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Titration (Neutralization) problem...is 2009 ADA wrong?

Discussion in 'DAT Discussions' started by Cofo, Aug 3, 2011.

  1. Cofo

    Cofo Super Cool Member
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    If 50.0 mL of H2SO4 required 50.0 mL of 1.0 M NaOH for a complete neutralization reaction, according to the equation shown below, what would be the molarity of an acid?
    2NaOH + H2SO4 → Na2SO4 + 2H2O

    (na)(Ma)(Va) = (nb)(Mb)(Vb)
    This is the formula Chad said to use. The answer is 2.0 M right?
    For some reason, the 2009 ADA test said 2 is wrong. :confused:
     
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  3. Golfguy

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    You would need half the molarity of the acid.
     
  4. FROGGBUSTER

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    Forget formulas for these type of problems. They're confusing and there's a simpler way to do them.

    Neutralization: moles H+ = moles OH-. If you just remember that, you can solve all of these.

    You know you require 50 mL of 1 M NaOH, which is 50 mmols. You also know you are using 50 mL of H2SO4, which is a di-protic acid.

    Ask yourself now, how do I produce 50 mmols of H+ from 50mL H2SO4? What must the molarity be? From there, it's a stoichiometry problem.

    50 mL H2SO4 * Molarity * 2 moles H+ per every mole of H2SO4 = 50 mmols H+.

    Solve and you get 0.5 Moles/L.
     
  5. LetsGo2DSchool

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    There's even an easier way to do this and it can be done with common sense.

    Given that both volumes are equal, knowing that you will get two H+ (from H2SO4) and one OH- (from NaOH), common logic would say the acid must be half as strong (or base twice as strong) for a neutralization to occur.

    Since there's 1 M NaOH, the molarity of H2SO4 must be 0.5 M
     
    #4 LetsGo2DSchool, Aug 3, 2011
    Last edited: Aug 3, 2011
  6. wired202808

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    you dont need their formula for the equation you just do:
    50ml H2SO4 * 2N of H2SO4* X = 50ml NaOH * 1M NaOH * 1N of NaOH
    solve for X = 0.5

    Just remember how do to normality and you're done!
     
  7. rah08e

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    Yeah, setting this up just like the destroyer/ how chad says works just fine.
    N1V1= N2V2 .... Taking normality into account with the molarity and you get 0.5... easiest way to solve this IMO
     
  8. rocky90

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    Can anyone please point which video in gen chem has chad discussed this in. I wanted to go over it again but cant fine it.
    Thanks
     
  9. Cofo

    Cofo Super Cool Member
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    Chem video 5.11. Just remember, if a question asks about a NEUTRALIZATION REACTION...use this normality equation:
    (nAcid)(MAcid)(VAcid)=(nBase)(MBase)(VBase).
    n stands for normality. M = molarity. V = volume
    nAcid = number of H^+
    nBase = number of OH^-
    BTW...Wow...this is extremely easy to do now that you mentioned normality was involved. Thank you, and to everyone else who chimed in.
     
  10. wired202808

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    no problem man! every titration problem should involve normality this way you can never get it wrong! plus the formula they give u is crap! its not even necessary to solve the problem and further confuses you. which is why I think I also got it wrong when I first did it.
     
  11. rah08e

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    Yeah, question# 108 on destroyer is a perfect example where you are given volumes and asked to find the Molarity of HBr given the molarity of Ca(OH)2. If you don't account for normality with Ca(OH)2, you will get the problem wrong.
     
  12. rah08e

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    I thought I understood this question but I'm confused again. How are the coefficients incorporated into solving this problem?

    Thanks!
     
  13. toothhornet88

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    You don't need the coefficients in order to solve this problem. Just solve it exactly how LetsGo2DSchool showed it
     
  14. rah08e

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    well I thought since the question tells you to refer to the equation, and since there are 2 Moles of NaOH, I thought that it would be used to solve the problem
     
  15. ScarletKnight24

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    given

    .05 L NaOH .05L H2SO4
    1M NaOH

    Find # of moles of NaOH
    .05L x 1M = .05 moles NaOH

    now convert .05 moles NaOH to moles of H2SO4.

    .05 moles NaOH x 1moleH2SO4/2moles NaOH = .025 moles H2SO4

    now plug this back into the formula M = #of moles/Liter solution
    .025 moles H2SO4/ .05L H2SO4=

    0.5 M of H2SO4------> ANWER
     
  16. wired202808

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    this is too much work for a simple problem. all you need to do is to factor in normality and you're done!

    50ml H2SO4 * 2N of H2SO4* X = 50ml NaOH * 1M NaOH * 1N of NaOH
    solve for X = 0.5

    Just remember how do to normality and you're done!
     
  17. ScarletKnight24

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    You are right, but I was trying to give a clear explanation :laugh:

    Just curious about what you did. What does the letter "N" in front of H2SO4 and NaOH mean? Are they the mole ratio? if so, shouldn't it be 2N NaOH and 1N H2SO4 based on the reaction?
     
    #16 ScarletKnight24, Aug 10, 2011
    Last edited: Aug 10, 2011
  18. wired202808

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    N is normality, meaning if you have 2 hydrogen atoms you would multiply the formula by 2, if you had one hydrogen atom then you multiply the volume and molarity by 1. same goes for OH's. Its a molar concentration amount that you have to use in order to get the right volume or molarity when doing titration problems.

    You can do it your way as well but this is how Chad showed it in his videos and how the Destroyer uses it. So on the other side you have NaOH hence its normality is 1, if it was Na(OH)2 then you would use 2 on the right hand side and have to include it in your multiplication.
     
  19. ScarletKnight24

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    So N= number of hydrogen atoms only ?
     
  20. es1113

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    It also applies to OH in the same way. As someone said, it it was something like NaOH, only one OH so N is 1
    For Ca(OH)2, N would be 2 because two OHs
     
  21. ScarletKnight24

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    Gotcha! Thanks I'll incorporate that into my arsenal lol.
     
  22. rah08e

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    I don't want to discredit your explanation but why would ADA give the equation with coefficients if it didn't matter? For problems like these, can you always just ignore the coefficients in the balanced equation? I thought since you had more moles of NaOH, it would factor into solving the problem.

    What I'm saying is: yes, you got the right answer, but will that work every time, even when they give a balanced equation?
     
  23. wired202808

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    as long as its a titration problem i dont see why it wouldnt work? the formula is simply telling you what the reaction looks like, but its not like you are using any values from it to determine the answer.

    Look through Destroyer or Chad there is never a formula for any titration problem. That is the first time i've ever seen one. Meanwhile using the NMV formula, I always get the correct answer because NMV is the proper way to calculate titration as explained per Chad and per Destroyer. Thats why I think a balanced formula is useless and confusing (in this sort of a problem)
     
  24. es1113

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    Yea, I feel like the formula is there as a trick? Either way most cases you don't need it (probably won't need it at all). Just use that equation and you should be good for these types of problems
     

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