3.6.3 Dilution of Solutions

Introduction

Sometimes, a chemist has a concentrated solution but needs a dilute solution for a particular experiment or procedure. The process of making a solution less concentrated by adding more solvent is called dilution.

Dilution is a common laboratory and industrial operation. For example:

• Concentrated acids are diluted before use.
• Stock solutions are diluted to prepare working solutions.
• Medicines are diluted to achieve the correct dosage.

Understanding dilution is important for:

• Laboratory work — preparing solutions of known concentration
• Industry — adjusting concentrations for specific processes
• Pharmaceuticals — ensuring correct drug concentrations
• Environmental science — preparing standards for analysis

Definition

Dilution: Dilution is a process by which you make a solution less concentrated by adding more of the liquid (usually water).

In simple words: Dilution means adding more solvent to a solution to make it weaker.

Example: Suppose you have a cup of strong lemonade. If you add more water, the lemonade becomes weaker or less strong. This means that dilution reduces the strength or concentration of a solution.

Main Idea

• Dilution reduces the concentration of a solution.
• The amount of solute remains the same before and after dilution.
• Only the volume of the solution increases.
• The relationship between concentration and volume before and after dilution is given by:

M₁ × V₁ = M₂ × V₂

Where:

• M₁ = concentration of the original (concentrated) solution
• V₁ = volume of the original solution used
• M₂ = concentration of the diluted solution
• V₂ = total volume of the final (diluted) solution

Core Concept

Key idea: When you dilute a solution, the moles of solute do not change. Only the volume changes.

Moles of solute before dilution = Moles of solute after dilution

M₁ V₁ = M₂ V₂

Simple explanation: If you have 100 mL of a 2 M solution, you have 0.2 moles of solute. If you add water to make the volume 200 mL, you still have 0.2 moles of solute, but now the concentration is 1 M. The moles of solute stayed the same; only the volume changed.

Detailed Explanation

Understanding Dilution

What happens during dilution:

• You start with a concentrated solution of known concentration (M₁) and known volume (V₁).
• You add more solvent (usually water).
• The total volume increases to a new value (V₂).
• The concentration decreases to a new value (M₂).
• The number of moles of solute remains constant.

Why moles remain constant:

• Adding solvent does not add or remove solute.
• The solute particles are simply spread out over a larger volume.
• The total number of solute particles is unchanged.

Using the Dilution Formula

Formula:

M₁ × V₁ = M₂ × V₂

Where:

• M₁ = concentration of the original (concentrated) solution
• V₁ = volume of the original solution you use
• M₂ = concentration you want after dilution
• V₂ = total volume of the final solution

How to use it:

• Plug the known values into the formula.
• Solve for the unknown value.
• Mix the calculated amount of the concentrated solution with enough water to reach the final volume you want.

Activity 3.1: Preparing a Dilute Solution

Objective: Prepare 250\ cm³ solution of 0.1\ M from concentrated hydrochloric acid.

Materials required:

• Conc. HCl
• 250\ cm³ volumetric flask
• Graduated pipette
• Distilled water

Steps:

Concentrated HCl is 12 M. Using the formula, calculate the volume of conc. HCl required to be diluted to 250\ cm³.

M₁ × V₁ = M₂ × V₂

Molarity of conc. HCl × its volume to be diluted = Required molarity of HCl × required volume of HCl solution

12 × V₁ = 0.1 × 250

V₁ = 0.1 × 25012

V₁ = 2.08\ cm³

Procedure:

• Transfer 2.08\ cm³ of conc. HCl with the help of a graduated pipette into a 250\ cm³ volumetric flask.
• Add distilled water up to the mark and shake well.
• This is the required 0.1\ M HCl solution.

Intermediate Level

Steps for dilution problems:

• Identify M₁, V₁, M₂, and V₂.
• Use the formula M₁ V₁ = M₂ V₂.
• Solve for the unknown value.
• Mix the calculated volume of concentrated solution with enough water to reach the final volume.

