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Help with TITRATIONS !

  1. Mar 26, 2008 #1
    Help with TITRATIONS URGENT!

    1. The problem statement, all variables and given/known data
    You need to neutralize 100 ml of 2.5 M HCl. You plan on using .25 M NaOH. How many liters of NaOH will you need to do this?


    2. Relevant equations
    What does the above mean?

    3. The attempt at a solution
    I could not attempt as I didn't attend the lesson due to absence. Could someone give me the 411 on how to do this using factor label and mole conversions. I googled this and got complex things that looked like physics,graphs and calculus jargon which I'm not on a high enough level of understanding to comprehend... :P
     
    Last edited: Mar 26, 2008
  2. jcsd
  3. Mar 26, 2008 #2
    Edit: Bad science, corrected in my next post.
     
    Last edited: Mar 26, 2008
  4. Mar 26, 2008 #3
    I don't know what that means, could you explain how I would apply this to something more complex such as "How many particles of HCl would be neutralized by 200 ml of .2 M NaOH?"
     
  5. Mar 26, 2008 #4
    Also if I have to use grams can I still use that equation? I just figured out what M and V meant, lol how silly of me...
     
  6. Mar 26, 2008 #5
    [tex]M= \frac {number of moles}{Volume}[/tex]

    You can calculate the number of moles of HCl, calculate how many moles of NaOH you need to neutralise this amount, then calculate the volume of NaOH you need.
     
  7. Mar 26, 2008 #6
    Whoops, okay, got off track on my last post, let's forget that. I was approaching the problem as a dilution rather than a titration, which just happened to work out almost correctly by coincidence.

    This problem is fairly simple; HCl and NaOH neutralize at a ratio of 1:1. You have a 100mL of 2.5M HCl. That's a tenth of a Liter, thus you have 0.25M HCl that needs to be neutralized. So you need to add 0.25M of NaOH to neutralize it.
     
  8. Mar 26, 2008 #7
    umm what if I have grams of something?
     
  9. Mar 26, 2008 #8
    Not hard to convert grams to moles using the molecular weight.
     
  10. Mar 26, 2008 #9
    I still have no idea on how to approach these... could you give me steps on how to solve a titration, I just asked a bunch of people in my class and they don't know.
     
  11. Mar 26, 2008 #10
    You've been given the tools on how to solve it, mate. Molarity = # of moles divided by the volume of the solution. With this, calculate the number of moles of HCl, with this number, calculate how many moles of NaOH you need to neutralise it if it's a 1:1 ratio, with this number, calculate how many litres of NaOH you need if you're given the molarity and you calculated the moles of NaOH needed for titration.
     
  12. Mar 26, 2008 #11
    Okay, I'll try it just the way you said with something else because I already did that problem...
    [​IMG]

    15 grams (KOH) times 1 Mol KOH divided by 56.01=.27 mols

    Now that I converted that to mols how do i find what volume of 6.0 M HCl would neutralize that? I have no idea... :(
     
  13. Mar 26, 2008 #12
    6.0 M means 6.0 moles per litre of HCl. How many litres would 0.27 moles of HCl be equal to?
     
  14. Mar 26, 2008 #13
    .27m*1L=6.0Mx(V2) .27=6x x=.045 can you confirm this snazzy? The original problem stated what volume of 6 M HCl would be needed to neutralize 15 grams of KOH.
     
  15. Mar 26, 2008 #14
    Looks right to me.
     
  16. Mar 26, 2008 #15
    thanks alot mate I really appreciate it now just 6 more problems to do :D
     
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