Lesson Explainer: Volumetric Analysis Chemistry • Third Year of Secondary School

In this explainer, we will learn how to describe volumetric analysis and use results from titration experiments to calculate the concentration of a solution.

There are many different quantitative analysis methods that a chemist can use to measure or quantify the amount of a substance in a sample. The substance that is investigated in chemical analysis is called the analyte.

The quantity of an analyte in a sample can be expressed in terms of mass, volume, concentration, moles, or relative abundance.

Volumetric analysis is a quantitative analytical method that measures the volume of the analyte directly or the concentration or volume of the analyte indirectly by measuring the volume of a second substance, called the titrant, that reacts with the analyte in a known proportion. The second method is called titrimetric or titration analysis.

Examples of titrimetric analysis (titrations) used in industry are

• determining concentrations of components in wastewater analysis,
• determining contamination levels in river waters and acid rain,
• determining levels of acids in juices and other foodstuff in nutrition,
• determining drug concentrations in medicine.

The diagram above shows that there are three broad categories of titration, based on the type of reaction that occurs between the analyte and the titrant. These are neutralization reactions, redox (oxidation–reduction) reactions, and precipitation reactions.

The type of titration is chosen depending on the nature of the analyte. Neutralization titrations can be used to quantify acids or bases, while substances that undergo redox reactions can be quantified by performing redox titrations, and substances that form sparingly soluble salts can be measured using precipitation titrations.

The diagram below shows the setup of a typical neutralization titration. Let us assume that the analyte is an acid and the titrant is a base.

An accurately known volume of the acid, whose concentration is unknown, is pipetted into a conical flask. The basic titrant solution has a concentration that is known exactly, and therefore, this solution is referred to as a standard solution. The titrant is added slowly to the analyte from the buret until the indicator changes color, at the end point, indicating that the reaction is complete and the neutralization of all the analyte in the conical flask has occurred.

The volume of the titrant added (the titer) is recorded. This process is repeated several times in different runs to ensure precision.

It is important to remember that the buret readings are taken at eye level to avoid parallax error.

Let us investigate how to calculate the concentration of an acid from the results of a titration experiment.

The three variables commonly used in titration calculations are

• , the number of moles (unit mol),
• , the volume of the solution (unit typically expressed as mL, L, or dm³),
• , the concentration of the solution (typically expressed as molarity: mol/L or mol/dm³).

The relationship between these three quantities is

The tables below show the data that is collected during a neutralization titration between an acid as an analyte, , and a standard base as a titrant, .

It is important to ensure that the acid–base reaction equation is balanced before doing calculations, since the stoichiometric ratio of the reactants influences the calculations. The schematic below shows the balanced reaction equation, the molar ratio of the reaction substances (from the stoichiometric coefficients), and the three steps for determining the concentration of the analyte:

In the first step, the number of moles of the titrant is calculated. Before doing this, however, the average titer must be calculated from the titer values of the three runs and the volume unit converted from millilitres to litres since the concentration of the solution is expressed in molarity, whose unit is mol/L. Remember that sometimes, molarity is expressed as mol/dm³, which is equivalent to the unit mol/L since .

The average volume of titrant is calculated by adding the volume of titrant delivered from the buret from each run and dividing by the number of runs:

Sometimes, the average volume of titrant is taken only from results that are concordant with each other (i.e., from titer results that differ from each other by 0.20 mL or less). In this example, the average of all the run results has been taken.

To convert a volume value in millilitres to litres, the value in millilitres is divided by 1‎ ‎000 according to this unit conversion calculation:

The mL units cancel with each other, giving a volume in litres of 0.01448 L. This titrant volume can be used to calculate the number of moles of base.

Now, steps 1 to 3 can be followed to determine the analyte concentration.

In step 1, the number of moles of the titrant can be calculated from its concentration, which is 0.0950 mol/L, and its average volume, which is 0.01448 L:

In step 2, the number of moles of the analyte is determined by relating it to the number of moles of the titrant from step 1 using the stoichiometric coefficients (molar ratio) from the balanced equation: 1 mole of reacts with 1 mole of (from the balanced equation); therefore, moles of will react with 0.0013756 moles .

Then, we can solve for :

In step 3, the concentration of the analyte can be calculated from its number of moles (from step 2) and its volume, which is 20.00 mL. This volume value needs to be converted to litres, which is done in the same manner as before, by dividing the value by 1‎ ‎000:

This value can then be used with the number of moles of to determine its concentration:

The concentration of the acid, in terms of molarity, is therefore 0.069 mol/L.

There is a simpler way to do this calculation, by combining the three steps into one step using the following equation: where is the stoichiometric coefficient of the acid from the balanced equation and is the stoichiometric coefficient of the base from the balanced equation.

When the concentration is expressed as a molarity, the symbol can be replaced by the symbol and the expression becomes

Example 1: Calculating the Concentration of Nitric Acid as an Analyte in a Titration

A 30 mL solution of nitric acid was titrated against a 0.1 M solution of potassium hydroxide. The addition of 26.6 mL of potassium hydroxide was found to neutralize the nitric acid. What is the concentration of the nitric acid? Give your answer to 2 decimal places.

