Using concentrations of solutions in mol/dm³ (chemistry only) (HT only)

What you'll learn

What concentration in mol/dm³ means, and why volume must usually be in dm³.
How to calculate moles of solute from concentration and volume.
How to calculate the mass of solute needed to make a solution.
How titration results can be used to find an unknown concentration.

The basic idea: what is in a solution?

A solution is a mixture where one substance is dissolved in another. The substance that dissolves is the solute. The liquid it dissolves in is the solvent.

For example, in sodium chloride solution, sodium chloride is the solute and water is the solvent.

Definition

Amount in moles

The amount of substance tells you how many particles are present. It is measured in moles, symbol mol. In calculations, we usually use the symbol $n$ for amount in moles.

The volume unit: dm³

Concentration in this topic is measured in moles per cubic decimetre, written as mol/dm³ or $dm−3\text{mol dm}^{-3}$ .

A cubic decimetre is the same volume as 1 litre.

cm31\text{ dm}^3 = 1000\text{ cm}^3

In practical work, volumes are often measured in cm³ using a pipette or burette, but the concentration equation needs volume in dm³.

Example

Converting cm³ to dm³

Convert 25.0 cm³ into dm³.

You are converting from cm³ to dm³, so divide by 1000.
Substitute the volume:
$\div 1000 = 0.0250\text{ dm}^3$
Keep the value as 0.0250 dm³ because the zeros show the precision of the measurement.

Common Mistake

Using cm³ directly

Do not put a volume in cm³ straight into the concentration equation. Always convert cm³ to dm³ first by dividing by 1000.

Concentration in mol/dm³

Definition

Concentration in mol/dm³

The concentration of a solution in mol/dm³ is the amount of solute, in moles, dissolved in 1 dm³ of solution.

The main equation is:

\frac{n}{V}

where:

$c$ is concentration in mol/dm³
$n$ is amount of solute in mol
$V$ is volume of solution in dm³

You can rearrange it depending on what you need:

n = c V

and

\frac{n}{c}

Key Idea

What concentration really compares

Concentration compares amount of solute with volume of solution. A solution is more concentrated if it has more moles of solute per dm³.

Calculating moles from concentration and volume

If you know the concentration and the volume, you can find the amount of solute using:

n = c V

Example

Calculating amount of solute

A sodium chloride solution has a concentration of 0.200 mol/dm³. Calculate the amount of sodium chloride in 250 cm³ of this solution.

Convert the volume into dm³:
$\div 1000 = 0.250\text{ dm}^3$
Use the equation $n = c V$ and substitute the values:
$\times 0.250$
Calculate the amount:
$0.0500\text{ mol}$

So there are 0.0500 mol of sodium chloride in 250 cm³ of the solution.

Tip

Sanity check

If the volume is less than 1 dm³, the number of moles should usually be smaller than the concentration value.

Linking concentration to mass

You also need to connect concentration with the mass of solute.

From earlier quantitative chemistry:

\frac{m}{M_r}

where:

$m$ is mass in grams
$M_r$ is relative formula mass, in g/mol for these calculations
$n$ is amount in mol

Since $n = c V$ , you can combine the equations:

m = cVM_r

This means the mass of solute depends on:

the concentration of the solution
the volume of solution being made
the relative formula mass of the solute

How mass, volume and concentration are related

For the same solute:

increasing the mass of solute in the same volume increases the concentration
increasing the volume while keeping the same mass decreases the concentration
halving the volume without changing the moles doubles the concentration

Another useful form is:

\frac{m}{M_rV}

Example

Finding the mass needed to make a solution

Calculate the mass of sodium carbonate, Na₂CO₃, needed to make 250 cm³ of a 0.100 mol/dm³ solution. The relative formula mass of Na₂CO₃ is 106.

Convert the volume into dm³:
$\div 1000 = 0.250\text{ dm}^3$
Use the equation $m = cVM_r$ :
$\times 0.250 \times 106$
Calculate the mass:
$2.65\text{ g}$

So you need 2.65 g of sodium carbonate.

Common Mistake

Volume of solution, not volume of water

If making a solution, the final volume must be 250 cm³ of solution. You do not simply add 250 cm³ of water to the solid, because the solid also affects the final volume.

Using concentrations in reacting solutions

This Higher Tier chemistry-only section also links concentration calculations to reacting ratios.

If two solutions react completely and you know:

the volume of both solutions
the concentration of one solution
the balanced symbol equation

then you can calculate the concentration of the other solution.

