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QUALITATIVE CHEMICAL ANALYSIS LAB
INTRODUCTION:
Chemical reactions can be grouped into different classes, such as precipitation reactions, acidbase (or neutralization) reactions, gas-formation reactions, and oxidation-reduction (or redox)
reactions. There are several other classifications, too, such as combination reactions,
decomposition reactions, oxidation reactions, combustion reactions, and so on. Some reactions
may belong to more than one classification. For example, the burning of carbon in the form of
coal,
C (s) + O2 (g) → CO2 (g)
can be classified as a combustion reaction because it depicts the burning of carbon, and it can be
classified as a combination reaction because two reactants combine to give one product. It can
also be classified as an oxidation reaction because carbon is oxidized during the reaction. It is
also a redox reaction because while carbon is oxidized, there is a concurrent reduction of oxygen.
Qualitative analysis is the identification of an unknown substance without regard to the amount or
concentration, and, by knowing how substances react, we can often identify unknowns very
easily. In this lab, you will be given seven colorless solutions that you will be required to identify
by carrying out various reactions with them.
In order to understand the procedure, let us first consider examples of the different kinds of
reactions that we will be carrying out in this lab:
I
Precipitation Reactions
Ionic compounds that dissolve in water dissociate into their component ions which move
about freely in solution. When two solutions are mixed, a precipitate may form if the cation
from one solution forms an “insoluble” (sparingly soluble) salt with the anion from the other
solution. A precipitate is defined as an insoluble salt formed during a reaction. The way we
will predict if a precipitate will form is to use the solubility rules:
Solubility Rules
1. All common compounds of Group I (the alkali metals), and ammonium ions are soluble.
2. All nitrates, acetates, and chlorates are soluble.
3. All binary compounds of the halogens (other than F) with metals are soluble, except those of
Ag, Hg(I), and Pb. (Pb halides are soluble in hot water.)
4. All sulfates are soluble, except those of barium, strontium, calcium, lead, silver, and mercury
(I).
5. Carbonates, hydroxides, oxides, silicates, and phosphates are insoluble. Exceptions are given
in Rule 1.
6. Sulfides are insoluble except for calcium, barium, strontium, magnesium, sodium, potassium,
and ammonium.
What is meant by “No Reaction”? For example, what would happen if we mixed a solution of
sodium chloride with a solution of potassium nitrate? In the first place, we know that sodium
chloride and potassium nitrate are soluble salts because of the first two solubility rules. We could
expect an exchange reaction (double-displacement reaction) to occur, forming sodium nitrate and
potassium chloride.
NaCl (aq) + KNO3 (aq) → KCl + NaNO3
But, from the solubility rules, we know that sodium nitrate and potassium chloride are also
soluble salts, and no precipitate would form.
NaCl (aq) + KNO3 (aq) → KCl (aq) + NaNO3 (aq)
Hence, the net ionic equation would look like this:
Na+ (aq) + Cl– (aq) + K+ (aq) + NO3– (aq) → K+ (aq) + Cl– (aq) + Na+ (aq) + NO3– (aq)
In other words, we would conclude that there is no reaction because all of the ions are spectator
ions and would cancel out.
What is meant by a precipitation reaction? On the other hand, what would be the result of
mixing a solution of silver nitrate and sodium chloride? A double-displacement reaction would
result in the formation of a precipitate of silver chloride, as in rule 3 of the solubility rules.
AgNO3 (aq) + NaCl (aq) → AgCl (s) + NaNO3 (aq)
Note : This is the molecular equation
(white, “curdy” ppt)
Na+ and NO3– being spectator ions since they remain soluble , the net ionic equation would be:
Ag+ (aq) + Cl– (aq) → AgCl (s)
Note : This is the net ionic equation where the spectator ions are absent .
Hence, if we suspected an unknown solution to be silver nitrate, we could identify it by adding
sodium chloride and observing the characteristic white, curd-like precipitate.
