiometry Chapter 4
Focus on reactions in water - the
most common and “universal”
solvent (aqueous solutions (aq))
with unique properties (mp, bp, etc...)
 Essential for life chemistry biochemistry (i.e., cellular
metabolism) and inorganic chemistry
(i.e., rusting, precipitation).
 Acids, bases and salts

Chapt. 4.1
Chem 106, Prof. J.T. Spencer
3
Solution Composition; Terms
Solution - homogeneous mixture of two or
more substances
 Solvent - substance of greater amount in
the homogeneous mixture (solution)
 Solute - compounds “dissolved” in the
solvent
 Concentration - the amount of solute
dissolved in a solvent. Expressed in
molarity (M)

Molarity (M) =
moles of solute
volume of solution (in liters)
Chapt. 4.1
Chem 106, Prof. J.T. Spencer
4
Molarity



5
Molarity - moles of solute per liter of solution.
(not L of solvent!)
A 1.0 M solution contains 1.00 moles of solute in
every liter of solution.
Calculate the molarity of a solution made by
dissolving 0.0575 mol NH4Cl in 400 mL of water.
Molarity (M) = 0.0575 mol NH4Cl = 0.144 M NH4Cl
0.400 L of H2O
Unit of Molarity is moles per liter
Molarity (M) =
moles of solute
volume of solution (in liters)
Chapt. 4.1
Chem 106, Prof. J.T. Spencer
Molarity

6
Calculate the molarity of a solution made by
dissolving 9.93 g of Na2SO4 in enough water to
form 650. mL.[MW = Na2SO4 = 119]
Moles of Na2SO4 = 9.93 g (1 mol Na2SO4) = 0.0834 mol
119 g
M = 0.0834 mol Na2SO4 = 0.128 M
0.650 L
Molarity (M) =
Chapt. 4.1
moles of solute
volume of solution (in liters)
Chem 106, Prof. J.T. Spencer
Molarity
7
(A) Calculate the number of moles of HNO3
in 2.0 L of a 0.200 M HNO3 solution.
 (B) What volume of a 0.30 M HNO3 solution
is required to supply 2.0 mol HNO3?

(A) Know: Molarity of solution
Volume of solution
Need: Moles of
Solute
(B) Know: Molarity of solution
Moles of solute
Need: Volume
Molarity (M) =
moles of solute
volume of solution (in liters)
Chapt. 4.1
Chem 106, Prof. J.T. Spencer
Molarity
8
(A) Calculate the number of moles of HNO3
in 2.0 L of a 0.200 M HNO3 solution.
 (B) What volume of a 0.30 M HNO3 solution
is required to supply 2.0 mol HNO3?

(A) Moles of HNO3 = 2.0 L soln (0.200 mol HNO3)
1 L soln
Moles of HNO3 = 0.40 mol HNO3
(B) L of HNO3 Soln. = 2.0 mol HNO3
L of HNO3 Soln. = 6.7 L
(1 L soln)
(0.30 mol HNO3)
Chapt. 4.1
Chem 106, Prof. J.T. Spencer
Molarity

Pure acetic acid is a liquid with a density of
1.049 g/mL at 25° C. Calculate the molarity of a
solution of acetic acid prepared by dissolving
10.00 mL of acetic acid at 25° C in enough water
to make 100.0 mL of solution. [Acetic acid
=C2H4O2, MW = 60.1 amu]
(A) Know:
Density = 1.049g.mL
Vol. pure AA used
Vol. final solution
Molarity (M) =
Need:
Molarity of
final solution
moles of solute
volume of solution (in liters)
Chapt. 4.1
Chem 106, Prof. J.T. Spencer
9
Molarity

10
Pure acetic acid is a liquid with a density of
1.049 g/mL at 25° C. Calculate the molarity of a
solution of acetic acid prepared by dissolving
10.00 mL of acetic acid at 25° C in enough water
to make 100.0 mL of solution. [Acetic acid
=C2H4O2, MW = 60.1 amu]
Moles of AA = 10 mL AA (1.049 g AA) (1 mol AA)
1 mL AA 60.1 g AA
= 0.175 mol AA
M = moles AA = 0.175 mol AA = 1.75 M AA soln.
V soln
0.100 L
Chapt. 4.1
Chem 106, Prof. J.T. Spencer
Molarity and Dilution
11

Volume of solution (L) = moles solute/molarity (M)

Moles of solute = (Vol of soln in L)(Molarity)

Dilution - adding solvent to a “stock” solution to
lower its concentration
moles(initial) = moles(final)
M(initial)V(initial) = M(final)V(final)
Molarity (M) =
Chapt. 4.1
moles of solute
volume of solution (in liters)
Chem 106, Prof. J.T. Spencer
Molarity; Additional Examples

