Which are the set of amphiprotic species?

H_{2} O, H_{2} P O_{4}^{-}, H P O_{4}^{2 -}

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H P O_{4}^{2 -}, H C O_{3}^{-}

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H_{2} P O_{2}^{-}, H_{2} P O_{3}^{-}, H C_{2} O_{4}^{-}

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H P O_{3}^{2 -}, H_{2} O, C O_{3}^{2 -}

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Solution

The correct options are
A $H_{2} O, H_{2} P O_{4}^{-}, H P O_{4}^{2 -}$
C $H P O_{4}^{2 -}, H C O_{3}^{-}$
$A m p h i p r o t i c s p e c i e s$ :-
A compound or ion which can both donate protons and also accept them is known as Amphiprotic compound or ions.

$2 H_{2} O ⇌ O H^{-} + H_{3} O^{+}$ (Amphiprotic)

The ions which are Amphiprotic are shown in the above figure.

Hence Option A and B are correct.

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Similar questions

Identify Bronsted acids in the reaction : $H_{2} C O_{3} + H P O_{4}^{2 -} ⇌ H_{2} P O_{4}^{-} + H C O_{3}^{-}$

H_{2} P O_{4}^{-}

H_{2} P O_{3}^{-}

H P O_{4}^{2 -}

H P O_{3}^{2 -}

C l O^{-}

C l O_{2}^{-}

N H_{3}

N H_{2}^{-}

H_{2} O + H_{3} P O_{4} ⇌ H_{3} O^{+} + H_{2} P O_{4}^{-}; p K_{1} = 2.15

H_{2} O + H_{2} P O_{4}^{-} ⇌ H_{3} O^{+} + H P O_{4}^{2 -}; p K_{2} = 7.20

p H

0.01 M N a H_{2} P O_{4}

H C_{2} O_{4}^{-} + P O_{4}^{3 -} ⇌ H P O_{4}^{2 -} + C_{2} O_{4}^{2 -}

H_{2} O + H_{3} P O_{4} + H_{3} O^{+} + H_{2} P O_{4}^{-}, p^{k_{1}} = 2.15

H_{2} O + H_{2} P O_{4}^{-} + H_{3} O^{+} + H P O_{4}^{2 -}, p^{k_{2}} = 7.20

p^{H}

N a H_{2} P O_{4}

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