An important fact to note about silver salts is that, in acidic solution, only the silver halides form
precipitates. This information gives us a good qualitative test for halides: If a precipitate is
obtained when dilute nitric acid and silver nitrate is added to an unknown solution containing a
simple salt (a salt containing one cation and one anion), then we can be sure that the anion is Cl–,
Br–, or I–. AgCl is a white, curd-like precipitate; AgBr is a pale-yellow, curdy precipitate; AgI is a
yellow, curdy precipitate. If dilute nitric acid is not added, many anions such as carbonate,
sulfate, sulfide, etc. will form a precipitate with the Ag+ from silver nitrate. These precipitates,
unlike the silver halides, are soluble in dilute nitric acid.
II Gas Formation Reactions
Test for carbonates
All carbonates give off carbon dioxide when a strong acid is added. An example is the reaction of
marble statuary with acid rain. Carbon dioxide, which is a colorless, odorless, acidic gas, may be
identified because it forms a white, milky precipitate of calcium carbonate when it is passed into
a solution of calcium hydroxide.
CaCO3 (s) + 2 HNO3 (aq) → Ca(NO3)2 (aq) + CO2 (g) + H2O (l)
(bubbles form)
CO2 (g) + Ca(OH)2 (aq) → CaCO3 (s) + H2O (l)
Test for sulfides
All sulfides give off hydrogen sulfide gas when mixed with a strong acid. H2S is an acidic gas
that is easily identified by its characteristic odor of rotten eggs. If H2S gas comes into contact
with lead acetate paper, the formation of lead(II) sulfide, PbS, turns the paper black. This
confirms the presence of sulfide in the original salt.
FeS (aq) + H2SO4 (aq) → FeSO4 (aq) + H2S (g)
(bubbles form)
Because H2S is a poisonous gas, we will identify the sulfide salt using a different qualitative
test:
The addition of a solution of copper (II) nitrate to a sulfide results in the formation of a black
precipitate of copper(II) sulfide.
K2S (aq) + Cu(NO3)2 (aq) → CuS (s) + 2KNO3 (aq)
(black ppt)
Note that this same test, the addition of a solution of copper (II) nitrate, Cu(NO3)2, to an iodide
solution results in the formation of a yellow precipitate of copper(I) iodide and red iodine
solution.
Cu(NO3)2 (aq) + 2NaI (aq) → I2 (aq) + CuI (s) + 2NaNO3 (aq)
(red solution)
(yellow ppt)
The formation of a yellow precipitate in a red solution makes the resulting mixture look brown.
This is a safe and effective test for iodide, but note that this is not a gas formation reaction.
Instead, it is an oxidation-reduction or “redox” reaction, because certain oxidation numbers
change during the reaction.
Test for sulfites
All sulfites give off sulfur dioxide gas when mixed with a strong acid.
Test for ammonium salts
All ammonium salts release ammonia gas when treated with a strong base. Ammonia has a
characteristic odor and is easily identified. It is also the only basic gas, and turns red litmus blue.
(NH4)2SO4 (s) + 2 KOH (aq) → 2 NH3 (g) + K2SO4 (aq) + 2 H2O (l)
III Single-Replacement Reactions or Displacement Reactions
Some metals are very reactive, some are not so reactive, and others are unreactive. For example,
sodium is so reactive with the components of air or with water that it has to be stored under
paraffin. Gold is a good example of a metal that is generally unreactive.
Magnesium reacts slowly with cold water but readily with steam:
Mg (s) + H2O (g) → H2 (g) + MgO (s)
Metals can be arranged in order of decreasing reactivity as shown in the table after this section.
The more reactive the metal, the higher it is placed in the table. Reactive metals, apart from
displacing hydrogen gas from acids and water or steam, have the ability to displace less reactive
metals from solutions of their salts. For example, if a strip of aluminum metal is added to a
solution of copper(II) sulfate, the aluminum will go into solution as the aluminum ion, while
displacing copper metal from the solution. The copper precipitate may be recognized by its
characteristic copper color, and the blue color of the copper sulfate solution will fade as the
copper ions are replaced by colorless aluminum ions.
2 Al (s) + 3 CuSO4 (aq) → Al2(SO4)3 (s) + 3 Cu (s)
The net ionic equation would be given by,
2 Al (s) + 3 Cu2+ (aq) → 2 Al3+ (aq) + 3 Cu (s)
As well as being a single-replacement reaction, this is an example of an oxidation-reaction or
redox reaction, as the oxidation numbers of the reactant species change during the reaction.
