When excess NaOH of solution is added in potash alum, the product is ?
 

Alum consists ${\mathrm{Al}}_2{\left({\mathrm{SO}}_4\right)}_3$ which reacts with NaOH.

${\mathrm{Al}}_2{\left({\mathrm{SO}}_4\right)}_3+6\mathrm{NaOH}\stackrel{ }{\to }2\mathrm{Al}{\left(\mathrm{OH}\right)}_3+3{\mathrm{Na}}_2{\mathrm{SO}}_4$

$\mathrm{Al}{\left(\mathrm{OH}\right)}_3+\mathrm{NaOH}\stackrel{ }{\to }{\mathrm{NaAlO}}_2+2{\mathrm{H}}_2\mathrm{O}$

The other sulphates ${\mathrm{K}}_2{\mathrm{SO}}_4$  and ${\mathrm{Na}}_2{\mathrm{SO}}_4$  are also soluble in water. Thus a clear solution is obtained.