1.Definitions
Solution - homogeneous mixture of two or
more substances
Aqueous - dissolved in water (aq)
Anion - negatively charged ion
Cation - positively charged ion

Electrode - conductor of electricity
- electrical conductors (wires, plates)
partially immersed in a solution and
connected to a source of electricity (e-)


Cathode- connected to the negative pole
Anode- connected to the positive pole

2. Arrhenius’ theory of electrolytic
dissolution

a. strong electrolyte

b. weak electrolyte

solute that is present in solution almost exclusively as
ions; good electrical conductor (ex. KCl, HBr, )
solute that is only partly ionized in solution; poor
conductor of electricity (ex. Acetic Acid (vinegar))
c. Nonelectrolyte 
Solute that is present in solution almost exclusively as
molecules; does not conduct electricity (ex. Sugar)

3. Solubility Rules

Memorize; refer to
back of stations lab

4. Dissolution equations Ex5.1 Write
dissolution equations for: NaCl(s),
Al(NO3)3(s), Na2SO4(s), and MgCl2(s)

1. Definitions

Double Replacement Reaction:


reaction between two ionic compounds in which the
cations and anions are exchanged to form two new ionic
compounds
Precipitate:


Insoluble ionic solid
Symbolized by: (s) or 

Full Ionic Equation



Net Ionic Equation



Equation in which all compounds are written as ions
REMEMBER Coefficients, Subscripts, etc.!
“net result” of the chemical reaction
Formation of solid, gas, H2O(l), etc.
Spectator Ions

Ions that do not take part in the reaction; present before and
after

Double Replacement Rxns
The positive ions from one solution interact with
the negative ions from the other solution to
form:

A precipitate that will settle out


A gas that will bubble out


Pb(NO3 )2(aq) + 2 NaCl (aq) ---> PbCl2(s) + 2 NaNO3(aq)
FeS (s) + 2HCl (aq) ---> H2S (g) + FeCl2 (aq)
A molecular compound, usually water

HCl (aq) + NaOH (aq) ---> NaCl (aq) + H2O (l)

What is happening in the solution?

The reactants are aqueous, which means that
ionic compounds are ionized.

Individual positive and negative ions are
present in the aqueous solutions

KCl
(s)
--H2O---> K+
(aq)
+ Cl- (aq)

Full (Complete) Ionic Equation

AgNO3 (aq) + NaCl (aq) ---> AgCl (s) + NaNO3(aq)

**Write the Complete Ionic Equation by separating
all aqueous compounds
Na+ (aq) + Cl- (aq) + Ag+ (aq) + NO3-(aq) ---> AgCl (s) + Na+ (aq) + NO3-(aq)

Spectator Ions


Ions that do not take part in the reaction
Ions are present before and after the reaction
Na+ (aq) + Cl- (aq) + Ag+ (aq) + NO3-(aq) ---> AgCl (s) + Na+ (aq) + NO3-(aq)
Spectator Ions

Net Ionic Equation

Remove the Spectator ions from the Complete
Ionic Reaction

N.I. = Cl- (aq) + Ag+ (aq) ---> AgCl (s)


2. Ex5.2 Predict whether or not the following pairs of reactants
will form precipitates. Then write the double replacement reaction,
the full ionic equation, and the net ionic equation and identify
spectator ions.
a. CuCl2 and (NH4)2SO4

b. Ba(NO3)2 and Na2CO3

c. MgCl2 and AgNO3

Ex5.3 What two aqueous solutions of
soluble compounds could be mixed to
produce a precipitate of CuCO3?
 Ex5.4 You suspect that a particular unlabeled aqueous solution is
one of the following: sodium sulfate, ammonia (NH3), or barium
nitrate. Explain how to use precipitation reactions on small samples
of the unknown solution to determine its identity.

1. Arrhenius Definition of acids and bases

Acids


Produces H+ in soln
Bases

Produces OH- in soln

2. Strong acids – strong electrolytes

HCl (g) -------> H+ (aq) + Cl- (aq)

Strong Acids to Memorize:
“Br I Cl SO NO ClO 4-3-4”

HCl, HBr, HI, HNO3, HClO4, H2SO4**

3. Weak acids – weak electrolytes

Any acid that isn’t “strong”
Majority of acids

HC2H3O2 (aq)

 
H+ (aq) + C2H3O2-1 (aq)

4. Strong bases – strong electrolytes

NaOH (s) --------> Na+ (aq) + OH- (aq)

Strong Bases to Memorize: (Group I, II Hydroxides,
except the 1st one in group I and the 1st two in group
II)


LiOH, NaOH, KOH, RbOH, CsOH
Ca(OH)2, Sr(OH)2, Ba(OH)2

5. Weak bases – weak electrolytes

Any base that isn’t strong
Watch for Nitrogen-containing compounds

NH3 (aq) + H2O (l) <-----> NH4+ (aq) + OH- (aq)


6. Oxides of metals and nonmetal

Oxides of nonmetals form acids


Ex. CO2 + H2O  H2CO3
Oxides of metals form bases

Ex. CaO + H2O  Ca(OH)2

1. Neutralization reaction and net ionic
equation

Neutralization Rxn:


Reaction between an acid and a base
**Most Rxns: Acid + Base  Water + Salt
Net Ionic Eqn:
H+ (aq) + OH-(aq)  H2O (l)

Ex5.5 Write the full neutralization reaction and net
ionic equations for the interactions between HNO3
and Ba(OH)2, and between H2SO4 and KOH.

