Concentration is moles of solute per cubic decimetre

Definition

Concentration

The amount of solute dissolved in a stated volume of solution, measured in g/dm³ or mol/dm³.

The relationship is: $(dm3)\text{concentration (mol/dm}^3\text{)} = \frac{\text{amount of solute (mol)}}{\text{volume of solution (dm}^3\text{)}}$
Convert a volume in $cm3\text{cm}^3$ into $dm3\text{dm}^3$ by dividing by 1000.
Rearranging gives amount in moles $=$ concentration $×\times$ $\times$ volume in $dm3\text{dm}^3$ $dm^{3}$ .
1. So $cm320.0\,\text{cm}^3$ of a $mol/dm30.100\,\text{mol/dm}^3$ solution contains $0.100 \times 0.0200 = 0.00200\,\text{mol}$ of solute.
Doubling the amount of solute in the same volume doubles the concentration, while doubling the volume for the same amount halves it.

Example

A solution contains $mol0.0500\,\text{mol}$ of sodium chloride in $cm3250\,\text{cm}^3$ ; calculate its concentration.
Convert the volume: $250 \div 1000 = 0.250\,\text{dm}^3$ .
Concentration $\dfrac{0.0500}{0.250} = 0.200\,\text{mol/dm}^3$ .

Relative formula mass links solute mass and concentration

Definition

Relative formula mass

The mass of one formula unit of a substance compared with one-twelfth of the mass of a carbon-12 atom, given the symbol Mr.

Convert a mass of solute into moles with amount $\dfrac{\text{mass (g)}}{M_r}$ .
Then divide the amount by the volume of solution in $dm3\text{dm}^3$ to find the concentration in $mol/dm3\text{mol/dm}^3$ .
Working backwards, mass of solute $=$ concentration $×\times$ $\times$ volume $×Mr\times M_r$ $\times M_{r}$ .
1. Making $cm3500\,\text{cm}^3$ of $mol/dm30.100\,\text{mol/dm}^3$ copper(II) sulfate, $M_r = 159.6$ , needs $0.100 \times 0.500 \times 159.6 = 7.98\,\text{g}$ of solid.
To change a concentration in $mol/dm3\text{mol/dm}^3$ into $g/dm3\text{g/dm}^3$ , multiply by $M_r$ , and to go the other way, divide by $M_r$ .

Example

Calculate the concentration in $mol/dm3\text{mol/dm}^3$ of a solution made by dissolving $g4.00\,\text{g}$ of sodium hydroxide to make $cm3250\,\text{cm}^3$ of solution ( $A_r$ : Na = 23.0, O = 16.0, H = 1.0).
$Mr(NaOH)=23.0+16.0+1.0=40.0M_r(\text{NaOH}) = 23.0 + 16.0 + 1.0 = 40.0$ .
Amount $\dfrac{4.00}{40.0} = 0.100\,\text{mol}$ .
Concentration $\dfrac{0.100}{0.250} = 0.400\,\text{mol/dm}^3$ .

The mole ratio links reacting volumes and concentrations

Definition

Mole ratio

The ratio of the amounts in moles of substances that react or form, given by the balancing numbers in the equation.

At the end point, the two substances have reacted completely, so their amounts in moles are in the ratio shown by the balanced equation.
The amount of the substance with a known concentration comes from concentration $×\times$ volume.
In $HCl(aq)+NaOH(aq)→NaCl(aq)+H2O(l)\text{HCl(aq)} + \text{NaOH(aq)} \rightarrow \text{NaCl(aq)} + \text{H}_2\text{O(l)}$ $HCl(aq) + NaOH(aq) \to NaCl(aq) + H_{2} O(l)$ , the ratio is 1 : 1, so equal volumes react when the concentrations are equal.
1. If the sodium hydroxide were twice as concentrated, only half the volume would be needed, because each $cm3\text{cm}^3$ would hold twice as many moles.
In $H2SO4(aq)+2NaOH(aq)→Na2SO4(aq)+2H2O(l)\text{H}_2\text{SO}_4\text{(aq)} + 2\text{NaOH(aq)} \rightarrow \text{Na}_2\text{SO}_4\text{(aq)} + 2\text{H}_2\text{O(l)}$ $H_{2} SO_{4} (aq) + 2 NaOH(aq) \to Na_{2} SO_{4} (aq) + 2 H_{2} O(l)$ , each mole of acid reacts with two moles of alkali.
1. So for solutions of equal concentration, the volume of sodium hydroxide needed is twice the volume of sulfuric acid.

