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Chemical Reactions II:
Reactions of Ions in Solution1
PRE-LAB ASSIGNMENTS:
To be assigned by your lab instructor.
STUDENT LEARNING OUTCOMES:




Learn how to classify chemical reactions by driving force (precipitation, neutralization,
oxidation-reduction).
Learn how to observe precipitation, neutralization, and oxidation-reduction reactions.
Learn how to write ionic and net ionic equations for double displacement reactions.
Learn that properties of ionic compounds are principally the properties of the ions that
they contain.
EXPERIMENTAL GOALS:
The goal of this experiment is to observe a series of precipitation and neutralization reactions
(double displacement reactions) and deduce some of the properties of the ions involved in those
reactions. You will also observe a series of oxidation-reduction reactions, and draw some
conclusions about them.
INTRODUCTION:
In this lab, you will examine properties and reactions of ionic compounds in aqueous
solution. The focus will be on double displacement (precipitation and neutralization)
reactions, but some attention also will be given to oxidation-reduction (or redox) reactions.
From this experiment, you will learn that many of the properties of an ionic compound are due to
the properties of the ions that make up the compound.
Ionic compounds consist of a positive ion (cation) and a negative ion (anion). Some ions
are monatomic, consisting of a single atom with a positive or negative charge (Na+, Ca2+, Cl-,
O2-, etc.), and some are polyatomic, consisting of a cluster of atoms having an overall positive or
negative charge (NH4+, NO3-, CO32-, SO42-, PO43-, etc.). Each ion has its own characteristic color
and other properties. Many ions are colorless in aqueous solutions, or appear to be white if they
are part of a dry salt. Other ions have a definite color either in solution or as a dry salt, although
the color may change when the salt is dissolved in water. (For example, some copper salts are
green in the solid form, but blue when dissolved in water. The color of an ion may also be
affected by the concentration of the solution.) If a compound consists of a red cation, for
Adapted from “Experiments in General Chemistry, Book 1,” 2nd ed., July 1993, Calvin College. Used with verbal
permission of Larry Louters, Chemistry Department Chair, given August 8, 2008.
1
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example, plus a colorless anion, the compound will usually be red. If both cation and anion are
colorless, the compound will be colorless in solution, or appear white as a solid. If both are
colored, the compound will be a mixture or blending of the two colors. The color of a salt,
therefore, usually depends on the colors of the ions composing it. However, there are situations
when a cation and anion which are usually colorless form a colored solid salt when they are
combined. This is a result of the intimate contact between ions in the solid.
In the Reactions I experiment, we used a classification scheme for chemical reactions
which is organized by counting the number of reactants and products, and examining the way
that groups are exchanged between reactants and products. Another classification scheme for
chemical reactions considers instead the driving force for chemical reactions: precipitation,
neutralization, and oxidation-reduction. This scheme provides us with another way of looking at
chemical reactions, and gives us an additional understanding of these reactions.
Classes of Chemical Reactions:
Classification by Driving Force
I. PRECIPITATION REACTIONS.
A precipitation reaction occurs when two ionic compounds react in an aqueous solution
to produce a precipitate, an insoluble substance which falls out of the solution. These reactions
are a type of double-displacement or metathesis reaction, because the cations and anions of the
reactants “change partners” in the products:
AB + CD  AD + CB
The cations and anions have the same charges on both sides of the reactions, and the formulas of
the products of the reaction can therefore be predicted using the rules for writing the formulas of
ionic compounds.
The formation of the insoluble precipitate is the driving force for these reactions: the
attractions between the cation and the anion in the precipitate are stronger than their attraction
for the solvent molecules. The other ions stay in solution as spectator ions, and do not
participate in the overall chemical change.
Just because we can write an equation for a precipitation reaction does not mean that the
reaction will actually take place. In order to predict whether a precipitation reaction occurs, we
use the solubility rules in Table 1 to determine whether any of the potential products of the
reaction are insoluble. If one of the products of the reaction is insoluble, the reaction occurs, and
must be balanced appropriately. (If one of the products is a nonelectrolyte like water, a weak
electrolyte, or a gas, the reaction is a neutralization reaction, which will be discussed in the next
section.) If no insoluble product (or weak electrolyte, gas, or nonelectrolyte) forms, then no
reaction (NR) takes place, and all of the ions remain in solution as spectator ions.
