Question

Zinc hydroxide is an amphoteric substance. Write equations that describe $\mathrm{Zn}(\mathrm{OH}{)}_2$ acting as a Brønsted-Lowry base toward ${\mathrm{H}}^{+}$ and as a Lewis acid toward ${\mathrm{OH}}^{-}$.

Short answer

Zinc hydroxide, $\mathrm{Zn}(\mathrm{OH}{)}_2$, can act as a Brønsted-Lowry base and a Lewis acid. In the Brønsted-Lowry base reaction, it accepts a proton (${\mathrm{H}}^{+}$): $$\mathrm{Zn}(\mathrm{OH}{)}_2+{\mathrm{H}}^{+}\to \mathrm{Zn}(\mathrm{OH}{)}_3^{+}$$ In the Lewis acid reaction, it accepts an electron pair from a hydroxide ion (${\mathrm{OH}}^{-}$): $$\mathrm{Zn}(\mathrm{OH}{)}_2+{\mathrm{OH}}^{-}\to \mathrm{Zn}(\mathrm{OH}{)}_3^{-}$$

Step-by-step solution

Acting as a Brønsted-Lowry base

Zinc hydroxide, $\mathrm{Zn}(\mathrm{OH}{)}_2$, can accept a proton (${\mathrm{H}}^{+}$) to form a new compound. In this case, the reaction would be: $$\mathrm{Zn}(\mathrm{OH}{)}_2+{\mathrm{H}}^{+}\to \mathrm{Zn}(\mathrm{OH}{)}_3^{+}$$

Acting as a Lewis acid

Zinc hydroxide, $\mathrm{Zn}(\mathrm{OH}{)}_2$, can accept an electron pair from a hydroxide ion (${\mathrm{OH}}^{-}$) to form a new compound. In this case, the reaction would be: $$\mathrm{Zn}(\mathrm{OH}{)}_2+{\mathrm{OH}}^{-}\to \mathrm{Zn}(\mathrm{OH}{)}_3^{-}$$ These equations show how zinc hydroxide can act as a Brønsted-Lowry base by accepting a proton when it reacts with a hydrogen ion, and as a Lewis acid by accepting an electron pair when it reacts with a hydroxide ion.