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

A
a bluish precipitate
B
clear solution
C
white precipitate
D
greenish precipitate

Solution:

Formula of potash alum is $K_{2} {SO}_{4} \cdot {Al}_{2} {({SO}_{4})}_{3} \cdot 24 H_{2} O$ . Potash alum consists of ${Al}_{2} {({SO}_{4})}_{3}$ which initially forms $Al {(OH)}_{3}$ , when reacts with NaOH. As NaOH is in excess, therefore $Al {(OH)}_{3}$ further react with NaOH to form NaAlO₂ which is soluble and a clear solution is obtained.
$\begin{array}{l} {Al}_{2} {({SO}_{4})}_{3} + 6 NaOH ⟶ 2 Al (OH)_{3} + 3 {Na}_{2} {SO}_{4} \\ Al (OH)_{3} + NaOH ⟶ \underset{Soluble}{{NaAlO}_{2}} + 2 H_{2} O \end{array}$

Solution:

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