12
An experiment requires 150 mL of a 2 M HCl
solution. All that is available in the laboratory is
a 10.0 M HCl stock solution. How would you
prepare the required solution?
M(initial)V(initial) = M(final)V(final)
M(initial) = 10 M HCl
V(initial) = ? L
M(final) = 2 M HCl
V(final) = 0.150 L
Vi = (2 M)(0.150 L) = 0.030 L of 10 M HCl
(10 M)
How??
30 mL of the 10 M HCl solution is volumetrically
pipetted into a 150 mL volumetric flask and diluted
to mark
Chapt. 4.1
Chem 106, Prof. J.T. Spencer
Molarity and Stiochiometry

13
How many mL of a 0.250 M HCl solution would be
required to do each of the following operations;
 prepare 50.0 mL of a 0.100 M HCl soln.
 react with the OH- in 15.0 mL of a 0.200 M
Ba(OH)2 soln.
 dissolve 0.500 g CaCO3 according to the
reaction; CaCO3 + 2H+
Ca+2 + H2O + CO2
Equations:
M(initial)V(initial) = M(final)V(final)
Molarity (M) =
Chapt. 4.1
moles of solute
volume of solution (in liters)
Chem 106, Prof. J.T. Spencer
Molarity and Dilution

14
How many mL of a 0.250 M HCl solution would be
required to do each of the following operations;
 prepare 50.0 mL of a 0.100 M HCl soln.
M(initial)V(initial) = M(final)V(final)
Vi = Mf Vf
Mi
M(initial) = 0.250 M HCl
V(initial) = ? L
M(final) = 0.100 M HCl
V(final) = 0.050 L
Vi = (0.100 M)(0.050 L) = 0.020 L
(0.250 M)
20 mL of the 0.250 M HCl solution is volumetrically
pipeted into a 50 mL volumetric flask and diluted to
mark
Chapt. 4.1
Chem 106, Prof. J.T. Spencer
Molarity and Dilution

15
How many mL of a 0.250 M HCl solution would be
required to do each of the following operations;
 react with the OH- in 15.0 mL of a 0.200 M
Ba(OH)2 soln.
2 HCl + Ba(OH)2
2 H2O + BaCl2
moles OH- = (0.200 m Ba(OH)2) (0.015L Ba(OH)2) (2 m HCl) = 0.006mol
liter
1 m Ba(OH)2
vol HCl (L) = (0.006 mol HCl req.) = 0.024 L or 24 mL
0.250 M
volume of solution (in liters)= moles of solute
Molarity (M)
Chapt. 4.1
Chem 106, Prof. J.T. Spencer
Molarity and Dilution

16
How many mL of a 0.250 M HCl solution would be
required to do each of the following operations;
 dissolve 0.500 g CaCO3 according to the
reaction; CaCO3 + 2H+
Ca+2 + H2O + CO2
moles CaCO3 = (0.50 g CaCO3) = 0.005 m CaCO3
100.1 amu
mol HCl req = (0.005 mol CaCO3) (2 HCl ) = 0.010 mol HCl
1 CaCO3
Vol HCl = 0.010 mol HCl = 0.040 L or 40 mL
0.250 M
volume of solution (in liters)= moles of solute
Molarity (M)
Chapt. 4.1
Chem 106, Prof. J.T. Spencer
Electrolytes
Bulb does not light
Nonelectrolyte Solution
[no electrical current carriers]
Bulb glows dimly
Weak Electrolyte Solution
[some current carriers]
solution
The Ions complete the
circuit by carrying
electrical charge between
the electrodes
Chapt. 4.2
Bulb glows brightly
Strong Electrolyte Solution
[many electrical current carriers]
Chem 106, Prof. J.T. Spencer
17
Electrolytes





Electrolytes - solutes that exist as ions in
solution and conduct electrical current.
Ionic compounds which dissociate into ions in
solution are electrolytes;
NaCl(s)
Na+1(aq) + Cl-1(aq)
Other solutes which are not ionic but dissolve to
form ions are electrolytes (HCl, H2SO4, etc...)
HCl(g)
H +1(aq) + Cl-1(aq)
Non ionizing compounds form nonconducting
solutions and are, therefore, nonelectrolytes.
Electrolyte solutes are characterized as either
strong or weak depending on the extent of
dissociation (and, therefore, current conduction).
Most ionic compounds are strong electrolytes
Chapt. 4.2
Chem 106, Prof. J.T. Spencer
18
Electrolytes


Strong electrolytes are completely dissociated in
solution
Weak electrolytes are only partially dissociated
into ions in solution [CHEMICAL EQUILIBRIUM]
HC2H3O2(aq)


