Ch 7: Reactions in Aqueous Solutions / Types of Reactions
7.1 Predicting Whether a Reaction Will Occur
1. Driving forces – or changes – that “pull” a reaction toward products:
a. The formation of a solid (a “precipitate”)
b. The formation of a gas
c. The formation of water
d. The transfer of electrons (oxidation-reduction reactions)
2. Many types of reaction occur in water. Why?
a. Organisms are mostly composed of water
b. Many reactions readily occur in water.
c. Water is the most universal liquid (making it a “universal solvent”).
3. States of matter in reactions:
(aq) = aqueous (something is dissolved in water, is in solution, or is soluble)
(s) = a solid, or any precipitate formed in an aqueous reaction
(l) = liquid
(g) = gas
7.2 Reactions in Which a Solid Forms
1. A solid formed in an aqueous solution of ionic compounds is called a precipitate.
a. Ionic compounds, which consist of a metal bonded to a nonmetal, are collectively known as salts.
b. Solutions of dissolved salts consist of separated cations and anions. Chemists know that separated
ions are present when ionic solids dissolve in water because the solutions become excellent conductors of
electricity. Pure water does not conduct electricity. (See Fig 7.2 in text.)
c. When solutions of dissolved salts are mixed together, the process of solid formation is called
precipitation. A solid forms because some salt products are insoluble in water.
d. The reaction is known as a precipitation (or a double-displacement) reaction.
e. Example: when a solution of potassium chromate is mixed with a solution of barium nitrate a
yellow precipitate forms.
K2CrO4(aq) + Ba(NO3)2(aq)  yellow precipitate
f. What is the composition of the yellow precipitate? How does it form?
2. To answer the above, let’s look at the nature of ionic compounds dissolved in water.
a. Ionic compounds (a metal bonded to a nonmetal) dissolve in water producing separate ions.
b. In the example above, potassium chromate dissolves in water producing potassium ions and
chromate ions,
K2CrO4  2K+ + CrO42likewise, barium nitrate dissolves to produce barium ions and nitrate ions.
Ba(NO3)2  Ba2+ + 2NO31
c. A more accurate equation shows the ions present when the two solutions are mixed:
2K+(aq) + CrO42-(aq) + Ba2+(aq) + 2NO3-(aq)  products
d. So, what product will form?
1) Determine all possible combinations of products by making a table, as shown below. Put the
cations on one side and the anions across the top. Pair up each cation with each anion.
{
CrO42-
NO3-
K+
K2CrO4
KNO3
Ba2+
BaCrO4
Ba(NO3)2
}
cation-anion
combinations
cations
{
anions
Method 1
2) Discard the possible products that are the same as the reactants. This leaves KNO3 and
BaCrO4 as possible products. Which one, in fact, is the yellow solid?
3) Use the Solubility Rules handout (or Table 7.1 in text) to determine the solid product.
a) A soluble solid is one that dissolves readily in water
b) An insoluble or slightly soluble solid does not dissolve, or dissolves very little, in water.
c) According to solubility rules, BaCrO4 is insoluble in water, whereas KNO3 is soluble.
Therefore, the solid product is BaCrO4. KNO3 remains as dissolved ions in solution.
d) Write the balanced molecular equation showing the appropriate states of matter:
K2CrO4(aq) + Ba(NO3)2(aq)  2KNO3(aq) + BaCrO4(s)
3. A precipitation (or double-displacement) reaction can also be described in the general form below,
AB
+
CD

AD
+
CB
Method 2
K2CrO4(aq) + Ba(NO3)2(aq)  2KNO3(aq) + BaCrO4(s)
a. Note that the cations switch anion partners to form the possible products. The potassium ion is
paired with the nitrate anion and the barium cation is paired with the chromate anion to produce the
possible products.
b. Think of couples at a dance switching partners. The “aqueous” partners stay at the dance, whereas
the “solid” partners could be said to have left the dance together – they “precipitated” out!
Summary: Steps for predicting the precipitate product in aqueous reactions.
Step 1: Write reactants as ions.
Step 2: Exchange the anions of the dissolved compounds (salts) to determine possible products.
Step 3: Use solubility rules to determine if a solid forms.
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Questions:
1. Use the solubility rules to predict which of the following substances will be soluble or insoluble in
water.
a. PbS
___________
i. potassium sulfide
________________
b. Mg(OH)2
________________
j. cobalt(III) nitrate
___________
c. Na2SO4
________________
k. ammonium phosphate
___________
d. (NH4)2S
___________
l. cesium sulfite
___________
e. BaCO3
________________
m. aluminum nitride
___________
f. AlPO4
________________
n. potassium hydroxide
___________
g. PbCl2
________________
m. manganese(II) iodide
___________
h. Au(NO3)3
________________
n. iron(II) nitride
___________
2. Apply what you’ve learned to determine the product formed in the precipitation reaction below. Note:
if no precipitate is formed, there was no reaction. All you have is a mixture of dissolved ions.
KNO3(aq)
+
BaCl2(aq)

Use the solubility rules to determine the product. Write the balanced molecular equation and show states
of matter.
3. Identify the precipitate for the following reaction. Write the balanced molecular equation for the
reaction and show states of matter.
AgNO3(aq)
+
KCl(aq) 
_____________________________________________________________________________________
In the space below and on back, or on your own paper, answer questions 17, 19, and 21 on page 202 in
your text.
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