Mark out the incorrect statements:
 

Statement – a – incorrect . 
$\begin{array}{l}{\mathrm{H}}_3{\mathrm{PO}}_4\stackrel{{\mathrm{K}}_{\mathrm{a}1}}{\rightleftharpoons }{\mathrm{H}}^{+}+{\mathrm{H}}_2{\mathrm{PO}}_4^{-}\\ {\mathrm{H}}_2{\mathrm{PO}}_4^{-}\stackrel{{\mathrm{K}}_{\mathrm{a}2}}{\rightleftharpoons }{\mathrm{H}}^{+}+{\mathrm{HPO}}_4^{2-}\\ {\mathrm{HPO}}_4^{2-}\stackrel{{\mathrm{K}}_{\mathrm{a}3}}{\rightleftharpoons }{\mathrm{H}}^{+}+{\mathrm{PO}}_4^{3-}\end{array}$
${\mathrm{p}}^{\mathrm{H}}\left({\mathrm{H}}_2{\mathrm{PO}}_4^{-}\right)=\frac{{\mathrm{P}}^{{\mathrm{K}}_{{\mathrm{a}}_1}}+{\mathrm{P}}^{{\mathrm{K}}_{{\mathrm{a}}_2}}}{2}$ this is valid only when H₃PO₄ during titration is completely converted to ${\mathrm{H}}_2{\mathrm{PO}}_4^{2-}$
${\mathrm{HPO}}_4^{2-}\stackrel{{\mathrm{K}}_{{\mathrm{a}}_3}}{\rightleftharpoons }{\mathrm{H}}^{+}+{\mathrm{PO}}_4^{3-}$
${\mathrm{p}}^{\mathrm{H}}\left({\mathrm{H}}_2{\mathrm{PO}}_4^{2-}\right)=\frac{{\mathrm{P}}^{{\mathrm{K}}_{{\mathrm{a}}_2}}+{\mathrm{P}}^{{\mathrm{K}}_{{\mathrm{a}}_3}}}{2}$ This again valid when starting with ${\mathrm{H}}_2{\mathrm{PO}}_4^{2-}$ and is completely converted to ${\mathrm{H}}_2{\mathrm{PO}}_4^{2-}$
Statement – b –correct .
Ka –value decreases successively. Since it is difficult to remove H⁺ - ion from an anion than a neutral compound. Similarly it is difficult to remove H⁺ - ion from an Ionian than and anion. Hence ${\mathrm{K}}_{{\mathrm{a}}_1}>{\mathrm{K}}_{{\mathrm{a}}_2}>{\mathrm{K}}_{{\mathrm{a}}_3}$.
Statement – c –correct .
From K_a - value, it is evident that, both ${\mathrm{H}}_3{\mathrm{PO}}_4\mathrm{\&}{\mathrm{H}}_2{\mathrm{PO}}_4^{-}$ are more acidic the ${\mathrm{H}}_2{\mathrm{PO}}_4^{2-}$
Statement – d – incorrect .
Both ${\mathrm{H}}_2{\mathrm{PO}}_4^{-}\mathrm{and} {\mathrm{HPO}}_4^{2-}$  are amphiprotic anions in the solution.
$\begin{array}{r}{\mathrm{H}}_3{\mathrm{P}}_4\stackrel{+{\mathrm{H}}^{\oplus }}{⟵}{\mathrm{H}}_2{\mathrm{P}}_4^{-}\stackrel{-{\mathrm{H}}^{\oplus }}{⟶}{\mathrm{HPO}}_4^{2-}+{\mathrm{H}}^{\oplus }\\ {\mathrm{H}}_2{\mathrm{P}}_4\stackrel{+{\mathrm{H}}^{\oplus }}{⟵}{\mathrm{HPO}}_4^{2-}\stackrel{-{\mathrm{H}}^{\oplus }}{⟶}{\mathrm{PO}}_4^{3-}+{\mathrm{H}}^{\oplus }\end{array}$