9.1: Reactions of
Ions in Solution
Making Predictions about Solubility
When aqueous ionic compounds are mixed
together, two things can occur:
1. There is no reaction
2. A double displacement reaction occurs
Factors Affecting the Solubility
of Ionic Substances
1. Ionic Charge
• Ionic compounds with small charges are
soluble
• However, as charge increases, the
strength of the bond holding the
molecule together increases making it
less soluble
2. Ion Size
• Small ions can bond tightly together
• Thus, the bond between small ions is
stronger than that between larger ions
making larger ions more soluble.
Predicting Solubility
Use the solubility table on page 424
Example:
Determine whether the following compounds
are soluble or insoluble in water.
a)Aluminum sulfate
b)Magnesium hydroxide
c) Nickel (II) carbonate
d)Silver nitrate
e) Barium sulfate
Equations for the
Reactions of Ions
 Recall, from Unit 2:
BaCl2(aq) + Na2SO4(aq)  BaSO4 (s) + 2 NaCl (aq)
But, sodium chloride (aq) does not actually exist
as NaCl in water – it dissociates into its ions: Na+
and Cl-
Reactions in Aqueous Solutions
Double displacement reactions can be observed
when one of the following is produced:
 A precipitate
 A gas
 Salt and water (Neutralization)
1. Double Displacement Reactions that
Produce a Precipitate
Using your solubility table you can
determine which products will form precipitates
(if any)
Example:
Pb(NO3)2 (aq) + 2KI(aq)  2KNO3 (aq) + 2PbI2 (aq)
2. Double Displacement Reactions that
Produce a Gas
Types of Gases that are produced:
1. Hydrogen gas – metal hydrides react with
water to form a base and hydrogen gas
Example:
LiH(s) + H2O(l)  LiOH(aq) + H2 (g)
2. Double Displacement Reactions that
Produce a Gas
2. Hydrogen sulfide gas – sulfides react with acids
to form hydrogen sulfide gas
Example:
K2S(aq) + 2HCl(aq)  2KCl(aq) + H2S (g)
2. Double Displacement Reactions that
Produce a Gas
3. Acids such as sulfuric and carbonic acids are so
strong that upon formation dissociate to form water and
the corresponding gas (sulfur dioxide and carbon
dioxide)
Example:
Na2SO3(aq) + 2HCl(aq)  2NaCl(aq) + H2SO3(aq)
Breaksdown
H20(l) + SO2(g)
Therefore, the new equation (NET EQUATION)
becomes:
Na2SO3(aq) + 2HCl(aq)  2NaCl(aq) + H20(l) + SO2(g)
2. Double Displacement Reactions that
Produce a Gas
Example:
Na2CO3(aq) + HCl(aq)  2NaCl(aq) + H2CO3(aq)
Breaksdown
H20(l) + CO2(g)
Therefore, the new equation (NET EQUATION)
becomes:
Na2CO3(aq) + HCl(aq)  2NaCl(aq) + H20(l) + CO2(g)
2. Double Displacement Reactions that
Produce a Gas
4. Ammonia gas – ammonia salts react with bases
Example:
NH4Cl(aq) + NaOH(aq)  NaCl(aq) + NH3(aq) + H2O(l)
Ammonia gas is very soluble in water. You can
detect it easily however by it’s sharp, pungent
smell.
2. Double Displacement Reactions that
Produce Water
Reactions that produce water are
Neutralization Reactions
a) Acid and base reactions
H2SO4(aq) + 2NaOH(aq)  Na2SO4(s) + 2 H2O(aq)
b) Metal oxide (are bases – recall from gr.10)
reacting with an acid neutralizes the reaction
2HNO3(aq) + MgO(s)  Mg(NO3)2(aq) + H2O(l)
a) Non-metal oxide (are acids – recall from gr.10)
reacting with a base neutralizes the reaction
2LiOH(aq) + CO2(aq)  Li2CO3(aq) + H2O(aq)
Representing Aqueous Ionic Reactions
with Net Ionic Equations
 A double displacement ionic reaction can
be represented by breaking the compounds
into their ionic form only if they are soluble in
water (aqueous).
 All soluble ionic compounds are written as
dissociated ions. (anything aqueous
dissociates)
 Must still be balanced (check the charges
AND the coefficients!)
Steps for Writing Total and Net Ionic
Equations
1. Write the balanced chemical equation with states
for the reaction (Use the solubility table and rules).
2. Write the total ionic equation by showing all of the
soluble compounds as ions.
3. Determine the spectator ions and cancel them out.
4. Write the net ionic equation.
**Note: can be done for both single and double
displacement reactions
Example
silver nitrate + sodium chloride  sodium nitrate + silver chloride
Balanced Equation:
AgNO3(aq) + NaCl(aq) NaNO3(aq) + AgCl(s)
Non-soluble
thus doesn’t
break apart in
sol’n
Total Ionic Equation:
Ag+(aq) + NO3-(aq) + Na+(aq) + Cl-(aq)  NO3-(aq) + Na+(aq) + AgCl(s)
At this point we can cancel out all of the ions that are the same on both sides of
the equation. These ions are called SPECTATOR IONS.
Example
Ag+(aq) + NO3-(aq) + Na+(aq) + Cl-(aq)  NO3-(aq) + Na+(aq) + AgCl(s)
The spectator ions for this example are:
NO3-(aq) and Na+(aq)
When the spectator ions are eliminated, the equation
becomes:
Ag+(aq) + Cl-(aq)  AgCl(s)
This is known as the NET IONIC EQUATION!
 Example: Identify the spectator ions and
the net ionic equation for the reaction of
solutions of lead (II) nitrate and potassium
iodide.
pg 427 # 1; pg 428 # 1, 3, 4, 5, 6, 9