The volume (in mL) of 0.1 N NaOH required to neutralise 10 mL of 0.1 N phosphinic acid is ______

Phosphoinic acid is ${\mathrm{H}}_3{\mathrm{PO}}_3$

$\mathrm{xNaOH}+{\mathrm{H}}_3{\mathrm{PO}}_3\to \underset{ }{{\mathrm{Na}}_{\mathrm{x}}{\mathrm{H}}_{3-\mathrm{x}}{\mathrm{PO}}_3}+{\mathrm{H}}_2\mathrm{O}$

$\begin{array}{l}{\mathrm{n}}_1{\mathrm{M}}_1{\mathrm{v}}_1={\mathrm{n}}_2{\mathrm{N}}_2{\mathrm{v}}_2 {\mathrm{n}}_2{\mathrm{M}}_2={\mathrm{N}}_2=0.1\mathrm{N}\\ 1\times 0.1\times {\mathrm{V}}_{\mathrm{NaOH}}=0.1\times 10\\ { \mathrm{V}}_{\mathrm{NaOH}}=10 \mathrm{m}1\end{array}$