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SECTION 13.1
Compounds in
Aqueous Solutions
Teacher Notes and Answers
SECTION 1 Compounds in Aqueous Solution
1.Dissociation is the separation of ions in a
dissolved ionic compound. Dissolution is the
process of a solvent separating formula units
of an ionic compound from a crystal to form a
uniform solution.
2.A double displacement reaction would not take
place because the ions are not all in solution.
The compounds would simply settle to the
bottom of the container.
3.solid
4.an uninvolved observer; a charged particle
formed when an atom gains or loses electrons
 
5.​H+
​ ​(aq) and SO​2−
  aq)
4   ​(
6.In ionization, ions form from neutral molecules.
In dissociation, ionic bonds are broken to form
ions.
7.nonelectrolyte, weak electrolyte, strong
electrolyte
8.Sodium nitrate is a strong electrolyte because
it is an ionic compound that will dissociate in
solution.
Practice
A.1 mol; 1 mol; 2 mol
​H​ ​O
3+
–
Ba.AlC​l3​ ​(s) ______
​  2 ⟶
  A​l​ ​(aq) + 3C​l​ ​(aq); 1 mol of
aluminum ions produced, 3 mol of chloride ions
produced, 4 mol total ions produced
​H​ ​O
+
2–
Bb.N​a2​ ​S(s) ______
​  2 ⟶
   2N​a​ ​(aq) + ​S​ ​(aq); 2 mol of
sodium ions produced, 1 mol of sulfide ions
produced, 3 mol total ions produced
​H​ ​O
2+
3–
Bc.B​a3​ ​​N​2​(s) ______
​  2 ⟶
   3B​a​ ​(aq) + 2​N​ ​(aq); 1.5 mol
of barium ions produced, 1 mol of nitride ions
produced, 2.5 mol total ions produced
Ca.N​i2+
​ ​(aq) + 2C​l–​ ​(aq)
Cb.​K+
​ ​(aq) + MnO​ 4​(   aq)
2−
Cc.C​u2+
​ ​(aq) + SO​ 
   aq)
4  ​(
−
2+
 
Cd.P​b​ ​(aq) + 2NO​3   ​(aq)
Ce.no dissociation
Cf.no dissociation
Da.soluble
Db.soluble
Dc.insoluble
Dd.insoluble
De.insoluble
Df.soluble
 
E.yes; B​a2+
​ ​(aq) + SO​2   aq) ⟶ BaS​O4​ ​(s)
4  ​(
F.does not apply
2−
G.yes; B​a2+
​ ​(aq) + SO​ 
   aq) ⟶ BaS​O4​ ​(s)
4  ​(
Review
1.Ionization is the formation of ions from solute
molecules by the action of the solvent.
​H​ ​O
−
2+
2a.Sr(N​O​ ​​)​ ​(aq) ______
​  2 ⟶
   Sr​ ​(aq) + 2NO​    ​(aq)
3 2
3
2b.0.5 mol of strontium ions, 1 mol of nitrate ions
3.No precipitate will form.
4.If the attraction of the polar solvent molecules is
stronger than a covalent bond in the solute, the
molecule will break into ions.
5.HCl is essentially 100% ionized in water
solution, while HF is only slightly ionized
because its hydrogen-fluoride covalent bond is
strong.
6a.0.10 M HCl
6b.0.10 M HCl
6c.0.10 M Ca​Cl​2​
I o n s i n A q u e o u s S o l u t i o n s a n d C o l l i g at i v e P r o p e r t i e s
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SECTION 13.1
Compounds in
Aqueous Solutions
Ionic compounds and molecular compounds have different
properties. Ionic compounds form a crystal lattice that
consists of a regular pattern of charged particles held
together by ionic attractions. Although molecules are held
together by strong covalent bonds, molecules within a
molecular compound are held together by relatively weak
intermolecular forces. This difference in nature causes the
two types of compound to behave differently when they
dissolve in water.
