To convert 12 g of NaH₂PO₄ completely into Na₃PO₄, the volume of 1molar NaOH required is

Method (I) 

${\mathrm{NaH}}_2{\mathrm{PO}}_4+2\mathrm{NaOH}\to {\mathrm{Na}}_3{\mathrm{PO}}_4+2{\mathrm{H}}_2\mathrm{O}$

 120g                    2x40

 12g                         X

 X= 8g

W_NaOH required = 8g

M_NaOH =  1M

$\therefore 1=\frac{8}{40}\mathrm{X}\frac{1000}{\mathrm{V}\left(\mathrm{ml}\right)}$

   V_NaOH = 200ml

Alternate method

${\mathrm{NaH}}_2{\mathrm{PO}}_4\to {\mathrm{Na}}_3{\mathrm{PO}}_4$

Basicity =2

At neutralization point

No.of gm equivalents of NaH₂PO₄ = No.of gm equivalents of NaOH

$\frac{\mathrm{W}}{\mathrm{GEW}}=\frac{{\mathrm{V}}_{\mathrm{b}}{\mathrm{N}}_{\mathrm{b}}}{1000}$

$\frac{12}{\left(\frac{120}{2}\right)}=\frac{{\mathrm{V}}_{\mathrm{b}}\left(1\right)}{1000}$

${\mathrm{V}}_{\mathrm{b}}=200\mathrm{ml}$

Vol. of NaOH required = 200ml