The balanced equation is essential because it tells you the mole ratio.

Titrations and unknown concentrations

A titration is a practical method used to find the concentration of a solution by reacting it with another solution of known concentration.

The solution of known concentration is usually placed in a burette, so its volume can be measured accurately. A known volume of the unknown solution is measured using a pipette and placed in a conical flask with an indicator.

Labelled titration setup with burette, pipette, conical flask, indicator and white tile

Definition

Titre

The titre is the volume delivered from the burette during a titration. Repeated titres that are very close together are called concordant results.

At the endpoint, the reacting solutions have reacted completely in the mole ratio shown by the balanced equation.

The general titration method for calculations

Use this reliable structure:

Convert all volumes from cm³ to dm³.
Calculate moles of the known solution using $n = c V$ .
Use the balanced equation to find the mole ratio.
Calculate moles of the unknown solution.
Use $\frac{n}{V}$ to find the unknown concentration.

Example

Calculating an unknown concentration from a titration

25.0 cm³ of sodium hydroxide solution is neutralised by 12.50 cm³ of 0.0800 mol/dm³ sulfuric acid. Calculate the concentration of the sodium hydroxide solution.

The equation is:

H₂SO₄(aq) + 2NaOH(aq) → Na₂SO₄(aq) + 2H₂O(l)

Convert the sulfuric acid volume into dm³:
$\div 1000 = 0.01250\text{ dm}^3$
Calculate the moles of sulfuric acid using $n = c V$ :
$\times 0.01250 = 0.00100\text{ mol}$
Use the equation ratio. 1 mol of H₂SO₄ reacts with 2 mol of NaOH, so:
$moln(\text{NaOH}) = 2 \times 0.00100 = 0.00200\text{ mol}$
Convert the sodium hydroxide volume into dm³:
$\div 1000 = 0.0250\text{ dm}^3$
Calculate the sodium hydroxide concentration:
$\frac{0.00200}{0.0250} = 0.0800\text{ mol/dm}^3$

The concentration of the sodium hydroxide solution is 0.0800 mol/dm³.

Common Mistake

The 1:1 shortcut is not always valid

For reactions like HCl(aq) + NaOH(aq) → NaCl(aq) + H₂O(l), the ratio is 1:1, so the moles of acid and alkali are equal at neutralisation. For equations like H₂SO₄(aq) + 2NaOH(aq), they are not equal.

Dilution: changing volume changes concentration

If you add more solvent to a solution, you dilute it. The moles of solute stay the same, but the volume increases, so the concentration decreases.

Key Idea

Dilution

During dilution, the amount of solute stays constant, so increasing the volume lowers the concentration.

Example

Calculating concentration after dilution

A solution contains 0.0500 mol of solute. It is diluted to a final volume of 500 cm³. Calculate the new concentration.

Convert the final volume into dm³:
$\div 1000 = 0.500\text{ dm}^3$
Use $\frac{n}{V}$ :
$\frac{0.0500}{0.500}$
Calculate the concentration:
$0.100\text{ mol/dm}^3$

The diluted solution has a concentration of 0.100 mol/dm³.

Exam technique

In the exam

Convert cm³ to dm³ before using concentration in mol/dm³.
Write down the balanced equation and use its mole ratio, especially if the reaction is not 1:1.
Carry units through your working so the examiner can see whether you are calculating moles, mass, volume or concentration.

Self review

Check yourself

Why must 25.0 cm³ be converted to 0.0250 dm³ before using $n = c V$ ?
How is concentration in mol/dm³ related to mass of solute and volume of solution?
In a titration calculation, why is the balanced equation needed?

Using concentrations of solutions in mol/dm³ (chemistry only) (HT only)

What you'll learn

What concentration in mol/dm³ means, and why volume must usually be in dm³.
How to calculate moles of solute from concentration and volume.
How to calculate the mass of solute needed to make a solution.
How titration results can be used to find an unknown concentration.

The basic idea: what is in a solution?

A solution is a mixture where one substance is dissolved in another. The substance that dissolves is the solute. The liquid it dissolves in is the solvent.

For example, in sodium chloride solution, sodium chloride is the solute and water is the solvent.

Definition

Amount in moles

The amount of substance tells you how many particles are present. It is measured in moles, symbol mol. In calculations, we usually use the symbol $n$ for amount in moles.

The volume unit: dm³

Concentration in this topic is measured in moles per cubic decimetre, written as mol/dm³ or $dm−3\text{mol dm}^{-3}$ .