3. Acid Base Titration

Titration


Lab procedure where acid and base are made to
combine in their stoichiometric proportions
Follows the same stoich process as in Chap 4

1. Definition

Reaction in which a gas is formed, sometimes
as the result of a decomposition of one of the
products

Ex. H2CO3

2. Metal and acid

PURE metals react with acid to produce H2 gas

Ex. Mg + HCl  MgCl2 + H2
3. Carbonate & acid
MgCO3 + HCl  MgCl2 + H2CO3

H2CO3 H2O + CO2
1.
General Classification:
Synthesis, Decomposition, Single and Double Rep.

2. Redox Reaction


Species lose/gain electrons as reaction occurs
COMING SOON Section 5.7

3. Product favored reactions

All of the preceding have been product-favored
reactions

Removing ions from solution to form a new product
(solid, gas, water, etc.)

1. Examples and categories

Uncombined Elements


Fe(s) + Cu+2(aq)  Fe+2(aq) + Cu(s)
Oxygen or Halogens

PbO + V+3 + H2O  PbO2 + VO + H+

2. Species which loses
electrons is said to be:



Oxidized
3. Species which gains
electrons is said to be:

GER!!!!!!
“LEO” the Lion
Reduced
Remember…


LEO the Lion Says---GER!
-or- OIL RIG
“OIL RIG”

4. Half-reaction examples
Fe(s) + Cu+2(aq)  Fe+2(aq) + Cu(s)
Fe  Fe+2 + 2e-
OXIDATION
Cu+2 + 2e-  Cu
REDUCTION
5.7 OXIDATION-REDUCTION
REACTIONS

5. Rules for determining oxidation numbers

a. Neutral species (isolated atom, molecule, formula
unit)



b. Monatomic ions, Grp 1A, 2A, 7A …


The total of all oxidation numbers is 0.
Ex. Fe = 0 Cl2 = 0 C6H12O6 = 0 NaCl = 0
+1, +2, +3, skip -3, -2, -1 (usually)
c. Oxygen


Most of the time, oxidation # = -2
Exception-Peroxides Oxidation # = -1
O2-2)
(O bonded to O 

d. Sum of oxidation numbers for neutral and charged
species:

Neutral Species- total of all oxidation #’s = 0

Charged Species- total of all oxidation #’s = total charge

Ex. NH4+ = +1
NO3-1 = -1
PO4-3 = -3
Cr+3 = +3

Ex5.6 Assign oxidation numbers for each element
in the following species:

CaC2O4
Cr2O72-

N2O
N2O4

ClO1-
ClO41-

HAsO42-
K2O

HCO31-
MgSO4

HIO3
Na2MoO4

NO21-
NH41+

S2O32-
TeF82-

6. REDUCTION IS…

7. OXIDATION IS…
1. Write all reactants and products EXCEPT H+ and H2O
in the form of separate oxidation and reduction halfreactions.
2. For each half-rxn, adjust coefficients for all atoms
except H and O.
3. For each half-rxn, if the number of oxygen atoms
differs between the left and right sides of the equation,
add water molecules to the side needing more oxygen
atoms.
4. For each half-rxn, if the number of hydrogen atoms
differs between the left and right sides, at H+ to the
side need hydrogen atoms.
5. For each half-rxn, add the number of e- need to
balance the charge. (Oxidation rxn will have e- on the
right and the Reduction will have e- on the left).
6. Multiply each half-rxn by the minimum factor required
to equalize the number of e- in each half-rxn.
7.Add the equations for the half-rxns together, canceling
e- and excess water molecules or hydrogen ions.
8. IF DONE IN BASIC SOLUTION:


Add OH- ions to BOTH sides of the final equation to convert
ALL of the H+ to water
Cancel any newly formed H2O molecules if necessary

See Steps on Previous Slide for Explanations!!
Ex. PbO + V+3 + H2O  PbO2 + VO + H+

Step 1:



Step 2: (all atoms except O & H are already balanced)



PbO  PbO2
V+3  VO
Step 3:



PbO  PbO2
V+3  VO
H2O + PbO  PbO2
H2O + V+3  VO
Step 4:

H O + PbO  PbO + 2H+


PbO + V+3 + H2O  PbO2 + VO + H+
Step 5:



Step 6:



H2O + PbO  PbO2 + 2H+ + 2e1 e- + H2O + V+3  VO + 2H+
H2O + PbO  PbO2 + 2H+ + 2e2e- + 2H2O + 2V+3  2VO + 4H+
Step 7:

