The volume of 0.025 M Ca(OH)₂ solution which can neutralize 100 mL of 10^-4 M H₃PO₄ is

Molarity of Calcium hydroxide (Ca(OH)₂) is 0.025 M and its valency factor is 2. Therefore, its normality is:

$\begin{array}{ccc}\mathrm{Normality}& =& \mathrm{valency} \mathrm{factor}\times \mathrm{molarity}\\ & =& 2\times 0.025 \mathrm{M}\\ & =& 0.05 \mathrm{N}\end{array}$

Molarity of H₃PO₄ is ${10}^{-4} \mathrm{M}$ and its valency factor is 3. Therefore, its normality is:

$\begin{array}{ccc}\mathrm{Normality}& =& 3\times \mathrm{molarity}\\ & =& 3\times {10}^{-4} \mathrm{N}\end{array}$

Volume of calcium hydroxide is calculated as:

$\begin{array}{ccc}\mathrm{Normality} \left(\mathrm{Ca}{\left(\mathrm{OH}\right)}_2\right)\times \mathrm{Volume} \left(\mathrm{Ca}{\left(\mathrm{OH}\right)}_2\right)& =& \mathrm{Normality} \left({\mathrm{H}}_3{\mathrm{PO}}_4\right)\times \mathrm{Volume} \left({\mathrm{H}}_3{\mathrm{PO}}_4\right)\\ 0.05 \mathrm{N}\times \mathrm{Volume} \left(\mathrm{Ca}{\left(\mathrm{OH}\right)}_2\right)& =& 3\times {10}^{-4} \mathrm{N}\times 100 \mathrm{mL}\\ \mathrm{Volume} \left(\mathrm{Ca}{\left(\mathrm{OH}\right)}_2\right)& =& 0.6 \mathrm{mL}\end{array}$