In the following reaction ${HC}_{2} O_{4}^{-} (a q) + {PO}_{4}^{3 -} (a q) ⇌ {HPO}_{4}^{2 -} (a q) + C_{2} O_{4}^{2 -} (a q)$ which are the two Bronsted bases?

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${PO}_{4}^{3 -}$ and $C_{2} O_{4}^{2 -}$

${HC}_{2} O_{4}^{-}$ and ${PO}_{4}^{3 -}$

${HC}_{2} O_{4}^{-}$ and ${HPO}_{4}^{2 -}$

${HPO}_{4}^{2 -}$ and $C_{2} O_{4}^{2 -}$

answer is D.

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Detailed Solution

${PO}_{4}^{3 -}$ and $C_{2} O_{4}^{2 -}$ are the two Bronsted bases.

Bronsted bases are proton $(H^{+})$ acceptor

$\begin{aligned} {PO}_{4}^{3 -} + H^{+} \to {HPO}_{4}^{2 -} \\ C_{2} O_{4}^{2 -} + H^{+} \to {HC}_{2} O_{4}^{-} \end{aligned}$

${HPO}_{4}^{2 -}$ and ${HC}_{2} O_{4}^{-}$ are Bronsted acids. They are proton $(H^{+})$ donor.

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