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G.C. Problem

Discussion in 'DAT Discussions' started by dwcardman, Jul 28, 2006.

  1. dwcardman

    dwcardman New Member

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    Opening myself up for some abuse here, but what the heck:

    (From Topscore)

    What volume of HCl was added if 20mL of 1.0M NaOH was titrated with 1.0M HCl to produce a pH=2?

    a) 10.2 mL
    b) 20.0 mL
    c) 30.4 mL
    d) 35.5 mL
    e) none of these

    ANSWER: e

    It's obviously over 20 mL, but how far? What am I missing?
     
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  3. thehipster

    thehipster Senior Member
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  4. dat_student

    dat_student Junior Member
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    V = Volume
    pH = 2
    -log [H+] = 2
    [H+] = 10 ^ -2

    (V * 1.0 M HCl) - (20 * 1.0 M OH-)
    ---------------------------------- = 10^-2
    V + 20

    V - 20
    ------- = 10 ^ -2
    V + 20

    0.99 V = 20.2

    V = ~20.4 mL
     
  5. allstardentist

    allstardentist All-Star
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    its 20.4 ml. A sign is wrong in the above explanation. it should be 1.0xV + 20 x 1.0M...
     
  6. dat_student

    dat_student Junior Member
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    i double checked my math. It's right. :)

    "1.0xV + 20 x 1.0M" makes no sense. You need to subtract the # of moles of the base from the # of moles of the acid & then divide by the total volume :)
     
  7. allstardentist

    allstardentist All-Star
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    didnt u have 20.2 before or i saw it wrong? we set up our equations differently, i set the V as the additional volume needed and also the numbers are different. my fault.
     
  8. dat_student

    dat_student Junior Member
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    no, 20.4. You can only get 20.4 with the following equation: :)

    V - 20
    ------- = 10 ^ -2
    V + 20
     
  9. allstardentist

    allstardentist All-Star
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    haha u wanna bet? ... try this...
    (1V + 10^-7)/(V+40) = 10^-2

    and see what u get... where v is the additional volume needed after the it reached pH 7. i know ur way might be simpler but i just did this by intuition.
     
  10. dat_student

    dat_student Junior Member
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    Why do you add 10^-7 to V ????!!!! :confused:
     
  11. allstardentist

    allstardentist All-Star
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    because i did this problem in 2 two steps. first step was to add 20 ml to get to ph=7 and then from there, i calculated how much hcl you need to get from ph=7 to ph=2. thats how i did it
     
  12. allstardentist

    allstardentist All-Star
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    dat student, do u go to ucla undergrad or ucla dental school alrdy, or you want to go to ucla dental school?
     
  13. dat_student

    dat_student Junior Member
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    "the 10^-7 part" makes no sense to me. If you want to do it in 2 steps you should use the following equation instead:

    (1V)/(V+40) = 10^-2

    (i.e. (1V )/( (V+20) + 20) = 10^-2)

    That gives you V = 0.4

    and then you simply have to add 20 to 0.4 to get 20.4

    UCLA School of Dentistry Class of 2010 :) I got in Dec of last yr :)
     
  14. allstardentist

    allstardentist All-Star
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    oh i forgot to multiply by 40ml(.040L) to 10^-7. but anyways, you have to add that because at that point the solution is neutral(pH=7). So there is 10^-7(.040) moles of H+ in the solution.

    Thats great, im an UCLA undergrad and UCLA is my dream school. hopefully i can get in but we'll see.
     
  15. allstardentist

    allstardentist All-Star
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    but yeah still get the same answer either way cuz 10^-7 is really small, doesnt affect the answer much.
     
  16. dat_student

    dat_student Junior Member
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    true but you can't add it just because it won't affect it much.

    (1V)/(V+40) = 10^-2

    or

    (1V)/((V+20) + 20) = 10^-2

    means

    (moles of H+ left) divided by the total volume = concentration of 10^-2 M

    V+20 is the volume of H+
    20 is the volume of OH-

    UCLA is a great school. I went to a party school for my other degrees. :) I partied too much. I'll probably have a hard time adapting to the UCLA lifestyle ;)
     
  17. allstardentist

    allstardentist All-Star
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    im not saying that... im saying 10^-7 and 10^-7x40ml arent much different since its so small. you have to add 10^-7x40ml term to be truly correct because according to ur equation there is no H+ moles in the solution at ph=7 which is not true. you have to consider the number H+ moles there are at ph=7.

    Although i hear ucla dental school is tough, you should be able to handle it..
     
  18. dat_student

    dat_student Junior Member
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    my 2nd equation is for pH = 2 (i.e. conc. = 10^-2) NOT pH = 7. At pH = 7, we have no extra H+


    I'll do my best
     
  19. allstardentist

    allstardentist All-Star
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    why is solution pH =7? its because at ph =7, the concentration of H+ in the solution is 10^-7M. This means that 40 ml of solution will have (10^-7M*40ml) moles of H+. Maybe u dont understand what i did exactly,

    #of moles of H+ at ph=7 in the solution-> (10^-7Mx40ml) + (1MxV)<-#of H+ that will be added ------------------------
    V + 40 <- u know that

    that term equals 10^-2M
     
  20. dat_student

    dat_student Junior Member
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    Do we also have this many moles of OH- in the neutral solution: 10^-7 * 40 ml? (i.e. pOH = 7 too) If we do doesn't that cancel 10^-7 * 40 ml H+. I think we need to find the extra H+ right?
     
  21. allstardentist

    allstardentist All-Star
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    You're right that there is 10^-7x40 ml of OH- in the solution. but you cant say they cancel out because equal number are present. ph=7 means that there are (10^-7x40ml) are dissociated (and same as for OH- as a result). So we need to include it. That's what i think. Does the topscore have your explanation? cuz ihavent taken topscore yet.
     
  22. dat_student

    dat_student Junior Member
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    I don't have topscore. Now, I think you are absolutely right. We usually ignore 10^-7 because it's usually a small number (relatively). In this case (and most other cases), 10^-7 doesn't affect the final answer much but technically & scientifically you're right. Thanks for the explanation. ;)

    Here is an interesting problem:

    We add 10^-8 moles of H+ to 1 L solution. What is the pH? (assuming volume stays at 1 L)
     
  23. allstardentist

    allstardentist All-Star
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    assuming the 1L solution is at ph=7, then the pH of the new solution would be -log (10^-7+10^-8)/1. if theres some trick to it, this would probably be wrong.
     
  24. dat_student

    dat_student Junior Member
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    :thumbup:
     
  25. DonExodus

    DonExodus Dentist in Virgin Islands
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    C1V1=C2V2 C=[] V=Volume.

    AT pH=7: C1=10e-7 V=40mL
    AT pH=2: C2=10e-2 V=X (20+x actually, just add 20 to x will suffice).

    Therefore, (10e-7)(40)=(.01)(x)

    4e-6 = .01x

    (4x10^-6)/(.01) = 4e4 = .0004L = .4mL = x

    20mL+.4mL = total volume of HCl added = 20.4mL

    Thats wonderful, but I goofed: Started with mL, ended with L... somehow Im off by 1k, any clue where?
     
  26. slayerdeus

    slayerdeus Member
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    We can also deduce that since it only takes .4 mL of HCl to drop the pH to 2 after reaching the isoelectric point at pH 7, the principle of a strong base titrated with strong acid. You know should know that it doesn't take much to lower the pH drastically after the isoelectric point is reached.

    [​IMG]
     

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