3.6.4 Calculating the Concentration of a Solution Through Titration

Introduction

In chemistry, we often need to find the concentration of an unknown solution. One of the most accurate and widely used methods for doing this is titration.

Titration is a laboratory technique in which a solution of known concentration (called the standard solution) is gradually added to a solution of unknown concentration until the reaction between them is just complete. The point at which the reaction is complete is called the end point, and it is usually detected by a colour change using an indicator.

Titration is important in:

• Laboratory work — to determine unknown concentrations
• Industry — to check the purity of chemicals
• Pharmaceuticals — to ensure correct drug concentrations
• Environmental science — to analyze pollutants
• Food industry — to determine acidity or nutrient content

Definition

Titration: Titration is a laboratory method used to determine the concentration of a solution by reacting it with a solution of known concentration.

In simple words: Titration is a way to find out how strong an unknown solution is by slowly adding a known solution until the reaction is complete.

Standard Solution: A solution of known concentration.

End Point: The point at which the reaction is complete, usually shown by a colour change.

Indicator: A substance that changes colour at the end point (e.g., phenolphthalein).

Main Idea

• In titration, a solution of known concentration (standard solution) is added to a solution of unknown concentration until the reaction is complete.
• The volume of the standard solution used is measured accurately using a burette.
• The concentration of the unknown solution is calculated using the titration formula.
• For a reaction between an acid and a base:

M₁ × V₁n₁ = M₂ × V₂n₂

Where:

• M₁ = molarity of solution 1 (known)
• V₁ = volume of solution 1
• n₁ = number of moles of solution 1 from the balanced equation
• M₂ = molarity of solution 2 (unknown)
• V₂ = volume of solution 2
• n₂ = number of moles of solution 2 from the balanced equation

Key point: Titration uses the stoichiometry of the balanced equation to relate the concentrations and volumes of the two solutions.

Core Concept

For the reaction:

HCl_(aq) + NaOH_(aq) → NaCl_(aq) + H₂O_(l)

The balanced equation shows:

• n₁ = 1 mole of HCl
• n₂ = 1 mole of NaOH

The titration formula becomes:

M₁ × V₁1 = M₂ × V₂1

M₁ V₁ = M₂ V₂

Simple explanation: At the end point, the moles of acid equal the moles of base (for a 1:1 reaction). So we can use the volumes and one known concentration to find the unknown concentration.

Detailed Explanation

Activity 3.2: Standardising a Solution of Hydrochloric Acid

Objective: To standardise the given solution of hydrochloric acid.

Note: Perform this activity in the chemistry laboratory.

You will need:

• Burette
• Pipette
• Burette stand
• Beakers
• Conical flask
• Glass rod
• Standard 0.1\ M NaOH solution
• Phenolphthalein

Chemical equation:

HCl_(aq) + NaOH_(aq) → NaCl_(aq) + H₂O_(l)

n₁ = 1 mole n₂ = 1 mole

Carry out the following:

• Fit up a clean burette in the burette stand vertically.
• Fill the burette with HCl solution up to zero mark.
• Take 10\ cm³ of NaOH solution in a conical flask with the help of a pipette.
• Add a few drops of phenolphthalein as an indicator.
• Note the initial reading on the burette.
• Run the acid solution in the conical flask drop by drop, and shake the flask constantly.
• Go on adding the acid solution till the pink colour just disappears.
• Note down the final reading from the burette.
• The difference between the final and initial readings gives the volume of the acid used to neutralize 10.0\ cm³ of NaOH solution.
• Repeat the experiment to get three concordant readings.
• Find the mean volume of the HCl solution used.

Observations and calculations:

Suppose the volume of HCl solution used = V₁ = 10\ cm³

Molarity of HCl solution = M₁ = ?

Volume NaOH solution used = V₂ = 10\ cm³

Molarity of NaOH solution = M₂ = 0.1\ M

No. of moles of HCl = n₁ = 1

No. of moles of NaOH = n₂ = 1

M₁ × V₁n₁ = M₂ × V₂n₂

M₁ × 101 = 0.1 × 101

M₁ = 0.1\ M

Result: Molarity of HCl solution is 0.1\ M

Strength of HCl solution:

Strength = Molarity × Molar mass

= 0.1 × 36.5

= 3.65\ g/dm³

Intermediate Level

Steps for titration calculations:

• Write the balanced chemical equation.
• Identify n₁ and n₂ (the coefficients of the reactants).
• Identify the known and unknown values.
• Use the formula:
• Solve for the unknown.

Key point: The units of volume must be the same on both sides (both in cm³ or both in dm³).

