Video Transcript
In this lesson, we will learn how
to identify aqueous negative ions based on their reactivity and the color and
solubility of their salts. In the laboratory, it’s useful to
know if we have substances that are dissolved in a sample of liquid water or not, in
other words whether the sample of water is pure. It may also be very useful to be
able to identify which substances are present in unknown solution.
Many substances dissolve in
water. In this lesson, we will focus on
substances called ionic salts. Many ionic salts dissolve easily in
water to make solutions. Ionic salts that dissolve easily in
water are therefore known as soluble salts. If we added a soluble ionic salt to
water and stirred well, we’d expect to get a solution of the ionic salt. Once the ionic salt is dissolved,
we would not be able to tell that the salt particles are in the solution. Very often, ionic salts have no
obvious color. This is particularly true for salts
of metals from group one and group two of the periodic table.
Many ionic salts also have no
obvious smell. And when they dissolve in small
amounts to make solutions, it seems that they have disappeared. However, ionic salts contain
positive ions and negative ions held together by strong ionic bonding in a crystal
lattice. The positive ions are known as
cations, and the negative ions are known as anions. In the case of the salt sodium
chloride, the cations are sodium ions, or Na+, and the anions are chloride ions, or
Cl−.
When these solid ionic crystals are
placed into water, the ions are surrounded by water molecules and they break free
from the crystal lattice. The ions become dissolved into a
solution, where they can move around freely. So, solutions of ionic salts will
contain cations and anions that are mobile in the solution. Very often, we wouldn’t know that
they’re there just by looking at the solution.
A sodium chloride solution will
contain positively charged sodium ions and negatively charged chloride ions. Both ions are colorless, so a
solution of sodium chloride looks just like a sample of pure water. In the laboratory, it may be very
important to know if these ions are present. We must assume that all unknown
substances in the lab are toxic or poisonous. So we must never taste unknown
samples. We need some simple chemical tests
to answer questions such as are there chloride ions in this solution.
Fortunately, there are a range of
reliable tests that can easily be performed in the lab with some test tubes and some
known test chemicals or reagents. These tests are quick and give an
obvious result. Simple observations that may be
noted during these tests might include color changes. This includes turning
colorless. Another observation we may notice
is the appearance of a solid from a mixture of solutions. This is known as a precipitate. We might notice gas bubbles or
fizzing, which is also known as effervescence, being produced from the reaction
mixture. The observations need to be unique
for the ions being tested so that the result does not cause confusion. It would not be a reliable test if
several different ions all produce the same result in a chemical test.
In this lesson, we’re focusing on
tests for anions, or negatively charged ions. We will see how the tests for
chloride, bromide, and iodide ions, collectively known as the halide ions, are
performed. We will also see the tests for
sulfate and sulfite ions. These are distinctly separate ions,
despite their similar names and chemical formulae. We will also see the tests for
carbonate and nitrate ions. We will see which observations
would be noted to confirm a positive test result for these ions.
We will begin with a chemical test
for carbonate ions. Carbonate ions are an example of a
base. If they are present in a solution,
they will react with acids to produce carbon dioxide gas. For example, sodium carbonate is a
soluble metal carbonate. When it’s dissolved into water, it
will release carbonate ions into the aqueous solution. When we add hydrochloric acid to
sodium carbonate, the salt sodium chloride and water and carbon dioxide are
produced. This is a neutralization
reaction. So if we add a few drops of dilute
hydrochloric acid to the carbonate solution, bubbles of carbon dioxide gas will be
observed. This observation may also be
described as fizzing or effervescence. We cannot see the carbon dioxide
produced as it’s colorless. It’s also odorless. The observation in this test is the
fizzing.
Of course, many other gases may
cause fizzing. So to prove that the gas is carbon
dioxide and that it originated from the carbonate ions, we bubble some of the gas
through a limewater solution. This may be done by collecting some
of the invisible gas from above the liquid in the test tube with a plastic
pipette. The air is squeezed out of the
pipette first and filled with carbon dioxide gas only. It is transferred immediately to
the limewater in a separate test tube and bubbled through it. Alternatively, if a lot of gas is
produced, a delivery tube may be used to bubble the gas across from the reaction
vessel.
