Calculating Concentration (VCE SSCE Chemistry): Revision Notes

Calculating Concentration

Understanding how to calculate and express the concentration of solutions is a fundamental skill in chemistry. Whether you're preparing medicines, analyzing water quality, or conducting laboratory experiments, knowing the exact amount of solute in a solution is essential.

What is concentration?

The concentration of a solution tells us the relationship between the amount of solute dissolved and the amount of solvent present. When we describe a solution as concentrated, we mean it has a high ratio of solute to solvent. Think of undiluted cordial - it has a lot of flavouring dissolved in a small amount of liquid. On the other hand, a dilute solution has a low ratio of solute to solvent, like a quarter teaspoon of sugar in a litre of water.

The accuracy of pharmaceutical concentrations can be a matter of life and death. Medications must contain the exact amount of active ingredient specified on the label to ensure patient safety and treatment effectiveness.

Understanding concentration units

Concentration measurements always have two components:

1. Information about the amount of solute
2. Information about the amount of solution

For example, if sodium chloride has a concentration of $17 \text{ g L}^{-1}$, this tells us that $17 \text{ g}$ of $\text{NaCl}$ (the solute amount) is dissolved in every $1 \text{ L}$ of the solution (the solution amount).

Chemists use different concentration units depending on the context and application. Let's explore the most common ones.

Concentration in grams per litre ($\text{g L}^{-1}$)

This concentration unit expresses the mass of solute, measured in grams, that is dissolved in one litre of solution. It's a straightforward way to describe concentration that directly relates mass to volume.

The formula for calculating concentration in grams per litre is:

$\text{concentration (g L}^{-1}\text{)} = \frac{\text{mass of solute (in g)}}{\text{volume of solution (in L)}}$

This unit is practical when you're working with solid solutes dissolved in liquid solvents. For instance, if seawater contains sodium chloride at a concentration of $20 \text{ g L}^{-1}$, every litre of seawater contains $20 \text{ g}$ of dissolved salt.

Concentration in parts per million (ppm)

When dealing with very dilute solutions, such as trace contaminants in water or food, we use parts per million. This unit is more convenient than using very small decimal numbers.

Fish sold commercially must meet strict safety standards. In Australia, mercury levels in fish cannot exceed $1 \text{ ppm}$ to be safe for consumption.

The formula for parts per million is:

$\text{concentration (ppm)} = \frac{\text{mass of solute (in mg)}}{\text{mass of solution (in kg)}}$

This works because there are one million milligrams in one kilogram. So if a solution contains $154 \text{ mg}$ of sodium chloride in $1 \text{ kg}$ of solution, the concentration is $154 \text{ ppm}$.

Threshold limit values

Electronic highway signs display ozone warnings when air quality becomes a health concern.

Other concentration units on consumer products

You'll often see different concentration units on everyday products in supermarkets and pharmacies. These units are chosen because they're practical and familiar to consumers.

Household cleaning products display various concentration units on their labels.

Percentage by volume: % (v/v)

This unit is used when both the solute and solution are measured by volume, typically for liquid solutes. The volumes must use the same units.

$\text{concentration \% (v/v)} = \frac{\text{volume of solute (in mL)}}{\text{volume of solution (in mL)}} \times 100\%$

Wine labels show alcohol content as a percentage by volume. A wine labeled $14.5\%$ (v/v) contains $14.5 \text{ mL}$ of alcohol in every $100 \text{ mL}$ of wine.

Percentage mass/volume: % (m/v)

This unit expresses the mass of solute in grams that is present in 100 mL of solution. Sometimes you'll see this written as % (w/v) on product labels, where "w" stands for weight.

$\text{concentration \% (m/v)} = \frac{\text{mass of solute (in g)}}{\text{volume of solution (in mL)}} \times 100$

A saline drip used in medical procedures typically contains sodium chloride at a concentration of $0.9\%$ (m/v), meaning $100 \text{ mL}$ of the solution contains $0.9 \text{ g}$ of dissolved salt.

