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

$3\mathrm{Ca}{\left(\mathrm{OH}\right)}_2+2{\mathrm{H}}_3{\mathrm{PO}}_4\to {\mathrm{Ca}}_3{\left({\mathrm{PO}}_4\right)}_2+6{\mathrm{H}}_2\mathrm{O}$

Ca(OH)₂                  H₃PO₄

V =in ml                 V =100ml

M = 0.025M         M =10^-4M

n=3                       n =2

$\frac{{\mathrm{M}}_1\times {\mathrm{V}}_1}{{\mathrm{n}}_1}=\frac{{\mathrm{M}}_2\times {\mathrm{V}}_2}{{\mathrm{n}}_2}$

$\frac{\mathrm{V}\times 0.025}{3}=\frac{100\times {10}^{-4}}{2}$

$\mathrm{V} =\frac{100\times {10}^{-4}\times 3}{0.025\times 2}$

V = 0.6 ml