Chemistry - Mole Concept Question with Solution | TestHub

${\mathrm{H}}_3\text{P}{\text{O}}_3$is a dibasic acid (containing two ionisable protons attached to O directly)
${\mathrm{H}}_3\text{P}{\text{O}}_3\mathrm{}\leftrightarrow \text{2}{\text{H}}^{+}\text{+HP}{\text{O}}_3^{2-}$
$\text{0}\text{.1}{\text{M H}}_{\text{3}}{\text{PO}}_{\text{3}}\text{=}\text{0}\text{.2}{\text{N H}}_{\text{3}}{\text{PO}}_{\text{3}}$
$\text{0}\text{.1}\text{M NaOH}\text{=}\text{0}\text{.1}\text{N NaOH}$
${\mathrm{N}}_1{\mathrm{V}}_1\text{=}{\text{N}}_2{\mathrm{V}}_2\mathrm{}$
$\left(\text{NaOH}\right)\left({\mathrm{H}}_3\mathrm{P}{\mathrm{O}}_3\right)$
$0.1\times {\mathrm{V}}_1=0.2\times 20$
${\text{V}}_{\text{1}}=40\text{mL}$
Alternative short method
${\mathrm{H}}_3\mathrm{P}{\mathrm{O}}_3+2\mathrm{N}\mathrm{a}\mathrm{O}\mathrm{H}\to \mathrm{N}{\mathrm{a}}_2\mathrm{H}\mathrm{P}{\mathrm{O}}_3+2{\mathrm{H}}_2\mathrm{O}$
$\frac{{\mathrm{M}}_{\mathrm{A}}{\mathrm{V}}_{\mathrm{A}}}{{\mathrm{x}}_{\mathrm{A}}}=\frac{{\mathrm{M}}_{\mathrm{B}}{\mathrm{V}}_{\mathrm{B}}}{{\mathrm{x}}_{\mathrm{B}}}$…(i)
Here${\mathrm{x}}_{\mathrm{A}}$and${\mathrm{x}}_{\mathrm{B}}$are balancing coefficient of acid and base respectively.
Now putting all the values in equation (i) we get
$\frac{20\times 0.1}{1}=\frac{0.1\times {\mathrm{V}}_{\mathrm{B}}}{2}$
${\text{V}}_{\text{B}}=40\text{ml}$