# Question 8a5f1

Jun 29, 2015

$N {a}_{2} Z n {O}_{2 \left(a q\right)} + 2 H C {l}_{\left(a q\right)} \rightarrow Z n {O}_{\left(s\right)} + {H}_{2} {O}_{\left(l\right)} + N a C {l}_{\left(a q\right)}$

#### Explanation:

The ions present in sodium zincate are $N {a}^{+}$ and zincate which is $Z n {O}_{2}^{2 -}$.

Zinc oxide is amphoteric, which means that it can react react with both acid and base.

Sodium zincate is formed if zinc oxide is treated with an alkali such as $N a O H$:

$Z n {O}_{\left(s\right)} + 2 N a O {H}_{\left(a q\right)} \rightarrow N {a}_{2} Z n {O}_{2 \left(a q\right)} + {H}_{2} {O}_{\left(l\right)}$

We can reverse the process by adding acid:

$N {a}_{2} Z n {O}_{2 \left(a q\right)} + 2 H C {l}_{\left(a q\right)} \rightarrow Z n {O}_{\left(s\right)} + {H}_{2} {O}_{\left(l\right)} + N a C {l}_{\left(a q\right)}$

The zincate ion is written more precisely as $Z n {\left(O H\right)}_{4 \left(a q\right)}^{2 -}$.

Adding acid would remove 2 $O {H}^{-}$ ions to leave a white precipitate of zinc hydroxide Zn(OH)_(2(s)#.