mass of NaOH = 15 g
molar mass of NaOH = 40 g/mol
moles of NaOH = 15 / 40
                        Â
= 0.375
moles of HNO3 = molarity x volume
                        Â
= 0.250 x 0.150
                         Â
= 0.0375
so
0.0375 moles of HNO3 react with 0.0375 moles of NaOH
remaining moles of NaOH = 0.375 - 0.0375 = 0.3375
final conentration of OH- = 0.3375 / volume
                                        Â
= 0.3375 / 0.150
                                         Â
=Â Â 2.25 M
