An excess of NaOH was added to 100 mL of a FeCl₃ solution which gives 2.14 g of Fe(OH)₃. Calculate the normality of FeCl₃ solution.

Chemical reaction between NaOH and FeCl₃ is:

$3\mathrm{NaOH}+{\mathrm{FeCl}}_3\to \mathrm{Fe}{\left(\mathrm{OH}\right)}_3+3\mathrm{NaCl}$

Above reaction indicates that NaOH is present in excess.

Moles of Fe(OH)₃ present in the solution is calculated as:

$\begin{array}{ccc}\mathrm{Number} \mathrm{of} \mathrm{moles}& =& \frac{\mathrm{Mass}}{\mathrm{Molar} \mathrm{mass}}\\ & =& \frac{2.14 \mathrm{g}}{107 \mathrm{g}/\mathrm{mol}}\\ & =& 2\times {10}^{-2} \mathrm{mol}\end{array}$

From balanced chemical reaction, one mole of Fe(OH)₃ forms from one mole of FeCl₃, similarly $2\times {10}^{-2} \mathrm{mol}$ of Fe(OH)₃ forms from $2\times {10}^{-2} \mathrm{mol}$ of FeCl₃.

Molarity of FeCl₃ is calculated as follows:

$\begin{array}{ccc}\mathrm{Molarity}& =& \frac{\mathrm{Number} \mathrm{of} \mathrm{moles} \mathrm{of} \mathrm{solute}}{\mathrm{Volume} \mathrm{of} \mathrm{solution} (\mathrm{in} \mathrm{L})}\\ & =& \frac{2\times {10}^{-2} \mathrm{mol}}{100 \mathrm{mL}\times {\displaystyle \frac{1 \mathrm{L}}{1000 \mathrm{mL}}}}\\ & =& 0.2 \mathrm{M}\end{array}$

Normality is calculated by multiplying molarity with valency factor. Charge on Fe in FeCl₃ is Fe³⁺. This implies that the valency factor of FeCl₃ is 3. Normality of FeCl₃ is calculated as:

$\begin{array}{ccc}\mathrm{Normality}& =& \mathrm{Molarity}\times \mathrm{Valency} \mathrm{factor}\\ & =& 0.2\times 3\\ & =& 0.6 \mathrm{N}\end{array}$