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

Expert Answers

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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

Approved by eNotes Editorial Team

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