A cubic decimetre is the same volume as 1 litre.

cm31\text{ dm}^3 = 1000\text{ cm}^3

In practical work, volumes are often measured in cm³ using a pipette or burette, but the concentration equation needs volume in dm³.

Example

Converting cm³ to dm³

Convert 25.0 cm³ into dm³.

You are converting from cm³ to dm³, so divide by 1000.
Substitute the volume:
$\div 1000 = 0.0250\text{ dm}^3$
Keep the value as 0.0250 dm³ because the zeros show the precision of the measurement.

Common Mistake

Using cm³ directly

Do not put a volume in cm³ straight into the concentration equation. Always convert cm³ to dm³ first by dividing by 1000.

Concentration in mol/dm³

Definition

Concentration in mol/dm³

The concentration of a solution in mol/dm³ is the amount of solute, in moles, dissolved in 1 dm³ of solution.

The main equation is:

\frac{n}{V}

where:

$c$ is concentration in mol/dm³
$n$ is amount of solute in mol
$V$ is volume of solution in dm³

You can rearrange it depending on what you need:

n = c V

and

\frac{n}{c}

Key Idea

What concentration really compares

Concentration compares amount of solute with volume of solution. A solution is more concentrated if it has more moles of solute per dm³.

Calculating moles from concentration and volume

If you know the concentration and the volume, you can find the amount of solute using:

n = c V

Example

Calculating amount of solute

A sodium chloride solution has a concentration of 0.200 mol/dm³. Calculate the amount of sodium chloride in 250 cm³ of this solution.

Convert the volume into dm³:
$\div 1000 = 0.250\text{ dm}^3$
Use the equation $n = c V$ and substitute the values:
$\times 0.250$
Calculate the amount:
$0.0500\text{ mol}$

So there are 0.0500 mol of sodium chloride in 250 cm³ of the solution.

Tip

Sanity check

If the volume is less than 1 dm³, the number of moles should usually be smaller than the concentration value.

Linking concentration to mass

You also need to connect concentration with the mass of solute.

From earlier quantitative chemistry:

\frac{m}{M_r}

where:

$m$ is mass in grams
$M_r$ is relative formula mass, in g/mol for these calculations
$n$ is amount in mol

Since $n = c V$ , you can combine the equations:

m = cVM_r

This means the mass of solute depends on:

the concentration of the solution
the volume of solution being made
the relative formula mass of the solute

How mass, volume and concentration are related

For the same solute:

increasing the mass of solute in the same volume increases the concentration
increasing the volume while keeping the same mass decreases the concentration
halving the volume without changing the moles doubles the concentration

Another useful form is:

\frac{m}{M_rV}

Example

Finding the mass needed to make a solution

Calculate the mass of sodium carbonate, Na₂CO₃, needed to make 250 cm³ of a 0.100 mol/dm³ solution. The relative formula mass of Na₂CO₃ is 106.

Convert the volume into dm³:
$\div 1000 = 0.250\text{ dm}^3$
Use the equation $m = cVM_r$ :
$\times 0.250 \times 106$
Calculate the mass:
$2.65\text{ g}$

So you need 2.65 g of sodium carbonate.

Common Mistake

Volume of solution, not volume of water

If making a solution, the final volume must be 250 cm³ of solution. You do not simply add 250 cm³ of water to the solid, because the solid also affects the final volume.

Using concentrations in reacting solutions

This Higher Tier chemistry-only section also links concentration calculations to reacting ratios.

If two solutions react completely and you know:

the volume of both solutions
the concentration of one solution
the balanced symbol equation

then you can calculate the concentration of the other solution.

The balanced equation is essential because it tells you the mole ratio.

Titrations and unknown concentrations

A titration is a practical method used to find the concentration of a solution by reacting it with another solution of known concentration.

Labelled titration setup with burette, pipette, conical flask, indicator and white tile

Definition

Titre

The titre is the volume delivered from the burette during a titration. Repeated titres that are very close together are called concordant results.

At the endpoint, the reacting solutions have reacted completely in the mole ratio shown by the balanced equation.

The general titration method for calculations

Use this reliable structure:

Convert all volumes from cm³ to dm³.
Calculate moles of the known solution using $n = c V$ .
Use the balanced equation to find the mole ratio.
Calculate moles of the unknown solution.
Use $\frac{n}{V}$ to find the unknown concentration.

