In $N H_{3} (a q) + H_{2} C O_{3} (a q) ⟶ N H_{4}^{+} (a q) + H C O_{3}^{-} (a q), N H_{4}^{+} (a q)$ acts as a\an:

indicator

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hydrate

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acid

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base

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salt

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Solution

The correct option is B acid
${N H}_{4}^{+} \to$ Since it is colorless so no indicator property only it can give proton to bases.
Thus it acts as a Bronsted acid.

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Similar questions

N H_{3} (a q) + H_{2} C O_{3} (a q) ⇋ N H_{4}^{+} (a q) + H C O_{3}^{-} (a q)

N H_{4}^{+}

{N H}_{4}^{+} (a q) + {C N}^{-} (a q) ⇌ H C N (a q) + {N H}_{3} (a q)

N H_{3 (a q)} + H_{2} O_{(l)} = N H_{4}^{+} (a q) + O H^{-} (a q)

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

Δ_{f} G^{o}

N H_{4} O H (a q) ⇌ N H_{4}^{+} (a q) + O H^{-} (a q)

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