4.2.5b Calculating quantities in titrations

Titration calculations link volumes through moles

Definition

Mole

The amount of a substance, measured in mol, where one mole has a mass in grams numerically equal to its relative formula mass.

Definition

Concentration

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

Start with the reactant whose concentration and volume are both known.
Find its moles using $cm31000\text{moles} = \dfrac{\text{concentration} \times \text{volume in cm}^3}{1000}$
Use the ratio in the balanced equation to find the moles of the other reactant.
Find the unknown concentration using $cm3\text{concentration} = \dfrac{\text{moles} \times 1000}{\text{volume in cm}^3}$

Key Idea

Always convert volumes from $cm3\text{cm}^3$ to $dm3\text{dm}^3$ , or keep the $÷1000\div 1000$ in the formula.
The mole ratio comes from the balanced equation, so balance it first.

Definition

Balanced equation

A chemical equation containing the same number of atoms of each element on both sides.

Example

In a titration, 24.0 $cm3\text{cm}^3$ of sulfuric acid exactly neutralises 25.0 $cm3\text{cm}^3$ of 0.100 $mol/dm3\text{mol/dm}^3$ sodium hydroxide, with equation $2NaOH+H2SO4→Na2SO4+2H2O2\text{NaOH} + \text{H}_2\text{SO}_4 \rightarrow \text{Na}_2\text{SO}_4 + 2\text{H}_2\text{O}$
Moles of $mol\text{NaOH} = \dfrac{0.100 \times 25.0}{1000} = 0.00250\ \text{mol}$ .
The ratio is $2 : 1$ , so moles of $mol\text{H}_2\text{SO}_4 = \dfrac{0.00250}{2} = 0.00125\ \text{mol}$ .
Concentration of $mol/dm3\text{H}_2\text{SO}_4 = \dfrac{0.00125 \times 1000}{24.0} = 0.0521\ \text{mol/dm}^3$ .

Exam technique

Set the work out in clear steps, because each stage earns method marks even if a later number is wrong.
Use the mean concordant titre, not the rough titre, as the volume in the calculation.

Converting between $mol/dm3\text{mol/dm}^3$ and $g/dm3\text{g/dm}^3$

Definition

Solute

The substance dissolved in a solvent to form a solution.

Convert using $mol/dm3×Mr\text{concentration in g/dm}^3 = \text{concentration in mol/dm}^3 \times M_r$
For sulfuric acid $M_r = 98$ , so $g/dm30.0521\ \text{mol/dm}^3 \times 98 = 5.11\ \text{g/dm}^3$ .
Reverse the conversion by dividing the $g/dm3\text{g/dm}^3$ value by the $M_r$ .

Self review

Write the formula linking moles, concentration and volume in $cm3\text{cm}^3$ .
In $2NaOH+H2SO42\text{NaOH} + \text{H}_2\text{SO}_4$ , how many moles of acid react with 0.00250 mol of alkali?
How do you convert a concentration in $mol/dm3\text{mol/dm}^3$ to $g/dm3\text{g/dm}^3$ ?
Why should you use the mean concordant titre in the calculation?
Find the concentration of an acid if 0.00200 mol reacts in 25.0 $cm3\text{cm}^3$ of solution.

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Titration setup showing sulfuric acid in a burette added to sodium hydroxide in a conical flask, with the measured volumes and concentration labelled

A titration links the volume and concentration of two reacting solutions through their amounts in moles. Start with the solution whose concentration and volume are known, then use the balanced equation to calculate the moles of the other substance.

For a volume measured in $cm3\text{cm}^3$ :

\text{moles} = \frac{\text{concentration} \times \text{volume}}{1000}

The factor of $1000$ converts $cm3\text{cm}^3$ to $dm3\text{dm}^3$ , because $\, \text{cm}^3 = 1 \, \text{dm}^3$ .

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Which reactant should you start with in a titration calculation?

4.2.5b Calculating quantities in titrations

Titration calculations link volumes through moles

Definition

Mole

The amount of a substance, measured in mol, where one mole has a mass in grams numerically equal to its relative formula mass.

Definition

Concentration

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

Start with the reactant whose concentration and volume are both known.
Find its moles using $cm31000\text{moles} = \dfrac{\text{concentration} \times \text{volume in cm}^3}{1000}$
Use the ratio in the balanced equation to find the moles of the other reactant.
Find the unknown concentration using $cm3\text{concentration} = \dfrac{\text{moles} \times 1000}{\text{volume in cm}^3}$

Key Idea

Always convert volumes from $cm3\text{cm}^3$ to $dm3\text{dm}^3$ , or keep the $÷1000\div 1000$ in the formula.
The mole ratio comes from the balanced equation, so balance it first.

