Question #0dc61

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Question #0dc61

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Answer 1 · 2016-06-28T23:28:14

${NH}_{4 (a q)}^{+} + {OH}_{(a q)}^{-} \to {NH}_{3 (g)} ↑ ⏐ ⏐ + H_{2} O_{(l)}$ $"NH"_ (4(aq))^(+) + "OH"_ ((aq))^(-) -> "NH"_ (3(g)) uarr + "H"_ 2 "O"_ ((l))$

Explanation:

The ammonium cation, ${NH}_{4}^{+}$ $"NH"_4^(+)$ , will react with the hydroxide anion, ${OH}^{-}$ $"OH"^(-)$ , which is delivered to the solution by the sodium hydroxide, $NaOH$ $"NaOH"$ , to form ammonia, ${NH}_{3}$ $"NH"_3$ , and water.

This is essentially a neutralization reaction that takes place between a weak acid, the ammonium ion, and a strong base, sodium hydroxide.

The balanced chemical equation that describes the reaction looks like this

${NH}_{4 (a q)}^{+} + {OH}_{(a q)}^{-} \to {NH}_{3 (g)} ↑ ⏐ ⏐ + H_{2} O_{(l)}$ $"NH"_ (4(aq))^(+) + "OH"_ ((aq))^(-) -> "NH"_ (3(g)) uarr + "H"_ 2 "O"_ ((l))$

Because ammonia gas is given off, this reaction is used as a confirmatory test for the presence of the ammonium cations.

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Question #0dc61

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