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[OH-] or [H+] ?

  1. Mar 3, 2004 #1
    __________HA <------> H+ + A-
    initial (mol) .1 ________ 0 ___ 0
    change __ -x _______ +x ___ +x
    final _____.1-x _______ x ___ x

    Ka = 10^-8
    find pH of .1 mol acid
    since ka X 100 = 10^-6 < .1 can use initial concentration: .1-x = .1

    Ka = [H+][A-]/[HA]
    10^-8=x^2/.1
    x=[H+]= 3.16x10^-5
    pH = 4.5

    now add .01 mols of OH- and see how it changes the pH

    Ka=x[3.16x10^-5 + .01]/(.1 - .01)
    10^-8 = .01x/.09
    x=9x10^-8

    -log(9x10^-8) = pH of 7 (also a pOH of 7)

    what is the 9x10^-8 ion concentration representing, the [OH-] or [H+] ions?

    if it's the [OH-] then shouldn't it be 14-pOH = pH
    14-7=7

    is this formula the best way find the pH in this case?
    Ka=x([H+] + [OH-])/([HA] - [OH-])
     
  2. jcsd
  3. Mar 30, 2004 #2
    A little confused- can you give some indication as to volume/amount. I dunno if I am reading that right but you seem to have derrived pH from moles rather than mol dm-3- I'm a little tired at the moment so I don't know though.
     
  4. Mar 31, 2004 #3
    now add .01 mols of OH- and see how it changes the pH

    Ka=x[3.16x10^-5 + .01]/(.1 - .01)
    10^-8 = .01x/.09
    x=9x10^-8

    I'm not quite sure what you did here could you post the actual problem and then explain what you did above
     
  5. Mar 31, 2004 #4
    what is the 9x10^-8 ion concentration representing, the [OH-] or [H+] ions?

    to answer this question the 9*10^-8 represents both the OH- and the H+ since your pH is 7 those two concentrations have to be equal
     
  6. Mar 31, 2004 #5
    Okay i got it see if ya follow me

    your gonna take 14-4.5=pOH then take 10^-pOH=[OH-] + the [.01] your adding, that equals .01 then take the -log.01=pOH 14-pOH=pH
     
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