Definition

Balanced equation

A chemical equation containing the same number of atoms of each element on both sides.

Example

In a titration, 24.0 $cm3\text{cm}^3$ of sulfuric acid exactly neutralises 25.0 $cm3\text{cm}^3$ of 0.100 $mol/dm3\text{mol/dm}^3$ sodium hydroxide, with equation $2NaOH+H2SO4→Na2SO4+2H2O2\text{NaOH} + \text{H}_2\text{SO}_4 \rightarrow \text{Na}_2\text{SO}_4 + 2\text{H}_2\text{O}$
Moles of $mol\text{NaOH} = \dfrac{0.100 \times 25.0}{1000} = 0.00250\ \text{mol}$ .
The ratio is $2 : 1$ , so moles of $mol\text{H}_2\text{SO}_4 = \dfrac{0.00250}{2} = 0.00125\ \text{mol}$ .
Concentration of $mol/dm3\text{H}_2\text{SO}_4 = \dfrac{0.00125 \times 1000}{24.0} = 0.0521\ \text{mol/dm}^3$ .

Exam technique

Set the work out in clear steps, because each stage earns method marks even if a later number is wrong.
Use the mean concordant titre, not the rough titre, as the volume in the calculation.

Converting between $mol/dm3\text{mol/dm}^3$ and $g/dm3\text{g/dm}^3$

Definition

Solute

The substance dissolved in a solvent to form a solution.

Convert using $mol/dm3×Mr\text{concentration in g/dm}^3 = \text{concentration in mol/dm}^3 \times M_r$
For sulfuric acid $M_r = 98$ , so $g/dm30.0521\ \text{mol/dm}^3 \times 98 = 5.11\ \text{g/dm}^3$ .
Reverse the conversion by dividing the $g/dm3\text{g/dm}^3$ value by the $M_r$ .

Self review

Write the formula linking moles, concentration and volume in $cm3\text{cm}^3$ .
In $2NaOH+H2SO42\text{NaOH} + \text{H}_2\text{SO}_4$ , how many moles of acid react with 0.00250 mol of alkali?
How do you convert a concentration in $mol/dm3\text{mol/dm}^3$ to $g/dm3\text{g/dm}^3$ ?
Why should you use the mean concordant titre in the calculation?
Find the concentration of an acid if 0.00200 mol reacts in 25.0 $cm3\text{cm}^3$ of solution.

Previous Next

How was this guide?

Teach Genie

Review 4.2.5b Calculating quantities in titrations by teaching Genie

Teach it back in your own words, spot gaps, and remember it better.

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Lesson

Recap your knowledge with an interactive lesson

7 minute activity

Start lesson

Titration setup showing sulfuric acid in a burette added to sodium hydroxide in a conical flask, with the measured volumes and concentration labelled

For a volume measured in $cm3\text{cm}^3$ :

\text{moles} = \frac{\text{concentration} \times \text{volume}}{1000}

The factor of $1000$ converts $cm3\text{cm}^3$ to $dm3\text{dm}^3$ , because $\, \text{cm}^3 = 1 \, \text{dm}^3$ .

Flashcards

Remember key concepts with flashcards

21 flashcards

Practice flashcards

Which reactant should you start with in a titration calculation?

4.2.5b Calculating quantities in titrations

Titration calculations link volumes through moles

Converting between mol/dm3\text{mol/dm}^3mol/dm3 and g/dm3\text{g/dm}^3g/dm3

Review 4.2.5b Calculating quantities in titrations by teaching Genie

Recap your knowledge with an interactive lesson

Remember key concepts with flashcards

4.2.5b Calculating quantities in titrations Revision Guide

4.2.5b Calculating quantities in titrations

Titration calculations link volumes through moles

Converting between mol/dm3\text{mol/dm}^3mol/dm3 and g/dm3\text{g/dm}^3g/dm3

Review 4.2.5b Calculating quantities in titrations by teaching Genie

Recap your knowledge with an interactive lesson

Remember key concepts with flashcards

4.2.5b Calculating quantities in titrations Revision Guide

Converting between $mol/dm3\text{mol/dm}^3$ and $g/dm3\text{g/dm}^3$

Converting between $mol/dm3\text{mol/dm}^3$ and $g/dm3\text{g/dm}^3$