50 mL of 0.1 M $NaOH$ is added to $60 mL$ of 0.15 M ${\mathrm{H}}_3{\mathrm{PO}}_4$ solution $\left(K_1,K_2\text{ and }{K}_3\text{ for }{\mathrm{H}}_3{\mathrm{PO}}_4\text{ are }{10}^{-3}\right.$, ${10}^{-8}\text{ and }{10}^{-13}\text{ respectively). }$The $pH$ of the mixture would be about (log 2 = 0.3) :