. What volume of 0.250 M nitric acid is needed to neutralize 17.35 mL of 0.195 M KOH solution?

Balanced equation for the neutralization:
HNO₃ + KOH → NaNO₃ + H₂O
Mole ratio  HNO₃ : KOH = 1 : 1

Millimoles of KOH reacted = (0.195 mmol/mL) × (17.35 mL) = 3.383 mmol
Millimoles of HNO₃ needed = 3.383 mmol
Volume of HNO₃ needed = (3.383 mmol) / (0.250 mmol/mL) = 13.5 mL

====
OR:

(0.195 mol KOH / 1000 mL KOH solution) × (17.35 mL KOH solution) × (1 mol HNO₃ / 1 mol KOH) × (1000 mL HNO₃ solution / 0.250 mol HNO₃)
= 13.5 mL HNO₃ solution