What volume of 3.3 m hno3 solution is needed to completely neutralize 175 ml of a 4.10 m koh solution?

HNO₃ + KOH = KNO₃ + H₂O

c (HNO₃) = 3.3 mol/L

c (KOH) = 4.10 mol/L

v (KOH) = 175 mL = 0.175 L

n (KOH) = n (HNO₃)

n (KOH) = c (KOH) v (KOH)

n (HNO₃) = c (HNO₃) v (HNO₃)

c (KOH) v (KOH) = c (HNO₃) v (HNO₃)

v (HNO₃) = c (KOH) v (KOH) / c (HNO₃)

v (HNO₃) = 4.10*0.175/3.3=0.217 L = 217 mL