What volume (in milliliters) of 0.180 M HNO₃ are required to neutralize 80.0 mL of 0.190 M NaOH?

Answer 1

First balance equation: 

NaOH + HNO₃ -> H₂O + NaNO₃

1 mole of NaOH to 1 mole HNO₃

moles of NaOH = 0.080L * 0.190M = 0.0152 moles

0.0152 moles of HNO₃

Divide by molarity to get volume in L

0.0152moles /0.180M= 0.084L or 84 mL