Limewater contains calcium
hydroxide. Limewater reacts with carbon
dioxide gas to produce a white milky precipitate. This precipitate is calcium
carbonate, which is very insoluble. This additional step to the test
confirms that the gas is carbon dioxide and that the original solution therefore
contained carbonate ions.
Halide ions are found in salts
containing the negatively charged ions of the halogens. Although halogens may produce
colored solutions in water, the halide ions are all colorless. To test for halide ions, we must
first acidify the test solution with dilute nitric acid. This step is done first to remove
other ions that may be present, such as carbonate ions and hydroxide ions. If present, these ions will give a
positive result in the second part of the test, which would be misleading. It’s important to use nitric acid
here and not hydrochloric acid, as hydrochloric acid contains chloride ions. We don’t want to add the ions that
we’re actually testing for.
After the addition of a few drops
of dilute nitric acid, a few drops of silver nitrate solution is added. This is usually stored as a dilute
solution in a brown bottle as it’s light-sensitive. If halide ions are present, some
solids will appear from the solutions. These solids are called
precipitates, and they have distinct colors.
Chloride ions will form silver
chloride, which is an insoluble white precipitate. Aqueous silver ions react with
aqueous chloride ions to give silver chloride, which is a white solid. Bromide ions will form silver
bromide, which is a cream-colored precipitate. Aqueous silver ions react with
aqueous bromide ions to produce solid silver bromide. Iodide ions will form a yellow
precipitate of silver iodide. Aqueous silver ions react with
aqueous iodide ions to give silver iodide, which is a solid.
Note that fluoride ions cannot be
identified using this test, as silver fluoride is soluble and does not produce an
observable precipitate at all. The white, cream, and yellow
precipitates formed in these tests can be hard to distinguish from each other on
occasions. For example, silver chloride can
darken on exposure to light. A further test using ammonia
solution can be used to confirm which silver halide is causing the observed
precipitate. Silver chloride dissolves easily if
some dilute ammonia is added to the precipitate. Silver bromide only dissolves in
concentrated ammonia solution. Silver iodide is insoluble in
dilute and concentrated ammonia solutions. If solid halide salts are
available, concentrated sulfuric acid may be added to a small quantity of the solid
salt to produce interesting observations.
As the likely products of these
reactions are harmful or toxic, these tests should be done in a fume cupboard. If we took four test tubes
containing solid samples of sodium fluoride, sodium chloride, sodium bromide, and
sodium iodide and added a few drops of concentrated sulfuric acid to each tube in
turn, we would expect the following observations.
Upon warming slightly, the sodium
fluoride will produce corrosive fumes of hydrogen fluoride gas. This will etch the glass on a glass
rod with a drop of water on the end of it. The sodium chloride will produce
acidic fumes of hydrogen chloride gas. These fumes will appear as steamy
white fumes. This gas will turn damp blue litmus
to red. The sodium bromide will produce
orange fumes of bromine and also sulfur dioxide gas, which is an acidic gas. Sulfur dioxide gas will turn fresh
acidified potassium dichromate paper from an orange to a green color. The sodium iodide will produce a
range of products. Most noticeable are a purple vapor
caused by iodine and a smell of rotten eggs caused by the hydrogen sulfide gas. The iodine may condense as a brown
or black solid inside the mouth of the tube.
Sulfate ions in solutions can be
identified using a combination of hydrochloric acid added first, followed by barium
chloride solution. The acidification step is needed
first to remove other ions, such as carbonate ions, that will interfere with the
test by giving a false-positive result. Any carbonate ions are removed as
carbon dioxide gas.
After acidification with
hydrochloric acid, not sulfuric ’cause this contains sulfate ions, which we’re
testing for, a few drops of barium chloride are added. Barium chloride is a very soluble
barium salt. If sulfate ions are present, a
dense white precipitate will be produced. The precipitate is barium sulfate,
which is a very insoluble barium salt. Barium sulfate is so insoluble it
can be used in medical applications, where it is swallowed to provide contrast on
X-ray images. It does not poison the patient as
it is not absorbed into the body.