Ions separate from each other when ionic compounds
are dissolved in water.
Key Terms
dissociation
net ionic equation
spectator ion
ionization
hydronium ion
strong electrolyte
weak electrolyte
Na+
H2O
Cl-
Dissociation is the separation of ions that occurs when an
ionic compound dissolves. For example, the dissociation
of sodium chloride in water can be represented by this
equation. Note that a formula unit for sodium chloride
produces two ions in solution.
NaCl
a formula unit of solid
sodium chloride
one dissolved
sodium ion
H​2​O
-
yields when added
to water
one dissolved
chloride ion
NaCl(s) N​a​+​(aq) + C​l​ ​(aq)
READING CHECK
1.
2
What is the difference between dissociation and dissolution?
C H A P TER 1 3
When NaCl dissolves in water, the
ions separate as they leave the crystal.
Moles of Dissolved Ions Chemical equations representing
dissolution reflect the number of moles of each substance that
are reacted and produced. So, for the dissolution of sodium
chloride in water, 1 mol of sodium chloride dissociates into
​ ​ions).
2 mol of ions (1 mol of N​a+
​ ions and 1 mol of C​lH​2​O
N​a+
​ ​(aq) + C​l​ ​(aq)
NaCl(s) 1 mol
1 mol
TIP
It is important not
to confuse the terms
­dissociation and dissolution.
Dissociation is the process
of splitting apart or separating.
Dissolution is the process
of dissolving.
1 mol
This relationship only holds if the compound undergoes
100% dissociation. In other words, one mole of sodium
chloride will only form one mole of each type of ion if every
formula unit in the mole of sodium chloride dissociates.
PRACTICE
A. Complete
the graphic below to show how many moles of
each ion will form from the 100% dissociation of CaC​l​2​.
H​2​O
C​a2+
​ ​(aq) + 2C​l​ ​(aq)
CaC​l2​ ​(s) B.
Write the equation for the dissolution of each sample in
water. Then determine the number of moles of each ion
produced as well as the total moles of ions produced.
a.
1 mol aluminum chloride
b.
1 mol sodium sulfide
c.
0.5 mol barium nitrate
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Precipitation Reactions
Every ionic compound is soluble to some degree. Some
compounds, such as sodium chloride, form clear solutions in
water. Others form solutions with a characteristic color. The
solutions are still homogeneous even though the solute
particles change the appearance of the water.
However, there are some ionic compounds that have a very
low solubility. Those compounds are considered insoluble for
practical purposes. The photograph shows some ionic
­compounds that are soluble and some that are insoluble.
(a) ​NiCl​2​forms a green solution in
water. (b) ​KMnO​4​forms a purple
solution in water. (c) ​CuSO​4​forms a
blue solution in water. (d) ​Pb(​NO​3​)2​ ​
forms a clear solution in water.
(e) AgCl and (f) CdS are insoluble
in water.
(a)
(c)
(b)
(d)
(e)
(f)
PRACTICE
C.
If possible, write dissociation equations for the six compounds
shown in the photograph. Otherwise write “no dissociation.”
a.
​NiCl​2​(aq) b.
​KMnO​4​(aq) c.
​CuSO​4​(aq) d.
​Pb(​NO​3​)2​ ​(aq) e.
AgCl(aq) f.
CdS(aq) 4
H​2​O
H​2​O
H​2​O
H​2​O
H​2​O
H​2​O
C H A P TER 1 3
Solubility Guidelines The table below gives some general
guidelines about whether an ionic compound is soluble in water
or not. According to the table, sodium carbonate, N
​ a​2​​CO​3​, is
soluble. Most carbonate compounds are insoluble according to
Rule 5. However, Rule 5 also states that the compounds listed
in Rule 1 are an exception. As another example, calcium
phosphate is insoluble. According to Rule 5, most phosphates
are insoluble and calcium is not one of the exceptions.