Common Mistake

Do not assume every acid reacts with an alkali in a 1 : 1 ratio; always read the ratio from the balanced equation.
Do not put a volume in $cm3\text{cm}^3$ into concentration $×\times$ volume, because the concentration is per $dm3\text{dm}^3$ .

Titration results give an unknown concentration

Definition

Titration

A technique that finds the volume of one solution needed to react completely with a measured volume of another solution.

The pipette gives the fixed volume of one solution and the mean titre gives the volume of the other.
Write the balanced equation first, so you know the mole ratio.
Find the amount in moles of the solution whose concentration is known.
Use the mole ratio to find the amount in moles of the other substance.
Divide that amount by its volume in $dm3\text{dm}^3$ to find the unknown concentration.

Example

Concentration of sodium hydroxide from a titration

$cm325.0\,\text{cm}^3$ of sodium hydroxide solution is neutralised by a mean titre of $cm320.00\,\text{cm}^3$ of $mol/dm30.100\,\text{mol/dm}^3$ sulfuric acid; find the concentration of the sodium hydroxide.
Amount of $mol\text{H}_2\text{SO}_4 = 0.100 \times 0.02000 = 0.002000\,\text{mol}$ .
The ratio $H2SO4:NaOH\text{H}_2\text{SO}_4 : \text{NaOH}$ is $1 : 2$ , so the amount of $mol\text{NaOH} = 2 \times 0.002000 = 0.004000\,\text{mol}$ .
Concentration of $mol/dm3\text{NaOH} = \dfrac{0.004000}{0.0250} = 0.160\,\text{mol/dm}^3$ .

Finding required volumes and mass concentrations

To find a volume needed, divide the amount in moles required by the concentration of that solution.
1. $cm325.0\,\text{cm}^3$ of $mol/dm30.100\,\text{mol/dm}^3$ sodium hydroxide contains $mol0.00250\,\text{mol}$ , so it needs $mol0.00250\,\text{mol}$ of hydrochloric acid.
2. With $mol/dm30.200\,\text{mol/dm}^3$ acid, the volume needed is $dm3\dfrac{0.00250}{0.200} = 0.0125\,\text{dm}^3$ , which is $cm312.5\,\text{cm}^3$ .
Convert a volume in $dm3\text{dm}^3$ back to $cm3\text{cm}^3$ by multiplying by 1000.
To give an answer in $g/dm3\text{g/dm}^3$ $g/dm^{3}$ , multiply the concentration in $mol/dm3\text{mol/dm}^3$ $mol/dm^{3}$ by $M_{r}$ .
1. The $mol/dm30.160\,\text{mol/dm}^3$ sodium hydroxide above has a concentration of $0.160 \times 40.0 = 6.40\,\text{g/dm}^3$ .

Exam technique

When a question asks for a concentration from titration results, write out each link of the chain, volume to moles, moles to moles by the ratio, then moles to concentration, with a unit on every value.
Check that the answer fits the equation, so in a 1 : 2 reaction the amount in moles of the second substance must be twice the amount of the first.

Self review

What is the concentration of $mol0.600\,\text{mol}$ of solute in $cm3250\,\text{cm}^3$ of solution?
How many moles of solute are in $cm330.0\,\text{cm}^3$ of a $mol/dm30.200\,\text{mol/dm}^3$ solution?
How do you convert a concentration in $mol/dm3\text{mol/dm}^3$ into $g/dm3\text{g/dm}^3$ ?
Why does the mole ratio in the balanced equation matter in a titration calculation?
$cm320.0\,\text{cm}^3$ of $mol/dm30.150\,\text{mol/dm}^3$ hydrochloric acid neutralises $cm325.0\,\text{cm}^3$ of potassium hydroxide solution; what is the concentration of the potassium hydroxide?

Concentration is moles of solute per cubic decimetre

Definition

Concentration

The amount of solute dissolved in a stated volume of solution, measured in g/dm³ or mol/dm³.

The relationship is: $(dm3)\text{concentration (mol/dm}^3\text{)} = \frac{\text{amount of solute (mol)}}{\text{volume of solution (dm}^3\text{)}}$
Convert a volume in $cm3\text{cm}^3$ into $dm3\text{dm}^3$ by dividing by 1000.
Rearranging gives amount in moles $=$ concentration $×\times$ $\times$ volume in $dm3\text{dm}^3$ $dm^{3}$ .
1. So $cm320.0\,\text{cm}^3$ of a $mol/dm30.100\,\text{mol/dm}^3$ solution contains $0.100 \times 0.0200 = 0.00200\,\text{mol}$ of solute.
Doubling the amount of solute in the same volume doubles the concentration, while doubling the volume for the same amount halves it.