It is important to note that in the case of the solution of a lead salt, the finely divided lead
that is displaced by a more reactive metal may not look metallic but appear as a black
precipitate.
IV Oxidation-reduction or Redox Reactions
Two reactions that you will carry out in this experiment can be classified as redox reactions.
Oxidation is the loss of electrons while reduction is the gain of electrons. For example, in the
reaction
Fe (s) + Cu(NO3)2 (aq) → Cu (s) + Fe(NO3)2 (aq)
Iron loses electrons to go from a charge of 0 since it is a neutral metal, to a charge of +2.
Therefore iron is oxidized in this reaction. The copper, on the other hand, gains electrons as it
goes from a +2 cation on the reactant side to a neutral metal with a charge of 0 on the product
side. The copper is reduced.
In this lab, a redox reactions you will see is the reaction of copper(II) nitrate with an iodide. This
is both a redox and a single-replacement reaction in which a metal replaces the cation of a salt in
solution.
This lab will be divided into two parts, each part being performed during a different lab period.
During the first lab period, you will perform each of the tests described in the introduction and
note the observations for yourself. You will also practice writing the correct, balanced molecular
and net ionic equation for each reaction. It is important to take careful notes on each of the
reactions.
During the second lab period, you will be given three unknowns which must be identified using
the tests given in the lab introduction. It will be your responsibility to construct a series of
experiments to identify each. As you identify each unknown, an explanation of the experimental
process which lead you to your conclusion will be required.
PRE-LAB QUESTIONS
1. Write the molecular and net ionic equations for the reaction between copper(II) nitrate and
potassium iodide. Identify which metal is oxidized in the reaction and which is reduced. Show
your reasoning.
2. In an experiment, you are given a set of seven colorless aqueous solutions in glass vials
labeled with numbers. The unknowns are barium chloride, sodium sulfate, potassium iodide, lead
(II) nitrate, sodium carbonate, sodium sulfide, and ammonium chloride. Write the formula for
each of the unknowns.
3. Which unknowns would give a white precipitate with sodium sulfate solution? Write
molecular and net equations for all of them.
4. Suppose you had added silver nitrate solution to all of the test tubes, which of the unknowns
would have reacted? Write molecular and net ionic equations for all of them.
5. Suppose you had added a magnesium strip to all of the test tubes, which of the unknowns
would have reacted? Write balanced chemical equations for all of them.
Activity Series of the metals from Section III: single Replacement Reactions or Displacement
Reactions
Higher metals on the series will displace metals lower in the series from aqueous solutions of
their salts.
For example, if Zn metal is reacted with Mg(NO3)2 no reaction will occur because Mg is above
Zn in the series. However is Zn metal is reacted with Fe(NO3)2 then a redox reaction will occur,
producing solid Fe metal and aqueous Zn(NO3)2
Li
K
React vigorously with acidic solutions
Ba
to give H2
Sr
React vigorously with liquid water
to give H2
Ca
Na
Mg
Al
React with acids to give H2
Zn
React slowly with liquid water but
Cr
readily with steam to give H2
Fe
Cd
Co
Ni
Do not react with water
Sn
Pb
H2
Cu
Do not react with acids to give H2
Hg
Ag
Au
Do not react with water
PROCEDURE and DATA SHEET for Part I of the Experiment
You will be given a set of seven colorless aqueous solutions in vials labeled with their identity.
The samples are barium chloride, sodium sulfate, potassium iodide, lead(II) nitrate, sodium
carbonate, sodium sulfide, and ammonium chloride.
Refer to the Introduction section to familiarize yourself with the chemistry of the test for
each of the seven compounds. Perform all tests in the order specified in these procedures.
Classify each reaction that you observe as precipitation, gas formation, single replacement or
redox. A reaction may also have more than one classification.
Use the white porcelain, 12-well spot plate to mix drops of the samples and reagents. Use a
pipette to add drops of unknown to the well plate. Avoid cross-contamination of the unknowns by
mixing up your droppers.
*Note: The number of drops to be added is approximate. Add fewer or more drops if needed. You
have to add just enough drops to effect a change, making sure that you mix the solution well after
each addition. You can use the tip of a clean glass rod to mix the drops in each well.