General Solubility Guidelines
1. Sodium, potassium, and ammonium compounds are soluble in water.
2. Nitrates, acetates, and chlorates are soluble.
3. Most chlorides are soluble, except those of silver, mercury(I), and lead.
Lead(II) chloride is soluble in hot water.
4. Most sulfates are soluble, except those of barium, strontium, lead,
calcium, and mercury.
5. Most carbonates, phosphates, and silicates are insoluble, except those
of sodium, potassium, and ammonium.
6. Most sulfides are insoluble, except those of calcium, strontium, sodium,
potassium, and ammonium.
PRACTICE
D. Use
the table above to predict whether each of the
following compounds is soluble or insoluble.
a.
KCl
d.
BaS​O4​ ​
b.
NaN​O3​ ​
e.
C​a3​ ​(P​O4​ ​​)​2​
c.
AgCl
f.
Pb(Cl​O3​ ​​)​2​
Formation of Precipitates The solubility guidelines listed
above are also important for determining what happens when
two different soluble compounds are mixed. Two sets of
cations and two sets of anions would be present in such a
mixture. If the attraction between the two types of ions is
greater than the attraction between the ions and the polar
water molecules, then the ions will bond and precipitation will
occur. This happens if a double-displacement reaction
between the compounds results in an insoluble compound. If
the attraction between the water molecules and the ions is
greater, then both compounds will remain dissolved.
Critical Thinking
2. Infer Assuming no additional
energy is supplied, what do you
expect will happen if you mix two
insoluble compounds in water,
and why?
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Before reaction
S2-
H2O
After reaction
NH+
4
(NH4)2S(aq)
Cd2+
NO3
CdS(s) + NH4NO3(aq)
CdS(s)
Cd(NO3)2(aq)
For example, consider the mixture shown in the photograph
above. The test tube contains a solution of ammonium sulfide
and the beaker contains a solution of cadmium nitrate. The
dissociation equations for the two substances are shown below.
H​2​O
(N​H4​ ​​)​2​ S(s) H​2​O
Cd(N​O​3​​)​2​(s)    ​(  aq) + ​S2 2NH​+
​ ​(aq)
4
 (aq)
C​d2+
​ ​(aq) + 2NO​3  
The following equation represents a potential doubledisplacement reaction between the two compounds. The
question marks indicate that states that are undetermined.
(N​H​4​​)​2​S(aq) + Cd(N​O3​ ​​)​2​(aq)
2N​H4​ ​N​O3​ ​(?) + CdS(?)
According to Rules 1 and 2 in the table on the previous
page, ammonium nitrate is soluble in water. However, according to Rule 6, cadmium sulfide is insoluble in water. Therefore,
the cadmium sulfide will precipitate out of solution. The
completed equation for the reaction in the photograph is
(N​H4​ ​​)​2​S(aq) + Cd(N​O3​ ​​)​2​(aq)
6
C H A P TER 1 3
2N​H4​ ​N​O3​ ​(aq) + CdS(s)
Ammonium sulfide and cadmium
nitrate are mixed, resulting in the
formation of the yellow precipitate
cadmium sulfide.
! Remember
A double-displacement reaction
is a reaction in which two ionic
compounds swap ions so that the
cation from one compound bonds
with the anion from the other
compound.
READING CHECK
3. If one product of a double­displacement reaction is a
precipitate, what is the state of
that product?
Net Ionic Equations
A reaction between ions in aqueous solutions is usually
represented by the component ions instead of by the ionic
compounds themselves. One reason is that the ions are the
important units in the reaction, not the formula units of the
compound’s crystal lattice. An equation that represents
aqueous solutions of ionic compounds as ions is called an
ionic equation.
A net ionic equation is a chemical reaction that includes
only those compounds and ions that undergo a chemical
change in a reaction in an aqueous solution. Ions that do
not participate in the reaction are not included. These
spectator ions are found in solution before and after the
reaction occurs.