19
H+(aq) + C2H3O2-(aq)
Nonelectrolytes are not dissociated into ions in
solution
Extent of dissolution does not dictate strong or
weak electrolyte solution (i.e., HC2H3O2 is very
soluble but is a weak electrolyte while Ba(OH)2 is
only slightly soluble is a strong electrolyte)
Chapt. 4.2
Chem 106, Prof. J.T. Spencer
Chemical Equilibria

When a reaction proceeds to completion, its
equation is written with a single arrow.
HNO3(aq)

H+(aq) + NO3-(aq)
When a reaction’s progress can be more
accurately considered to be a balance between
the forward and back reactions, then the two
processes (forward and backward) are
involved in an equilibrium which is written
as a double arrow.
HC2H3O2(aq)
H+(aq) + C2H3O2-(aq)
Chapt. 4.2
Chem 106, Prof. J.T. Spencer
20
Chemical Equilibria
Backward Reaction
Forward Reaction
Chapt. 4.2
Chem 106, Prof. J.T. Spencer
21
Metathesis Reactions
 Precipitation
22
Reactions - formation of an
insoluble solid (less than 0.01 mol/L)
AgNO3(aq) + NaCl(aq)
NaNO3(aq)
 Solubility
AgCl(s) +
Rules - allow predictions of
whether a ppt will form. Rules not based
upon simple physical properties (i.e., charge,
conductivity, etc...)
– Rules developed by observations, but many
generalizations are possible
Chapt. 4.5
Chem 106, Prof. J.T. Spencer

Soluble
Solubility
23
– all Nitrates (NO3-) and acetates (C2H3O2-)
– all chlorides, bromides and iodides (except of Ag, Hg, and
Pb)
– all sulfates (except (Ca, Sr, Ba, Pb, Hg, and Ag)
Chapt. 4.5
Chem 106, Prof. J.T. Spencer

Soluble
Solubility
24
– all Nitrates (NO3-) and acetates (C2H3O2-)
– all chlorides, bromides and iodides (except of Ag, Hg, and
Pb)
– all sulfates (except (Ca, Sr, Ba, Pb, Hg, and Ag)

Insoluble
– all sulfides (S2-) except those of groups 1 (1A) and 2 (2A) and
ammonium
– all carbonates (CO32-) except those of group 1 (1A) and
ammonium
– all phosphates except those of group 1 (1A) and ammonium
– all hydroxides except those of group 1 (1A) and Ba, Sr and Ca

Use Rules to Predict Metathesis Reactions
Chapt. 4.5
Chem 106, Prof. J.T. Spencer
Metathesis Reactions
25
Metathesis Reactions - a reaction of ionic
species in which partners exchange (metathesis
is Greek for “transpose”).
AX + BY
AY + BX
 These reactions require a “driving force” by
removing ions from solution by;

– formation of an insoluble solid
– formation of either a soluble weak electrolyte or
soluble nonelectrolyte
– formation of a gas which escapes from solution
Chapt. 4.5
Chem 106, Prof. J.T. Spencer
Metathesis Reactions

26
Driving force can also be the formation of weak
electrolytes or nonelectrolytes
– acid base neutralizations (even water insoluble
hydroxides can act as bases in these reactions)
Mg(OH)2(s) + 2 HCl
MgCl2(aq) + 2 H2O
– metal oxides with acids (oxide can react with 2H+ to
yield water)
FeO(s) + H2SO4(aq)
Fe(SO4)(aq) + H2O(l)
FeO(s) + 2H+(aq)
Fe2+(aq) + H2O(l)
– ionic reactions occur when ions are removed to form a
weak electrolyte
HCl(aq) + NaC2H3O3(aq)
HC2H3O3(aq) + NaCl(aq)
Chapt. 4.5
Chem 106, Prof. J.T. Spencer
Metathesis Reactions

27
Driving force may also be the formation of a gas as a
product which has a low water solubility.
– 2HCl(aq) + Na2S(aq)
H2S(g) + 2 NaCl(aq)
2H+(aq) + S2-(aq)
H2S(g) (net ionic)
– HCl(aq) + NaHCO3(aq)
NaCl(aq) + [H2CO3(aq)]
UNSTABLE INITIAL PRODUCT
H2O(l) + CO2(g)
gas formation to drive
reaction
Chapt. 4.5
Chem 106, Prof. J.T. Spencer
Metathesis Reactions

28
Reaction of PbS with HCl
Pb+2(aq) + S2-(aq) + 2H+(aq) + 2Cl-(aq)
PbCl2(s) + H2S(g)
complete equation, net ionic equation
driving force = precipitation and gas evol.