Answer

In this titration, the analyte is the nitric acid, , whose concentration is unknown but whose volume is known to be 30 mL. The titrant is the standard solution potassium hydroxide, , whose known concentration is given as a molarity of 0.1 M, which is 0.1 mol/L. The volume of the base, which is the titer, is 26.6 mL.

The balanced acid–base reaction equation is

It is convenient to write down the data in a format that makes it easy to use. One such way is to write the data underneath the corresponding substance in the balanced equation as follows:

There are different ways to set out the solution to this problem. One way is to follow the three-step procedure as in the diagram below:

Before starting step 1 of the calculation, make sure that the volume of the base is expressed in units compatible with the base concentration units. Concentration is given in moles per litre (mol/L), so the volume can be expressed in litres, and the unit conversion from millilitres to litres is as follows:

Now we can proceed with step 1.

Step 1: Calculate the number of moles of the base from its concentration and its volume:

Step 2: Relate the number of moles of the base to the number of moles of the acid according to the stoichiometric coefficients (molar ratio) from the balanced equation to determine the number of moles of acid: 1 mole of reacts with 1 mole of (from the balanced equation); therefore, moles of will react with 0.00266 moles of .

We can then solve for :

Step 3: Calculate the concentration of the acid from its number of moles (from step 2) and its volume in litres:

Then,

The concentration of the nitric acid, in terms of molarity, is therefore 0.09 mol/L.

Sometimes, the titrant is an acid and the analyte is a base. The same calculations can be performed to determine the concentration of a base as an analyte.

Example 2: Calculating the Concentration of Lithium Hydroxide Base as an Analyte in a Titration

A standard solution of 0.25 M  is used to determine the concentration of a 220 mL  solution. The addition of 143 mL of resulted in complete neutralization. What is the concentration of the solution? Give your answer in units of millimolars.

Answer

The question refers to a neutralization titration reaction between an acid and a base. The reaction between an acid and a base produces a salt and water; hence, the equation for the reaction is

The equation needs to be balanced as follows:

The concentration of the analyte, , can be determined in two ways:

Method 1: A three-step process according to the steps shown in this diagram.

Step 1: The number of moles of the titrant, which is in this question, can be calculated from its concentration, which is 0.25 mol/L, and its volume, which is 143 mL. The volume of 143 mL must first be converted to a volume in litres.

Step 2: The number of moles of (from step 1) and the molar ratio (the stoichiometric coefficients) of the two reactants from the balanced equation are used to determine the number of moles of the base, .

Step 3: The concentration of the base can be calculated from its number of moles (from step 2) and its volume of 220 mL. The volume of 220 mL must first be converted to a volume in litres.

Method 2: By using the following equation where

is the stoichiometric coefficient of the acid from the balanced equation and

is the stoichiometric coefficient of the base from the balanced equation.

Let us use method 2. The values we will substitute are

• ,
• ,

which needs to be converted to litres by multiplying it by the conversion factor of and then cancelling the mL units:

• (from the stoichiometric coefficient in the balanced equation)
• , which needs to be converted to litres by multiplying it by the conversion factor of and then cancelling the mL units:
• (from the stoichiometric coefficient in the balanced equation)

Putting these values into the equation, we get which can then be solved for the concentration of the base:

The question asks that the answer be expressed in millimolars, which is mol/L. To convert the answer of 0.325 mol/L to millimoles per litre (mmol/L), we can multiply by the conversion factor of :

Since a mole per litre is a molar, the answer can be written as 325 mmol/L or 325 mM.

Neutralization titrations are also useful for determining the percentage of an acid or a base in a mixture.

An example is when a mixture of 1.8 g of solid and is titrated against 0.25 M . If 25 mL of the acid is required to completely neutralize all the base in the mixture, we can determine the percentage of base in the mixture and also the percentage of the salt.

First, the reaction equation needs to be balanced as follows:

Then, the steps shown in the diagram above can be followed to determine the percentage of the base in the original solid mixture.

Step 1: Use the concentration (0.25 M) and volume (25 mL) of the acid to calculate the number of moles of acid that was required to neutralize all the base.

The volume value of 25 mL must be converted to a value in litres by multiplying by the conversion factor of and then canceling the mL units:

Then, the number of moles of acid can be calculated:

Step 2: The number of moles of acid (from step 1) and the molar ratio (the stoichiometric coefficients) of the two reactants from the balanced equation are used to determine the number of moles of the base.

The balanced equation above shows that 2 moles of reacts with 1 mole of ; therefore, 0.00625 moles of reacts with moles of .

We can then solve for :

This value of 0.003125 mol  is also the number of moles of base in the original solid mixture, before the titration.

Step 3: The mass of the base can be calculated from its number of moles and molar mass. The molar mass of is which means that there is 74 g of in every 1 mole of ; therefore, there are g of in 0.003125 mol of .

Solving for , we get

This value of 0.23125 g  is also the mass of base in the original solid mixture, before the titration.

Step 4: The mass of the base in the original solid mixture can be converted to a percentage. To calculate the percentage of the base in the / mixture, we can use the formula or

The total mass of the mixture was 1.8 g, of which 0.23125 g was . Substituting in these values, we can solve

We can go one step further and use this value to calculate the percentage of the salt . The total mass of the original solid mixture is equivalent to ; therefore, we can determine the percentage of by subtracting the percentage of the base from as follows:

There was of the salt in the original solid mixture.