Dilution

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

    \[\frac{{{M_1}{V_1}}}{{1000}} = \frac{{{M_2}{V_2}}}{{1000}}\]
    or
    \[{M_1}{V_1} = {M_2}{V_2}\]
    M1 = Molarity before dilution
    M2 = Molarity after dilution
    V1 = Volume before dilution
    V2 = Volume after dilution

    Example 1
    100cm3 of 0.5 mol dm-3 sodium chloride solution is diluted with distilled water to produce 250 cm3 of solution. Calculate the concentration (in mol dm-3) of the sodium chloride solution after the dilution.

    Answer:
    M1 = 0.5 mol dm-3
    M2 = ?
    V1 = 100cm3
    V2 = 250 cm3
    \[\begin{gathered}
    {M_1}{V_1} = {M_2}{V_2} \hfill \\
    (0.5)(100) = {M_2}(250) \hfill \\
    {M_2} = \frac{{(0.5)(100)}}{{(250)}} \hfill \\
    {M_2} = 0.1mold{m^{ – 3}} \hfill \\
    \end{gathered} \]

    Example 2
    Find the volume of 2 mol/dm3 nitric acid that needs to be diluted with distill water to produce 500cm3 of 0.05 mol/dm3 nitric acid.

    Answer:
    M1 = 2 mol dm-3
    M2 = 0.05 mol/dm3
    V1 = ？
    V2 = 500 cm3
    \[\begin{gathered}
    {M_1}{V_1} = {M_2}{V_2} \hfill \\
    (2){V_1} = (0.05)(500) \hfill \\
    {V_1} = \frac{{(0.05)(500)}}{{(2)}} \hfill \\
    {V_1} = 12.5c{m^{ – 3}} \hfill \\
    \end{gathered} \]