What volume of 0.120 M HNO3 (qq) is needed to completely neutralize 150.0 milliliters of 0.100 M NaOH (aq)

Balanced equation for the neutralisation reaction is as follows

NaOH + HNO₃ - - > NaNO₃ + H₂O

stoichiometry of NaOH to HNO₃ is 1:1

number of NaOH moles to be neutralised is - 0.100 mol/L x 0.1500 L = 0.015 mol

therefore number of HNO₃ moles reacted - 0.015 mol

molarity of HNO₃ solution is 0.120 M

volume of solution in which there 0.120 mol - 1 L

then 0.015 mol are in - 0.015 mol / 0.120 mol/L = 125 mL

volume of HNO₃ required - 125 mL