Question #56cc5

In water (or solutions), `"ZnCl"_2` dissolves.

`"ZnCl"_2 (s) -> "Zn"^{2+} (aq) + 2"Cl"^{-} (aq)`

The chloride ions do not participate in the following reactions.

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For the sodium hydroxide reaction, the `"Zn"^{2+}` will react with the `"OH"^-` ions (from NaOH) in the solution.

`"Zn"^{2+} (aq) + 2 "OH"^{-} (aq) -> "Zn"("OH")_2 (s)`

`"Zn"("OH")_2` is amphoteric, so it can reacts with the excess NaOH.

`"Zn"("OH")_2 (s) + 2 "OH"^{-} (aq) -> "Zn"("OH")_4^{2-} (aq)`

Zinc hydroxide will dissolve because the ion is normally surrounded by water ligands; when there is excess sodium hydroxide, the hydroxide ions displace the water ligands and the complex will acquire a `-2` charge, making it soluble.

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For the ammonia reaction, the `"NH"_3` first dissociates in water.

`"NH"_3 (aq) + "H"_2"O" (l) rightleftharpoons "NH"_4^+ (aq) + "OH"^{-} (aq)`

The `"Zn"^{2+}` will then react with the `"OH"^-` ions in the solution.

`"Zn"^{2+} (aq) + 2 "OH"^{-} (aq) -> "Zn"("OH")_2 (s)`

Zinc hydroxide also dissolves in excess aqueous ammonia to form a colorless, water-soluble ammine complex.

`"Zn"("OH")_2 (s) + 4"NH"_3 (aq) -> "Zn"("NH"_3)_4^{2+} (aq) + 2"OH"^{-} (aq)`

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The rest of the compounds provided will also dissolve in the solutions, but their hydroxides will precipitate out, rendering them "insoluble".