### Video Transcript

In this video, we will learn what volumetric analysis is and how to use titration
results to calculate the concentration of a solution.

What is volumetric analysis? In many areas of science and industry, chemical compounds need to be analyzed. They can be identified and quantified. A compound under investigation is called an analyte. Examples of analytes a chemist might study include compounds in wastewater,
pollutants and their levels in river water, the amounts of nutrients and other
compounds in foods, and the concentrations of drugs and medicines in research as
well as in body fluids, such as blood. All these different analytes and many more can be quantified in terms of mass, moles,
volume, concentration, or relative abundance. We are going to focus on quantifying an analyte by volume, and this is volumetric
analysis.

Volumetric analysis is a quantitative analytical method which measures the volume of
analyte directly or the volume of a second substance called the titrant which reacts
with the analyte in a known proportion. This second method is an indirect method of analysis and is known as titration or
titrimetric analysis. In a titration, it is usually the concentration of the analyte which we determine
using the volume of a second substance called the titrant whose concentration is
accurately known.

Three titration methods commonly used depend on the type of reaction between the
analyte and the titrant. These are neutralization titrations, when the analyte is an acid or a base; redox
titrations, which are used to quantify analytes which undergo redox reactions; and
precipitation titrations, which are used to measure substances which react to form a
precipitate or a sparingly soluble salt. Now, let’s look at an example of how to calculate the concentration of an analyte by
titration. Let’s use a neutralization reaction in our example. In another video, the details of how to perform a titration are discussed in depth,
but let’s quickly recap before we do the calculation.

A burette containing the titrant solution — in our example, let’s make it a base,
sodium hydroxide — is clamped to a retort stand, and its initial volume is
recorded. The analyte — in this example is an acid solution, and let’s make it hydrochloric
acid — is pipetted into an Erlenmeyer flask or a conical flask. The accurately known volume of the analyte is recorded. The concentration of the titrant is accurately known. The volume of the analyte is accurately known, but its concentration is unknown. Next, we add a few drops of suitable indicator to the analyte. And then, the tap or stopcock on the burette is opened, and titrant is slowly added
to the analyte.

A point is reached where the indicator in the analyte changes color. This is called end point. At the end point of the reaction, the moles of analyte have been exactly neutralized
by the moles of titrant added. The tap is closed, and the final volume of the titrant is recorded. We can then subtract the final and initial volumes of the titrant to get the titer,
which is the volume of titrant added. And we now know this value. Often when we do a titration, we repeat this whole process several times. If three runs are performed, three titer values will be obtained. We then find the average titer of the three and use this value in our
calculations.

So, to find the concentration of the analyte, the three pieces of information that we
need are the volume of the analyte, the concentration of the titrant, and an average
titer value. Let’s put some numbers in. If the concentration of the base is 0.0950 moles per liter and the average titer
14.48 milliliters and the pipetted volume of the analyte exactly 20 milliliters, we
can determine the concentration of the analyte as follows. First, we need the balanced equation for the reaction between the titrant and
analyte. The molar ratio of the two reactants is one as to one. We get this from the stoichiometric coefficients of the balanced equation. We can use a handy table for our data. When we transfer the data that we know, we get the following.

We can now use the key equation number of moles is equal to concentration times
volume to solve this problem. There are three steps to solving this calculation: step one, calculate the number of
moles of titrant from its concentration and average titer using the key
equation. Note, however, that first we must convert the average titer from a value in
milliliters to a value in liters. This is because its concentration is given in moles per liter. Let’s clear some space to do this. We can convert from milliliters to liters by multiplying by this conversion
factor. The milliliter units cancel out, and we are left with liters as the unit. And we get 0.01448 liters. Let’s put that in our table.

Now, we can use our key equation to determine the number of moles of titrant, 𝑛
equals 𝑐𝑉. Substituting in the concentration of the base and its volume in terms of liters, we
get 0.0013756 moles of titrant reacted. We can put this value into the table. Note that I have kept all the decimal places at this point. We will only round off at the very end of the calculation to ensure that we get the
correct answer, and we will then express our answer to the correct number of
significant figures.

Step two is to calculate the number of moles of analyte that reacted from the number
of moles of titrant. And we do this using the stoichiometric coefficients from the balanced equation. We know that one mole of sodium hydroxide reacts with one mole of hydrochloric
acid. Therefore, 0.0013756 moles of sodium hydroxide reacts with 𝑥 moles of hydrochloric
acid. And, of course, because it is a one-as-to-one ratio, 𝑥 is equal to 0.0013756 moles
of hydrochloric acid, and we can put this value into the table.

The third and last step is to calculate the concentration of the analyte using its
volume and number of moles. Once again, we’ll have to convert its volume in milliliters to a value in liters so
that we can express its concentration in terms of a molarity, moles per liter. I have condensed step two to create some more space. So, step three is first convert the analyte volume from a milliliter value to a liter
value by multiplying by a suitable conversion factor, and we get 0.02000 liters. We can write this in the table.

