${\mathrm{H}}_3{\mathrm{AsO}}_3+{\mathrm{I}}_2+{\mathrm{H}}_2\mathrm{O}={\mathrm{H}}_3{\mathrm{AsO}}_4+2\mathrm{HI}$ 0.01 equivalent of the mentioned acid will neutralize how many
grams of NaOH?

$$\begin{gathered} {\mathrm{H}}_3{\stackrel{+3}{\mathrm{As}}\mathrm{O}}_3+{\mathrm{I}}_2+{\mathrm{H}}_2\mathrm{O}={\mathrm{H}}_3{\stackrel{+5}{\mathrm{As}}\mathrm{O}}_4+2\mathrm{HI} \\ \therefore \mathrm{x} 2 \\ \therefore \text{ Equiv. wt. of }{\mathrm{H}}_3{\mathrm{AsO}}_3=\frac{\text{ Mol. wt. of }{\mathrm{H}}_3{\mathrm{AsO}}_3}{2}=\frac{126}{2}=63 \\ \begin{array}{l}\therefore 0.01\text{ equiv. of }{\mathrm{H}}_3{\mathrm{AsO}}_3=0.63{\mathrm{gH}}_3{\mathrm{AsO}}_3\\ {\mathrm{H}}_3{\mathrm{AsO}}_3+2\mathrm{NaOH}={\mathrm{Na}}_2{\mathrm{HasO}}_3+2{\mathrm{H}}_2\mathrm{O}\end{array} \\ 126{\mathrm{gH}}_3{\mathrm{AsO}}_3\text{ reacts with }80\mathrm{g}\text{ of }\mathrm{NaOH} \\ 0.63{\mathrm{gH}}_3{\mathrm{AsO}}_3\text{ reacts with }\frac{80}{126}\times 0.63\mathrm{g}\text{ of }\mathrm{NaOH} \\ = 0.4 \mathrm{g} \mathrm{of} \mathrm{NaOH}. \end{gathered}$$