\(\mathrm{Zn}(\mathrm{OH})_{2}\) is an amphoteric hydroxide. Write balanced ionic equations to show its reaction with (a) \(\mathrm{HCl},\) (b) \(\mathrm{NaOH}\) [the product is \(\left.\mathrm{Zn}(\mathrm{OH})_{4}^{2-}\right]\).

Short Answer

Expert verified
The net ionic equations for the reactions are: \(\mathrm{H^+} + \mathrm{OH^-}\rightarrow \mathrm{H}_{2}\mathrm{O}\) (for the reaction with \(\mathrm{HCl}\)) and \(\mathrm{Zn^{2+}} + 4\mathrm{OH^-}\rightarrow \mathrm{Zn}(\mathrm{OH})_{4}^{2-}\) (for the reaction with \(\mathrm{NaOH}\))

Step by step solution

01

Reaction with HCl

Write the balanced equation for the reaction of zinc hydroxide with hydrochloric acid: \(\mathrm{Zn}(\mathrm{OH})_{2} + 2\mathrm{HCl}\rightarrow \mathrm{ZnCl}_{2} + 2\mathrm{H}_{2}\mathrm{O}\). This is a typical acid-base reaction, with zinc hydroxide acting as a base.
02

Write the ionic equation for the reaction with HCl

Now break the compounds into their ions to write the ionic equation: \(\mathrm{Zn^{2+}} + 2\mathrm{OH^-} + 2\mathrm{H^+} + 2\mathrm{Cl^-}\rightarrow \mathrm{Zn^{2+}} + 2\mathrm{Cl^-} + 2\mathrm{H}_{2}\mathrm{O}\). Note that \(\mathrm{Zn^{2+}}\) and \(\mathrm{Cl^-}\) appear on both sides, so these are spectator ions that can be removed from the ionic equation.
03

Simplify the ionic equation for the reaction with HCl

Take out the spectator ions to get the net ionic equation: \(2\mathrm{H^+} + 2\mathrm{OH^-}\rightarrow 2\mathrm{H}_{2}\mathrm{O}\). This can be simplified to \(\mathrm{H^+} + \mathrm{OH^-}\rightarrow \mathrm{H}_{2}\mathrm{O}\).
04

Reaction with NaOH

Write the balanced equation for the reaction of zinc hydroxide with sodium hydroxide: \(\mathrm{Zn}(\mathrm{OH})_{2} + 2\mathrm{NaOH}\rightarrow \mathrm{Na}_{2}\mathrm{Zn}(\mathrm{OH})_{4}\). This is a typical acid-base reaction, but here zinc hydroxide acts as an acid.
05

Write the ionic equation for the reaction with NaOH

Break the compounds into their ions to write the ionic equation: \(\mathrm{Zn^{2+}} + 2\mathrm{OH^-} + 2\mathrm{Na^+} + 2\mathrm{OH^-}\rightarrow 2\mathrm{Na^+} + \mathrm{Zn}(\mathrm{OH})_{4}^{2-}\). Here, \(\mathrm{Na^+}\) is a spectator ion and can be eliminated from the equation.
06

Simplify the ionic equation for the reaction with NaOH

Take out the spectator ions to get the net ionic equation: \(\mathrm{Zn^{2+}} + 4\mathrm{OH^-}\rightarrow \mathrm{Zn}(\mathrm{OH})_{4}^{2-}\). This equation can't be simplified further, because it is already balanced.

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