Accepcting the definition that an acid is a proton donor, the acid in the following reaction, $N H_{3} + H_{2} O \to N H_{4}^{+} + O H^{-} i s :$

N H_{3}

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H^{+}

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N H_{4}^{+}

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H_{2} O

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Solution

The correct option is D $H_{2} O$
In this reaction, $H_{2} O$ acts as a acid.

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Similar questions

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

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

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

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

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

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