Reaction of Cr(C2H3O2)2 with HCl
Cr+2(aq) + 2C2H3O3-(aq) + 2H+(aq) + 2Cl-(aq)
Cr+2(aq) + 2HC2H3O3(aq) + 2Cl-(aq)
2C2H3O3-(aq) + 2H+(aq)
2HC2H3O3(aq) (net ionic)

Reaction of Co(CO3) with HCl
Co+2(aq) + CO3-2(aq) + 2H+(aq) + 2Cl-(aq)
Co+2(aq) +CO2(g) + H2O(l) +2Cl-(aq)
Chapt. 4.5
Chem 106, Prof. J.T. Spencer
Ionic Equations

Molecular equation - shows the complete
chemical formulas.
Ba(OH)2(s) + 2 HNO3(aq)
H2O(l)

29
Ba(NO3)2 (aq) + 2
Complete Ionic Equation - in solutions, where
the reactants and products are ionized, it is more
useful (and accurate) to indicate ions.
Ba2+(aq) + 2OH-(aq) + 2H+(aq) + 2NO3- (aq)
Ba2+(aq) + 2NO3- (aq) + 2 H2O(l)

Net Ionic Equation - omits ions which appear
unchanged (spectator ions) on both sides of eqn.
2OH-(aq) + 2H+(aq)
Chapt. 4.4
2 H2O(l)
Chem 106, Prof. J.T. Spencer
Ionic Equations

Net Ionic Equations greatly simplify equations
and points out similarities and differences
between reactions more easily. (i.e., ANY
neutralization reaction appears the same);
2OH-(aq) + 2H+(aq)
2 H2O(l)
Chapt. 4.4
Chem 106, Prof. J.T. Spencer
30
Ionic Equations

Net Ionic Equations greatly simplify equations
and points out similarities and differences
between reactions more easily. (i.e., EVERY
neutralization reaction appears the same);
2OH-(aq) + 2H+(aq)

2 H2O(l)
To write an ionic equation;
–is the compound soluble?
–if its is soluble, is it a strong electrolyte?
» Only if both answers are yes then an ionic
equation is good. (otherwise write molecular
equation)
Chapt. 4.4
Chem 106, Prof. J.T. Spencer
31
Acids
 Acid - a substance able to ionize to form H+
ion, “proton”, thereby increasing the H+
concentration in solution.
ACIDS ARE PROTON DONORS

Type of acid named by how many protons it
can donate (i.e., monoprotic, diprotic,....).
H3PO3(aq)
H2PO3-1(aq)
HPO3-2(aq)
H+(aq) + H2PO3-1(aq)
H+(aq) + HPO3-2(aq)
H+(aq) + PO3-3(aq)
Chapt. 4.3
Chem 106, Prof. J.T. Spencer
32
Bases
33
 Base - a substance able to ionize to form OH-
ions, thereby increasing the OHconcentration in aqueous solutions.
 Base - a substance which reacts with or
accepts H+ ions [Neutralization Reactions]:
H+(aq) + OH-(aq)
H2O(l)
BASES ARE PROTON ACCEPTORS
Chapt. 4.3
Chem 106, Prof. J.T. Spencer
Bases
 Kinds
of Bases -
 Ionic
Hydroxides - dissociate to form
OH- and M+
 Proton Acceptors w/out Hydroxide (OH-)
NH3(aq) + H2O(l)
NH4+(aq) + OH-(aq)
only a small portion of the ammonia (ca. 1%)
reacts with the water to form the ammonium
ion, it is a weak electrolyte (weak base).
Note: Compounds, such as CH3OH, which contain an OH
group, are essentially not dissociated, are not electrolytes,
and are very weak bases.
Chapt. 4.3
Chem 106, Prof. J.T. Spencer
34
Acids and Bases
35
Increasing concentration of acid (H+)
decreases concentration of base (OH-).
 Increasing concentration of base (OH-)
decreases the concentration of acid (H+)
 Strong and Weak Acids and Bases - Strong
indicates complete dissociation (strong
electrolyte) and weak refers to only partial
dissociation into ions (weak electrolyte)
 Strong acids and bases are generally more
reactive than weak acids and bases when the
reactivity depends only on H+ or OHconcentration

Chapt. 4.3
Chem 106, Prof. J.T. Spencer
A cids and Bases

Strong Acids HCl
HBr
HI
HNO3
H2SO4
HClO3
HClO4

36
hydrochloric acid
hydrobromic acid
hydroiodic acid
nitric acid
sulfuric acid
chloric acid
perchloric acid
H+ with a
halide ion
Most Acids
(others) Are
Weak
Strong Bases
Group 1 (or 1A) metal hydroxides (i.e., LiOH)
Group 2 (or 2A) heavy metal hydroxides (i.e., Ba(OH)2)
Chapt. 4.3
Chem 106, Prof. J.T. Spencer
Salts