Key point: The moles of solute before and after dilution are the same.

Moles before = Moles after

M₁ V₁ = M₂ V₂

Example: 150 cm³ of 4 M KNO₃ → prepare 500 cm³ of 1 M:

4 × V₁ = 1 × 500

V₁ = 125\ cm³

Formula / Mathematical Form

M₁ × V₁ = M₂ × V₂

Where:

• M₁ = initial concentration (mol/dm³ or M)
• V₁ = initial volume (dm³ or cm³)
• M₂ = final concentration (mol/dm³ or M)
• V₂ = final volume (dm³ or cm³)

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

Solved Example

Example 3.15

If you have 150\ cm³ of a 4\ M potassium nitrate (KNO₃) solution, what volume of this solution will be needed to prepare 500\ cm³ of a 1\ M potassium nitrate solution?

Solution:

Given: M₁ = 4\ M (initial concentration) V₁ = ? (initial volume to find) M₂ = 1\ M (final concentration) V₂ = 500\ cm³ (final volume)

Formula:

M₁ × V₁ = M₂ × V₂

Substitution:

4\ M × V₁ = 1\ M × 500\ cm³

Calculation:

V₁ = 1 × 5004

V₁ = 125\ cm³

Final Answer: To prepare 500\ cm³ of a 1\ M potassium nitrate solution, you need to use 125\ cm³ of the 4\ M potassium nitrate solution.

Procedure: Mix 125\ cm³ of the 4\ M solution with enough water to make a total of 500\ cm³ to achieve the desired concentration.

Concept Assessment Exercise 3.11 (Part 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 (Note: The question asks for the volume of the original solution needed to make 200 cm³ of 2 M solution, so V₂ is the final volume.)

Formula:

M₁ × V₁ = M₂ × V₂

Substitution:

6 × V₁ = 2 × 200

Calculation:

V₁ = 2 × 2006

V₁ = 4006 = 66.67\ cm³

Final 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

• - Dilution = making a solution less concentrated by adding solvent.
• - The moles of solute remain constant during dilution.
• - Only the volume changes.
• - Formula:
• M₁ × V₁ = M₂ × V₂
• - M₁ = initial concentration; V₁ = initial volume.
• - M₂ = final concentration; V₂ = final volume.
• - The units of volume must be the same on both sides.
• - Dilution is used to prepare working solutions from stock solutions.
• - Example: 12 M HCl → 0.1 M HCl by dilution.
• - Example: 4 M KNO₃ → 1 M KNO₃ by dilution.

Remember

• - Dilution = adding solvent to reduce concentration.
• - Moles of solute stay the same.
• - Only volume changes.
• - M₁ V₁ = M₂ V₂.
• - Units of volume must match.
• - M₁, V₁ = initial (concentrated).
• - M₂, V₂ = final (diluted).
• - Example: 12 M HCl → 0.1 M HCl.
• - Example: 4 M KNO₃ → 1 M KNO₃.
• - Dilution is based on conservation of moles.

Common Mistakes

Quick Revision

• Dilution:
• Dilution is a process by which you make a solution less concentrated by adding more of the liquid (usually water).

Example: Suppose you have a cup of strong lemonade. If you add more water, the lemonade becomes weaker or less strong. This means that dilution reduces the strength or concentration of a solution.

Formula:

• M₁ × V₁ = M₂ × V₂

Where:

• M₁ is the concentration of the original (concentrated) solution.
• V₁ is the volume of the original solution you use.
• M₂ is the concentration you want after dilution.
• V₂ is the total volume of the final solution.

• Activity 3.1:
• Prepare 250\ cm³ solution of 0.1\ M from concentrated hydrochloric acid.

• Conc. HCl = 12 M
• 12 × V₁ = 0.1 × 250
• V₁ = 2.08\ cm³

• Example 3.15:
• If you have 150\ cm³ of a 4\ M KNO₃ solution, what volume of this solution will be needed to prepare