Each identification test will be performed on all seven of the compounds so that you can observe
and record the results with the exception of the first test for a sulfide.
Test #1: It would be prudent to first identify the sodium sulfide using a “safe” test so that
we can eliminate it from our pool of unknowns. This will avoid generating poisonous
H2S gas when we add acid in subsequent tests. Add a few drops of Cu(NO3)2 to each
unknown in the well plate. Identify the unknown that forms a black precipitate. This
sample is sodium sulfide.
You will also see a brownish mixture form in one well, and you will see a pale blue
precipitate form with another sample
Obsdervations
______
Balanced molecular equation for the reaction of sodium sulfide and Cu(NO3)2:
Therefore net ionic equation:
Classification of this reaction:
Dispose of all the solutions from the well-plate into a waste beaker. Place the waste beaker
in the fume hood and obtain another waste beaker for the rest of the experiment. Rinse the
well plate well and dry with a paper towel.
Test #2: Identify the unknown that formed a brownish mixture with Cu(NO3)2. Add a few
drops of this sample to a well and add to it a drop of dilute nitric acid. Then add two or
three drops of silver nitrate.
Observations
______
Identify the compound and record your observations in the data table.
Balanced molecular equation
Therefore net ionic equation:
Classification of this reaction:
Test #3: Identify the sample that formed a pale blue precipitate with Cu(NO3)2. Remove
the stopper from the apparatus shown below and add about 20 drops of this sample to the
clean test tube. The Beral pipette and tubing is attached to the stopper. Squeeze the bulb
of the Beral pipette and dip the end of it into the vial of dilute HCl. Release the bulb,
drawing some HCl into the Beral pipette. Stopper the test tube tightly.
Beral pipette
containing
dilute HCl
Poly vinyl tubing
Stopper
Unknown solution
dip end of tubing into
saturated calcium hydrooxide
Collect about 10 drops of Ca(OH)2 solution in another test tube and dip the end of the tubing
into the calcium hydroxide. Keeping the end of the poly vinyl tubing dipped in the Ca(OH)2,
squeeze the bulb of the Beral pipette, releasing the HCl into the sample solution. Observe
what happens in both test tubes.
Observations
Balanced molecular equation for the reaction of dilute HCl and the sample:
Therefore net ionic equation:
Classification of this reaction:
Molecular equation for the reaction that occurred in the test tube containing Ca(OH)2:
Test #4A: Place a few drops of the remaining four unknowns in clean wells in the well
plate, and add a few drops of dilute HCl to each well. You should observe a change in
only one well.
Observations
Balanced molecular equation for the reaction of dilute HCl with the unknown:
Therefore net ionic equation:
Classification of this reaction:
Test #4B: To confirm this sample, add a few drops of it into a clean well, and stir it with
a magnesium strip.
Observation
Balanced molecular equation for the reaction of Mg with the sample:
Therefore net ionic equation:
Classification of this reaction:
Test #5: Place a few drops of the remaining three samples in clean wells in the well plate,
and add a few drops of NaOH solution to each well. Moisten three strips of red litmus
with distilled water, and hold a moistened end above each well so that any gas that results
from a reaction in the well will rise and make contact with the litmus. Do not let the
litmus touch the solution in the well, which has NaOH in it.
Observation with the litmus paper
Balanced molecular equation for the reaction of NaOH(aq) with the reacting sample:
Therefore net ionic equation:
Classification of this reaction:
Test #6: Place a few drops of the remaining two samples in clean wells in the well plate
and add a few drops of a solution of sodium sulfate into each sample. You should observe
a reaction with only one sample.
Observation
______
Balanced molecular equation for the reaction of sodium sulfate with the sample:
Therefore net ionic equation:
Classification of this reaction:
Test #7: Place a few drops of the remaining sample in a clean well and add a few drops
of a solution of barium chloride.
Observations
Balanced molecular equation for the reaction of barium chloride with the sample:
Therefore net ionic equation:
Classification of this reaction:
Test #8: Place a few drops of the lead (II) nitrate in a clean well and add a few drops of
the potassium iodide. Write your observation and the equations.
Observation
Balanced molecular equation for the reaction of lead (II) nitrate with potassium iodide:
Therefore net ionic equation:
Classification of this reaction: