Question

How many mL of a 0.4 M nitric acid solution are required to neutralize 200 mL of a 0.16 M potassium hydroxide

Answer 1

Answer:

80 ml.

Explanation:

The balanced equation is:

KOH + HNO3 ---->  KNO3 + H2O

So 1 mole of KOH reacts with 1 mole of HNO3

200ml of 0.16M KOH contains 0.16 * 200/1000

= 0.16 * 0.2

= 0.032 M KOH.

So we need the volume (V) of 0.4 M HNO3 which contains 0.032 moles.

0.4 * V = 0.032

V = 0.032 / 0.4

V = 0.08 liters

= 80 mls.