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Titration of Trimethylamine with HCl (g)

  1. Oct 31, 2009 #1
    1. The problem statement, all variables and given/known data
    Trimethylamine has a Kb of 6.3 x 10-5. How many liters of HCl gas, measured at 1.20 bar and 298 K, must be added to 250.0 mL of 0.500 M trimethylamine to give a pH of 10.30?

    2. Relevant equations

    3. The attempt at a solution
    I don't know where to start because I don't know how the gaseous HCl affects the pH.
     
  2. jcsd
  3. Oct 31, 2009 #2

    Borek

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    Staff: Mentor

    It is gaseous just to make you think it is a difficult question. Just solve for number of moles of HCl that have to be added to get given pH, then it is simple pV=nRT.

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  4. Oct 31, 2009 #3
    How do I set up the expression for Ka for this reaction?

    Is Ka = [(CH3)3N][H3O+] / [(CH3)3NH]?

    Since Kb = 6.3*10^-5, I found Ka by dividing Kw by Kb, and that turned out to be 1.6*10^-10
    Also, since pH = 10.3, [H3O+] = 5*10^-11

    I'm not sure what to do from that point on...

    I tried to set an equation up:
    Ka = (x)(5*10^-11 + x) / (0.500 - x)
    But this can't be right...

    Am I supposed to use the equation pH = pKa + log [A-]/[HA]?
     
    Last edited: Oct 31, 2009
  5. Oct 31, 2009 #4

    symbolipoint

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    The form of your equation looks reasonable, but you might be much more comfortable using the Kb value and any other corresponding changes.
     
  6. Nov 1, 2009 #5

    Borek

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    Staff: Mentor

    That sounds like the simplest approach, although part under the log would take

    [tex]\frac {} {[BH^+]}[/tex]

    form. But that's just a minor detail.

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