H2O + PbO + 2e- + 2H2O + 2V+3  PbO2 + 2H+ + 2e- + 2VO + 4H+
FINAL: PbO + 3H2O + 2V+3  PbO2 + 2VO + 6H+

9. Ex5.7 Balance in acidic solution:
MnO41- + Fe2+  Mn2+ + Fe3+
Cr2O72- + H2SO3  Cr3+ + SO42-

10. Ex5.8 Balance in Basic Solution
I1- + MnO41-  I2 + MnO2
S2O32- + I2  SO42- + I1-

1. Molarity – definition, equation, and notation
(M and [ ])

Unit of concentration for solutions

Molarity = M = mol-solute
L-soln

[ ] = molar concentration
Ex. [Mg+2] [NaCl]

Ex5.9 What is the molarity of a solution made by
dissolving 20.0g of NaCl(s) in enough water to make 1.00L
of solution? In enough water to make 300.mL of solution?

Ex5.10 How many liters of 0.430M solution of Na2SO4 can
be made starting with 10.0g of solid?

Ex5.11 How many grams of solid solute can be
recrystallized from 10.0mL of 0.020M CuCl2?

5. Dissolution equations and molarity

Concentration of compound is related to
concentration of ions via equation stoich!

Ex. 0.500 M MgCl2

Ex5.12 State the concentration of each ion in a. 0.250M
Na2CO3 b. 0.023M barium phosphate c. 0.380M
glucose

7. Dilution problems and equation

Start with conc. soln.  dilute with H2O to lower
conc.

M1V1 = M2V2

8. Ex5.13 What is the new concentration of a solution
made by adding 150.mL of water to 23.0mL of a
2.50M solution of NaCl?

9. Ex5.14 A student starts with100mL of a 0.500M solution of
HCl. How much water needs to be added to obtain a solution of
HCl with a concentration of 0.0250M. What is the final volume of
the solution?

1. pH scale – definitions and diagram

pH scale: ranges from 0 (acid) to 14 (base); measures
“the Power of Hydrogen”

pH = -log [H+]

[H3O+] = 10-pH

Ex5.15 State the pH of the following solutions with
given hydrogen ion concentrations:

a. [H+] = 0.0230M
b. [H+] = 0.000560M
c. [H+] = 3.00M

Ex5.16 What is the hydrogen ion concentration for
each of the following solutions with given pH?

a. pH = 4.30
b. pH = 11.2
c. pH = 6.30

Precipitation Reaction Ex5.17 A precipitate forms when solutions of
silver nitrate and scandium (III) chloride are mixed. What volume of
0.0385M scandium (III) chloride is needed to react completely with
22.00mL of 0.130M silver nitrate? What is the mass of the precipitate
formed?

Precipitation Reaction Ex5.18 What is the mass percent of NaCl in a
mixture of sodium chloride and sodium nitrate if a 0.9056g sample of the
mixture yields 0.9372g of AgCl(s) when allowed to react with excess
AgNO3(aq)?

Acid Base Reaction Ex5.19 How many milliliters of
0.0195M HCl are required to titrate 10.00mL of 0.0116M
Ca(OH)2?

Acid Base Reaction Ex5.20 A Vitamin C capsule (ascorbic acid) is analyzed by
titrating is with 0.250M sodium hydroxide. It is found that 10.30mL of the base is
required to react with a capsule with a mass of 0.518g. What is the percentage of
vitamin C, C8H8O6, in the capsule if the acid and base react in a 1 to 1 molar
ratio?
 Gas Forming Reaction Ex5.21 If 38.55mL of HCl are used to react with
2.150g of Na2CO3, what is the molarity of the HCl solution?
 Redox Reaction Ex5.22 To analyze an iron containing compound, you
convert all the iron to Fe2+ in aqueous solution and then titrate the solution
with a known concentration of KMnO4 according to the following
balanced equation:

MnO41-(aq) + 5Fe2+(aq) + 8H1+(aq)  Mn2+(aq) + 5Fe3+(aq) +
4H2O(l)

A 0.598g sample of the iron containing compound requires 22.25mL of 0.0123M
KMnO4 for titration to the equivalence point. What is the mass percent of iron in the
sample?

#1 – 21, 23, 25, 27, 29, 105, 107, 117


#2 – 31, 33, 35, 37, 39, 41, 43



Molar concentrations, preparing solutions
#6 – 73, 75, 77, 109, 87, 89, 91, 93, 101


Gas forming and Predicting Products
#4 – Worksheet – Balancing Redox Equations
#5 – 59, 61, 63, 64, 67, 69, 71, 113


Net Ionic Equations & Acid-Base
#3 – 45, 47, 49, 95, 97, 53, 55, 57


Predicting Solubilities
pH & Acid-Base Titration
#7 – 79, 83, 99, 1-3, 111, 115, 125