A.P. Chemistry
Chapter 4: Reactions in Aqueous
Solutions
Part 1. 4.1-4.7
4.1 Water, the Common Solvent
Definition-solution prepared by dissolving
substances in water (aqueous solutions); can
be classified as non-electrolyte or electrolyte,
depending on their ability to conduct
electricity.
Water is not a linear molecule; it is bent at an
angle of about 105o. Electrons are not evenly
distributed around the atoms in the water,
making the molecule polar. Draw it.
4.2 The Nature of Aqueous Solutions: Strong & Weak Electrolytes
Like dissolves like. (p. 133) The following classes of molecules are
miscible:
Polar and ionic
Polar and polar
Nonpolar and nonpolar
Ionic salts dissolve in water
Example:
NaCl(s)  Na+1(aq) + Cl-1(aq)
CaCl2(s) 
K2Cr2O7(s) 
Ba(OH)2(s) 
Compounds that contain only carbon and hydrogen are nonpolar.
Example: Predict whether each pair of substances will mix and indicate
why or why not:
a. NaNO3 and H2O (y)
c. C6H14 (hexane) and H2O (n)
b. I2 and C6H14 (y)
d. I2 and H2O (n)
(What does immiscible mean?)
Solution- a homogeneous mixture of two or more
substances
Solute- the substance present in the smaller amount; or,
if it changes phase upon making the solution (p. 134)
Solvent- substance present in the larger amount (p. 134)
Electrolyte- a substance that, when dissolved in water,
results in a solution that can conduct electricity (p.
134)
Non-electrolyte- does not conduct electricity when
dissolved in water (p. 134)
Pure water contains very few ions- poor conductor
Water w/ NaCl- because salt dissolves in water, a salt
solution is electrolytic
Electrolyte
Strong
Weak
Non
Conductivity o of Dissociation
Examples
high
total
strong acids
(HCl); many salts;
strong bases
(NaOH); other
Group 1 and
Group 2
hydroxides
low-moderate partial
weak
organic acids;
weak bases
none
close to zero sugar, AgCl,
Fe2O3
Example: Which of these would be a strong,
weak or non-electrolyte?
HClO4 (S)
C6H12 (N)
LiOH (S)
NH3 (W)
CaCl2 (S)
HC2H3O2 (W)
Problem: Identify the following as strong, weak,
or nonelectrolyte:
a. CH3OH (methyl alcohol) (W)
b. CH4
(N)
c. Na2CO3 (S)
4.3 The Composition of Solutions
Concentration
Percent by Mass (p. 514)
Percent solute = mass of solute x 100%
Mass of solution
Percent =
mass of solute
x 100%
Mass of solute + mass of solvent
Molarity- number of moles of solute divided by the volume of solution
in liters; M
(Molarity may change due to change in temperature as volume
expands or contracts)(p. 139)
Molarity = number of moles of solute
Number of liters of solution
Example: Calculate the molarity of a solution prepared by dissolving
11.5 g of solid NaOH in enough water to make 1.50 L of solution.
11. 5 g NaOH x 1 mol/40g = 0.2875 mol
0.2875 mol/1.50 L = 0.192 M
Problem: Calculate the molarity of a solution prepared by dissolving
1.56 g of gaseous HCl in enough water to make 26.8 mL of solution.
1.56 g HCl x 1 mol/36.5 g = 0.0427 mol
0.0427 mol/0.0268 L = 1.59 M
Molality- number of moles of solute per kilogram of solvent; m
(molality will not change due to change in temperature; mass is not variable
w/ temp. Therefore, use this concentration when working with colligative
properties)(p. 514, 551)
Molality = number of moles of solute
Number of kilograms of solvent
Example: A solution is prepared by mixing 1.00 g of ethanol with 100.0 g of
water to give a final volume of 101.mL. Calculate the molality, molarity,
and mass percent.
Molar mass ethanol = 46 g/mol
1.00 g x 1 mol/46 g = 0.0217 mol
0.0217 mol/ 0.101 L = 0.215 M
0.0217 mol/0.100 kg = 0.217 m
1.00 g/101 g x 100% = 0.99 %
Problem: The electrolyte in automobile lead storage batteries is a 3.75 M
sulfuric acid solution that has a density of 1.230 g/mL. Calculate the mass
percent and molality of the sulfuric acid.
Mole Fraction (XA, XB)- a dimensionless quantity, no
unit
XA =
number of moles of A
# of moles of A + # of moles B
Standard Solution (p. 142) –a solution whose
concentration is accurately known.
Dilution (p. 144-5) –the process of adding solvent to
lower the concentration of solute in a solution
4.4 Types of Chemical Reactions (read textbook
only; practice in class)
4.5 Precipitation Reactions
Precipitate (p. 147) –an insoluble substance that may
form when two solutions are mixed together.
Basic Solubility Rules: Learn these!! (Table 4.1, p. 150)
All Group I metallic salts, ammonium salts, nitrates, acetates,
chlorates, and perchlorates are soluble.
All chlorides, bromides, and iodides are soluble EXCEPT when
with silver, lead (II), or mercury (I)
All sulfates are soluble EXCEPT when with lead (II), Ca, Sr, or
Ba.
Group I, Ca, Sr, Ba metallic oxides are soluble (basic
anhydrides) as well as their resulting hydroxides
Insoluble (except w/ cations in Rule #1): carbonates,
phosphates, sulfides, and bisulfites (hydrogen sulfites).
4.6 Describing Reactions in Solution
Full Formula Unit Equation: show complete formulas for all
compounds (aka molecular equation)
AgNO3(aq) + NaCl(aq) AgCl(s) + NaNO3(aq)
Complete Ionic Equation: show the predominant form in
which each substance exists when it is in contact with
aqueous solution (p. 152)
Ag+ + NO3- + Na+ + Cl-  AgCl(s) + Na+ + ClNet Ionic Equation: only show the species that react;
eliminate spectator ions.(p. 152)
Ag+ + Cl-  AgCl(s)
Spectator Ions (p. 152)
Na+ , NO3-
4.7 Stoichiometry of Precipitation Reactions
Individual Ion Concentrations when an ionic compound
is dissolved in water
Example: Give the concentration of each type of ion in
the following solutions:
0.50 M Co(NO3)2 (0.50 M Co2+, 1.0 M NO3-1)
1 M Fe(ClO4)3 (1 M Fe3+, 3 M ClO4-)
Problem: Calculate the number of moles of chloride ions
in 1.75 L of 1.0 x 10-3 M ZnCl2.
ZnCl2  Zn2+ + 2Cl1.75 L x 1.0 x 10-3 mol/liter x 2 mol Cl-/1 mol ZnCl2 =
0.0035 mol Chloride ion
Example: A 0.5662 g sample of an ionic
compound containing chloride and an
unknown metal is dissolved in water and
treated with an excess of AgNO3 solution. If
1.0882 g of AgCl precipitate forms, what is the
percent by mass of Cl in the original
compound.
% Cl (in AgCl) = 35.45 g/143.25 g x 100% = 24.7%
0.247 x 1.0882 g AgCl = 0.2693 g Cl
0.2693 g/0.5662 g x 100% = 47.6%