pH when 100 mL of $0.1 \mathrm{M} {\mathrm{H}}_3{\mathrm{PO}}_4$ is titrated with $150 \mathrm{mL} 0.1 \mathrm{M} \mathrm{NaOH}$ solution will be :

 

The reaction takes place as,

$\underset{\underset{-}{10}}{{\mathrm{H}}_3{\mathrm{PO}}_4}+\underset{\underset{5}{15}}{\mathrm{NaOH}}⟶\underset{\underset{10}{-}}{{\mathrm{NaH}}_2{\mathrm{PO}}_4}+{\mathrm{H}}_2\mathrm{O}$

$\underset{\underset{-}{10}}{{\mathrm{NaH}}_2{\mathrm{PO}}_4}+\underset{\underset{5 5\}\mathrm{pH}=p{K}_{a_2}}{15}}{\mathrm{NaOH}}⟶{\mathrm{Na}}_2{\mathrm{HPO}}_4+{\mathrm{H}}_2\mathrm{O}$

This is the required pH.