37
Salts are ionic compounds formed by replacing
one or more H+ of an acid by a different cation.
HCl becomes NaCl by replacing H+ with Na+
H2SO4 becomes CuSO4 by replacing 2H+ with
Cu+2 HNO3 becomes CoNO3 by replacing H+ with
Co+1 HClO4 becomes KClO4 by replacing H+ with
K+1
 Salts are typically strong electrolytes [with the
exception of salts of heavy metal cations such as
mercury and lead].
 A neutralization reaction between an acid and a
metal hydroxide produces water and a salt;
HCl(aq) + NaOH(aq)
Chapt. 4.3
H2O(l) + NaCl(aq)
Chem 106, Prof. J.T. Spencer
Acids and Bases
38
Identifying Strong and Weak Acids and Bases
 Most
salts are strong electrolytes[EXCEPT
heavy metal salts, i.e., HgCl2].
 Most acids are weak electrolytes [EXCEPT
HCl, HBr, HI, HNO3, H2SO4, HClO3 and
HClO4].
 The common strong bases are the hydroxides
of the alkali metals and the heavy alkaline
earths.
 Ammonia is a weak electrolyte.
 Most other compounds are nonelectrolytes.
Chapt. 4.3
Chem 106, Prof. J.T. Spencer
Acids, Bases, and Salts
39
Identify each of the following as a strong electrolyte,
a weak electrolyte or a non electrolyte AND (where possible)
as a weak or strong acid, base or salt (in aqueous solution);
LiOH
BaBr2
H3PO4
H2SO4
CH3CN
Ba(OH)2
CuCl2
C6H12O6
NH3









strong base (strong electrolyte)
salt (strong electrolyte)
strong acid (strong electrolyte)
strong acid (strong electrolyte)
nonelectrolyte
strong base (strong electrolyte)
salt (strong electrolyte)
nonelectrolyte
weak base (weak electrolyte)
Chapt. 4.3
Chem 106, Prof. J.T. Spencer
Acids, Bases, and Salts
40
Complete and balance each of the following equations

(1) Barium hydroxide reacting with nitric acid
Ba(OH)2(s) + 2 HNO3(aq)
Ba(NO3)2 (aq) + 2 H2O(l)
(2) Phosphoric acid reacting with lithium hydroxide

H3PO4(aq) + 3 LiOH(aq)
Li3PO4(aq) + 3 H2O(l)
(3) Acetic acid is neutralized with sodium hydroxide

CH3CO2H(aq) + NaOH(aq)
CH3CO2Na(aq) + H2O(l)
(4) Ammonia(aq) is added to hydrochloric acid(aq)

NH3(aq) + HCl(aq)
NH4+ + Cl-(aq)
Chapt. 4.3
Chem 106, Prof. J.T. Spencer
 Metals
Reactions with Metals
41
react with many compounds -
corrosion reaction is specifically the reaction of a
metal with an “environmental” (naturally
occurring
 Oxidation
Reactions - loss of electrons by a
substance (the substance is oxidized).
4 Na(s) + O2(g)
2 Na2O(s)
Na goes from 0 charge to +1 charge
(oxidized)
O goes from 0 charge to -2
charge (reduced)
4 Na0
+
O0
O0
2 Na+1
O-2
Na+1
Chapt. 4.6
Chem 106, Prof. J.T. Spencer
Reactions with Metals

4 Na(s) + O2(g)
42
2 Na2O(s)
Na goes from 0 charge to +1 charge (oxidized)
O goes from 0 charge to -2 charge (reduced)
 Reduction
Reactions - gain of electrons by
a substance (the substance is reduced).
 When one reactant loses electrons
(oxidized) some other reactant must gain
the same number of electrons (reduced)
[net electron transfer]
Chapt. 4.6
Chem 106, Prof. J.T. Spencer
Oxidation Numbers


H
F
43
EN = 4.0 - 2.1 = 1.9
Polar Covalent Bond
partial - charge
partial + charge
 May formally assign charges to atoms in polar covalent
compounds (bookkeeping purposes).
 Charges assigned are called Oxidation Numbers.
 H ox. no. = +1 and F ox. no. = -1
 Oxidation numbers DO NOT correspond with actual
charges within a molecule (except for ionic
compounds).
Chem 106, Prof. J.T. Spencer
Oxidation Numbers

44
Rules to determining Oxidation Number:
– Ox. No. of an element is 0.
– Ox. No. of a monatomic ion is the same as its charge.
– For binary compounds, the element with the greater
EN is assigned the negative Ox. No. (same charge as
found in ionic compounds for the anion).
– The sum of the oxidation numbers (total) must equal
the charge of the molecule (including the neutral
case).
Chem 106, Prof. J.T. Spencer
Formal Charge vs. Oxidation Number

