Question

Three reactions involving $H_2{PO}_4^{-}$ are given below:
(i) $H_3{PO}_4+H_{2}O\to H_3{O}^{+}+H_2{PO}_4^{-}$
(ii) $H_2{PO}_4^{-}+H_{2}O\to H{PO}_4^{2-}+H_3{O}^{+}$
(iii) $H_2{PO}_4^{-}+{OH}^{-}\to H_3{PO}_4+O^{2-}$

Answer 1

ii) $H_{2}PO{4}^{-}+H_{2}O\to HPO{4}^{2-}+H_3{O}^{+}$

In second reaction $H_{2}PO{4}^{-}$ acts as

an acid because acid gives $H^{+}$ in

aqueous medium.

$H_{2}PO{4}^{-}\left(aq\right)+H_{2}O\left(l\right)\to H_3{O}^{+}+HPO{4}^{2-}\left(aq\right)$

i) $H_{3}PO4+H_{2}O\to H_3{O}^{+}+H_{2}PO{4}^{-}$

$H_{2}PO{4}^{-}$ is a con jugged base of $H_{3}PO4$

iii) $H_{2}PO{4}^{-}+O{H}^{-}\to H_{3}PO4+O_2^{-}$

In this reaction $H_{2}PO{4}^{-}$ acts as a

conjugate base which goined one hydrogen

and made it parent acid.