Question

$H_{3}P{O}_4\rightleftharpoons H^{\oplus }+H_{2}P{O}_4^{⊝};K_{a_1}:$
$H_{3}P{O}_4^{⊝}\rightleftharpoons H^{\oplus }+HP{O}_4^{2-};K_{a_2}:$
$HP{O}_4^{2-}\rightleftharpoons H^{\oplus }+P{O}_4^{3-};K_{a_3}:$
Mark out the incorrect statements

• A. $K_{a_1}>K_{a_2}>K_{a_3}$
• B. $pH\left(H_{2}P{O}_4^{⊝}\right)=\frac{p{K}_{a_1}+p{K}_{a_2}}{2}$
• C. Both $H_{3}P{O}_4$ and $H_{2}P{O}_4^{⊝}$ are more acidic than $HP{O}_4^{2-}$
• D. Only $HP{O}_4^{2-}$ is amphiprotic anion in the solution

Answer 1

Answer: (B), (D)

The correct options are
B $pH\left(H_{2}P{O}_4^{⊝}\right)=\frac{p{K}_{a_1}+p{K}_{a_2}}{2}$
D Only $HP{O}_4^{2-}$ is amphiprotic anion in the solution

After losing one $H^{+}$ ion, the tendency to lose another $H^{+}$ ion decreases due to the existing charge on the molecule and incomplete octet.
Hence, $K_{a_1}>K_{a_2}>K_{a_3}$, which implies Both $H_{3}P{O}_4$ and $H_{2}P{O}_4^{⊝}$ are more acidic than $HP{O}_4^{2-}$.
Amphiprotic molecules are those which can either donate or accept a proton (H+)
$HP{O}_4^{2-}$ and $H_{2}P{O}_4^{-}$ are amphiprotic anion in the solution.
Further, the Ka values cannot be directly related.