45
DO NOT CONFUSE FORMAL CHARGE WITH
OXIDATION NUMBERS!!
FORMAL CHARGE
0
OXIDATION NUMBER
+2
Cl
H C Cl
Cl
0
0
+1
-1
Chem 106, Prof. J.T. Spencer
Oxidation Numbers

46
Determine Oxidation Numbers:
Cl
H C Cl
Cl
Cl
S Cl
H +1
H N H
H
O
H N O
F
Cl F
F
Chem 106, Prof. J.T. Spencer
Oxidation Numbers

47
Determine Oxidation Numbers:
Cl
H C Cl
Cl
H +1
H N H
H
O
H N O
H = +1
N = -3
H = +1
N = +3
O = -2
H = +1
C = +2
Cl = -1
Cl
SI Cl
S = +2
Cl = -1
F
Cl F
F
F = -1
Cl = +3
Chem 106, Prof. J.T. Spencer
Oxidation of Metals


48
Many metals react with acids to form salts and
hydrogen gas (the metal is oxidized!)
metal + acid
salt + hydrogen
Zn(s) + 2HCl(aq)
Fe(s) + H2SO4(aq)
ZnCl2(aq) + H2(g)
FeSO4(aq) + H2(g)
(H+ is reduced and the metal is oxidized)
Can we predict when reactions involving oxidation and
reduction will occur?
– Different metals vary in the ease with which they are oxidized
– Activity Series
Chapt. 4.4
Chem 106, Prof. J.T. Spencer
Activity Series of Metals





49
Activity Series lists metals in decreasing order of ease
of oxidation (how easily a metal loses electrons).
Metals at the top are most easily oxidized and are
referredactive metals (i.e., alkali and alkaline earth
metals)
Metals at the bottom are most stable WRT oxidation (i.e.,
noble metals)
Series can be used to predict reactions;
ANY METAL ON THE LIST CAN BE OXIDIZED BY
IONS OF THE ELEMENT BELOW IT
Only those metals ABOVE hydrogen on the list can react
with acids to form hydrogen gas.
Chapt. 4.4
Chem 106, Prof. J.T. Spencer
Activity Series
Lithium
Potassium
Barium
Calcium
Sodium
Magnesium
Aluminum
Zinc
Chromium
Iron
Nickel
Hydrogen
Copper
Silver
Platinum
Gold
M(s)
Easier
Oxidation
More
Difficult
Oxidation
M+n + n e-
Alkali Metals
Noble Metals
Chapt. 4.6
Chem 106, Prof. J.T. Spencer
50
Activity Series
Lithium
Potassium
Barium
Calcium
Sodium
Magnesium
Aluminum
Zinc
Chromium
Iron
Nickel
Hydrogen
Copper
Silver
Platinum
Gold
51
ANY METAL ON THE LIST CAN
BE OXIDIZED BY IONS OF THE
ELEMENT BELOW IT
EXAMPLES
 Iron ions cannot oxidize
Copper
Easier
Oxidation  Copper ions can oxidize Iron
 Copper does not react with H+
More
 Nickel reacts with H+ to form
Difficult
hydrogen
Oxidation
 Nickel is oxidized by Copper
ions
 Gold is not oxidized by other
metal ions
Chapt. 4.6
Chem 106, Prof. J.T. Spencer
Activity Series



52
Note: Copper should not be oxidized by acid (H+) to
form hydrogen (from activity series table data)
Copper does not react with HCl(aq)
HOWEVER, copper does react with nitric acid. WHY?
Copper reacts with the nitrate ion of the acid NOT the H+
(in agreement with the activity series)
Cu(s) + 4HNO3(aq)
Cu(NO3)(aq) + 2H2O(l) + 2NO2(g)
Chapt. 4.6
Chem 106, Prof. J.T. Spencer
Activity Series
Lithium
Potassium
Barium
Calcium
Sodium
Magnesium
Aluminum
Zinc
Chromium
Iron
Nickel
Hydrogen
Copper
Silver
Platinum
Gold
M(s)
53
M+n + n e-
EXAMPLES
Which of the following metals
will be oxidized by Zn(SO4)?






