# Question 56cc5

Feb 8, 2016

${\text{ZnCl}}_{2}$

#### Explanation:

In water (or solutions), ${\text{ZnCl}}_{2}$ dissolves.

${\text{ZnCl"_2 (s) -> "Zn"^{2+} (aq) + 2"Cl}}^{-} \left(a q\right)$

The chloride ions do not participate in the following reactions.

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For the sodium hydroxide reaction, the ${\text{Zn}}^{2 +}$ will react with the ${\text{OH}}^{-}$ 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 ${\text{NH}}_{3}$ first dissociates in water.

${\text{NH"_3 (aq) + "H"_2"O" (l) rightleftharpoons "NH"_4^+ (aq) + "OH}}^{-} \left(a q\right)$

The ${\text{Zn}}^{2 +}$ will then react with the ${\text{OH}}^{-}$ 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.

${\text{Zn"("OH")_2 (s) + 4"NH"_3 (aq) -> "Zn"("NH"_3)_4^{2+} (aq) + 2"OH}}^{-} \left(a q\right)$

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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".