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

Expert Answers
ndnordic eNotes educator| Certified Educator

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

blackeye | Student
12 mL of KOH needed

Access hundreds of thousands of answers with a free trial.

Start Free Trial
Ask a Question