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Table 1. Solubility Rules for Ionic Compounds in Aqueous Solution
IONS
SOLUBLE COMPOUNDS
INSOLUBLE COMPOUNDS
Salts that are Mostly Soluble
Group I cations
All
None
NH4+
All
None
Cl-, I-, Br-
Most
Ag+, Hg22+, Pb2+, Cu+
F-
Most
Mg2+, Ca2+, Sr2+, Ba2+, Pb2+
NO3-, ClO3-, ClO4-,
C2H3O2- (CH3COO-)
All
None
SO42-
Most
Sr2+, Ba2+, Pb2+, Hg22+
CO32-, PO43oxalates (C2O42-)
chromates (CrO42-)
Group I cations, NH4+
Most
S2-
Group I & II cations, NH4+
Most
OH-, O2-
Group I, Sr2+, Ba2+, Ra2+, NH4+
Most
Salts that are Mostly Insoluble
For example, if we mix solutions of NaNO3 and KI, all we get is a mixture of dissolved
ions; no net reaction takes place because none of the potential products are insoluble in water:
NaCl(aq) + KI(aq)  KCl(aq) + NaI(aq): NR
(All of the ions in this reaction are spectator ions.) On the other hand, if we mix Pb(NO3)2 and
KI, a reaction does take place, producing a solid precipitate of PbI2:
Pb(NO3)2(aq) + 2KI(aq)  2KNO3(aq) + PbI2(s)
There are three types of equations which are used to represent precipitation reactions. In
the overall reaction, all of the species are written in their usual ionic formulas (e.g., sodium
chloride is written as NaCl). In the complete ionic equation, all soluble strong electrolytes
(soluble ionic compounds, and strong acids) are dissociated into their separate cations and anions
(e.g., NaCl is written as Na+ and Cl-), while insoluble precipitates, weak acids, and molecular
compounds are left intact. In the net ionic equation, ions that are the same on both sides of the
equation are canceled out; these ions are called spectator ions, because they do not participate in
the overall chemical reaction that is taking place.
For example, consider the reaction between an aqueous solution of sodium chloride and
silver nitrate. One of the products of the reaction is insoluble silver chloride, so the reaction does
take place:
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Overall reaction in aqueous solution:
NaCl(aq) + AgNO3(aq)  AgCl(s) + NaNO3(aq)
According to the solubility rules, NaCl, AgNO3, and NaNO3 are soluble in water, so in the
complete ionic equation, they are dissociated into their ions. However, AgCl is insoluble, and is
left intact:
Complete ionic equation: [phase labels omitted for clarity]
Na+ + Cl+ + Ag+ + NO3-  AgCl(s) + Na+ + NO3Spectator ions: Na+ and NO3-.
The spectator ions in the reaction are Na+ and NO3-, since they are the same on both sides of the
equation. The Ag+ and Cl- are different, because they are dissociated on the left side, but in an
insoluble precipitate on the right side. In the net ionic equation, Na+ and NO3- are canceled out,
which emphasizes the actual chemical change that takes place: the formation of insoluble silver
chloride:
Net ionic equation:
Ag+(aq) + Cl-(aq)  AgCl(s)
Similarly, for the reaction of lead(II) nitrate and potassium iodide:
Overall reaction in aqueous solution:
Pb(NO3)2(aq) + 2KI(aq)  2KNO3(aq) + PbI2(s)
Complete ionic equation: [phase labels omitted for clarity]
Pb2+ + 2NO3- + 2K+ + 2I-  PbI2(s) + 2K+ + 2NO3Spectator ions: K+ and NO3-.
Net ionic equation:
Pb2+(aq) + 2I-(aq)  PbI2(s)
Notice that the complete ionic equations and net ionic equations must be balanced, with respect
to the number of atoms, and with respect to the total charge. It is important to remember that the
cations and anions have the same charges on both sides of the reaction: precipitation reaction do
not involve changes in oxidation number (see section III).
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II. NEUTRALIZATION REACTIONS.
Acid-base reactions or neutralization reactions occur when an acid and base react to
form water, a weak electrolyte, or a gas. Like precipitation reactions, these reactions are a type
of double-displacement reaction.
