Ch. 9: Introduction to
Solutions and Aqueous
Reactions
Dr. Namphol Sinkaset
Chem 200: General Chemistry I
I. Chapter Outline
I. Introduction
II. Solution Concentrations
III. Solution Calculations
IV. Aqueous Solutions
V. Precipitation Reactions
VI. Acid/Base Reactions
VII. Gas-Evolution Reactions
VIII. Oxidation-Reduction Reactions
I. Aqueous Chemistry
•
•
Water-based chemistry is the most well
studied – why?
In this chapter, we will focus on
reactions that take place in water and
look at:
1) Solution stoichiometry
2) Common aqueous reactions
II. Solution Concentration
• There are two parts of a solution.
 solute: substance present in smaller
amount
 solvent: substance present in larger
amount
• For stoichiometry, the important aspect
of a solution is its concentration.
• concentration: amount of solute present
in a certain volume of solution
II. Concentrated vs. Dilute
• Concentrated
solutions have a lot
of solute relative to
solvent.
• Dilute solutions
have a little solute
relative to solvent.
II. Quantitative Concentrations
• The most common concentration unit is
molarity, which is moles solute per L of
solution.
In a solution, solute is evenly dispersed in the solvent!!
II. Solution Preparation
• Preparing solutions
requires a series of
exacting steps.
III. Solution Calculations
•
There are 3 main types of solution-based
calculations.
1) What is the concentration?

Use definition of molarity.
2) Solution creation/dilution.

Use definition of molarity / M1V1 = M2V2
3) Stoichiometry.

Use molarity as a conversion factor.
• Start w/ problems using definition of molarity.
III. Sample Problem
• e.g. Calculate the molarity of a solution
formed when 24.2 g NaCl is dissolved in
water to make 124.1 mL of solution.
III. Sample Problem
• e.g. How many grams of Na2HPO4 are
needed to make 1.50 L of a 0.500 M
Na2HPO4 solution?
III. Solution Dilution
• A stock solution is a
solution of high
concentration.
• Lower concentration
solutions can be
made from the stock
via dilution.
III. Sample Problem
• e.g. How many mL of a 2.0 M NaCl
solution are needed to make 250 mL of
a 0.50 M NaCl solution?
III. Solution Stoichiometry
• In the stoichiometry we’ve done before,
amounts were converted between
grams and moles.
• In solutions, amounts are converted
between volumes and concentrations.
• The key is remembering that molarity is
a conversion factor between moles and
volume!
III. Sample Problem
• e.g. How many mL of 0.10 M HCl reacts
with 0.10 g Al(OH)3 according to the
reaction below?
Al(OH)3(s) + 3HCl(aq)  AlCl3(aq) + 3H2O(l)
III. Sample Problem
• e.g. How much PbCl2 forms when 267
mL 1.50 M lead(II) acetate reacts with
125 mL 3.40 M sodium chloride
according to the reaction below?
Pb(CH3COO)2(aq) + 2NaCl(aq)  PbCl2(s) + 2NaCH3COO(aq)
IV. Chemical Reactions
• There are countless reactions, but only
a few categories of reactions.
• With experience, it becomes easier to
identify what will happen in a reaction.
• First, we take a close look at the solute
and solvent in a solution.
IV. Forming a Solution
• Attractive forces hold
solute together.
• When solute is added
to a solvent, new
potentially attractive
forces arise.
• Competition between
these forces occurs.
IV. Aqueous Solutions
• Water is a particularly
“active” solvent.
• As a solvent, water
has one important
characteristic: it is
polar!
IV. Interactions in Aqueous
Solutions
• The polar nature of water allows it to interact
with charged species in solution.
IV. Water-Ion > Na+Cl-
IV. Electrolytes
• Compounds that dissociate in water and lead to
electrical conductivity are called electrolytes.
IV. Strong/Weak Electrolytes
• Strong electrolytes dissociate
completely in water.
H 2O
 NaCl(s)  Na+(aq) + Cl-(aq)
• Logically, weak electrolytes do not
dissociate completely in water.
H 2O
 CH3COOH(aq)  H+(aq) + CH3COO-(aq)
IV. Sample Problem
• e.g. How many moles of each ion are in
a solution formed by dissolving 354 g of
magnesium hydroxide in water?
V. Precipitation Reactions
• The formation of a
precipitate (ppt) is a
strong driving force
for a reaction.
• Precipitate is a
fancy word for solid.
• The attractions in
these solids are too
strong for H2O to
break up.
V. Predicting Precipitates (95%)
1) Li+, Na+, K+, NH4+ salts are soluble.
2) NO3-, CH3COO-, ClO4- salts are
soluble.
3) Ag+, Pb2+, Hg22+ salts are insoluble.
4) Cl-, Br-, I- salts are soluble.
5) CO32-, S2-, O2-, OH- salts are insoluble.
6) SO42- salts are soluble except for
CaSO4,SrSO4, and BaSO4.
7) If none of these apply, it’s insoluble.
V. Sample Problem
•
e.g. Predict the precipitates in the
following aqueous reactions.
a)
b)
c)
d)
sodium hydroxide + cadmium(II) nitrate
magnesium bromide + potassium acetate
ammonium sulfate + barium chloride
sodium iodide + lead(II) nitrate
V. Writing Reactions
No
spectator
ions
V. Sample Problem
• Write balanced molecular, total ionic,
and net ionic equations for the reaction
between strontium chloride and lithium
phosphate.
VI. Acids/Bases
• There are many definitions of acids and
bases, but we will use the Arrhenius
definitions for now.
• Acids are molecular compounds that
produce H+ ions in aqueous solution.
• Bases are substances that produce OHions in aqueous solution.
VI. Aqueous Acids
• What is H+ comprised of?
• Water interacts so strongly w/ H+, that it
forms a bond with it.
 e.g. HCl(g) + H2O(l)  (H2O)H+(aq) + Cl-(aq)
• This is called the hydronium ion, and it’s
usually written as H3O+.
 Note that H+(aq) = H3O+(aq)
• Polyprotic acids have more than one
ionizable H+, e.g. H2SO4.
VI. Common Acids & Bases
VI. Acid Nomenclature
•
There are two categories of acids that
have different naming rules.
1) Binary acids
2) Oxoacids
• You can easily recognize acids
because their formula has H as the
first element.
VI. Naming Binary Acids
• Binary acids contain a nonmetal anion.




