Video Transcript
In this video, we will learn how to
determine the mass or composition of an analyte using precipitation gravimetry.
Some chemical reactions that take
place in water or in aqueous conditions produce in soluble products. There are many colorful examples of
this. For example, if a solution of lead
nitrate is mixed with a solution of potassium iodide, a bright yellow solid is
immediately formed. This is a completely unexpected
result in this reaction, since the two starting solutions are colorless. Both lead nitrate and potassium
iodide are soluble salts. Most common nitrates are soluble in
water. Most salts of group one metals are
also soluble.
Lead nitrate and potassium iodide
are ionic compounds. And when dissolved, the ions are
free to move around and interact with each other. A double decomposition reaction
takes place here, and the ion swap over. The two products formed are
potassium nitrate and lead iodide. Lead iodide does not dissolve in
water. It has very limited solubility, and
it’s responsible for the dense yellow solid observed in this reaction. The yellow lead iodide solid is
known as a precipitate. And the reaction could be described
as a precipitation reaction. A precipitate is simply a solid
formed in a reaction from a mixture of solutions.
So, how could we collect a pure dry
sample of this lead iodide precipitate? Firstly, we would need to separate
it from the surrounding liquid that contains other dissolved ions. A piece of filter paper and a
filter funnel would be used for this task. This procedure is called
filtration, not funneling as is often believed. The yellow lead iodide solid will
not pass through the filter paper, whereas any solution will drop into the flask
that supports the funnel. The yellow lead iodide that remains
on the filter paper will be contaminated with other dissolved ions. These solid filtrate should
therefore be rinsed with distilled or deionized water whilst it’s in the filter
paper.
Now that the solid precipitate has
been separated from the solution containing other dissolved ions, it must be dried
fully to remove traces of water. This can be achieved by placing the
filter paper and the solid in a warm dry place. Ideally, a thermostatic drying oven
is used set to 110 degrees Celsius. This temperature is plenty hot
enough to remove the water but not too hot to risk burning the paper or melting or
decomposing the product. To ensure that all traces of water
are truly removed, the paper and the filtrate can be removed and its mass recorded
on an accurate balance. The paper and the filtrate is
returned to the drying oven for a few more hours, and then its mass is
rerecorded. If the dry masses are the same, we
can be convinced that all the water has been removed. This technique is known as drying
to constant mass.
Now that we’ve seen how a pure dry
sample of a precipitate can be collected, we can explore how this idea can be used
to quantitatively analyze solutions of ions. A quantitative chemical analysis is
one where the amount of one or more constituents in a sample are determined. As an example, the result of the
chemical analysis may be expressed in concentration units or as a percentage by
mass. Beyond simply confirming the
presence of various ions, precipitates can be used to perform quantitative chemical
analysis. The exact mass of a precipitate
formed must be established. This can be achieved by taking mass
measurements before and after the filtration and drying processes described
earlier.
For example, the mass of a dry
clean filter paper is recorded on a three-decimal-place balance. Balances that are accurate to the
nearest 0.1 of a milligram are often referred to as analytical balances. They measure mass to four decimal
places. As these balances are very
sensitive, they usually incorporate a draft shield to stop airflow. The clean dry filter paper would
then be used to collect all of the precipitate from a chemical reaction. The filter paper and the
precipitate would be washed and dried, as described earlier. Having been dried to constant mass,
the mass of the filter paper and the precipitate is then recorded. The difference in mass is
equivalent to the mass of precipitate that was formed in the chemical reaction.
Sometimes the precipitate is dried
in an ignition crucible using ashless filter paper. So we do not need to worry about
preweighing the filter paper first. When a substance is analyzed based
upon its mass, the method of analysis is known as gravimetry. The quantitative analytical method
can be described as a gravimetric technique. In a chemical analysis, the
substance being analyzed is often referred to as the analyte. In precipitation gravimetry, the
analyte may be a soluble compound that is converted to an insoluble precipitate
during the course of a simple reaction.
As an example, we could take a
solution of sodium chloride which will be the analyte that is to be determined
here. To this analyte solution we could
add an excess of silver nitrate solution. The chemical reaction would produce
a white precipitate of silver chloride and a solution of sodium nitrate. Using an analytical balance and a
piece of filter paper, the mass of silver chloride precipitate can be established as
described earlier. If we want to answer the question
“What mass of sodium chloride was in the analyte solution?,” we need to look
carefully at the balanced equation for the reaction that occurred.
From the balanced chemical
equation, we can see that one mole of sodium chloride reacts with one mole of silver
nitrate. Since the silver nitrate is in
excess, all of the chloride ions are reacted. One mole of chloride ions,
therefore, produces one mole of silver chloride as a solid precipitate. To convert the mass of silver
chloride precipitated to the moles of silver chloride precipitated, we would take
its mass and divide by the molar mass for silver chloride. The molar mass for silver chloride
is found to be 143.4 grams per mole. The moles of silver chloride formed
is therefore 0.139 grams divided by 143.4 grams per mole. This equates to 0.000969 moles of
silver chloride.