For a strong acid (HA2) and base (MOH), this can be written as
HA(aq) + MOH(aq)  MA(aq) + H2O(l)
and the ionic equation:
H+ + A- + M+ + OH-  M+ + A- + H2O(l)
Canceling the spectator ions M+ and A- gives the net ionic equation:
H+(aq) + OH-(aq)  H2O(l)
We will see the same net ionic equation for the reaction of any strong acid with any strong base.
The driving force for this reaction is the formation of a very stable water molecule. When one of
the species is a weak acid or base, one of the products is usually still water, but the net ionic
equation will be slightly different, since weak acids and bases do not dissociate:
HF(aq) + NaOH(aq) → NaF(aq) + H2O(l)
Net ionic equation: HF + OH- → F- + H2O(l)
We’ve already seen one neutralization reaction: the reaction between acetic acid and
sodium hydroxide in the Titration of Vinegar lab is a neutralization:
HC2H3O2(aq) + NaOH(aq)  NaC2H3O2(aq) + H2O(l)
Some acid-base reactions result in the formation of gases, either directly or in the
decomposition of an unstable intermediate. This removes ions from the solution, driving the
reaction to occur. For example, carbonic acid, H2CO3, is unstable, and falls apart to give water
and carbon dioxide gas, so any time carbonic acid is produced in a neutralization reaction, the
reaction takes place because of the formation of the carbon dioxide:
2HCl(aq) + K2CO3(aq)  2KCl(aq) + H2CO3(aq)
2HCl(aq) + K2CO3(aq)  2KCl(aq) + H2O(l) + CO2(g)
Other gases that can be produced in neutralization reactions include H2S(g), SO2(g) (from the
decomposition of H2SO3), HCN(g), and NH3(g) (from the dissociation of NH4OH).
2
Sometimes chemistry can be funny. This isn’t one of those times.
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Neutralization reactions can also result in the formation of weak electrolytes, which also do not
dissociate. Usually, these will be weak acids3:
HCl(aq) + NaF(aq  HF(aq) + NaCl(aq)
In the reaction classification scheme from the Reactions I lab, precipitation reactions and
neutralization reactions are types of double-displacement or metathesis reactions. In these
reactions, the cations and anions in two ionic compounds “change partners”:
AB + CD  AD + CB
In order for a double-displacement to occur, one of four things must occur, otherwise, there is no
reaction (NR):
1. In precipitation reactions, an insoluble ionic compound forms (check the solubility
rules to see if there is a precipitate).
2. In neutralization (acid-base) reactions, an acid and a base form a salt and water, a gas
(CO2, SO2, H2S, HCN, or NH3), or a weak electrolyte (a weak acid).
III. OXIDATION-REDUCTION (REDOX) REACTIONS
An oxidation-reduction (redox) reaction is a process in which electrons are transferred
from one substance to another. Oxidation is the loss of electrons, and reduction is the gain of
electrons.4
For example, in the reaction between elemental iron (Fe) and copper(II) ions (Cu2+), two
electrons are transferred from Fe to Cu2+, producing Fe2+ and elemental Cu. These processes
may be represented by half-reactions that show the separate oxidation and reduction processes:
Oxidation:
Fe  Fe2+ + 2e-
Reduction:
Cu2+ + 2e-  Cu
Overall:
Fe + Cu2+  Fe2+ + Cu
Notice that when the oxidation and reduction half-reactions are added together, the electrons
cancel out.
3
Remember, the six strong acids are HCl, HBr, HI, HNO3, HClO4, and H2SO4. Any other acid is (for now)
considered to be a weak acid.
4
A common mnemonic device for this is LEO GER (a lion growling): Loss of Electrons is Oxidation, Gain of
Electrons is Reduction.
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oxidation
Fe(s) + Cu2+(aq)  Fe2+(aq) + Cu(s)
reduction
The oxidizing agent is the substance that causes oxidation to occur by accepting
electrons. The oxidizing agent itself becomes reduced. The reducing agent is the substance
that causes reduction to occur by losing electrons. The reducing agent itself becomes oxidized.