HCl = hydrochloric acid
HBr = hydrobromic acid
H2Se = hydroselenic acid
HI = hydroiodic acid
VI. Naming Oxoacids
Oxoacids contain an oxoanion.
• Set 1
• Set 2








HNO3 = nitric acid
H2SO4 = sulfuric acid
HClO3 = chloric acid
HClO4 = perchloric acid
HNO2 = nitrous acid
HClO2 = chlorous acid
HClO = hypochlorous acid
H2SO3 = sulfurous acid
VI. Acid/Base Reactions
• The driving force for this
reaction is the formation
of water; other product is
a salt.
• The net ionic equation for
acid/base reactions is
always the same!!
 H+(aq) + OH-(aq)  H2O(l)
VI. Acid/Base Titrations
• The concentration of
an acid or base can
be determined
experimentally.
VI. Titration Terminology
• titration: procedure in which one
solution of known [ ] is used to
determine the [ ] of another solution
• indicator: a substance used to visualize
the end of a reaction
• The equivalance point occurs when
moles acid = moles base.
• The endpoint occurs when the solution
changes color due to the indicator.
VI. Titration Problems
• The first step in solving a titration
problem is writing the titration reaction!!
• After identifying the reaction, it becomes
a solution stoichiometry problem!
• Again, use unit labels on the numbers to
guide your calculation.
VI. Sample Problem
• e.g. To determine the concentration of a
solution of H2SO4, you titrate a 50.00
mL sample of it with 0.250 M NaOH. If
it takes 22.35 mL of the NaOH solution
to reach the endpoint, what’s the
concentration of the H2SO4?
VII. Gas-Evolution Reactions
• Many gas-evolution reactions are also acidbase reactions.
• The formation of the gas could be direct or
through decomposition of a product.
 H2CO3(aq)  H2O(l) + CO2(g)
VII. It Makes Bubbles
VII. Common Gas Products
VII. Sample Problem
• Write balanced molecular, total ionic,
and net ionic equations for the reaction
between aqueous solutions of
hydrobromic acid and potassium sulfite.
VIII. Reactions Involving ein Motion
• The movement of e- from one atom to
another is another driving force for
reactions.
• Oxidation is the loss of e-.
• Reduction is the gain of e-.
• They are coupled processes; one
cannot occur w/out the other.
VIII. Reduction-Oxidation
Reactions
• These are also known as redox reactions.
• Formation of NaCl is an example.
 Na  Na+ + e ½ Cl2 + e-  Cl-
• Cl2 oxidizes Na; Cl2 is the oxidizing agent.
• Na reduces Cl2; Na is the reducing agent.
VIII. NaCl Formation via Redox
VIII. Nonpolar to Polar
• You don’t need
complete e- transfer
for a redox reaction.
• Can also have just a
shift of e- density.
• Consider H2(g) + Cl2(g)
 2HCl(g)
VIII. Oxidation States
• Oxidation states (oxidation numbers)
allow us to keep track of which atoms
are gaining/losing e- in a reaction.
• oxidation state (number): the charge an
atom would have if e- are transferred
completely and not shared
• Note that in ionic compounds, we
consider e- as totally transferred, so the
ionic charge is the oxidation state.
VIII. Rules for Assigning O.N.
1)
2)
3)
4)
5)
6)
Atoms in elemental form have O.N. = 0.
Charge on a monatomic ion equals its O.N.
The sum of all O.N. must equal the total charge.
For Group 1, O.N. = +1.
For Group 2, O.N. = +2.
For H, O.N. = +1 w/ nonmetals, -1 w/ metals and
B.
7) For F, O.N. = -1.
8) For O, O.N. = -1 in peroxides and -2 in all others.
9) For Group 17, typically O.N. = -1.
VIII. Assigning O.N.
•
e.g. Determine O.N. for all atoms in
the following.
a)
b)
c)
d)
KMnO4
NH4+
IF3
ZnCl2
VIII. Sample Problem
• e.g. Determine the substances that are
oxidized and reduced in the reaction
below.
5CO(g) + I2O5(s)  I2(s) + 5CO2(g)