LOOKING CLOSER
4. Write separate definitions for
these terms in your own words.
spectator:
ion:
The first step in writing a net ionic equation is to write a
complete ionic equation. Consider the reaction that was
discussed on the previous page. The double-displacement
reaction can be written as follows, by writing all of the substances in an aqueous state as dissociated ions.
  ​(   aq) + ​S2    ​(aq)
​ ​(aq) + C​d2+
​ ​(aq) + 2NO​2 NH​+
4
3
  ​(   aq) + 2NO​    ​(aq) + CdS(s)
2NH​+
4
3
  ​,  
This ionic equation shows that aqueous ammonium ion, NH​+
4
appears on both sides of the equation. The aqueous nitrate
    ​, also appears on both sides of the equation. These
ion, NO​+
3
ions are spectator ions. To form the net ionic equation, cancel
the spectator ions from both sides of the equation.
  ​(   aq) + ​S2   ​(  aq)
​ ​(aq) + C​d2+
​ ​(aq) + 2NO​2 NH​+
4
3
+
 
2NH​ 
 
 
 
(
​
aq)
+
2NO​
4
3   ​(aq) + CdS(s)
​ ​(aq)
C​d2+
​ ​(aq) + ​S2–
CdS(s)
This net ionic equation applies to other reactions besides the
one between ammonium sulfide and cadmium nitrate. It
applies to any reaction in which ions are combined in solution
and cadmium sulfide forms. For example, the reaction of
​ ​2​S, also has
cadmium sulfate, CdS​O​4​, and hydrogen sulfide, H
the same net ionic equation.
Critical Thinking
5. Apply What are the two
spectator ions in the reactions
of cadmium sulfate and
hydrogen sulfide?
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A molecular compound ionizes in a polar solvent.
Ionic compounds form ions in solution through the process
called dissociation. Molecular compounds can also form ions
in solution. However, molecular compounds undergo a
different process. Ions are formed from solute molecules by
the action of the solvent in a process called ionization.
The extent to which a solute ionizes depends on the
strength of the bonds within the molecule of the solute and
the strength of the attraction between the solute and the
solvent molecules. If the strength of the bonds in the solute
molecule are weaker, then the covalent bond breaks and the
molecule separates into ions. The ions form because the more
electronegative atom holds on to the electrons that were
shared in the covalent bond.
READING CHECK
6. How is ionization different
from dissociation?
One example of ionization is in the dissolution of hydrogen
chloride, HCl. The hydrogen-chlorine bond is highly polar. The
attraction between the polar HCl molecule and the polar
water molecules is strong enough to break the covalent bond.
The chlorine atom keeps the shared electrons and forms a
chloride ion, while the hydrogen atom forms a hydrogen ion.
HCl(g) H​2​O
​H+
​ ​(aq) + C​l​ ​(aq)
The Hydronium Ion
The hydrogen ion, ​H+
​ ​, is equivalent to a proton, a
concentrated area of positive charge. This ion is attracted to
negatively charged particles so strongly that it does not
normally exist alone. In an aqueous solution, the hydrogen
atom will be attracted to the negative end of a water molecule.
The result is the formation of an H
​ 3​ ​​O​+​ion, known as a
hydronium ion.
The ionization of hydrogen chloride is better represented
by the transfer of a proton from a hydrogen chloride molecule
to a water molecule. The chemical reaction is shown below.
+
+
+
​H2​ ​O HCl ​H3​ ​​O​+​ C​l​​
8
C H A P TER 1 3
-
! Remember
A hydrogen-1 atom consists of a
single electron that orbits a nucleus
consisting of a single proton. If
the electron is stripped away to
form a hydrogen ion, only the
proton remains.
SAMPLE PROBLEM
Write the net ionic equation for the reaction of aqueous
solutions of zinc nitrate and ammonium sulfide.
SOLUTION
1 ANALYZE
Determine the information that is given and unknown.