The reaction occurring in this test
is often represented by the simplified ionic equation where spectator ions have been
removed. Aqueous barium ions react with
aqueous sulfate ions to produce barium sulfate, which is a white solid.
Sulfite ions have the chemical
formula SO32−. The word sulfite is sometimes
spelled with a p and an h instead of an f. The sulfite ion is also sometimes
known as the sulfate one V or sulfate(IV) ion. This is easily confused with the
sulfate or sulfate(VI) ion, which has the formula SO42−. Sulfite salts and sulfite solutions
are less commonly encountered in the laboratory. They’re sometimes used as
preservatives in food products.
There is an interesting test to see
if they are present in solution. When hydrochloric acid is added to
a solution containing sulfites, sulfur dioxide gas is produced. Although sulfur dioxide gas is very
soluble in water, its presence can be confirmed by using fresh acidified potassium
dichromate paper. The damp paper will turn from an
orange color to a green color.
Alternatively, a drop of potassium
permanganate solution acidified with sulfuric acid may be held in the gas using the
tip of a glass rod. The purple solution containing
manganate(VII) ions will become colorless as manganese(II) sulfate is produced.
The presence of nitrate ions in a
solution can be confirmed by adding sodium hydroxide solution first. Then, small pieces of aluminum foil
or aluminum powder are added. The mixture is then heated
carefully, and ammonia gas will be produced. The ammonia gas can easily be
detected by placing a piece of damp red litmus paper above the mouth of the test
tube. It will immediately turn a blue
color as ammonia gas is less dense than air. The presence of ammonia gas
generated in this reaction proves that nitrate ions were present in the original
solution. As all common nitrate salts are
soluble, it’s not possible to use a simple test forming a precipitate for the
nitrate ion.
We will now look at a question to
test your understanding of testing for anions.
What is the net ionic equation for
the reaction between barium nitrate and a metal sulfate, which is used as a test for
the sulfate anion? (A) NO32− aqueous plus SO42−
aqueous gives SO3− aqueous plus NO2 gas plus O2 gas. (B) Ba solid plus SO42− aqueous
gives BaSO4 solid. (C) Ba2+ aqueous plus SO42− aqueous
gives BaSO4 solid. (D) Ba2+ aqueous plus SO42− aqueous
gives SO2 gas plus O2 gas plus Ba solid. (E) Ba(NO3)2 solid plus H2SO4
aqueous gives BaSO4 solid plus two HNO3 aqueous.
In this question, we need to
consider the reaction between barium nitrate, which is a soluble barium salt, and a
metal sulfate, which can be represented as MSO4 (aq), since it also must be
soluble. Since we are going to show a net
ionic equation for this reaction, only the ions reacting will be shown. Any spectator ions will be
removed. Spectator ions are those that do
not change in the reaction at all.
When one mole of barium nitrate is
dissolved in water, it yields one mole of barium two plus ions and two moles of
nitrate ions. These are aqueous ions. When our soluble metal sulfate,
represented as MSO4, is added to water, it will yield one mole of M2+ ions aqueous
and one mole of SO42− ions, also aqueous.
Now we have a complete list of
reactant ions, we can consider what happens to them during the course of this
reaction. The barium ions react with the
sulfate ions to form insoluble barium sulfate. This is written BaSO4 brackets
s. The metal ions and nitrate ions
form a soluble nitrate salt. As these ions are soluble, they can
be added to the list of products in our ionic equation. The nitrate ions and metal ions are
in fact spectator ions. And they can be removed from this
ionic equation. This produces the net ionic
equation which exactly matches answer (C) with the correct state symbols.
Now, we can summarize the key
points in this lesson. Carbonate ions react with acids to
produce carbon dioxide gas, which turns limewater milky. Halide ions react with silver
nitrate solution to give silver halide precipitates that can be identified from
their colors. Sulfate ions react with soluble
barium salts to give a white precipitate of barium sulfate.
