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gchem question

Discussion in 'DAT Discussions' started by dontbam, Jul 18, 2006.

  1. dontbam

    dontbam Member
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    Can anyone help me with this question?

    What volume of HCl was added if 20mL of 1M NaOh is titrated with 1M of HCl to produce a pH=2 solution?

    according to topscore the answer is 20.4 mL
     
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  3. tissy

    U.S. Public Health Service 10+ Year Member

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    Think this is right..if not sorry...:p
     
  4. thehipster

    thehipster Senior Member
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    The answer doesn't happen to be 200 mL, does it?
     
  5. asckwan

    asckwan Member
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    I got it to be .2L. Anyone else?

    (forgot my dot there :oops: )
     
  6. lovechild404

    lovechild404 Member
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    i got the answer to be .2L = 20 mL. i'm not sure why the answer would be 20.4mL...where did the .4 come from?
     
  7. dontbam

    dontbam Member
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    hey can you guys show me how you calculated the 20ml? in top score they set it up where it's like 0.01 * (40 + x) = x

    solving, that's where the .4 comes from, how did you guys solve it?
     
  8. Chga

    Chga Member
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    you have to convert to normality
    1 mole NaOH/1L = x mole NaOH/.02L
    x= .02 moles
    (.02 moles)(1 equiv of -OH)= .02 N
    since OH and H in HCl are 1 to 1 to equalout
    there is .02N of H+ ions meaning .02M of HCl
    since its 1M HCl there are .02L of it.
    .02L = 20ml
    have no idea how some people got .2L or where the 20.4 comes in.
    it doesnt make sense
     
  9. djeffreyt

    djeffreyt Senior Member
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    first...you have 20mL of a 1M Strong Base that will take only 1 proton per molecule (NaOH). To neutralize that, add an equal amount of a strong acid (HCl) at 1 M. So to neutralize the 20mL of 1 M NaOH, we add 20 mL of 1 M HCl.

    That was step 1, and you do that in your head just by looking at the problem. Then the question is how much more do we have to add of 1M HCl to our now 40mL of neutral solution to get the pH = 2

    You need to know this formula below and how to work with it
    pH = 2
    pH = -log [H+]
    2 = -log [H+]
    10^-2 = [H+]
    or
    [H+] = 0.01 M

    if we call X the amount of HCl we need to add to our 40ml solution to reach a pH = 2 then you can put it together like this. It's kind of a dilution question. You start with 1 M HCl. You are diluting it by adding some of this 1M HCl to a 40mL neutral solution. When will the solution's pH drop of 0.01 M HCL 9which is pH =2)

    (1 M HCl) * (X mL) = (0.01 M HCl) * (40 + X ml)

    I'm not gonna do the math here...but the answer comes out right.
     
  10. asckwan

    asckwan Member
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    Oh dang! I see now, 20mL is the amount to neutralize the solution, but then you have to take it one step further to get the pH = 2.

    Achiever also has a question like this:

    How many milliliters of 0.05 M HCl are required to turn 20 ml of 0.05 M KOH into a solution of pH 2.00?

    Practice practice practice!!
     
  11. dontbam

    dontbam Member
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    thanks guys!
     
  12. fancymylotus

    fancymylotus A Whole New World
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    Can someone do this problem out? I'm having a ****** moment.
     
  13. cocokeala

    cocokeala katonk!
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    finishing up Dr. Jeffs math hope its right cuz i came out with the right answer :)

    V= .4+.01V
    .99V= .4
    V= .4

    .4 is the extra amount of HCL you need to add to the neutralized soln which was 40 (remember 20 HCL + 20NaOH)

    well we only want how much HCL was added
    so we know that 20 HCL was added just to neutralize the NaOH, and we added .4 HCL to get it to pH=2.

    so 20+.4= 20.4
     
  14. lovechild404

    lovechild404 Member
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    BRILLIANT!...we're all on our way to dental school now! :laugh:
     
  15. tom_servo_dds

    tom_servo_dds Senior Member
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    I came up with 30 mL for HCl needed:

    (0.05M HCl)(X mL) = (0.01)(40mL + X mL)
    X = 10 mL + 20mL to neutralize = 30 mL total
     
  16. fancymylotus

    fancymylotus A Whole New World
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    huh. 40mL :confused:
     
  17. tom_servo_dds

    tom_servo_dds Senior Member
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  18. Envision

    Envision Envisioning...
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    Hey tom,

    doing that problem i got 28. is your 30 round off? or is my math incorrect? thanx.
     
  19. dat_student

    dat_student Junior Member
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    ok, I solve it again:

    V is the total volume of HCl

    1 * V - 1 * 20
    __________ = 10^-2
    V + 20

    V = 20.2 / 0.99

    V = 22.4

    or you can do the following:

    20 ml of HCl is needed to neutralize 20 ml of 1 M NaOH. So, total volume of the final solution is V + 20 + 20 (or V + 40). V is the extra volume of HCl that is needed to produce a pH=2 solution.

    1 * V
    _______ = 10^-2
    V + 40

    V = 0.4

    20 (to neutralize) + 0.4 (extra) = 20.4

    *Again, I assumed that at pH = 7 we have little H+.
     
  20. tom_servo_dds

    tom_servo_dds Senior Member
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    You're right, 28 - thanks for catching that!
     
  21. Envision

    Envision Envisioning...
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    no wait, your right! i did it again and i got thirty! haha :D
     
  22. tom_servo_dds

    tom_servo_dds Senior Member
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    Looks like we both made the same mistake :laugh: - it is 30, which is good since the options were:

    A. 10
    B. 15
    C. 20
    D. 25
    E. 30
     

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