Concentration in moles per litre ($\text{mol L}^{-1}$)

While units like $\text{g L}^{-1}$ and percentages appear on consumer products, chemists prefer a different unit for laboratory work: moles per litre, or $\text{mol L}^{-1}$. This unit is also called molarity and given the symbol M.

The reason chemists favour this unit is that it allows direct comparison of the number of particles (atoms, molecules, or ions) in solutions. This makes quantitative calculations much easier.

The formula for molar concentration is:

$\text{concentration (mol L}^{-1}\text{)} = \frac{\text{amount of solute (in mol)}}{\text{volume of solution (in L)}}$

Using symbols, we write this as:

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

where $c$ is the concentration in $\text{mol L}^{-1}$, $n$ is the amount in moles, and $V$ is the volume in litres.

Laboratory bottles of dilute acids showing their concentrations in $\text{mol L}^{-1}$ (or M).

Understanding molarity

A solution with a concentration of $1 \text{ mol L}^{-1}$ (or $1 \text{ M}$) contains one mole of solute dissolved in one litre of solution. We can describe this solution in several equivalent ways:

• It has a concentration of one mole per litre
• It has a concentration of $1 \text{ mol L}^{-1}$
• It is $1$ molar
• It is $1 \text{ M}$
• It has a molarity of $1 \text{ M}$

For instance, a $0.1 \text{ M}$ solution of hydrochloric acid contains $0.1$ moles of $\text{HCl}$ in $1 \text{ L}$ of solution.

Calculating molarity from mass of solute

Often you'll know the mass of a solute and need to calculate the molarity. This requires two main steps:

1. Calculate the number of moles of solute from its mass using $n = \frac{m}{M}$
2. Calculate the concentration using the number of moles and volume using $c = \frac{n}{V}$

Calculating moles of solute in a solution

We can rearrange the molarity formula $c = \frac{n}{V}$ to find the number of moles of solute in a solution of known concentration and volume:

$n = c \times V$

where $n$ is the amount in moles, $c$ is the concentration in $\text{mol L}^{-1}$, and $V$ is the volume in litres.

Dilution of solutions

Many commercial products, from pesticides to cleaning products, are sold as concentrates. This reduces transportation costs and gives users flexibility to prepare solutions of different strengths as needed.

Agricultural pesticide solutions are prepared by diluting concentrated stock solutions.

Common examples of dilution include:

• Adding water to cordial before drinking
• A laboratory technician preparing $1 \text{ M}$ hydrochloric acid from concentrated acid
• A gardener diluting fertiliser concentrate for lawn application
• A farmer diluting weedkiller for crop spraying
• Kitchen staff diluting concentrated detergent for dishwashing

Understanding the dilution process

Dilution is the process of adding more solvent to a solution, which decreases its concentration. The crucial point to understand is that dilution doesn't change the amount of solute present - it only changes how spread out the solute particles are.

When we dilute a solution by adding more solvent, the solute particles become more spread out, but their total number remains the same.

For example, if $50 \text{ mL}$ of water is added to $50 \text{ mL}$ of $0.10 \text{ mol L}^{-1}$ sugar solution, the amount of sugar stays the same, but the volume doubles. The concentration therefore halves to $0.050 \text{ mol L}^{-1}$.

The dilution formula

Because the amount of solute remains constant during dilution, we can derive a useful relationship. If we start with volume $V_1$ at concentration $c_1$, the amount of solute is:

$n_1 = c_1 \times V_1$

After dilution to volume $V_2$ at concentration $c_2$, the amount of solute is:

$n_2 = c_2 \times V_2$

Since $n_1 = n_2$ (the amount of solute hasn't changed):

$c_1V_1 = c_2V_2$

Safety when diluting strong acids

The golden rule: always add acid to water. Adding water to concentrated acid can cause violent boiling and splashing of dangerous acid.

The reason for adding acid to water (not the reverse) is that concentrated acid is denser than water. When acid is added to water, it sinks and mixes naturally. The heat generated disperses throughout the water. If water were added to acid, the heat would be concentrated at the interface between the liquids, potentially causing violent boiling and ejection of hot, concentrated acid.

Converting between concentration units

Sometimes we need to convert a concentration from one unit to another. This can be done systematically by working through a series of steps.