$_{HA(aq)+NaOH(aq)\to NaA(aq)+{{H}_{2}}O(l)}^{\left[ HA \right]=\left[ {{H}_{3}}{{O}^{+}} \right]={{10}^{-PH}}={{10}^{-3}}mol.{{L}^{-1}}}$  $?molNaOH=10LHA\times \frac{{{10}^{-3}}molHA}{1LHA}\times \frac{1molNaOH}{1molHA}={{10}^{-2}}molNaOH$