Consider the following reaction:
$N a N H_{2 (s)} + H_{2} O_{(l)} ⇌ N a_{(a q)}^{+} + O H_{(a q)}^{-} + N H_{3 (a q)}$
What is the conjugate acid in this reaction?

H_{2} O

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N a^{+}

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O H^{-}

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N H_{3}

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Solution

The correct option is A $H_{2} O$
$N a N H_{2 (s)} + H_{2} O_{(l)} ⇋ N a_{(a q)}^{+} + O H_{(a q)}^{-} + N H_{3 (a q)}$
When a base accepts a proton, it results into its conjugate acid.
In the above equation, in backward reaction, the base $O H^{-}$ accepts a proton to give $H_{2} O$ . Hence, $H_{2} O$ is the conjugate acid in this reaction.

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

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

N O_{3}^{-} (a q)

H_{2} O_{(l)} + H_{2} O_{(l)} ⇌ H_{3} O_{(a q)}^{+} + O H_{(a q)}^{-}

H_{2} S_{(g)} + {H_{2} O}_{(l)} ⇌ {H_{3} O}_{(a q)}^{+} + {H S^{-}}_{(a q)}

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

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