Earlier, you learnt how to calculate the molar concentration of a solution. The molar concentration of a solution is the number of moles of solute per litre of solvent (\(\text{mol·L$^{-1}$}\)). This is more commonly given as moles of solute per cubic decimetre of solution (\(\text{mol·dm$^{-3}$}\)).

To calculate concentration we use \(C = \frac{n}{V}\), where \(C\) is the molar concentration, \(n\) is the number of moles and \(V\) is the volume of the solution.

Calculating molar concentrations is useful to determine how much solute we need to add to a given volume of solvent in order to make a standard solution.

A standard solution is a solution in which the concentration is known to a high degree of precision. When we work with standard solutions we can take the concentration to be constant.


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 a volume in \(\text{cm$^{3}$}\) by \(\text{1 000}\) will give you the equivalent volume in \(\text{dm$^{3}$}\).

Example: Concentration Calculations


How much sodium chloride (in g) will one need to prepare \(\text{500}\) \(\text{cm$^{3}$}\) of a standard solution with a concentration of \(\text{0.01}\) \(\text{mol·dm$^{-3}$}\)?

Step 1: Convert all quantities into the correct units for this equation.

The volume must be converted to \(\text{dm$^{3}$}\): \begin{align*} V & = \frac{\text{500}}{\text{1 000}} \\ & = \text{0.5}\text{ dm$^{3}$} \end{align*}

Step 2: Calculate the number of moles of sodium chloride needed.

\begin{align*} C &= \frac{n}{V} \\ \text{0.01} & = \frac{n}{\text{0.5}} \\ n & = \text{0.005}\text{ mol} \end{align*}

Step 3: Convert moles of \(\text{NaCl}\) to mass.

To find the mass of \(\text{NaCl}\) we need the molar mass of \(\text{NaCl}\). We can get this from the periodic table (recall from grade 10 how to calculate the molar mass of a compound).

\begin{align*} m & = nM \\ & = (\text{0.005})(\text{58})\\ & = \text{0.29}\text{ g} \end{align*}

The mass of sodium chloride needed is \(\text{0.29}\) \(\text{g}\)

We will now look at another use of concentration which is for titration calculations.


A titration is a technique for determining the concentration of an unknown solution. Titrations can be done using many different types of reactions. Acid-base reactions and redox reactions are both commonly used for titrations.

In grade 10 you did a simple acid-base titration. Now we will look at how to calculate the concentration of an unknown solution using an acid-base titration.


When performing a titration we say that the substance of unknown concentration is titrated with the standard solution. A pipette is a measuring device that is used to measure an exact amount of a liquid. If you use a pipette to add liquid to a flask then you would say that the liquid was pipetted into a flask.

We can reduce the number of calculations that we have to do in titration calculations by using the following: \begin{align*} \frac{C_{A}V_{A}}{a} & = \frac{C_{B}V_{B}}{b} \end{align*}

The \(a\) and \(b\) are the stoichiometric coefficients of compounds A and B respectively.

Example: Titration Calculation


Given the equation:

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

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

Step 1: Write down all the information you know about the reaction, and make sure that the equation is balanced.

\(\text{NaOH}\): \(V =\text{15}\text{ cm$^{3}$}\)

\(\text{HCl}\): \(V=\text{25}\text{ cm$^{3}$}\); \(C= \text{0.2}\text{ mol·dm$^{-3}$}\)

The equation is already balanced.

Step 2: Convert the volume to \(\text{dm$^{3}$}\)

\begin{align*} V_{\text{NaOH}} & = \frac{\text{15}}{\text{1 000}} \\ & = \text{0.015}\text{ dm$^{3}$} \end{align*}\begin{align*} V_{\text{HCl}} & = \frac{\text{25}}{\text{1 000}} \\ & = \text{0.025}\text{ dm$^{3}$} \end{align*}

Step 3: Calculate the concentration of the sodium hydroxide

\begin{align*} \frac{C_{A}V_{A}}{a} & = \frac{C_{B}V_{B}}{b} \\ \frac{(\text{0.2})(\text{0.025})}{\text{1}} & = \frac{(C_{\text{NaOH}})(\text{0.015})}{\text{1}} \\ \text{0.005} & = (\text{0.015})C_{\text{NaOH}}\\ C_{\text{NaOH}} & = \text{0.33}\text{ mol·dm$^{-3}$} \end{align*}

The concentration of the \(\text{NaOH}\) solution is \(\text{0.33}\) \(\text{mol·dm$^{-3}$}\)

Example: Titration Calculation


\(\text{4.9}\) \(\text{g}\) of sulfuric acid is dissolved in water and the final solution has a volume of \(\text{220}\) \(\text{cm$^{3}$}\). Using an acid-base 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 sodium hydroxide solution. Calculate the concentration of the sodium hydroxide in \(\text{mol·dm$^{-3}$}\).

Step 1: Write a balanced equation for the titration reaction.

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

Step 2: Calculate the concentration of the sulfuric acid solution.

First convert the volume into \(\text{dm$^{3}$}\):

\begin{align*} V & = \frac{\text{220}}{\text{1 000}} = \text{0.22}\text{ dm$^{3}$} \end{align*}

Then calculate the number of moles of sulfuric acid:

\begin{align*} n & = \frac{m}{M} \\ & = \frac{\text{4.9}}{\text{98}}\\ & = \text{0.05}\text{ mol} \end{align*}

Now we can calculate the concentration of the sulfuric acid:

\begin{align*} C & = \frac{n}{V} \\ & = \frac{\text{0.05}}{\text{0.22}}\\ & = \text{0.227}\text{ mol·dm$^{-3}$} \end{align*}

Step 3: Calculate the concentration of the sodium hydroxide solution.

Remember that only \(\text{20}\) \(\text{cm$^{3}$}\) or \(\text{0.02}\) \(\text{dm$^{3}$}\) of the sulfuric acid solution is used.

\begin{align*} \frac{C_{1}V_{1}}{n_{1}} & = \frac{C_{2}V_{2}}{n_{2}} \\ \frac{(\text{0.227})(\text{0.02})}{\text{1}} & = \frac{(C_{\text{NaOH}})(\text{0.01})}{\text{2}} \\ \text{0.00454} & = (\text{0.005})C_{\text{NaOH}}\\ C_{\text{NaOH}} & = \text{0.909}\text{ mol·dm$^{-3}$} \end{align*}

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