Example

A solution contains $mol0.0500\,\text{mol}$ of sodium chloride in $cm3250\,\text{cm}^3$ ; calculate its concentration.
Convert the volume: $250 \div 1000 = 0.250\,\text{dm}^3$ .
Concentration $\dfrac{0.0500}{0.250} = 0.200\,\text{mol/dm}^3$ .

Relative formula mass links solute mass and concentration

Definition

Relative formula mass

The mass of one formula unit of a substance compared with one-twelfth of the mass of a carbon-12 atom, given the symbol Mr.

Convert a mass of solute into moles with amount $\dfrac{\text{mass (g)}}{M_r}$ .
Then divide the amount by the volume of solution in $dm3\text{dm}^3$ to find the concentration in $mol/dm3\text{mol/dm}^3$ .
Working backwards, mass of solute $=$ concentration $×\times$ $\times$ volume $×Mr\times M_r$ $\times M_{r}$ .
1. Making $cm3500\,\text{cm}^3$ of $mol/dm30.100\,\text{mol/dm}^3$ copper(II) sulfate, $M_r = 159.6$ , needs $0.100 \times 0.500 \times 159.6 = 7.98\,\text{g}$ of solid.
To change a concentration in $mol/dm3\text{mol/dm}^3$ into $g/dm3\text{g/dm}^3$ , multiply by $M_r$ , and to go the other way, divide by $M_r$ .

Example

Calculate the concentration in $mol/dm3\text{mol/dm}^3$ of a solution made by dissolving $g4.00\,\text{g}$ of sodium hydroxide to make $cm3250\,\text{cm}^3$ of solution ( $A_r$ : Na = 23.0, O = 16.0, H = 1.0).
$Mr(NaOH)=23.0+16.0+1.0=40.0M_r(\text{NaOH}) = 23.0 + 16.0 + 1.0 = 40.0$ .
Amount $\dfrac{4.00}{40.0} = 0.100\,\text{mol}$ .
Concentration $\dfrac{0.100}{0.250} = 0.400\,\text{mol/dm}^3$ .

The mole ratio links reacting volumes and concentrations

Definition

Mole ratio

The ratio of the amounts in moles of substances that react or form, given by the balancing numbers in the equation.

At the end point, the two substances have reacted completely, so their amounts in moles are in the ratio shown by the balanced equation.
The amount of the substance with a known concentration comes from concentration $×\times$ volume.
In $HCl(aq)+NaOH(aq)→NaCl(aq)+H2O(l)\text{HCl(aq)} + \text{NaOH(aq)} \rightarrow \text{NaCl(aq)} + \text{H}_2\text{O(l)}$ $HCl(aq) + NaOH(aq) \to NaCl(aq) + H_{2} O(l)$ , the ratio is 1 : 1, so equal volumes react when the concentrations are equal.
1. If the sodium hydroxide were twice as concentrated, only half the volume would be needed, because each $cm3\text{cm}^3$ would hold twice as many moles.
In $H2SO4(aq)+2NaOH(aq)→Na2SO4(aq)+2H2O(l)\text{H}_2\text{SO}_4\text{(aq)} + 2\text{NaOH(aq)} \rightarrow \text{Na}_2\text{SO}_4\text{(aq)} + 2\text{H}_2\text{O(l)}$ $H_{2} SO_{4} (aq) + 2 NaOH(aq) \to Na_{2} SO_{4} (aq) + 2 H_{2} O(l)$ , each mole of acid reacts with two moles of alkali.
1. So for solutions of equal concentration, the volume of sodium hydroxide needed is twice the volume of sulfuric acid.

Common Mistake

Do not assume every acid reacts with an alkali in a 1 : 1 ratio; always read the ratio from the balanced equation.
Do not put a volume in $cm3\text{cm}^3$ into concentration $×\times$ volume, because the concentration is per $dm3\text{dm}^3$ .

Titration results give an unknown concentration

Definition

Titration

A technique that finds the volume of one solution needed to react completely with a measured volume of another solution.

The pipette gives the fixed volume of one solution and the mean titre gives the volume of the other.
Write the balanced equation first, so you know the mole ratio.
Find the amount in moles of the solution whose concentration is known.
Use the mole ratio to find the amount in moles of the other substance.
Divide that amount by its volume in $dm3\text{dm}^3$ to find the unknown concentration.