Example

Calculating an unknown concentration from a titration

25.0 cm³ of sodium hydroxide solution is neutralised by 12.50 cm³ of 0.0800 mol/dm³ sulfuric acid. Calculate the concentration of the sodium hydroxide solution.

The equation is:

H₂SO₄(aq) + 2NaOH(aq) → Na₂SO₄(aq) + 2H₂O(l)

Convert the sulfuric acid volume into dm³:
$\div 1000 = 0.01250\text{ dm}^3$
Calculate the moles of sulfuric acid using $n = c V$ :
$\times 0.01250 = 0.00100\text{ mol}$
Use the equation ratio. 1 mol of H₂SO₄ reacts with 2 mol of NaOH, so:
$moln(\text{NaOH}) = 2 \times 0.00100 = 0.00200\text{ mol}$
Convert the sodium hydroxide volume into dm³:
$\div 1000 = 0.0250\text{ dm}^3$
Calculate the sodium hydroxide concentration:
$\frac{0.00200}{0.0250} = 0.0800\text{ mol/dm}^3$

The concentration of the sodium hydroxide solution is 0.0800 mol/dm³.

Common Mistake

The 1:1 shortcut is not always valid

Dilution: changing volume changes concentration

If you add more solvent to a solution, you dilute it. The moles of solute stay the same, but the volume increases, so the concentration decreases.

Key Idea

Dilution

During dilution, the amount of solute stays constant, so increasing the volume lowers the concentration.

Example

Calculating concentration after dilution

A solution contains 0.0500 mol of solute. It is diluted to a final volume of 500 cm³. Calculate the new concentration.

Convert the final volume into dm³:
$\div 1000 = 0.500\text{ dm}^3$
Use $\frac{n}{V}$ :
$\frac{0.0500}{0.500}$
Calculate the concentration:
$0.100\text{ mol/dm}^3$

The diluted solution has a concentration of 0.100 mol/dm³.

Exam technique

In the exam

Convert cm³ to dm³ before using concentration in mol/dm³.
Write down the balanced equation and use its mole ratio, especially if the reaction is not 1:1.
Carry units through your working so the examiner can see whether you are calculating moles, mass, volume or concentration.

Self review

Check yourself

Why must 25.0 cm³ be converted to 0.0250 dm³ before using $n = c V$ ?
How is concentration in mol/dm³ related to mass of solute and volume of solution?
In a titration calculation, why is the balanced equation needed?

Using concentrations of solutions in mol/dm³ (chemistry only) (HT only)

What you'll learn

The basic idea: what is in a solution?

The volume unit: dm³

Concentration in mol/dm³

Calculating moles from concentration and volume

Linking concentration to mass

How mass, volume and concentration are related

Using concentrations in reacting solutions

Titrations and unknown concentrations

The general titration method for calculations

Dilution: changing volume changes concentration

Recap questions

You've reached the end

Using concentrations of solutions in mol/dm³ (chemistry only) (HT only)

What you'll learn

The basic idea: what is in a solution?

The volume unit: dm³

Concentration in mol/dm³

Calculating moles from concentration and volume

Linking concentration to mass

How mass, volume and concentration are related

Using concentrations in reacting solutions

Titrations and unknown concentrations

The general titration method for calculations

Dilution: changing volume changes concentration

Recap questions

You've reached the end

Using concentrations of solutions in mol/dm3 (chemistry only) (HT only)

What you'll learn

The basic idea: what is in a solution?

The volume unit: dm³

Concentration in mol/dm³

Calculating moles from concentration and volume

Linking concentration to mass

How mass, volume and concentration are related

Using concentrations in reacting solutions

Titrations and unknown concentrations

The general titration method for calculations

Dilution: changing volume changes concentration

Recap questions

You've reached the end

Using concentrations of solutions in mol/dm<sup>3</sup> (chemistry only) (HT only) Revision Guide

Using concentrations of solutions in mol/dm3 (chemistry only) (HT only)

What you'll learn

The basic idea: what is in a solution?

The volume unit: dm³

Concentration in mol/dm³

Calculating moles from concentration and volume

Linking concentration to mass

How mass, volume and concentration are related

Using concentrations in reacting solutions

Titrations and unknown concentrations

The general titration method for calculations

Dilution: changing volume changes concentration

Recap questions

You've reached the end

Using concentrations of solutions in mol/dm<sup>3</sup> (chemistry only) (HT only) Revision Guide

Using concentrations of solutions in mol/dm³ (chemistry only) (HT only)

Using concentrations of solutions in mol/dm³ (chemistry only) (HT only)