Lithium
Silver
Chromium
Nickel
Hydrogen
Magnesium






Yes
No
No
No
No
Yes
ANY METAL
ON THE
LIST CAN BE
OXIDIZED
BY IONS OF
THE
ELEMENT
BELOW IT
Chapt. 4.6
Chem 106, Prof. J.T. Spencer
Solution Stoichiometry

54
How do we determine (experimentally) the
concentration of an unknown solution?
– USE CHEMICAL REACTIONS AND
SOLUTION STOICHIOMETRY INVOLVING
ONE SOLUTION OF KNOWN
CONCENTRATION
BUT WHAT REACTIONS?
Acid Base Reactions (previously discussed)
Metathesis Reactions
Oxidation-Reduction Reactions
Chem 106, Prof. J.T. Spencer
Solution Stoichiometry

55
How do we determine (experimentally) the
concentration of an unknown solution?
– use a second solution (standard solution) of KNOWN
CONCENTRATION which reacts with the first solution,
then use molar/molarity conversions and the balanced
equation to determine the concentration of the unknown
solution (TITRATION).
Grams of
Compound A
Grams of
Compound B
use
MW
use
MW
Moles of
Compound A
use
bal. eqn.
Moles of
Compound B
Volume or Molarity
of Compound A
use
M=mol/V
Volume or Molarity
of Compound B
use
M=mol/V
Chapt. 4.7
Chem 106, Prof. J.T. Spencer
Titrations
56

add
add
1 drop
Equivalence
Point
Add volume of a
known conc.
standard
solution until
equivalence
point [reaction of
first “excess” of
standard
solution with
“indicator” to
give a visible
color change (i.e.,
litmus)].
Chapt. 4.7
Chem 106, Prof. J.T. Spencer
Titrations: Indicators
57
Indicator - A compound that changes color and is
used to mark the end point of a titration. The color
change occurs when either there is an excess of acid
or base in solution.
Phenolphthalein
OH
-H+
O
C
COLORLESS
O
OH
OH
C
O
O
C
O-
RED
Chem 106, Prof. J.T. Spencer
Titrations: Indicators
58
Indicator - A compound that changes color and is
used to mark the end point of a titration. The color
change occurs when either there is an excess of acid
or base in solution.


Acid-Base Indicator from Red Cabbage
Aqueous Carbon Dioxide
QuickT ime™ and a
Graphics decompressor
are needed to see this picture.
S Tape 1:1 (5:44 + 3:28)
Chem 106, Prof. J.T. Spencer
Titrations
M(strd)V(strd) = M(unk)V(unk)
known
59
M(strd) = Molarity of strd
V(strd) = vol. to equiv. point
M(unk) = determine
V(unk) = vol. of unknown
solution
volume of solution (in liters)= moles of solute
Molarity (M)
Five Example Problems Follow To Illustrate
Chapt. 4.7
Chem 106, Prof. J.T. Spencer
Titrations

60
What volume of 0.827 M KOH solution is required to
completely neutralize 35.00 mL of 0.737 H2SO4 solution?

2 KOH + H2SO4
M(strd)V(strd) = M(unk)V(unk)
2 H2O + K2SO4
M(unk) = 0.8270 M KOH
V(unk) = ? L KOH
M(strd) = 0.737 M H2SO4
V(strd) = 0.0350 L H2SO4
V(KOH) = (0.737 M)(0.0350 L) 2 = 0.0624 L
(0.827 M)
V(KOH) = 62.4 mL
Chem 106, Prof. J.T. Spencer
Titrations

61
What volume of 0.827 M KOH solution is required to
completely neutralize 35.00 mL of 0.737 H2SO4 solution?

2 KOH + H2SO4
2 H2O + K2SO4
V(KOH) = (0.737 M)(0.0350 L) 2 = 0.0624 L
(0.827 M)
V(KOH) = (0.737 mol H2SO4)(0.0350 L H2SO4)(2 mol KOH)
1 L H2SO4 soln
1 mol H2SO4
0.827 mol KOH
1 L KOH soln
Chem 106, Prof. J.T. Spencer
Titrations

A 17.5 mL sample of acetic acid (CH3CO2H) required
29.6 mL of 0.250 M NaOH to neutralize it. What was
the molarity of the acetic acid solution?
EXAM 1
Chem 106, Prof. J.T. Spencer
62
Titrations

63
A 17.5 mL sample of acetic acid (CH3CO2H) required
29.6 mL of 0.250 M NaOH to neutralize it. What was
the molarity of the acetic acid solution?
M(strd)V(strd) = M(unk)V(unk)
M(aa) = (0.250 M)(0.0296 L)
(0.0175 L)
M(strd) = 0.250 M NaOH
V(strd) = 0.0296 L NaOH
M(unk) = ?
V(unk) = 0.0175 L acetic acid
M(aa) = 0.42 M
EXAM 1
Chem 106, Prof. J.T. Spencer
Titrations

Example - Acetic acid (HC2H3O2) reacts with sodium
64
hydroxide in a neutralization reaction. If 8.30 mL of an
acetic acid soln. requires 42.4 mL of 0.1050 M NaOH to reach
the equivalence point, how many grams of acetic acid are in
a 500 mL sample? HC2H3O2 + NaOH
H2O + NaC2H3O2
Chapt. 4.7
Chem 106, Prof. J.T. Spencer
Titrations