In general, metals tend to act as reducing agents (lose electrons), and nonmetals tend to act as
oxidizing agents (gain electrons). In the previous example, Fe is the reducing agent, and Cu 2+ is
the oxidizing agent.
We can keep track of whether a reaction is a redox reaction, and what specifically is
being oxidized or reduced, by using oxidation numbers (aka oxidation state). The rules for
assigning oxidation numbers are listed in Table 2. (Note that oxidation number is not the same
thing as charge!!!!!)
Table 2. Rules for Assigning Oxidation Numbers
These rules are hierarchical. If any two rules conflict,
follow the rule that is higher on the list.
1. The oxidation number of an atom in its elemental state is zero.
2. The oxidation number of a monatomic ion is the same as its charge.
3. The sum of the oxidation numbers is 0 for a neutral compound and is equal to
the net charge for a polyatomic ion.
4. In their compounds, metals have positive oxidation numbers:
a. In compounds, Group 1A metals always have oxidation numbers of +1.
b. In compounds, Group 2A metals always have oxidation numbers of +2.
5. In their compounds, nonmetals are assigned oxidation states according to the
table below. Entries at the top of the table take precedence over entries at the
bottom of the table.
a. Fluorine
-1
b. Hydrogen
+1
(-1 when bonded with a metal)
c. Oxygen
-2
(-1 in peroxides, O22-)
d. Group 7A
-1
(except in compounds with O)
e. Group 6A
-2
(except in compounds with O)
f. Group 5A
-3
(except in compounds with O)
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Examples:
Na
Na is zero because it is an atom in its elemental state.
H2
H is zero because it is an atom in its elemental state.
Cl2
Cl is zero because it is an atom in its elemental state.
+
Na
Na is +1 because it is a monatomic ion, which has the same oxidation number as its
charge.
Ca2+
Ca is +2 because it is a monatomic ion.
2-
O
O is -2 because it is a monatomic ion.
Cl-
Cl is -1 because it is a monatomic ion.
H2SO4
H is +1, O is -2, so for the oxidation numbers to add up to 0, S must be +6.
ClO4-
O is -2, so for the oxidation numbers to add up to -1 (the charge on the polyatomic
ion), Cl must be +7.
ClO2-
O is -2, so for the oxidation numbers to add up to -1 (the charge on the polyatomic
ion), Cl must be +3.
ClO3
O is -2, so for the oxidation numbers to add up to 0 (the charge on the molecule), Cl
must be +6.
HCl
H is +1, so Cl must be -1.
CaH2
Ca is +2, so H must be -1. (H is bonded to a metal in this case.)
H2O
H is +1, and O is -2.
H2O2
H is +1, and O is -1 since this is a peroxide. (This can be seen if you try to apply
rules 5b and 5c: with H = +1 and O = -2, the oxidation numbers add up to -2, which
is not the charge on H2O2. Since rule 5b takes priority over rule 5c, H must be +1,
and since rule 3 takes priority over rule 5c, this forces O to be -1 for the oxidation
numbers to add up to 0.)
NaCl
Na is +1 and Cl is -1. (This is an ionic compound, so the oxidation numbers of
monatomic ions are the same as their charges.)
Cl2O
O is -2, which forces Cl to be +1 for the oxidation numbers to add up to 0.
HOBr
H is +1 and O is -2, which forces Br to be +1.
SO2
O is -2, which makes S +4.
SO42-
O is -2, which makes S +6 for the oxidation numbers to add up to -2.
NH3
H is +1, which makes N -3.
NO2
NO3
O is -2, so N must be +4 for the oxidation numbers to add up to 0.
-
NaNO3
O is -2, so N must be +5 for the oxidation numbers to add up to -1.
Na is +1, O is -2, so N must be +5 for the oxidation numbers to add up to 0. (Notice
that the oxidation number on N is the same as for the previous example, since this is
the same polyatomic ion.)
K2Cr2O7 K is +1, and O is -2; each Cr must have an oxidation number of +6.
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The oxidation number concept gives us another way to define redox reactions: A redox reaction
is one in which the oxidation numbers of species change. Oxidation is an increase in oxidation
number, and reduction is a decrease in oxidation number. Any reaction in which oxidation
numbers change is a redox reaction. If no oxidation numbers change, the reaction is not a redox
reaction.
