Bauxite ore is made up of ${A l}_{2} O_{3} + S i O_{2} + T i O_{2} + {F e}_{2} O_{3}$ . This ore is treated with conc. $N a O H$ solution at $500 K$ and $35$ bar pressure for few hours and filtered, the species present are:

N a A l {(O H)}_{4}

only

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{N a}_{2} T i {(O H)}_{6}

only

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N a A l {(O H)}_{4}

and

{N a}_{2} S i O_{3}

both

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{N a}_{2} S i O_{3}

only

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Solution

The correct option is C $N a A l {(O H)}_{4}$ and ${N a}_{2} S i O_{3}$ both
$T i O_{2}$ and ${F e}_{2} O_{3}$ are basic oxides. So they will not react with $N a O H$ . ${A l}_{2} O_{3}$ is amphoteric oxide and $S i O_{2}$ is acidic oxide. So they will react with $N a O H$
${A l}_{2} O_{3} + N a O H ⟶ N a A l {(O H)}_{4}$
$S i O_{2} + N a O H ⟶ {N a}_{2} S i O_{3}$

Suggest Corrections

Similar questions

c o n c . N a O H

{A l}_{2} O_{3} (s) + 2 O H^{-} (a q) \to 2 A l O_{2}^{-} (a q) + H_{2} O (l)

{F e}_{2} O_{3}

S i O_{2}

A l {(O H)}_{4}^{-}

2 A l {(O H)}_{4}^{-} ⟶ {A l}_{2} O_{3} \cdot 3 H_{2} O (s) + 2 O H_{(a q)}^{-}