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

Explanation:

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