Given: reactants are zinc nitrate and ammonium sulfide
Unknown: net ionic equation
2 PLAN
Describe how to write the net ionic equation.
First, determine the possible double-displacement reaction
between the two reactants. Then determine if either of the
products forms a precipitate. Then write the ionic equation
for the reaction and cancel the particles that appear
unchanged on both sides.
3 SOLVE
Use the plan above to write the equation for the reaction.
+
  ​  to form Zn(N​O​ ​​)​ ​. NH​ 
Z​n2+
​ ​combines with NO​3 2
3
4  ​ combines
with ​S​2–​to form (N​H4​ ​​)​2​S. The balanced double-displacement
reaction is as follows.
Zn(N​O3​ ​​)​2​(aq) + (N​H4​ ​​)​2​S(aq)
ZnS(?) + 2N​H4​ ​N​O3​ ​(?)
According to the “Solubility Guidelines” table earlier in this
chapter, zinc sulfide is insoluble, so it forms a precipitate in this
reaction. The completed equation for the reaction is as follows.
Zn(N​O​3​​)​2​(aq) + (N​H4​ ​​)​2​S(aq)
ZnS(s) + 2N​H4​ ​N​O3​ ​(aq)
The aqueous compounds in this reaction are written as ions to
produce the complete ionic equation. Canceling the spectator
ions yields the net ionic equation for the reaction.
  ​(   aq) + 2NH​+
  ​(   aq) + S
​ 2​ ​(aq)
​n​2+​(aq) + 2NO​Z
3
4
+
- ​(
    aq) + ZnS(s)
+ 2NO​ 
3    aq) + 2NH​4 ​(
​ ​(aq)
Z​n​2+​(aq) + ​S2–
4 CHECK
YOUR
WORK
ZnS(s)
Check the final equation to see if it is reasonable.
The chemical formulas for all of the particles that appear on
both sides of the equation have been canceled. The
precipitate is an insoluble ionic compound, and it contains
ions from both of the reactant compounds.
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PRACTICE
E. Will
a precipitate form if solutions of potassium sulfate
and barium nitrate are combined? If so, write the net
ionic equation for the reaction.
Chemical formulas: potassium sulfate:
barium nitrate:
Which of the products are soluble?
Complete double-displacement reaction:
Complete ionic equation:
Net ionic equation:
F. Will
a precipitate form if solutions of potassium nitrate
and magnesium sulfate are combined? If so, write the
net ionic equation for the reaction.
Chemical formulas: potassium nitrate:
magnesium sulfate:
Which of the products will be soluble?
Complete double-displacement reaction:
Complete ionic equation:
Net ionic equation:
G.Will
a precipitate form if solutions of barium chloride
and sodium sulfate are combined? If so, write the net
ionic equation for the reaction.
10
C H A P TER 1 3
TIP
If both ionic
compounds
are soluble, the
complete ionic
equation has the
same four ions on
both sides of the
equation. In any
of these practice
problems in which
this is the case,
write “does not
apply” for the net
ionic equation.
An electrolyte’s strength depends on how many
dissolved ions it contains.
As discussed in the chapter “Solutions,” substances that
yield ions and conduct an electric current in solution are
electrolytes. However, not all electrolytes have the same
ability to conduct electricity. Electrolytes that conduct
electricity well are called strong electrolytes. Electrolytes that
conduct electricity poorly are called weak electrolytes.
The factor that determines whether a substance is a strong
or a weak electrolyte is the proportion of the substance that
is found as aqueous ions in solution. A strong electrolyte is
composed of all, or nearly all, ions in solution. A weak
electrolyte mostly consists of molecules in solution, with just
a few molecules that are ionized.
Consider the group of substances known as hydrogen
halides. Hydrogen halides are polar molecules that are
gases at room temperature and soluble in water. Hydrogen
chloride, hydrogen bromide, and hydrogen iodide are strong
electrolytes. However, hydrogen fluoride is a weak electrolyte.
