Titration of Borax: Tincal : Suhaga :

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.