Titration Calculations

Titration calculations

So how exactly can a titration be carried out to determine an unknown concentration? Look at the following steps to help you to understand the process.

  1. A carefully measured volume of the solution with unknown concentration is put into a conical flask.

  2. A few drops of a suitable indicator is added to this solution (bromothymol blue and phenolphthalein are common indicators).

  3. The conical flask is placed on a white tile or piece of paper (to make colour changes easier to see).

  4. A volume of the standard solution (known concentration) is put into a burette (a measuring device) and is slowly added to the solution in the flask, drop by drop.


    It is better to have the acid in the burette as many bases will clog the stopcock and damage the burette. It is much easier to clean a burette that has contained acid only. It is important to make sure the burette is perpendicular to the desk and clamped securely.

    When filling the burette the stopcock should be closed and the funnel should be lifted slightly to allow air to escape, otherwise it will bubble through the liquid in the funnel and splash acid. Once the burette is filled a beaker should be placed underneath and a few centimeters of the solution should be drained through the stopcock. This ensures that the entire burette is full. There should be no air bubbles, as a bubble bursting can affect the volume measurement and disrupt the accuracy of the experiment.

  5. At some point, adding one more drop will change the colour of the unknown solution to the colour of the end-point of the reaction. Remember the colour changes from the previous lesson on pH scale.

  6. Record the volume of standard solution that has been added up to this point.

  7. Use the information you have gathered to calculate the exact concentration of the unknown solution. Worked examples are given to walk you through this step.

  8. Note that adding more solution once the end-point has been reached will result in a colour change from the end-point colour to that of the acid (if the solution in the conical flask is a base) or of the base (if the solution in the conical flask is an acid).

When you are busy with these calculations, you will need to remember the following:

\(\text{1}\) \(\text{dm$^{3}$}\) = \(\text{1}\) \(\text{L}\) = \(\text{1 000}\) \(\text{mL}\) = \(\text{1 000}\) \(\text{cm$^{3}$}\), therefore dividing \(\text{cm$^{3}$}\) by \(\text{1 000}\) will give you an answer in \(\text{dm$^{3}$}\).

Some other terms and equations which will be useful to remember are shown below:

  • concentration of a solution is measured in \(\text{mol.dm$^{-3}$}\)

  • moles (mol) \(=\) concentration (\(\text{mol·dm$^{-3}$}\)) \(\times\) volume (\(\text{dm$^{3}$}\))

  • \begin{align*} \text{concentration } = \frac{\text{moles}}{\text{volume}} \\ C (\text{mol·dm$^{-3}$}) = \frac{n (\text{mol})}{V (\text{dm$^{3}$})} \end{align*}
  • remember to make sure all the units are correct in your calculations



Given the equation:

\(\text{NaOH}(\text{aq}) + \text{HCl}(\text{aq})\) \(\to\) \(\text{NaCl}(\text{aq}) + \text{H}_{2}\text{O}(\text{l})\)

\(\text{25}\) \(\text{cm$^{3}$}\) of sodium hydroxide solution was pipetted into a conical flask and titrated with \(\text{0.2}\) \(\text{mol.dm$^{-3}$}\) hydrochloric acid. Using a suitable indicator, it was found that \(\text{15}\) \(\text{cm$^{3}$}\) of acid was needed to neutralise the base. Calculate the concentration of the sodium hydroxide.

Step 1: Make sure that the equation is balanced

There are equal numbers of each type of atom on each side of the equation, so the equation is balanced.

Step 2: Write down all the information you know about the reaction, converting to the correct units

NaOH: V = \(\text{25}\) \(\text{cm$^{3}$}\) \(\times \dfrac{\text{0.001} {\text{ dm}}^{3}}{\text{1} {\text{ cm}}^{3}}\) = \(\text{0.025}\) \(\text{dm$^{3}$}\)

HCl: V = \(\text{15}\) \(\text{cm$^{3}$}\) \(\times \dfrac{\text{0.001} {\text{ dm}}^{3}}{\text{1} {\text{ cm}}^{3}}\) = \(\text{0.015}\) \(\text{dm$^{3}$}\)

\(\phantom{\rule{25pt}{0ex}}\)C = \(\text{0.2}\) \(\text{mol.dm$^{-3}$}\)

Step 3: Calculate the number of moles of HCl that are added

\({\text{C}} = \dfrac{\text{n}}{\text{V}}\)

Therefore, n\((\text{HCl})\) = C \(\times\) V

n(HCl) = \(\text{0.2}\) \(\text{mol.dm$^{-3}$}\) \(\times\) \(\text{0.015}\) \(\text{dm$^{3}$}\) = \(\text{0.003}\) \(\text{mol}\)

\(\text{0.003}\) \(\text{moles}\) of \(\text{HCl}\) are required to neutralise the base.

Step 4: Calculate the number of moles of sodium hydroxide in the reaction

Look at the equation for the reaction: the molar ratio of \(\text{HCl}\):\(\text{NaOH}\) is \(\text{1}\):\(\text{1}\).

So for every mole of \(\text{HCl}\), there is one mole of \(\text{NaOH}\) that is involved in the reaction. Therefore, if \(\text{0.003}\) \(\text{mol}\) \(\text{HCl}\) is required to neutralise the solution, \(\text{0.003}\) \(\text{mol}\) \(\text{NaOH}\) must have been present in the sample of the unknown solution.