Example: 10 cm³ NaOH titrated against 9.8 cm³ of 0.1 M HCl:

0.1 × 9.81 = M₂ × 101

M₂ = 0.098\ M

Formula / Mathematical Form

M₁ × V₁n₁ = M₂ × V₂n₂

Where:

• M₁ = molarity of solution 1
• V₁ = volume of solution 1
• n₁ = number of moles of solution 1 from balanced equation
• M₂ = molarity of solution 2
• V₂ = volume of solution 2
• n₂ = number of moles of solution 2 from balanced equation

For a 1:1 reaction (like HCl + NaOH):

M₁ V₁ = M₂ V₂

Solved Example

Example: Titration of NaOH with HCl

When 10\ cm³ solution of NaOH solution is titrated against 9.8\ cm³ of 0.1\ M HCl solution for complete neutralization:

HCl_(aq) + NaOH_(aq) → NaCl_(aq) + H₂O_(l)

Calculate:

(i) The molarity of NaOH solution (ii) The amount of NaOH present per cm³ of solution

Solution:

Given: V₁ = 9.8\ cm³ (HCl) M₁ = 0.1\ M (HCl) n₁ = 1 V₂ = 10\ cm³ (NaOH) M₂ = ? (NaOH) n₂ = 1

Formula:

M₁ × V₁n₁ = M₂ × V₂n₂

Substitution:

0.1 × 9.81 = M₂ × 101

Calculation:

0.98 = 10 M₂

M₂ = 0.9810 = 0.098\ M

(i) Molarity of NaOH solution = 0.098\ M

(ii) Amount of NaOH present per cm³ of solution:

Molar mass of NaOH:

= 23 + 16 + 1 = 40\ g/mol

Mass of NaOH in 1 dm³ (1000 cm³) of solution:

Mass = Molarity × Molar mass

= 0.098 × 40 = 3.92\ g

Mass of NaOH per cm³:

= 3.921000 = 0.00392\ g/cm³

Final Answer:

• (i) Molarity of NaOH = 0.098\ M
• (ii) Amount of NaOH per cm³ = 0.00392\ g/cm³

Concept Assessment Exercise 3.11

1. When 10\ cm³ solution of NaOH solution is titrated against 9.8\ cm³ of 0.1\ M HCl solution for complete neutralization.

HCl_(aq) + NaOH_(aq) → NaCl_(aq) + H₂O_(l)

Calculate: (i) The molarity of NaOH solution (ii) The amount of NaOH present per cm³ of solution

Solution:

(i) Molarity of NaOH:

M₁ × V₁n₁ = M₂ × V₂n₂

0.1 × 9.81 = M₂ × 101

M₂ = 0.9810 = 0.098\ M

Answer: 0.098\ M

(ii) Amount of NaOH per cm³:

Mass in 1 dm³ = 0.098 × 40 = 3.92\ g

Mass per cm³ = 3.921000 = 0.00392\ g/cm³

Answer: 0.00392\ g/cm³

2. A chemist has a 200\ cm³ solution of hydrochloric acid with a concentration of 6 M. He wants to dilute it to a concentration of 2 M. How much of the original solution should be mixed with water to achieve the desired concentration?

Solution:

Given: M₁ = 6\ M V₁ = ? M₂ = 2\ M V₂ = 200\ cm³

Formula:

M₁ V₁ = M₂ V₂

Substitution:

6 × V₁ = 2 × 200

Calculation:

V₁ = 4006 = 66.67\ cm³

Answer: The chemist should mix 66.67\ cm³ of the 6\ M solution with enough water to make a total of 200\ cm³ of solution.

Summary Table

Important Points

• - Titration = method to find unknown concentration.
• - Standard solution = solution of known concentration.
• - End point = point at which reaction is complete (shown by colour change).
• - Indicator = substance that changes colour at the end point (e.g., phenolphthalein).
• - Formula:
• M₁ × V₁n₁ = M₂ × V₂n₂
• - For a 1:1 reaction, M₁ V₁ = M₂ V₂.
• - n₁ and n₂ come from the balanced equation.
• - Titration is used in laboratory, industry, pharmaceuticals, environmental science, and food industry.
• - Example: 10 cm³ NaOH titrated against 9.8 cm³ of 0.1 M HCl → 0.098 M NaOH.
• - Example: Strength of HCl = 0.1 × 36.5 = 3.65\ g/dm³.

Remember

• - Titration = find unknown concentration.
• - Standard solution = known concentration.
• - End point = reaction complete (colour change).
• - Indicator = phenolphthalein (for acid-base titrations).
• - Formula: M₁ V₁n₁ = M₂ V₂n₂.
• - For 1:1 reaction: M₁ V₁ = M₂ V₂.
• - n₁, n₂ from balanced equation.
• - Units