Question

The reaction ${B(OH)_3 + NaOH <=> Na[B(OH)_4]}$ ${ B(OH)_3 + NaOH <=> NaBO_2 + 2 \, H_2O}$ can be made to proceed to forward direction by the addition of

• A. borax
• B. cis-l, 2 diol
• C. trans-1, 2 diol
• D. ${Na_2HPO_4}$

Answer 1

Answer: (B)

cis-l, 2 diol

Answer 2

${B(OH)_3}$ behaves as a weak monobasic acid.
However, when cis-diol is added to the titration solution, it forms a very stable complex with ${[B(OH)_4]^-}$ formed by the forward reaction, thus effectively removing it from solution. This upsets the balance of the equation and the reaction proceeds completely to the right. Thus all the ${B(OH)_3}$ reacts with NaOH and it acts as a strong acid in the presence of the cis-diol