`H_(3)PO_(4)hArr H^(o+) +H_(2)PO_(4)^(Theta), K_(a_(1))`:
`H_(2)PO_(4)^(Theta) hArr H^(o+) +HPO_(4)^(2-),' K_(a_(2))`:
`HPO_(4)^(2-) hArr H^(o+) + PO_(4)^(3-), K_(a_(3)):`
Mark out the incorrect statements:

Options

A. `K_(a_(1)) gt K_(a_(2)) gt K_(a_(3))`
B. `pH (H_(2)PO_(4)^(Theta)) = (pK_(a_(1))+pK_(a_(2)))/(2)`
C. Both `H_(3)PO_(4)` and `H_(2)PO_(4)^(Theta)` are more acidic than `HPO_(4)^(2-)`
D. Only `HPO_(4)^(2-)` is amphiprotic anion in the solution.

Answer

B, D

Explanation

a. Statement (a) is correct.
`K_(a)` values decreases successively, since it is difficult to remove `H^(o+)` ion from an anion than a neutral compound. Similarly, it is difficult to remove `H^(o+)` ion from an dianion than and anion.
Hence `K_(a_(1)) gt K_(a_(2)) gt K_(a_(3))`
b. Statement (b) is incorrect.
`H_(3)PO_(4) overset(K_(a_(1)))hArr H^(o+) +H_(2)PO_(4)^(Theta)`
`H_(3)PO_(4)^(Theta) overset(K_(a_(2)))hArrH^(o+) +HPO_(4)^(2-)`
`pH(H_(2)PO_(4)^(Theta)) = (pK_(a_(1))+pK_(a_(1)))/(2)`
This is valid only when `H_(3)PO_(4)` during titration is completely converted to `H_(2)PO_(4)^(Theta)`.
`HPO_(4)^(2-) overset(K_(a_(3)))hArr H^(o+) +PO_(4)^(3-)`
`pH(HPO_(4)^(2-)) = (pK_(a_(2))+pK_(a_(3)))/(2)`
This is again valid when starting with `H_(2)PO_(4)^(2-)` and is completely converted to `HPO_(4)^(2-)`
c. Statement (c) is correct.
From the `K_(a)` values, it is evident that both `H_(3)PO_(4)` and `H_(2)PO_(4)^(Theta)` are more acidic than `HPO_(4)^(2-)`.
d. Statement (d) is incorrect.
Both `H_(2)PO_(4)^(Theta)` and `HPO_(4)^(2-)` are amphiprotic anion in the solution.
`H_(3)PO_(4) overset(+H^(o+))larrH_(2)PO_(4)^(Theta)overset(-H^(o+))rarrHPO_(4)^(2-)+H^(o+)`
`H_(2)PO_(4)^(Theta)overset(+H^(o+))larrHPO_(4)^(2-)overset(-H^(o+))rarrPO_(4)^(3-)+H^(o+)`