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