If solutions of these four substances are made at the same
concentration, the first three will conduct electricity strongly
and hydrogen fluoride will conduct electricity poorly.
Recall that a nonelectrolyte is a substance
that does not have the ability to conduct
electricity at all when it is in solution. Because
all of the substance remains in molecular
form, no charge can be transmitted by the
movement of charged particles in the solution.
The diagram at the right shows a molecular
view of a sucrose solution. Sucrose, C​12​​H​22​​O​11​,
is a nonelectrolyte. Compare this diagram to
the diagrams of strong and weak electrolytes
on the next page.
TIP
TIP
Be careful not to confuse
the classification of an
electrolyte as strong or weak with
the concentration of a solution.
Strong and weak electrolytes differ
in the percentage of dissolved
solute ionization or dissociation.
Concentration is a measure of
the percentage of solute that is
dissolved. A dilute solution can
still be a strong electrolyte.
A hydrogen halide forms
from a hydrogen atom
and an atom from Group 17, the
halogens. The term halide is
derived in the same way as that
of a negative ion. The ending -ide
is appended to the root of the
term halogen.
Nonelectrolyte
C12H22O11
C12H22O11(aq)
READING CHECK
7.
Order the following substances from lowest to highest
percentage of ions in solution: a nonelectrolyte, a strong
electrolyte, and a weak electrolyte.
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Strong Electrolytes
The distinguishing feature of a strong
electrolyte is that it exists in ionic form
when dissolved in water. This feature is
observable for dilute concentrations as
well as high concentrations of the solute.
Even if the compound has a low solubility,
the amount that does dissolve is only
present in ionic form. The diagram at the
right shows the molecular view of sodium
chloride, which is a strong electrolyte.
Strong electrolyte
Na+
Cl-
NaCl(aq)
All soluble ionic compounds are strong electrolytes.
Strong electrolytes also include the halides hydrogen chloride,
hydrogen bromide, and hydrogen iodide. Several other acids
are strong electrolytes as well.
Weak Electrolytes
Some molecular compounds form
aqueous solutions that contain not only
dissolved ions, but also dissolved
molecules that are not ionized. For
example, hydrogen fluoride dissolves in
water to form hydrofluoric acid. The
diagram at the right shows that
hydrofluoric acid consists of both ions
and molecules. The hydrogen-fluorine
bond, the strongest of the polar bonds in
the hydrogen halides, is difficult for the
polar water molecules to break.
Weak electrolyte
HF
H3O+
F-
Weak electrolytes exist in a state of equilibrium in
which molecules are being ionized and other molecules
are re-forming at equal rates. For example, the equilibrium
reaction for hydrofluoric acid is shown below.
HF(aq) + ​H2​ ​O(l)
​H3​ ​​O​+​(aq) + ​F–​ ​(aq)
READING CHECK
8.
12
Is sodium nitrate a strong electrolyte, a weak electrolyte, or
a nonelectrolyte? Explain.
C H A P TER 1 3
HF(aq)
SECTION 13.1 REVIEW
VOCABULARY
1. What is ionization?
REVIEW
2. a. Write the equation for the dissolution of Sr(N​O​3​​)​2​in water.
b.How many moles of strontium ions and nitrate ions are produced by
dissolving 0.5 mol of strontium nitrate?
3. Will a precipitate form if solutions of magnesium acetate and strontium
chloride are combined? If so, what is the identity of the precipitate?
4. What determines whether a molecular compound will be ionized
in a polar solvent?
5. Explain why HCl is a strong electrolyte and HF is a weak electrolyte.
Critical Thinking
6. PREDICTING OUTCOMES For each of the following pairs, tell which solution
contains the larger total concentration of ions.
a. 0.10 M HCl and 0.05 M HCl
b. 0.10 M HCl and 0.10 M HF
c. 0.10 M HCl and 0.10 M CaC​l2​ ​
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