Identify the Bronsted acid in the following equation:

$P O_{4}^{3 -} + H_{2} O (l) \to H P O_{4}^{2 -} (a q) + O H^{-} (a q)$

P O_{4}^{3 -}

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H P O_{4}

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H_{2} O

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O H^{-}

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Solution

The correct option is D $H_{2} O$
In the reaction:
$P O_{4}^{3 -} + H_{2} O (l) \to H P O_{4}^{2 -} (a q) + O H^{-} (a q)$

Only $H_{2} O$ acts as Bronsted acid among the given options as it donate the proton $(H^{+})$ to $O H^{-}$ ions.

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Similar questions

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

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

{NH}_{3} (aq) + H_{2} O (l) \to {NH}_{4}_{4}^{+} (aq) + {OH}^{-} (aq)

H C O_{3}^{-} (a q .) + O H^{-} (a q .) \to C O_{3}^{2} (a q .) + H_{2} O

H_{3} P O_{4}

H_{3} P O_{4} + O H^{-} \to H_{2} P O_{4}^{-} + H_{2} O

H_{3} P O_{4} + 2 O H^{-} \to H P O_{4}^{2 -} + 2 H_{2} O

H_{3} P O_{4} + 3 O H^{-} \to P O_{4}^{3 -} + 3 H_{2} O

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