Video Transcript
Sodium chloride can be prepared by
reacting sodium hydroxide with hydrochloric acid. In an experiment, a fixed volume of
sodium hydroxide and a few drops of phenolphthalein were added into an Erlenmeyer
flask. Hydrochloric acid was then added,
from a buret, until the reaction was complete. The reaction was then repeated,
without the indicator but using the exact volume of hydrochloric acid determined
from the previous experiment. The resulting solution was then
heated to produce white crystals of sodium chloride. Why were a few drops of
phenolphthalein added to the solution at the start? (A) To increase the solubility of
sodium chloride, (B) to increase the basicity of sodium hydroxide, (C) to remove any
impurities in the solution, (D) to act as an indicator and inform us when the
reaction is complete, or (E) to act as a catalyst in the reaction.
Let’s first review the methods used
in this experiment before answering the first part of our question. First, a fixed volume of sodium
hydroxide, which is a basic solution and can be represented using its chemical
formula NaOH with the state symbol aq representing that it is aqueous, was added to
an Erlenmeyer flask, which can also be called a conical flask. In this same Erlenmeyer flask, a
few drops of phenolphthalein were also added. Then, hydrochloric acid was reacted
with sodium hydroxide using a buret.
Hydrochloric acid can be
represented using the chemical formula HCl with the state symbol aq. Once this first reaction was
completed and the exact volume of hydrochloric acid needed was determined, the
process was repeated without the indicator. From this reaction, an aqueous
solution is produced of sodium chloride, which can be represented using the chemical
formula NaCl. This solution was then heated to
drive off the second product of the reaction, which was liquid water or H2O with the
state symbol l. With the water driven off, the
remaining white crystals of sodium chloride could be collected.
The question asks us to consider
why the phenolphthalein was added at the start of this experiment. Phenolphthalein is a compound whose
color is determined by the pH of the solution to which it is added. In solutions with a pH higher than
approximately eight, phenolphthalein appears pink. But in solutions with pHs lower
than approximately eight, phenolphthalein is colorless. In this reaction between sodium
hydroxide, which we can see from the diagram is a colorless solution, and it is a
basic solution, which means it has a pH of above seven. Sodium hydroxide is reacted with
hydrochloric acid, which we can see from the diagram is also a colorless solution
and being that it is an acid has a pH of below seven.
In this experiment, it was critical
to determine the exact volume of hydrochloric acid needed to completely react with
the fixed volume of sodium hydroxide so that the resulting solution contained only
the desired salt and water. At the start, phenolphthalein
appeared pink as it was in a basic solution with a pH of above approximately eight
but turned colorless when the reaction was complete, producing a resulting solution
with a pH of below approximately eight. In this experiment, phenolphthalein
served as a pH indicator, which is a substance that changes color within certain pH
ranges and in this case informed us when the reaction was complete. Therefore, the reason why a few
drops of phenolphthalein were added to the solution at the start is answer choice
(D) to act as an indicator and inform us when the reaction is complete.
Why would this method not work for
producing crystals of copper sulfate from copper(II) oxide and sulfuric acid at room
temperature? (A) Copper sulfate precipitates out
of solution and so filtration would be required. (B) Copper oxide is insoluble and
would not form an aqueous solution. (C) Copper oxide is acidic and
would not react with sulfuric acid. (D) Copper sulfate only forms when
copper oxide reacts with sulfur dioxide. Or (E) copper oxide does not react
with sulfuric acid.
Keeping the method of our
experiment in mind, let’s clear some space on screen so we can answer this part of
the question. In the method used to produce
sodium chloride crystals, a base which is aqueous was reacted with an acid that is
also aqueous. So, in this method, both the base
and acid reacted were aqueous solutions, which produced an aqueous salt and liquid
water. This question features copper
sulfate as the desired salt product, which would be soluble in water. We can represent this using the
chemical formula CuSO4 aqueous in the products of this reaction. The acidic reactant in this
question is sulfuric acid whose chemical formula is H2SO4 aqueous.
Copper(II) oxide is the other
reactant of this reaction whose chemical formula is CuO. According to water solubility
rules, copper(II) oxide is an insoluble compound, which we can represent using the
state symbol s for solid. While this reaction could feasibly
occur to produce the salt copper sulfate and water as products, the method used
required both reactants to be aqueous solutions and would not be appropriate for a
reaction involving a solid insoluble reactant. Other methods could be used such as
adding excess solid reactant to the acid, then stirring and gently warming the
mixture, where once the reaction is complete, excess solid reactant can be removed
through filtration. The solution can be heated to drive
off the water and the salt crystals can be collected.
Therefore, the answer choice which
explains why this method used to collect sodium chloride crystals would not work for
producing crystals of copper sulfate from copper(II) oxide and sulfuric acid at room
temperature is answer choice (B). Copper oxide is insoluble and would
not form an aqueous solution.
Why could the sodium chloride not
be obtained by filtering the solution? (A) Any insoluble impurities would
not be separated through filtration. (B) Only the sodium ions would be
filtered and not the chloride ions. (C) Any excess hydrochloric acid
would burn the filter paper. Or (D) sodium chloride is soluble
and would therefore not be separated through filtration.
In our reaction, sodium chloride
was produced as an aqueous solution, which means in water, the ions which make up
this salt disassociate and move freely from each other. However, they are extremely small
that if in an attempt to separate the sodium and chloride ions from the water using
filtration, the ions would be small enough to pass through the filter paper and
remain in the water. Filtration, which is a technique
that separates solid and liquid substances based on particle size, can be used to
collect insoluble substances when poured through a filter paper in a funnel
positioned above a container to collect the soluble or liquid substances, which can
be referred to as the filtrate, while the substance remaining separated by the
filter paper is referred to as the residue.
This specific type of filtration is
called gravity filtration. Because our salt sodium chloride is
water-soluble, it would not be separated using this method. Therefore, the answer choice that
best describes why sodium chloride could not be obtained by filtering the solution
is answer choice (D). Sodium chloride is soluble and
would therefore not be separated through filtration.
