The reaction, $\mathrm{B}(\mathrm{OH}{)}_3+\mathrm{NaOH}⟶\mathrm{Na}\left[\mathrm{B}(\mathrm{OH}{)}_4\right]$ can be made to proceed in forward direction by:

 

Solution:

Given reaction is reversible reaction because sodium metaborate, Na[B(OH)_4​] formed by the reaction between B(OH)₃​ and NaOH gets hydrolysed to regenerate B(OH)₃​ and NaOH.

$\mathrm{Na}\left[\mathrm{B}{\left(\mathrm{OH}\right)}_4\right] \stackrel{\mathrm{hydrolysis}}{\to } \mathrm{NaOH} + \mathrm{B}{\left(\mathrm{OH}\right)}_3$
If, some quantity of poly hydroxy compounds like cis- 1, 2-diol, catechol, glycerol etc is added to the reaction mixture then the B(OH)_3​ combines with such poly hydroxy compounds to give chelated complex compound. Due to complex compound formation, stability increases and due to higher stability of complex, reaction moves in forward direction.