Question

In the following unbalanced reaction, the product formed is:
NaAlO$_2$ + Al(OH)$_3$ + H$_2$O $\rightarrow$ ?

• A. Na$_3$[Al(OH)$_6$]
• B. Na$_2$[Al(OH)$_4$]
• C. Na[Al(OH)$_4$]
• D. Na[Al(OH)$_6$]

Hint:

Recognize complex ion formation from amphoteric Al(OH)$_3$
Tetrahydroxoaluminate is a key species in amphoteric behavior

Answer 1

The Correct Option is C

- Sodium aluminate reacts with aluminum hydroxide and water: \[ \text{NaAlO}_2 + \text{Al(OH)}_3 + \text{H}_2\text{O} \rightarrow \text{Na[Al(OH)}_4\text{]} \] - This forms a soluble complex ion: tetrahydroxoaluminate (Na[Al(OH)$_4$])