How can the following reaction be made to proceed in forward direction?

$\mathrm{B}(\mathrm{OH}{)}_3+\mathrm{NaOH}\rightleftharpoons \mathrm{Na}\left[\mathrm{B}(\mathrm{OH}{)}_4\right]$

Solution:

An aqueous solution of boric acid undergoes titration with NaOH using phenolphthalein as an indicator. Reaction is:

 $\mathrm{B}{\left(\mathrm{OH}\right)}_3+\mathrm{NaOH}\rightleftharpoons \mathrm{Na}\left[\mathrm{B}{\left(\mathrm{OH}\right)}_4\right]$

In cis-1,2-diol, both -OH groups present on the same side and forms chelate complex with B(OH)₃. This chelate complex enhances the acidic strength of B(OH)₃, which consequently increases the rate of reaction in the forward direction. In trans-1,2-diol, two -OH groups present on the opposite direction, as a result, chelate formation do not take place with B(OH)₃.