S + O2  SO2
0
0
0  +4
0  -2
S:
O:
+4
-2
oxidized
reduced
H2(g) + Cl2(g)  2HCl(g)
0  +1
0  -1
H:
Cl:
oxidized
reduced
MnO4-(aq) + 5Fe2+(aq) + 8H+(aq)  Mn2+(aq) + 5Fe3+(aq) + 4H2O
Fe: +2  +3
Mn: +7  +2
oxidized
reduced
Na2SO4(aq) + Pb(NO3)2(aq)  2NaNO3(aq) + PbSO4(s)
Na:
Pb:
S:
N:
O:
+1  +1
+2  +2
+6  +6
+5  +5
-2  -2
None of the oxidation
numbers change, so this
is not a redox reaction
The single displacement reactions discussed in Reactions I are examples of redox
reactions:
A + BC  AC + B
The more active metal A is oxidized (it is the reducing agent), while the less active metal B is
reduced (it is the oxidizing agent). The species C, usually either a monatomic anion or a
polyatomic anion, is a spectator ion in the process.
Double displacement reactions — whether they are precipitation or neutralization
reactions — are never redox reactions, because the cations and anions are simply “changing
partners,” and their charges do not change.
Combustion reactions are also examples of redox reactions; the carbon atoms in the
hydrocarbons are oxidized to the +4 state in carbon dioxide, while molecular oxygen is reduced.
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Combination and decomposition reactions may be redox reactions, although not all of
them are: check to see if the oxidation numbers change to determine whether or not the reaction
is a redox reaction.
Simple redox reactions can be balanced by inspection, but it is important to remember
that not only do the atoms have to be balanced, but the electrons that are being transferred need
to be balanced as well. This means that complex redox reactions in many cases cannot be
balanced by inspection, and a more complex system is needed to ensure that everything is
balanced properly. (There are two commonly used methods for balancing redox reactions: the
oxidation number method, and the half-reaction method, but these will not be discussed here.)
Redox reactions are important in a number of other applications:

Bleaching uses redox reactions to decolorize or lighten colored materials such as hair,
clothes, paper, etc.

Batteries use spontaneous redox reactions which occur in separated compartments; the
electrons that move between the oxidation compartment (anode) and the reduction
compartment (cathode) can be used to deliver electrical energy.

Metallurgy is the processes involved in extracting and purifying metals from their ores.
Roasting ores with carbon (coke) carries away oxygen in the form of CO or CO2, yielding
the pure metal:
Fe2O3(s) + 3C(s) → 2Fe(s) + 3CO(g)

Corrosion is the deterioration of a metal by oxidation, such as the rusting of iron in moist
air.
4Fe(s) + 3O2(g) + H2O  2Fe2O3•H2O(s)
PROCEDURE:
A. Observing Ionic Compounds in Solution.
All the salts listed in Part A of the report sheet are water soluble. Solutions of these salts have
been prepared, and are on display in the laboratory. Observe the color of each solution, if any,
and by deduction determine if the color (for colored compounds) is due to the cation, the anion,
or both. (For example, the NaCl solution is colorless, so if you see some other ionic solution
containing Na+ that has a color, the color is not due to the presence of Na+.) Record the formula
and charge for the cation and anion in each of the compounds (e.g., Na+), and its color.
B. Precipitation Reactions, Part I.
In part A, all the compounds were solutions of soluble salts. All ions have some salts (that is,
combinations with other ions) that are soluble. Nearly all ions also have some salts that are
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insoluble (although, as you will see, there are some exceptions). The solubility of a given salt,
therefore, depends on both ions. In the two exercises below you will produce precipitates by
combining ions. With the knowledge of the colors of ions you have gained in part A, and by
observing the colors of the precipitates and solutions in Part B, you should be able to determine
which ions have precipitated and which have remained in solution.
For this part, use solutions in dropping bottles. The centrifuge may be used to aid in the settling
of the precipitate. Its use will be explained by the instructor.
1. To 5 drops of silver nitrate solution in a test tube, add 5 drops of copper(II) chloride solution
and shake vigorously. Let settle, or centrifuge. Note the colors of the precipitate and the
resulting solution. From part A deduce which ions have gone to the precipitate and which
have remained in solution. Fill in the required information on the report sheet. Discard the
waste from this procedure in the “B” waste bottle.