Example

Concentration of sodium hydroxide from a titration

$cm325.0\,\text{cm}^3$ of sodium hydroxide solution is neutralised by a mean titre of $cm320.00\,\text{cm}^3$ of $mol/dm30.100\,\text{mol/dm}^3$ sulfuric acid; find the concentration of the sodium hydroxide.
Amount of $mol\text{H}_2\text{SO}_4 = 0.100 \times 0.02000 = 0.002000\,\text{mol}$ .
The ratio $H2SO4:NaOH\text{H}_2\text{SO}_4 : \text{NaOH}$ is $1 : 2$ , so the amount of $mol\text{NaOH} = 2 \times 0.002000 = 0.004000\,\text{mol}$ .
Concentration of $mol/dm3\text{NaOH} = \dfrac{0.004000}{0.0250} = 0.160\,\text{mol/dm}^3$ .

Finding required volumes and mass concentrations

To find a volume needed, divide the amount in moles required by the concentration of that solution.
1. $cm325.0\,\text{cm}^3$ of $mol/dm30.100\,\text{mol/dm}^3$ sodium hydroxide contains $mol0.00250\,\text{mol}$ , so it needs $mol0.00250\,\text{mol}$ of hydrochloric acid.
2. With $mol/dm30.200\,\text{mol/dm}^3$ acid, the volume needed is $dm3\dfrac{0.00250}{0.200} = 0.0125\,\text{dm}^3$ , which is $cm312.5\,\text{cm}^3$ .
Convert a volume in $dm3\text{dm}^3$ back to $cm3\text{cm}^3$ by multiplying by 1000.
To give an answer in $g/dm3\text{g/dm}^3$ $g/dm^{3}$ , multiply the concentration in $mol/dm3\text{mol/dm}^3$ $mol/dm^{3}$ by $M_{r}$ .
1. The $mol/dm30.160\,\text{mol/dm}^3$ sodium hydroxide above has a concentration of $0.160 \times 40.0 = 6.40\,\text{g/dm}^3$ .

Exam technique

When a question asks for a concentration from titration results, write out each link of the chain, volume to moles, moles to moles by the ratio, then moles to concentration, with a unit on every value.
Check that the answer fits the equation, so in a 1 : 2 reaction the amount in moles of the second substance must be twice the amount of the first.

Self review

What is the concentration of $mol0.600\,\text{mol}$ of solute in $cm3250\,\text{cm}^3$ of solution?
How many moles of solute are in $cm330.0\,\text{cm}^3$ of a $mol/dm30.200\,\text{mol/dm}^3$ solution?
How do you convert a concentration in $mol/dm3\text{mol/dm}^3$ into $g/dm3\text{g/dm}^3$ ?
Why does the mole ratio in the balanced equation matter in a titration calculation?
$cm320.0\,\text{cm}^3$ of $mol/dm30.150\,\text{mol/dm}^3$ hydrochloric acid neutralises $cm325.0\,\text{cm}^3$ of potassium hydroxide solution; what is the concentration of the potassium hydroxide?

5.1.4 Concentration calculations and titration results

Concentration is moles of solute per cubic decimetre

Relative formula mass links solute mass and concentration

The mole ratio links reacting volumes and concentrations

Titration results give an unknown concentration

Concentration of sodium hydroxide from a titration

Finding required volumes and mass concentrations

Review 5.1.4 Concentration calculations and titration results by teaching Genie

Recap your knowledge with an interactive lesson

Remember key concepts with flashcards

5.1.4 Concentration calculations and titration results

Concentration is moles of solute per cubic decimetre

Relative formula mass links solute mass and concentration

The mole ratio links reacting volumes and concentrations

Titration results give an unknown concentration

Concentration of sodium hydroxide from a titration

Finding required volumes and mass concentrations

Review 5.1.4 Concentration calculations and titration results by teaching Genie

Recap your knowledge with an interactive lesson

Remember key concepts with flashcards

5.1.4 Concentration calculations and titration results

Concentration is moles of solute per cubic decimetre

Relative formula mass links solute mass and concentration

The mole ratio links reacting volumes and concentrations

Titration results give an unknown concentration

Concentration of sodium hydroxide from a titration

Finding required volumes and mass concentrations

Review 5.1.4 Concentration calculations and titration results by teaching Genie

Recap your knowledge with an interactive lesson

Remember key concepts with flashcards

5.1.4 Concentration calculations and titration results Revision Guide

5.1.4 Concentration calculations and titration results

Concentration is moles of solute per cubic decimetre

Relative formula mass links solute mass and concentration

The mole ratio links reacting volumes and concentrations

Titration results give an unknown concentration

Concentration of sodium hydroxide from a titration

Finding required volumes and mass concentrations

Review 5.1.4 Concentration calculations and titration results by teaching Genie

Recap your knowledge with an interactive lesson

Remember key concepts with flashcards

5.1.4 Concentration calculations and titration results Revision Guide