Question #8a5f1

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Question #8a5f1

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Answer 1 · 2015-06-29T16:34:06

$N a_{2} Z n O_{2 (a q)} + 2 H C l_{(a q)} \to Z n O_{(s)} + H_{2} O_{(l)} + N a C l_{(a q)}$ $Na_2ZnO_(2(aq))+2HCl_((aq))rarrZnO_((s))+H_2O_((l))+NaCl_((aq))$

Explanation:

The ions present in sodium zincate are $N a^{+}$ $Na^(+)$ and zincate which is $Z n O_{2}^{2 -}$ $ZnO_2^(2-)$ .

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 $N a O H$ $NaOH$ :

$Z n O_{(s)} + 2 N a O H_{(a q)} \to N a_{2} Z n O_{2 (a q)} + H_{2} O_{(l)}$ $ZnO_((s))+2NaOH_((aq))rarrNa_2ZnO_(2(aq))+H_2O_((l))$

We can reverse the process by adding acid:

$N a_{2} Z n O_{2 (a q)} + 2 H C l_{(a q)} \to Z n O_{(s)} + H_{2} O_{(l)} + N a C l_{(a q)}$ $Na_2ZnO_(2(aq))+2HCl_((aq))rarrZnO_((s))+H_2O_((l))+NaCl_((aq))$

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

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

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Question #8a5f1

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