- Dilution is a process of adding water to the standard solution lowered the concentration of the solution.
- In dilution of solution, we should take note that mole of solute before dilution is equal to the mole of solute after dilution.

\[\frac{{{M_1}{V_1}}}{{1000}} = \frac{{{M_2}{V_2}}}{{1000}}\]

or

\[{M_1}{V_1} = {M_2}{V_2}\]

M_{1} = Molarity before dilution

M_{2} = Molarity after dilution

V_{1} = Volume before dilution

V_{2} = Volume after dilution

**Example 1**

100cm^{3} of 0.5 mol dm^{-3} sodium chloride solution is diluted with distilled water to produce 250 cm^{3} of solution. Calculate the concentration (in mol dm^{-3}) of the sodium chloride solution after the dilution.

**Answer**:

M_{1} = 0.5 mol dm^{-3}

M_{2} = ?

V_{1} = 100cm^{3}

V_{2} = 250 cm^{3}

\[\begin{gathered}

{M_1}{V_1} = {M_2}{V_2} \hfill \\

(0.5)(100) = {M_2}(250) \hfill \\

{M_2} = \frac{{(0.5)(100)}}{{(250)}} \hfill \\

{M_2} = 0.1mold{m^{ – 3}} \hfill \\

\end{gathered} \]

**Example 2**

Find the volume of 2 mol/dm^{3} nitric acid that needs to be diluted with distill water to produce 500cm^{3} of 0.05 mol/dm^{3} nitric acid.

**Answer**:

M_{1} = 2 mol dm^{-3}

M_{2} = 0.05 mol/dm^{3}

V_{1} = ？

V_{2} = 500 cm^{3}

\[\begin{gathered}

{M_1}{V_1} = {M_2}{V_2} \hfill \\

(2){V_1} = (0.05)(500) \hfill \\

{V_1} = \frac{{(0.05)(500)}}{{(2)}} \hfill \\

{V_1} = 12.5c{m^{ – 3}} \hfill \\

\end{gathered} \]

## Preparing Standard Solutions

- A standard solution is a solution in which its concentration is known.
- The steps taken in preparing a standard solution are:
- Determine the voloume and concentration that you want to prepare.
- Calculate the mass of solute needed to give the required volume and concentration.
- Weigh the solute
- Dissolve the solute completely dissolved in distilled water and then transfer it to a volumetric flask partially filled with distilled water.
- Add distilled water to the calibration mark of the volumetric flask.
- Invert the flask and shake it to make sure thorough mixing.