Now, we can rearrange the key equation to get concentration as the subject of the
formula. Concentration is equal to number of moles divided by volume. Now, we can substitute in our values for hydrochloric acid, 0.0013756 moles divided
by 0.02000 liters, and we get a concentration of acid analyte of 0.06878 moles per
liter. If we round that off to three significant figures, we get 0.0688 moles per liter, and
we can put this value in the table.

Neutralization titrations can also be used to determine the percentage of an acid or
base in a mixture. Let’s look at an example.

A 1.8-gram solid mixture of CaOH2 and CaCl2 is dissolved then titrated against
0.25-molar HCl aqueous. If 25 milliliters of the acid is required to completely neutralize all the base,
determine the percentage of base in the mixture.

And we are given the molar masses of calcium, oxygen, hydrogen, and chlorine. We are told we have a solid mixture of CaOH2 and CaCl2. CaOH2 is calcium hydroxide and CaCl2 is calcium chloride. The combined total mass is 1.8 grams. We are then told that it is dissolved and titrated against 0.25-molar HCl, or
hydrochloric acid. So, HCl is the titrant. And although both calcium hydroxide and calcium chloride are dissolved in the
Erlenmeyer flask, the analyte is only the calcium hydroxide because this base will
react with the titrant which is an acid.

This problem is solved in a similar manner to a normal titration calculation. We start with the balanced equation so that we can get the stoichiometric
coefficients or the molar ratio in which the analyte and titrant react. We can then draw a handy table and fill in all the information that we know about
concentration, volume, and number of moles for the two reactants. The concentration of the titrant is 0.25 molar. Its titer value is 25 milliliters, and that’s all the information we know. In this example, we are not asked to calculate the concentration of the analyte and
we are not told the volume of the analyte. We are asked to determine the percentage of analyte in the original solid
mixture.

There are four steps to this calculation: step one, calculate the number of moles of
titrant from its concentration and volume. Let’s clear some space to do this. First, we need to convert the volume in milliliters to a volume in liters so that its
unit is compatible with the concentration unit. Multiplying by a suitable conversion factor, we can get rid of the milliliter units
and we get the titrant volume in liters, which we put into the table. We can then use the key equation 𝑛 equals 𝑐 times 𝑉 and substitute in the
concentration and volume of the acid. Remember molar or molarity is moles per liter. And we get 0.00625 moles of HCl. This is the moles of titrant needed to neutralize all of the calcium hydroxide. We will round off at the very end of the calculation to the correct number of
significant figures.

In step two, we calculate the number of moles of analyte from the number of moles of
titrant using the stoichiometric coefficients from the balanced equation. One mole of calcium hydroxide reacts with two moles of hydrochloric acid. Therefore, 𝑥 moles of calcium hydroxide will react with 0.00625 moles of the
acid. We can solve for 𝑥. Since the moles of the acid are double the moles of the base, to get 𝑥, we can
divide the moles of acid that we have by two, and we get 0.003125 moles of calcium
hydroxide. This is the moles of base neutralized by the acid titrant and also the moles of base
in the original solid mixture.

In step three, we calculate the mass of calcium hydroxide, the analyte, from its
number of moles and molar mass. We use the key equation number of moles is equal to mass divided by molar mass. We want mass as the subject of the formula. So, when the equation is rearranged, we get mass is equal to number of moles
multiplied by molar mass. We can substitute in our number of moles. Now, I’m going to determine the molar mass of calcium hydroxide directly within the
equation.

There is one atom of calcium in CaOH2, and its molar mass is 40 grams per mole. So, we write 40 plus two oxygens with molar mass 16 grams per mole each. So, we write two times 16 plus two hydrogens with molar mass one gram per mole. So, we write two multiplied by one. This gives us a molar mass of calcium hydroxide of 74 grams per mole, which we
multiply by its number of moles. Solving, we get 0.23125 grams. This is the mass of calcium hydroxide in the titration as well as in the original
solid mixture.

There is one last step. We need to convert the mass of calcium hydroxide to a percentage. We can determine the percentage of calcium hydroxide in the original solid mixture as
follows. The percentage of calcium hydroxide is equal to the mass of calcium hydroxide divided
by the total mass of the original solid mixture multiplied by 100 percent. We can put in our values. The mass of calcium hydroxide is 0.23125 grams, and 1.8 grams is the mass of the
total combined original solid mixture. The grams units cancel, and we get 12.847 percent.

Now, let’s round off to the correct number of significant figures. The smallest number of significant figures that we were given from the original data
is two significant figures. So, we can round up to 13 percent, which is two significant figures. This means of the 1.8 grams of the total original solid mixture, only 13 percent was
contributed by the calcium hydroxide.

Now, it’s time to summarize the main points we have learnt. We learnt that volumetric analysis is a quantitative analytical method which measures
the volume of the analyte directly or the volume of a second substance called the
titrant which reacts with the analyte in a known proportion. This second method is called titration. There are different types of titration, but they all involve an analyte reacting with
a titrant. Typically, in a titration, the concentration of the analyte is unknown. However, its volume is known as well as the concentration and volume of the
titrant. These three pieces of information as well as the key equation number of moles is
equal to concentration times volume can be used to determine the concentration of
the analyte. We can also use a titration to determine the percentage of analyte in a mixture.