# Dilution

This is part of Year 11 HSC Chemistry course under the topic of Molarity

### HSC Chemistry Syllabus

• Manipulate variables and solve problems to calculate concentration, mass or volume using:
• Dilutions (number of moles before dilution = number of moles in sample after dilution.
• Conduct an investigation to make a standard solution and perform a dilution.

### Dilution

Definition:

Dilution refers to the process of decreasing the concnetration of a solution by adding more solvent to the existing stock solution. In most cases, this solvent is water. The primary equation governing this process is known as the dilution equation.

Dilution Equation

The dilution equation can be expressed as

C_1 \times V_1 = C_2 \times V_2

Where

C_1 : Concentration of the stock solution (before dilution) in mol L^-1

V_1 : Volume of stock solution before dilution in L

C_2 : Concentration of the new diluted solution

V_2 : Total volume of the new solution in L

### Examples

• Calculate the new concentration of a solution if water is added to a 100 mL solution of 0.25 M sodium chloride up till 1.5 L.

Solution:

Since C_1 = 0.25 M, V_1 = 0.1 L, and V_2 = 1.5 L, we can rearrange the equation to solve for C_2

$$C_2 = \frac{C_1 \times V_1}{V_2} = \frac{0.25\, \text{mol/L} \times 0.1\, \text{L}}{1.5\, \text{L}} = 0.017\, \text{mol/L}$$

• Calculate the volume in litres to which a 500 mL solution of 0.1 M hydrochloric acid solution must be diluted to make a new solution with a concentration of 0.001 M

Solution:

Given C_1 = 0.1M, V_1 = 0.5 L, and C_2 = 0.001 M, we can rearrange the formula to get:

$$V_2 = \frac{C_1 \times V_1}{C_2} = \frac{0.1\, \text{mol/L} \times 0.5\, \text{L}}{0.01\, \text{mol/L}} = 50\, \text{L}$$

• Calculate the original concentration of a copper sulfate standard solution if 30 mL of it was used to make a 300 mL dilute solution with a concentration of 0.1 M.

Solution:

Given V_1 = 0.03L, V_2 = 0.3 L, and C_2 = 0.1 M, we can rearrange the formula to get:

$$C_1 = \frac{V_1}{C_2 \times V_2} = \frac{0.03\, \text{L}}{0.1\, \text{mol/L} \times 0.3\, \text{L}} = 1\, \text{mol/L}$$

### Standard Solution

A standard solution is a solution whose concentration is known accurately. It's prepared by dissolving a primary standard, a specific solute, in an appropriate solvent

Primary Standard

A primary standard is a solute used to prepare a standard solution. It must fulfil the following criteria:

• High level of purity
• Accurately known composition
• Stable and unaffected by air when weighing
• Readily soluble (in water for high school-level studies)
• High molar weight
• Quickly and completely reacting

Certain substances, such as hydrochloric acid, are not suitable as primary standards due to their tendencies to fume, absorb moisture in the air, or effervesce when in contact with water.

### Summary

• Dilution is the process of reducing the concentration of a solution by adding more solvent.
• Standard solutions are known concentration prepared using a primary standard, which meets specific criteria
• Example problems demonstrate the application of the dilution equation for calculating new concentrations or volumes.