# Concentration and Standard Solution – HSC Chemistry

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

### HSC Chemistry Syllabus

• Conduct practical investigations to determine the concentrations of solutions and investigate the different ways in which concentrations are measured
• Manipulate variables and solve problems to calculate concentration, mass or volume using:
– c = \frac{n}{V} (molarity formula (ACSCH063))

### What is Concentration?

Concentration describes the amount of solute present in a specified volume of solvent.

A solution is a homogeneous mixture wherein a solute is dissolved in a solvent. For instance, salt water consists of the salt solute dissolved in water as the solvent.

• The descriptors 'dilute' and 'concentrated' indicate the relative concentrations of solutions.
• A solution is termed 'unsaturated' if it can dissolve more solute at a given temperature.
• Conversely, a 'saturated' solution cannot dissolve any more solute at a specified temperature.

### Concentration as Molarity (mol L–1)

Molarity (mol/L or M) is a common way to express the concentration of a solution, given by the number of moles in a litre.

Molarity is given by:

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

where:

• c = concentration of solution in mol L–1
• n = moles of substance being dissolved
• V = volume of solution in litres (L)

### Molarity Calculation Examples

1. Calculate the molarity of a 0.50 L solution containing 0.125 moles of NaCl.

$$c = \frac{0.125}{0.50} = 0.25 \text{ mol/L (2 s.f.)}$$

2. Calculate the number of moles of HCl in 300.0 mL of a 2.00 mol L–1 HCl solution.

$$n = c \times V$$
$$n = 2.00 \times 0.3 = 0.6 \text{ mol (3 s.f.)}$$

3. Calculate the volume of water that needs to be added to 0.30 moles of NaOH to make a 0.50 M solution.

$$V = \frac{n}{c}$$
$$V = \frac{0.30}{0.50} = 0.60 \text{ L}$$

### Concentration Units

Since there are numerous ways to describe the quantity of solutes and the amount of solvent, the unit for concentration can also be expressed in various ways.

 Concentration: amount of solute present in solution Concentration term Amount of solute Amount of solution Concentration units Mass/volume Mass Volume g L-1 Percent by mass % w/w or %m/m Mass Mass g /100 g Mass/volume percentage w/v% or m/v% Mass Volume g /100 mL Volumes/volume percentage v/v% Volume Volume mL /100 g Parts per million ppm Mass Mass mg kg-1, mg g-1 Mass Volume mg L-1, mg mL-1 Mole percent % Moles Moles % Molarity M Moles Volume mol L-1 Molar, M

### Concentration in Grams Per Litre (g L–1)

The concentration of a solution in grams per litre (g L–1) indicates the mass of the solute in grams dissolved in one litre of the solution.

Here's a practical example:

If the concentration of sodium chloride in seawater is 20 g L–1, this means that there are 20 g of sodium chloride in 1 L of seawater

To calculate concentration in g L–1 we simply divide the mass of solute by the volume in litres.

$$c= \frac{\text{mass of solute (g)}}{\text{volume of solution (L)}}$$

### Converting Between Molarity and g L–1

Recall that the number of moles of a substance is given by:

$$n = \frac{m}{MM}$$

where M is the molar mass of the substance.

To convert from molarity (mol L–1) to g L–1, we simply need to divide the concentration by the molar mass.

To convert from g L–1 to mol L–1, we multiply the concentration by the molar mass.

#### Examples

1. 10.0 g of NaOH is added to 250.0 mL of water. Calculate the molarity of the solution.

$$n = \frac{m}{MM}$$

$$n = \frac{10.0}{22.99 + 16.00 +1.008} = 0.25 \text{ mol}$$

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

$$c = \frac{0.25}{0.25} = 1.00 \text{ mol/L (3 s.f.)}$$

2. Convert 2.00 mol L–1 of HCl solution to g L–1.

$$MM = 1.008 + 35.45 = 36.458 \text{ g/mol}$$

$$c = 2.00 \times 36.458 = 72.9 \text{ g/L (3 s.f.)}$$

### Concentration in Parts Per Million (ppm)

Parts per million (ppm) is often used for very dilute solutions, and it can be expressed as weight/weight (w/w) or weight/volume (w/v):

• Weight/volume:  1 ppm = 1 mg L–1 = 1 µg L–1
• Weight/weight: 1 ppm = 1 mg kg–11 µg kg–1

#### Examples

• 2 L–1 to ppm:

Convert grams to milligrams: 2 g = 2000 mg

The concentration is 2000 ppm

• 0.33 g kg–1 to ppm:

Convert grams to milligrams: 0.33 g = 330 mg

The concentration is 330 ppm

### What is a 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.