The moles of silver chloride is now
linked to the moles of sodium chloride in the analyte solution via the mole ratio in
this balanced equation. This will enable us to find the
mass of sodium chloride in this analyte solution. Since the mole ratio of silver
chloride to sodium chloride in this balanced equation is one to one, there’s an
equal number of moles of sodium chloride present in the original analyte
solution. The molar mass of sodium chloride
equates to 58.5 grams per mole. The mass of sodium chloride in the
analyte solution is therefore equal to the moles of sodium chloride multiplied by
its molar mass. This equates to 0.000969 moles
multiplied by 58.5 grams per mole. The mass of sodium chloride in the
analyte solution was therefore 0.0567 grams to three significant figures.
We will now look at a question to
further develop the concept of precipitation gravimetry.
A magnesium halide salt has the
formula MgX2. A 0.593-gram sample of MgX2 was
dissolved in 100 milliliters of deionized water, followed by the addition of excess
NaOH. The precipitate of MgOH2 was
filtered, washed, and dried. The precipitate was found to have a
mass of 0.187 grams. What is the identity of X? (A) I, (B) F, (C) Br, (D) Cl.
In this question, we’re being asked
to find the identity of X which will be the halogen that’s responsible for the
halide ion in the ionic salt MgX2. The halogen present as the halide
ion in MgX2 could be iodine, fluorine, bromine, or chlorine. With this in mind, the actual
formula of MgX2 could be MgI2, MgF2, MgBr2, or MgCl2. Firstly, we need to look at the
reaction that occurs between the magnesium halide salt and NaOH, which is sodium
hydroxide. Both are dissolved in water, so
these are aqueous solutions. We are told in this question that
one of the products is MgOH2. This is magnesium hydroxide, which
is an insoluble precipitate. The other product of this reaction
will be a combination of the sodium ions and the halide ions from the two reactants
concerned. Sodium halides are soluble, so this
product is an aqueous solution.
We now need to balance this
equation. We can see from the formula for
magnesium hydroxide that one mole of magnesium hydroxide contains two moles of
hydroxide ions. This makes sense as the magnesium
ion is present as a two plus ion and the hydroxide ions are one minus ions. The charges must balance up, so we
need two hydroxide ions for every magnesium two plus ion. We therefore need two moles of
sodium hydroxide. This will provide two moles of
sodium ions and two moles of hydroxide ions on the left side of our equation, which
will balance with the hydroxide ions on the right side. We therefore need to form two moles
of NaX, our sodium halide, which will contain two moles of sodium ions. Two moles of NaX also contains two
moles of X−, or halide ions, which balance with the two moles of halide ions in the
magnesium halide salt we started with.
This equation is now balanced, and
we can use it with the quantitative data provided in the question. We know in this question that the
mass of magnesium hydroxide precipitate collected was 0.187 grams. Since we know the precise
composition of magnesium hydroxide from its formula, we can find the molar mass of
magnesium hydroxide. This equates to 58.3 grams per
mole. The moles of magnesium hydroxide
produced can then be found by taking its mass in grams and dividing it by the molar
mass in grams per mole. The moles of magnesium hydroxide is
therefore 0.0032.
We can link the moles of magnesium
hydroxide formed to the moles of magnesium halide salt in the analyte using the mole
ratio in the balanced equation. One mole of magnesium hydroxide
originates from one mole of magnesium halide salt. There must therefore have been
0.0032 moles of the magnesium halide salt in the sample. Since the 0.0032 moles of magnesium
halide salt was found in 0.593 grams of sample, we can use this information to find
the molar mass.
Since we now have the moles and the
mass for the magnesium halide salt, we can use this information to find the molar
mass of the magnesium halide salt itself. The molar mass of MgX2 is the mass
of the sample divided by the number of moles that it contains. The molar mass of MgX2 is therefore
185.3 grams per mole. Using the atomic mass from
magnesium, we can find the atomic mass for the element X in this magnesium halide
salt. By subtracting the atomic mass for
magnesium from the molar mass of the magnesium halide salt, we have the mass of
twice the halogen present. Twice the atomic mass of the
halogen present equates to 161. The atomic mass of the halogen X is
therefore 161 divided by two, which equals 80.5.
We need to bear in mind that this
will not be the precise atomic mass of our halogen X, as it’s derived from
experimental data where there are errors. The only halogen with an atomic
mass close to 80.5 is bromine. The atomic mass of bromine is
79.9. The correct answer is therefore
bromine.
Now it’s time to review the key
points. Some chemical reactions produce
precipitates. A precipitate is a solid product
formed from a mixture of reacting solutions. A precipitate may be collected by
filtration, washing, and then drying. Gravimetric analysis is a
quantitative technique where the analysis is based upon mass measurements. The mass of a precipitate formed
can be linked to the mass of a reacting analyte using a balanced equation.