Question

In the following reaction:

${H{C}_2{O}_4}^{-}\left(aq\right)+P{O}_4^{3-}\left(aq\right)\rightleftharpoons {HPO}^{-}\left(aq\right)+C_2{O}_4^{2-}\left(aq\right)$, which are the two bronsted bases?

• A. ${H{C}_2{O}_4}^{-}$ and $P{O}_4^{3-}$
• B. $HP{O}_4^{2-}$ and ${C_2{O}_4}^{-}$
• C. ${H{C}_2{O}_4}^{-}$ and $HP{O}_4^{2-}$
• D. $P{O}_4^{3-}$ and $C_2{O}_4^{2-}$

Answer 1

The correct option is D $P{O}_4^{3-}$ and $C_2{O}_4^{2-}$

$P{O}_4^{3-}and{C}_2{O}_4^{2-}$ are the two Bronsted bases.
Bronsted bases are proton $\left(H^{+}\right)$ acceptor.
$P{O}_4^{3-}+H^{+}\to HP{O}_4^{2-}$
$C_2{O}_4^{2-}+H^{+}\to H{C}_2{O}_4^{-}$
$HP{O}_4^{2-}$ and $H{C}_2{O}_4^{-}$ are Bronsted acids. They are proton $\left(H^{+}\right)$ donor.