Question

On acidification of aqueous borax solution, it gives white flakes. This is due to formation of

• A. $B(OH{)}_3$
• B. $B_2{O}_3$
• C. $B_2{H}_6$
• D. $NaB{O}_2$

Answer 1

The correct option is A $B(OH{)}_3$

Borax dissolve in water to give both $B\left(OH{)}_3\text{and}\right[\left(B\right(OH{)}_4{]}^{-}$. This solution contains equal amounts of weak acid and its salt, thus it can act as a buffer.
On acidification of this solution, $\left[\right(B(OH{)}_4{]}^{-}$ reaction with $H^{+}$ ions to form orthoboric acid. $\left[B_4{O}_5\right(OH{)}_4{]}^{2-}+5H_{2}O\rightleftharpoons 2B\left(OH{)}_3\right(weakacid)+2[B\left(OH\right)4{]}^{-}salt$
$2\left[B\right(OH{)}_4{]}^{-}+2H_3{O}^{+}\to 2B(OH{)}_3+4H_{2}O$