When Borax dissolved in water, it
is subject to completely dissociate, and then, hydrolyzes to orthoboric acid (B(OH)3 and [B(OH)4]- anions, and resulting PH of solution about 9.13.
According to the equation:
Na2[B4O5(OH)4]⋅8H2O(s)⟶2Na+(aq) + B4O5(OH)42−
(aq) + 8H2O(l)
B4O5(OH)42−(aq)
+ 5H2O(l) ⟶ 4 B(OH)3(aq) + 2OH−(aq)
The liberated B(OH)3(aq) acts as a Lewis acid on OH− ions
from auto ionization of water, and hence, in equilibrium with water:
B(OH)3(aq)+2H2O(l) ⟶ B(OH)4−(aq) + H3O+(aq)
Therefore, one mole of borax reacts with two moles of strong acid
solutions. This is because when Borax dissolved
in water Both B(OH)3(aq) and B(OH)4−(aq) are formed but
only
B(OH)4−(aq) react with acid.
For example:
Na2[B4O5(OH)4].8H2O(aq)
+ 2HCl(aq) ⟶ 4B(OH)3(aq)+ 2NaCl (aq) + 5 H2O
ILLUSTRATIVE EXAMPLE:
Aqueous solution of borax reacts with two moles
of acids.
This is
because of:
(A) Formation of 2mol of B(OH)3
only
(B) Formation of 2mol of Na[B(OH)4]¯
only
(C) Formation of 1mol each of B(OH)3
and Na[B(OH)4]¯
(D) Formation of 2mol each of Na[B(OH)4]¯and B(OH)3 of which only Na[B(OH)4]¯ reacts with acid.
Answer: (D)
Solution:
Na2B4O7
+ 7H2O ⟶ 2B(OH)3
+ 2Na[B(OH)4]
B(OH)3
or H3BO3 is an acid and does not react with acid. Hence Na[B(OH)4]
reacts with acid.