Step 5: Calculate the concentration of the sodium hydroxide solution

\({\text{C(NaOH)}} = \dfrac{\text{n}}{\text{V}} = \dfrac{\text{0.003}\text{ mol}}{\text{0.025}\text{ dm$^{3}$}} = \text{0.12}\text{ mol.dm$^{-3}$}\)

The concentration of the \(\text{NaOH}\) solution is \(\text{0.12}\) \(\text{mol.dm$^{-3}$}\)



\(\text{10}\) \(\text{g}\) of solid sodium hydroxide is dissolved in \(\text{500}\) \(\text{cm$^{3}$}\) water. Using titration, it was found that \(\text{20}\) \(\text{cm$^{3}$}\) of this solution was able to completely neutralise \(\text{10}\) \(\text{cm$^{3}$}\) of a sulfuric acid solution. Calculate the concentration of the sulfuric acid.

Step 1: Write a balanced equation for the titration reaction

The reactants are sodium hydroxide (\(\text{NaOH}\)) and sulfuric acid (\(\text{H}_{2}\text{SO}_{4}\)). The base has a hydroxide anion (\(\text{OH}^{-}\)), therefore the products will be a salt and water.

The cation for the salt (\(\text{Na}^{+}\)) will come from the base. The anion for the salt (\(\text{SO}_{4}^{2-}\)) will come from the acid. There must be \(\text{2}\) \(\text{Na}^{+}\) cations for every one \(\text{SO}_{4}^{2-}\) and the salt will be \(\text{Na}_{2}\text{SO}_{4}\).

\(\text{H}_{2}\text{SO}_{4}(\text{aq}) + \text{NaOH}(\text{aq})\) \(\to\) \(\text{Na}_{2}\text{SO}_{4}(\text{aq}) + \text{H}_{2}\text{O}(\text{l})\)

To balance the equation we need to multipy the number of sodium hydroxide molecules and the water molecules by two.

\(\text{H}_{2}\text{SO}_{4}(\text{aq}) + 2\text{NaOH}(\text{aq})\) \(\to\) \(\text{Na}_{2}\text{SO}_{4}(\text{aq}) + 2\text{H}_{2}\text{O}(\text{l})\)

Step 2: Calculate the concentration of the sodium hydroxide solution

The total volume that the \(\text{10}\) \(\text{g}\) was dissolved in must be used to calculate the concentration.

V = \(\text{500}\) \(\text{cm$^{3}$}\) \(\times \dfrac{\text{0.001} {\text{ dm}}^{3}}{\text{1} {\text{ cm}}^{3}}\) = \(\text{0.5}\) \(\text{dm$^{3}$}\)

M(NaOH) = \(\text{23.0}\) + \(\text{16.0}\) + \(\text{1.01}\) = \(\text{40.01}\) \(\text{g.mol$^{-1}$}\)

n(NaOH) = \(\dfrac{\text{m}}{\text{M}} = \dfrac{\text{10}\text{ g}}{\text{40.01}\text{ g.mol$^{-1}$}} = \text{0.25}\text{ mol}\)

C(NaOH) = \(\dfrac{\text{n}}{\text{V}} = \dfrac{\text{0.25}\text{ mol}}{\text{0.5}\text{ dm$^{3}$}} = \text{0.50}\text{ mol.dm$^{-3}$}\)

Step 3: Calculate the number of moles of sodium hydroxide that were used in the neutralisation reaction

Remember that only \(\text{20}\) \(\text{cm$^{3}$}\) of the sodium hydroxide solution is used:

V = \(\text{20}\) \(\text{cm$^{3}$}\) \(\times \dfrac{\text{0.001} {\text{ dm}}^{3}}{\text{1} {\text{ cm}}^{3}}\) = \(\text{0.02}\) \(\text{dm$^{3}$}\)

\(\text{C} = \dfrac{\text{n}}{\text{V}}\), therefore \(\text{n} = {\text{C}} \times {\text{V}}\)

\(\text{n} = \text{0.50}\text{ mol.dm$^{-3}$} \times \text{0.02}\text{ dm$^{3}$} = \text{0.01}\text{ mol}\)

Step 4: Calculate the number of moles of sulfuric acid that were neutralised

According to the balanced chemical equation, the mole ratio of \(\text{NaOH}\) to \(\text{H}_{2}\text{SO}_{4}\) is \(\text{2}\):\(\text{1}\). There are \(\text{2}\) moles of \(\text{NaOH}\) for every \(\text{1}\) mole of \(\text{H}_{2}\text{SO}_{4}\).

n(\(\text{H}_{2}\text{SO}_{4}\) = \(\dfrac{\text{0.01}\text{ mol}}{\text{2}}\) = \(\text{0.005}\) \(\text{mol}\).

Step 5: Calculate the concentration of the sulfuric acid solution

Remember that \(\text{10}\) \(\text{cm$^{3}$}\) of the sulfuric acid solution is neutralised.

V(\(\text{H}_{2}\text{SO}_{4}\)) = \(\text{10}\) \(\text{cm$^{3}$}\) \(\times \dfrac{\text{0.001} {\text{ dm}}^{3}}{\text{1} {\text{ cm}}^{3}}\) = \(\text{0.01}\) \(\text{dm$^{3}$}\)

\(\text{C} = \dfrac{\text{n}}{\text{V}} = \dfrac{\text{0.005}\text{ mol}}{\text{0.01}\text{ dm$^{3}$}} = \text{0.5}\text{ mol.dm$^{-3}$}\)

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