Example - Acetic acid (HC2H3O2) reacts with sodium
65
hydroxide in a neutralization reaction. If 8.30 mL of an
acetic acid soln. requires 42.4 mL of 0.1050 M NaOH to reach
the equivalence point, how many grams of acetic acid are in
a 500 mL sample? HC2H3O2 + NaOH
H2O + NaC2H3O2
M(strd)V(strd) = M(unk)V(unk)
M(aa) = (0.1050 M)(0.0424 L)
(0.00830 L)
M(strd) = 0.1050 M NaOH
V(strd) = 0.0424 L NaOH
M(unk) = ?
V(unk) = 0.00830 L acetic acid
M(aa) = 0.536 M
mol(aa) in 500 mL = (0.537 M aa)(0.5 L) = 0.268 mol aa
g(aa) = (0.268 mol aa)(60 amu) = 16.1 g aa in 500 mL of soln
Chapt. 4.7
Chem 106, Prof. J.T. Spencer

Titrations
Example - In the laboratory, 6.67 g of Sr(NO3)2 was
66
dissolved to make 750 mL of solution. A 100 mL sample of
this solution was titrated with a 0.0460 M Na2CrO4 solution.
What volume of the Na2CrO4 solution is required to
precipitate all the Sr2+ as SrCrO4? (MW Sr(NO3)2 =211.6)
Sr(NO3)2(aq) + Na2CrO4(aq)
SrCrO4(s) +
Na(NO3)(aq)
(6.67 g Sr(NO3)2) = 0.0315 mol Sr+2
(211.6 g/mol)
(0.0315 mol Sr+2) = 0.042 M and (0.042 M)(0.100 L) = 0.0042 mol
(0.750 L)
0.0042 Sr+2 = 0.0042 mol CrO4-2
0.042 M (0.100) = 0.0460(V) V = 0.042(0.100)
0.0460
V = 0.0913 L
Chapt. 4.7
Chem 106, Prof. J.T. Spencer
Titrations

67
Example - The arsenic in a 1.22 gram sample of
pesticide was chemically converted to AsO43-. It was
then titrated with Ag+1 to form Ag3AsO4 as a precipitate.
If it took 25.0 mL of a 0.102 M Ag+1 solution to reach the
equivalence point, what is the percentage arsenic is the
pesticide?
 What is the balanced equation?
AsO43-(aq) + 3Ag+1(aq)
Ag3AsO4(s)
 How
many moles of AsO43- are in the sample?
 How many moles As in sample ?
 How many grams of As in sample?
Chapt. 4.7
Chem 106, Prof. J.T. Spencer

Titrations
Example - The arsenic in a 1.22 gram sample of
pesticide was chemically converted to AsO43-. It was
then titrated with Ag+1 to form Ag3AsO4 as a precip. If it
took 25.0 mL of a 0.102 M Ag+1 solution to reach the
equivalence point, what is the percentage arsenic is the
pesticide?
 How many moles of AsO43- are in the sample?
AsO43-(aq) + 3Ag+1(aq)
Ag3AsO4(s)
Molarity = moles therefore
moles = (M)(V)
V
moles Ag1+ = (0.102M)(0.025L)
= 0.00244 mol Ag1+
mol AsO43- = (0.00255 mol Ag1+)(1 AsO43-) = 0.00085 mol
(3 Ag1+)
AsO43Chapt. 4.7
Chem 106, Prof. J.T. Spencer
68

Titrations
Example - The arsenic in a 1.22 gram sample of
pesticide was chemically converted to AsO43-. It was
then titrated with Ag+1 to form Ag3AsO4 as a precip. If it
took 25.0 mL of a 0.102 M Ag+1 solution to reach the
equivalence point, what is the percentage arsenic is the
pesticide?
 How many moles and3-grams As in sample ?
moles AsO4 = moles As
g As = (mole As)(AW of As) = (0.00085 mol As)(74.9 amu)
= 0.0637 grams As in sample
% As = g As (100) = (0.0637 g As) (100) = 5.22% As
g total
(1.22 g sample)
Chapt. 4.7
Chem 106, Prof. J.T. Spencer
69
End Chapter Four










Molarity
Strong Electrolytes, Weak Electrolytes and
Nonelectrolytes
Acids, Bases and Salts
Neutralization Reactions
Metathesis Reaction
Solubility rules
Reactions of Metals
Activity Series and oxidations and
reductions
Solution Stoichiometry
Titrations
Chapt. 4.1
Chem 106, Prof. J.T. Spencer
70