2. Repeat, using 10 drops of copper(II) sulfate solution and 5 drops of barium chloride solution.
If you are uncertain about your observations, repeat the test with 15 drops CuSO4 and 5 drops
BaCl2. Discard the waste from this procedure in the “B” waste bottle.
C. Precipitation Reactions, Part II.
You now know that certain combinations of ions are soluble (part A), and that at least two
combinations are insoluble (part B). On the report sheet are listed pairs of solutions. Using the
information obtained in Parts A and B, and some logical thinking, try to predict the formula of
the precipitate that forms (if any), the color of the solution, and the color of the precipitate. If
Parts A and B provide you with no information with which to make a prediction, you should
write “no prediction possible.”
Check your predictions by mixing together about 5 drops of each solution in a test tube. If the
results do not confirm your predictions, experiment a little by trying the reaction with different
relative amounts of the two solutions (for example, 3 drops of one with 10 of the other, or vice
versa). Record your observations of what happened in the Observation column on the report
sheet. Dispose of the waste in as directed on the data sheet.
D. Neutralization Reactions.
1. In a test tube, combine ~2 cm of a solution of 2.0 M HCl with ~2 cm of a solution of 2.0 M
NaOH. Record the chemical changes you observe during the reaction on the data sheet.
2. In a test tube, combine ~2 cm of a solution of 2.0 M HCl with a pinch of sodium bicarbonate,
NaHCO3 (about enough to fill the tip of a spatula). Record the chemical changes you
observe during the reaction on the data sheet.
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E. Oxidation-Reduction Reactions.
1. Place a pinch of Cu metal powder into a small test tube and heat it over a Bunsen burner.
When you are finished, put the used test tube in the waste beaker near the Cu metal
powder and solid Cu(NO3)2.
2. Place a few crystals of copper(II) nitrate in a test tube and heat strongly. Observe
carefully and note any gases produced or new solids formed. Can you identify the gas?
Put the used test tube in the waste beaker near the Cu metal powder and solid Cu(NO3)2.
3. Place a small piece of zinc into a test tube containing 2 mL of 6 M HCl. Test for H2 gas
by holding a burning splint near the mouth of the test tube; a popping sound indicates the
presence of H2. If the test is negative, heat the test tube gently to increase the production
of H2, which is indicated by increased bubble formation, and retest for the presence of H2.
Put the solution in the solution waste container labeled with a “C”. After rinsing in tap
water, dispose of any residual zinc in the waste basket.
4. Place an inverted test tube, filled with water, into a 100-mL beaker of water. Your
instructor will demonstrate this procedure. Now place a piece of calcium metal in the
water and collect the gas that is evolved in the inverted test tube. Bring the test tube to a
lighted Bunsen burner, and test it with a burning wooden splint.
5. Place a piece of copper wire into 2 mL of concentrated HCl; carefully observe for signs
of reaction. Place the same piece of copper wire into 2 mL of concentrated HNO3;
carefully observe for signs of reaction. Put the solutions in the solution waste container
labeled with a “C”.
6. Place a piece of copper wire into a solution of mercury(II) chloride, HgCl2. Wait a few
minutes; taking care not to touch the end of the wire with your hands, wipe off the
wire with a paper towel. Report any changes that you have seen on the surface of the
wire before or after the procedure. Place the Hg-contaminated materials in the beaker
labeled for that purpose. Wash your hands before proceeding to the next experiment.
7. Place an iron nail into copper(II) sulfate solution. Observe any color changes that occur.
The copper coated nail can be disposed of in the waste basket, or kept as a souvenir.
8. Place copper wire into an iron(III) chloride solution. Observe what happens. Rinse the
copper wire and return it to beaker from which it came.
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LAB REPORT
Chemical Reactions II: Reactions of Ions in Solution
Name ________________________________
Date _________
Partner ________________________________
Section _________
Report Grade ______
A. Observing Ionic Compounds in Solution.
Record the color of the following solutions and ions.
