April 28, 2024, 07:10:19 PM
Forum Rules: Read This Before Posting


Topic: Need help with this molarity problem that involves a neutralization reaction.  (Read 4167 times)

0 Members and 1 Guest are viewing this topic.

mummer43

  • Guest
I have no idea how to do this problem!

What volume (L) of 0.0250 M HNO3 is required to neutralize a solution prepared by disolving 17.5g of NaOH in 350 mL of water?

HNO3 (aq) + NaOH  -----> H2O (l) + NaNO3

Any help would be greatly appreciated!!

Offline mike

  • Retired Staff
  • Sr. Member
  • *
  • Posts: 1245
  • Mole Snacks: +121/-35
  • Gender: Male
How many moles of NaOH are there in the solution?
There is no science without fancy, and no art without facts.

metalriffzz

  • Guest
Well to get you started out this is a problem where you're going to have to neutralize the moles of NaOH you have with an equal number of moles of HNO3. Your molar weight for NaOH is roughly 40 g/mol and rember that molarity = moles per liter, meaning that for .0250 M HNO3 there's .0250 moles of acid per liter of solution you add.

17.5/40 = ?

? = ??(.0250)

Good luck.

Offline Borek

  • Mr. pH
  • Administrator
  • Deity Member
  • *
  • Posts: 27665
  • Mole Snacks: +1801/-410
  • Gender: Male
  • I am known to be occasionally wrong.
    • Chembuddy
Quite a lot - almost two buckets!

Are you sure you have to use such diluted acid?
ChemBuddy chemical calculators - stoichiometry, pH, concentration, buffer preparation, titrations.info

Sponsored Links