How much 6.0 M HNO2 is needed to neutralize 39 mL of 2.0 M KOH?

To solve a problem like this just remember that mL * M =
mL * M.

Then write a balanced chemical equation for this
reaction:

HNO2 + KOH --> KNO2 +
HOH

In this case it takes one mole of AHON2 to react with
one mole of KOH, producing one mole of KNO2 + one mole of water
(HOH)

So you take the know
quantities:

M of HNO2 = 6.0

M
of KOH = 2.0

mL of KOH =
39

Now solve for the unknown, the mL of
HNO2

mL * 6.0 = 39 * 2.0

So mL
of KOH = (39 *2)/6 = 12 mL of KOH