Salt
Sodium chloride,
NaCl
Copper(II) chloride,
CuCl2
Cobalt(II) Nitrate,
Co(NO3)2
Silver nitrate,
AgNO3
Potassium nitrate,
KNO3
Copper(II) nitrate,
Cu(NO3)2
Potassium chromate,
K2CrO4
Lead nitrate,
Pb(NO3)2
Potassium iodide,
KI
Barium chloride,
BaCl2
Potassium sulfate,
K2SO4
Zinc nitrate,
Zn(NO3)2
Copper(II) sulfate,
CuSO4
Nickel(II) nitrate,
Ni(NO3)2
Iron(III) nitrate,
Fe(NO3)3
Chromium(III) nitrate,
Cr(NO3)3
Manganese(II) nitrate,
Mn(NO3)2
Potassium
permanganate, KMnO4
Sodium hydrogen
sulfite, NaHSO3
Color of Solution
Cation & Its Color
Anion & Its Color
Colorless
Na+, colorless
Cl-, colorless
pale pink
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B. Precipitation Reactions, Part I.
1. Silver nitrate and copper(II) chloride
The color of the precipitate is __________, and it has the following formula _____________.
The color of the solution is __________, and the ions in the solution are ________________.
Overall equation for the reaction (include phase labels):
Complete ionic equation (omit phase labels):
Net ionic equation (include phase labels):
2. Copper(II) sulfate and barium chloride
The color of the precipitate is __________, and it has the following formula _____________.
The color of the solution is __________, and the ions in the solution are ________________.
Overall equation for the reaction (include phase labels):
Complete ionic equation (omit phase labels):
Net ionic equation (include phase labels):
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C. Precipitation Reactions, Part II.
Pair
KNO3 and
AgNO3
(Waste B)
CoCl2 and
AgNO3
(Waste B)
KNO3 and
CuCl2
(Waste C)
AgNO3 and
NaCl
(Waste B)
K2SO4 and
Cu(NO3)2
(Waste C)
Pb(NO3)2
and
K2CrO4
(Waste B)
Pb(NO3)2
and KI
(Waste B)
Predicted
Formula of
Precipitate
Predicted Color
of Precipitate
Predicted
Color of
Solution
Observations
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Questions on Parts A-C. Ionic Compounds and Precipitation Reactions.
1. List all the colored cations that were encountered in the entire experiment.
2. List all the colorless cations.
3. Draw a general conclusion about the place of colored cations in the periodic table.
4. At least one of the cations in this experiment is an apparent exception to the rule. Which one
is it?
5. There are two colored anions in this experiment. Which are they, and do they follow the
same rule?
D. Neutralization Reactions.
1. Write the molecular, ionic, and net ionic equations for the reaction between HCl and NaOH.
2. What evidence did you observe to indicate there was a reaction between HCl and NaOH?
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3. Write the molecular, ionic, and net ionic equations for the reaction between HCl and
NaHCO3.
4. What evidence did you observe to indicate there was a reaction between HCl and NaHCO3?
E. Oxidation-Reduction Reactions.
1. What physical evidence was there that a chemical reaction occurred when the copper metal
was heated? Write a balanced equation for the reaction.
2. Describe the results of heating copper nitrate, Cu(NO3)2. Can you identify any of the
products? (Recall the reactions of Cu with nitric acid.) Write a balanced equation for the
reaction.
3. Describe the results of adding zinc to HCl. Write the equation for this reaction.
4. Write an equation for the reaction of calcium metal with water. What was the appearance of
the products? Is the solution acidic or basic? How do you know?
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5. Does copper wire react with HCl? What part of the HNO3 does the Cu react with?
6. What do you observe when you place the copper wire in the solution of HgCl2? Write the
reaction (if any) for what happens.
7. What do you observe when you place a sample of iron into the CuSO4 solution? Write the
reaction (if any) for what happens.
8. What do you observe when you place the copper wire in the solution of FeCl 3? Write the
reaction (if any) for what happens.
9. List the elements that were studied in these reactions (Cu, Hg, Fe, H, and Zn) in order
decreasing chemical activity: an element listed on the left will replace those listed on the
right, and not vice versa. Compare to the activity series in the procedure. [If you cannot
decide the order for two specific elements, you may need to design and perform an additional
test to resolve the problem – check your procedure with your instructor before proceeding.]