# Dilution – HSC Chemistry

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.

### What is Dilution?

Dilution refers to the process of decreasing the concentration of a solution by adding more volume of solvent to the existing solution. In most cases, this solvent is water.

The most important principle to understand in dilution of a solution is that the number of moles of solute remains unchanged with the addition of solvent.

Therefore, the number of moles of solute before dilution equals the number of moles of solute after dilution, as given by:

$$n_{\text{before}} = n_{\text{after}}$$

Since the number of moles of a substance is given by:

$$n = c \times V$$

We get:

c_1 \times V_1 = c_2 \times V_2

where

• c_1 is the concentration of the original solution before dilution in mol L–1
• V_1 is the volume of the original solution before dilution in litres (L)
• c_2 is the concentration of the new diluted solution  in mol L–1
• V_2 is the new volume of the diluted solution in litres (L)

### Example 1

Water is added to 100.0 mL of 0.250 M sodium chloride solution until the volume reaches 1.50 L. Calculate the new concentration of this solution.

Solution:

The number of moles of sodium chloride before dilution is given by:

$$n = 0.250 \times 0.100 = 0.0250 \text{ mol}$$

After dilution, the number of moles of NaCl remains the same.

Therefore,

$$c_{\text{new}} = \frac{0.0250}{1.50} = 0.0167 \text{ mol/L (3 s.f.)}$$

﻿Alternatively, we can use the dilution equation:

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:

$$c_2 = \frac{c_1 \times V_1}{V_2} = \frac{0.25\, \text{mol/L} \times 0.1\, \text{L}}{1.5\, \text{L}} = 0.0167\, \text{mol/L}$$

### Example 2

How many litres of water needs to be added to a 500.0 mL solution of 0.10 M hydrochloric acid solution to make a 0.010 M solution?

Solution:
The number of moles of HCl is given by:

$$n = c \times V = (0.10)(0.5)$$
$$n = 0.050 \text{ mol}$$

The new final volume of solution is given by:

$$V = \frac{n}{c}$$
$$V = \frac{0.050}{0.010}$$
$$V = 5.0 \text{ L}$$

$$\text{Volume added} = 5.0 - 0.5 = 4.5 \text{ L (2 s.f.)}$$

### Example 3

200.0 mL of water is added to 30.0 mL of a copper sulfate solution. The final concentration of the solution is 0.120 mol L–1.

What is the original concentration of the copper sulfate solution?

Solution:
The number of moles of copper sulfate is given by:

$$n = c \times V = (0.120)(0.200 + 0.030)$$
$$n = 0.0276 \text{ mol}$$
The original solution would have contained the same number of moles of copper sulfate.

$$c = \frac{n}{V} = \frac{0.0276}{0.030}$$

$$c = 0.920 \text{ mol/L (3 s.f.)}$$

### Example 4

2.00 g of solid NaOH is added to 300.0 mL of water. 150.0 mL of this solution was transferred to a new beaker and diluted with 250.0 mL of water.

What is the final molarity of the NaOH solution?

Solution:

Let's first find the number of moles of NaOH added:

$$n = \frac{m}{MM} = \frac{2.00}{22.99 + 16.00 + 1.008}$$

$$n = 0.0500 \text{ mol}$$

Initial molarity:

$$c = \frac{n}{V} = \frac{0.0500}{0.300}$$

$$c = 0.167 \text{ mol/L}$$

The number of moles in the 150.0 mL of solution that is transferred to a new beaker is not the same as the original solution. However, this portion of the solution has the same concentration.

$$n_{\text{transferred}} = c \times V = 0.167 \times 0.150$$

$$n_{\text{transferred}} = 0.025 \text{ mol}$$

Lastly, during dilution with 250.0 mL of water, the number of moles of NaOH remains unchanged. The final volume = 150 + 250 = 400.0 mL = 0.400 L.

$$c = \frac{n}{V}$$

$$c = \frac{0.025}{0.400}$$

$$c = 0.0625 \text{ mol/L (3 s.f.)}$$